{"pdfData":[[{"bbox":[0,0,595,423],"type":"image","angle":0,"index":0,"img_path":"/ced86103fb59e0b03674d3fbef57fa32abd2c451b2a16f5c0f33f48c8f0d7d70.jpg","color":{"line":"rgba(89, 92, 220, 1)","fill":"rgba(89, 92, 220, 1)"},"img_ratio":1,"page_idx":0,"id":"fa30342a-35b6-4bf2-8eaa-4925078f37c5","page_size":[595,841],"block_position":"0-0"},{"bbox":[20,518,172,547],"type":"title","angle":0,"index":1,"level":2,"text":"## 数学分析笔记","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":0,"id":"d1da859a-5bc4-4975-ad58-38d835438996","page_size":[595,841],"block_position":"0-1"},{"bbox":[30,590,104,604],"type":"text","angle":0,"index":2,"merge_prev":false,"text":"作者：Reichtum ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":0,"id":"a96f12b4-5387-4597-adc1-f915c6f4dc66","page_size":[595,841],"block_position":"0-2"},{"bbox":[30,608,145,623],"type":"text","angle":0,"index":3,"merge_prev":false,"text":"组织：ZheJiang University ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":0,"id":"6dfd8f34-2977-4353-a223-98ac879ef00f","page_size":[595,841],"block_position":"0-3"},{"bbox":[30,626,145,641],"type":"text","angle":0,"index":4,"merge_prev":false,"text":"时间：September 22, 2022 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":0,"id":"ebf4483f-d4de-4851-b25e-f89412542cab","page_size":[595,841],"block_position":"0-4"},{"bbox":[31,645,77,659],"type":"text","angle":0,"index":5,"merge_prev":false,"text":"版本：0.1 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":0,"id":"17c3b333-eff2-4a11-900f-e5fa469228a1","page_size":[595,841],"block_position":"0-5"},{"bbox":[420,617,537,731],"type":"image","angle":0,"index":6,"img_path":"/f8e34492d364de6248e0c95112dac34ceb37621665dd12e77bb7f21e0c6e21aa.jpg","color":{"line":"rgba(89, 92, 220, 1)","fill":"rgba(89, 92, 220, 1)"},"img_ratio":0.2,"page_idx":0,"id":"6857ffde-20bb-4ea3-b2ac-0cc75c7341ed","page_size":[595,841],"block_position":"0-6"},{"bbox":[95,739,499,768],"type":"text","angle":0,"index":7,"merge_prev":false,"text":"There is nothing in the understanding which has not come from the senses, except the understanding itself, or the one who understands ——Gottfried Wilhelm Leibniz ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":0,"id":"e8ddc5ee-5fe3-472c-b5c0-af333af5cd6c","page_size":[595,841],"block_position":"0-7"}],[{"bbox":[279,74,316,95],"type":"title","angle":0,"index":0,"level":2,"text":"## 目录","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":1,"id":"25ddedcd-17e1-4ec6-a44d-678d502b349e","page_size":[595,841],"block_position":"1-0"},{"bbox":[55,126,542,760],"type":"text","angle":0,"index":1,"text":"第1章 数学基础 1\n1.1 三角函数 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1\n1.1.1 和差公式、万能公式、倍角公式、半角公式. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1\n1.1.2 和差化积与积化和差. . . .\n1.1.3 交替三角函数. . .\n1.1.4 重要的三角恒等式. . .\n1.1.5 反三角函数. . .\n1.1.6 三角不等式. . .\n1.2 恒等式. . .\n1.2.1 立方和、四次方和. . .\n1.2.2 最大值、最小值. . .\n1.2.3 次方和. . .\n1.3 不等式. . .\n1.3.1 基本不等式. . .\n1.3.2 Jensen 不等式. . .\n1.3.3 调和级数不等式. . .\n1.4 几何学基础. . .\n1.4.1 常见立体几何体积. . .\n第2章 数列极限 6\n2.1 数列极限的概念与性质. . .\n2.1.1 数列极限的概念. . .\n2.1.2 收敛数列的性质. . .\n2.2 一些重要极限结论. . .\n2.2.1 几个基本极限. . .\n2.2.2 Stirling 公式与 Wallis 公式. . .\n2.2.3 Euler 常数. . .\n2.3 夹逼准则. . .\n2.4 平均值定理. . .\n2.4.1 有限项幂次根号平均. . .\n2.4.2 平均值定理：无穷项算术平均与几何平均. \n2.4.3 技巧：一招解决 n 次根号与除以 n 10\n2.5 使用定积分定义. 10\n2.5.1 使用定积分计算极限. 10\n2.5.2 利用定积分定义构造等价无穷小 12\n2.6 单调有界定理 12\n2.6.1 单调有界定理及其推广 12\n2.6.2 相减法 13\n2.6.3 相除法 13\n2.6.4 不等式放缩 14\n2.7 递推问题 14 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":1,"id":"60635dbd-d302-4996-875e-f17baa357e22","page_size":[595,841],"block_position":"1-1"}],[{"bbox":[518,47,539,58],"type":"header","angle":0,"index":-99999,"text":"目录 ","color":{"line":"rgba(164, 164, 164, 1)","fill":"rgba(164, 164, 164, 1)"},"page_idx":2,"is_discarded":true,"id":"9b29f138-305a-4457-8fa4-2ca577da5036","page_size":[595,841],"block_position":"2-0"},{"bbox":[61,69,542,759],"type":"text","angle":0,"index":1,"text":"2.7.1 极限压缩定理 14 \n2.7.2 压缩映射定理 15 \n2.8 Stolz 定理 15 \n2.9 积分形式极限：分段法 16 \n2.9.1 定积分形极限 16 \n2.9.2 反常积分形极限 16 \n2.9.3 周期函数积分极限 17 \n\n第3章 函数极限与连续性 18 \n3.1 函数的极限 18 \n3.1.1 函数极限定义 18 \n3.1.2 函数极限的性质 18 \n3.1.3 无穷大与无穷小 19 \n3.1.4 使用定义直接计算 19 \n3.2 函数极限的计算 19 \n3.2.1 等价无穷小公式 19 \n3.2.2 Taylor 展开公式 20 \n3.3 函数的连续性 20 \n3.3.1 函数连续的定义 20 \n3.3.2 间断点的概念 21 \n3.3.3 介值性定理 22 \n3.3.4 介值性定理的无连续推广 23 \n3.4 连续性的应用 23 \n3.5 一致连续 24 \n3.5.1 一致连续的概念 24 \n3.5.2 Cantor 定理及其推广 25 \n3.5.3 Lip 连续与一致连续 26 \n3.5.4 区间合并问题 26 \n3.5.5 四则运算、复合的一致连续性 27 \n3.6 一致连续的经典问题 27 \n\n第4章 一元微分学 28 \n4.1 导数与微分 28 \n4.1.1 导数概念与可导性 28 \n4.1.2 微分的概念 29 \n4.1.3 常用的导数公式 29 \n4.1.4 隐函数、反函数求导 30 \n4.1.5 高阶导数 31 \n4.1.6 使用导数证明恒等式 31 \n4.2 微分中值定理 32 \n4.2.1 三大中值定理 32 \n4.2.2 中值问题：构造辅助函数 33 \n4.2.3 带积分的微分中值问题 35 \n4.2.4 多个中值点的中值问题 35 \n4.2.5 中值等式问题 36 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":2,"id":"c532fbb3-75fc-4d9a-810a-4756e1bf6717","page_size":[595,841],"block_position":"2-1"},{"bbox":[293,803,301,812],"type":"page_number","angle":0,"index":1000001,"text":"ii ","color":{"line":"rgba(164, 164, 164, 1)","fill":"rgba(164, 164, 164, 1)"},"page_idx":2,"is_discarded":true,"id":"bdbf5faa-9a6f-4b99-bd21-28bb23cb2ff3","page_size":[595,841],"block_position":"2-2"}],[{"bbox":[518,47,539,58],"type":"header","angle":0,"index":-99999,"text":"目录 ","color":{"line":"rgba(164, 164, 164, 1)","fill":"rgba(164, 164, 164, 1)"},"page_idx":3,"is_discarded":true,"id":"19501142-8005-4174-8776-d63acbed6e5a","page_size":[595,841],"block_position":"3-0"},{"bbox":[61,69,542,744],"type":"text","angle":0,"index":1,"text":"4.2.6 中值不等式 36 \n4.2.7 导数极限定理与导数介值定理 37 \n4.3 洛必达法则与Taylor公式 38 \n4.3.1 洛必达法则 38 \n4.3.2 Taylor公式 38 \n4.4 凸函数 39 \n4.4.1 函数的凹凸性 39 \n4.4.2 凸函数的基本性质 40 \n4.4.3 利用凹凸性证明不等式 41 \n4.5 一元极值与导数零点问题 41 \n4.5.1 一元极值的概念与判定 41 \n4.5.2 导数零点问题 41 \n\n第5章 不定积分 43 \n5.1 常用不定积分公式 43 \n5.2 不定积分基本方法 43 \n5.3 不定积分表选例 45 \n5.3.1 三角函数：重点 45 \n5.3.2 对数与指数 48 \n5.3.3 分式问题 48 \n5.3.4 根式问题 49 \n5.3.5 配对积分法 51 \n5.3.6 递推问题 51 \n\n第6章 定积分 52 \n6.1 定积分理论 52 \n6.1.1 可积性 52 \n6.1.2 Newton-Leibniz公式 53 \n6.1.3 四则运算与复合 54 \n6.1.4 定积分的保号性 55 \n6.1.5 积分中值定理 56 \n6.2 三角函数定积分计算 57 \n6.2.1 $[0,\\frac{\\pi}{2}]$ 上三角函数积分公式 57 \n6.2.2 $[0,\\pi],[0,2\\pi]$ 上三角函数积分 58 \n6.2.3 区间再现公式 59 \n6.2.4 Euler积分 60 \n6.2.5 Dirichlet积分 60 \n6.2.6 其他几个常见三角积分 61 \n6.3 积分(不)等式 61 \n6.3.1 整体放缩 61 \n6.3.2 同区域单调性放缩 61 \n6.3.3 利用Newton-Leibniz公式 62 \n6.3.4 利用Cauchy-Schwarz不等式 63 \n6.3.5 利用微分中值定理 63 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":3,"id":"65699078-3230-4061-a664-627a0a12a613","page_size":[595,841],"block_position":"3-1"},{"bbox":[54,753,135,766],"type":"text","angle":0,"index":2,"text":"第7章 反常积分 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":3,"id":"ad24412b-027d-49f2-94a4-a3fcf9d89645","page_size":[595,841],"block_position":"3-2"},{"bbox":[526,755,539,765],"type":"page_number","angle":0,"index":1000002,"text":"65 ","color":{"line":"rgba(164, 164, 164, 1)","fill":"rgba(164, 164, 164, 1)"},"page_idx":3,"is_discarded":true,"id":"f29a136d-1b11-4935-80fe-235cbdfe0c97","page_size":[595,841],"block_position":"3-3"},{"bbox":[292,803,302,812],"type":"page_number","angle":0,"index":1000003,"text":"iii ","color":{"line":"rgba(164, 164, 164, 1)","fill":"rgba(164, 164, 164, 1)"},"page_idx":3,"is_discarded":true,"id":"91e08249-119b-464c-9792-c30a00b3fdbc","page_size":[595,841],"block_position":"3-4"}],[{"bbox":[518,47,539,58],"type":"header","angle":0,"index":-99999,"text":"目录 ","color":{"line":"rgba(164, 164, 164, 1)","fill":"rgba(164, 164, 164, 1)"},"page_idx":4,"is_discarded":true,"id":"ad1defe5-b67e-443e-bb08-a76731a77c7d","page_size":[595,841],"block_position":"4-0"},{"bbox":[58,69,542,770],"type":"text","angle":0,"index":1,"text":"7.1 反常积分理论 65\n7.1.1 无穷积分与瑕积分概念 65\n7.1.2 反常积分比较判别法 66\n7.1.3 Dirichlet与Abel判别法 67\n7.1.4 函数在无穷远处的性质 69\n7.2 反常积分的计算 70\n7.2.1 两个常用的基础反常积分 70\n7.2.2 富如兰尼积分 70\n7.2.3 Dirichlet积分 71\n第8章 数项级数 73\n8.1 数项级数基本理论 73\n8.2 正项级数 74\n8.2.1 比较原则 74\n8.2.2 比式判别法与根式判别法 75\n8.2.3 积分判别法和拉贝判别法 77\n8.2.4 放缩证明敛散性 77\n8.3 任意项级数 78\n8.3.1 交错项级数 78\n8.3.2 Dirichlet与Abel判别法 79\n8.4 数项级数理论 79\n8.4.1 级数的加括号 79\n8.4.2 级数的重排 80\n8.4.3 正部与负部 81\n8.4.4 级数的乘积 82\n8.4.5 级数的相互控制 82\n第9章 函数列与函数项级数 84\n9.1 函数列的一致收敛 84\n9.1.1 一致收敛的概念 84\n9.1.2 一致收敛的基本判别法 84\n9.1.3 一致收敛判别练习 85\n9.1.4 四则运算、复合极限的一致收敛性 86\n9.2 函数列极限的性质 87\n9.2.1 极限函数的连续性 87\n9.2.2 极限函数的可积性 88\n9.2.3 极限函数的可微性 89\n9.2.4 连续+单调：推出一致收敛 90\n9.3 函数项级数的一致收敛性 90\n9.3.1 一致收敛的基本性质与判别 90\n9.3.2 Dirichlet与Abel判别法 91\n9.4 函数项级数极限函数的性质 92\n第10章 幂级数 93\n10.1 幂级数概念及其收敛区间 93\n10.1.1 收敛区间与收敛半径 93 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":4,"id":"8bf45759-ede1-4fe6-a568-aa7287a4fbf9","page_size":[595,841],"block_position":"4-1"},{"bbox":[292,803,302,812],"type":"page_number","angle":0,"index":1000001,"text":"iv ","color":{"line":"rgba(164, 164, 164, 1)","fill":"rgba(164, 164, 164, 1)"},"page_idx":4,"is_discarded":true,"id":"037b2af0-a976-4972-bf7e-45e76b19c82d","page_size":[595,841],"block_position":"4-2"}],[{"bbox":[518,47,539,58],"type":"header","angle":0,"index":-99999,"text":"目录 ","color":{"line":"rgba(164, 164, 164, 1)","fill":"rgba(164, 164, 164, 1)"},"page_idx":5,"is_discarded":true,"id":"3ae1b861-ff46-4880-83a6-b65becb50540","page_size":[595,841],"block_position":"5-0"},{"bbox":[57,69,542,764],"type":"text","angle":0,"index":1,"text":"10.1.2 幂级数的一致收敛性 ..... 94\n10.1.3 缺项幂级数的收敛区域 ..... 95\n10.2 幂级数展开式求和 ..... 95\n10.2.1 等比级数 ..... 96\n10.2.2 arctan x 级数 ..... 96\n10.2.3 $e^x$ 级数 ..... 97\n10.3 幂级数求和式展开 ..... 97\n10.3.1 基本理论与直接展开 ..... 97\n10.3.2 求导与求积 ..... 98\n\n第 11 章 傅里叶级数 ..... 99\n11.1 以 $2\\pi$ 为周期的傅里叶级数 ..... 99\n11.2 以 $2l$ 为周期的傅里叶级数 ..... 99\n11.3 Parseval 等式 ..... 100\n\n第 12 章 多元极限与连续 ..... 101\n12.1 累次极限与重极限 ..... 101\n12.1.1 基本概念 ..... 101\n12.1.2 重极限的计算 ..... 102\n12.2 多元函数连续性 ..... 103\n12.2.1 多元函数连续性 ..... 103\n12.2.2 多元函数的一致连续性 ..... 103\n\n第 13 章 多元微分学 ..... 105\n13.1 多变量导数与微分 ..... 105\n13.1.1 全微分与偏导数 ..... 105\n13.1.2 可微性的证明 ..... 105\n13.1.3 梯度与方向导数 ..... 106\n13.1.4 可微、偏导、方向导数、连续关系总结 ..... 107\n13.1.5 复合求导：链式法则 ..... 108\n13.1.6 混合偏导 ..... 108\n13.2 多元中值定理与泰勒展开 ..... 109\n13.2.1 多元中值定理 ..... 109\n13.2.2 多元 Taylor 展开 ..... 110\n13.3 隐函数定理 ..... 110\n13.3.1 隐函数存在定理与可微定理 ..... 110\n13.3.2 隐函数组存在定理与微分定理 ..... 111\n13.4 多元无条件极值问题 ..... 112\n13.4.1 无条件极值 ..... 113\n13.4.2 最值问题：极值点 + 边界 ..... 114\n13.4.3 多元极值理论证明 ..... 114\n13.5 条件极值 ..... 115\n13.5.1 Lagrange 乘数法 ..... 115\n\n第 14 章 含参积分 ..... 117\n14.1 含参正常积分 ..... 117 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":5,"id":"b775bcdb-9e96-4df4-80e4-7ad8c04eb86f","page_size":[595,841],"block_position":"5-1"},{"bbox":[293,803,300,811],"type":"page_number","angle":0,"index":1000001,"text":"V ","color":{"line":"rgba(164, 164, 164, 1)","fill":"rgba(164, 164, 164, 1)"},"page_idx":5,"is_discarded":true,"id":"d7f33043-9baf-41c6-93ee-3279f1695be8","page_size":[595,841],"block_position":"5-2"}],[{"bbox":[518,47,539,58],"type":"header","angle":0,"index":-99999,"text":"目录 ","color":{"line":"rgba(164, 164, 164, 1)","fill":"rgba(164, 164, 164, 1)"},"page_idx":6,"is_discarded":true,"id":"a4731d73-96a8-4fa3-bea8-b3065b805ec9","page_size":[595,841],"block_position":"6-0"},{"bbox":[54,70,542,770],"type":"text","angle":0,"index":1,"text":"14.1.1 含参正常积分基本概念与定理 ..... 117\n14.1.2 常用含参正常积分的计算 ..... 118\n14.2 含参反常积分与一致收敛 ..... 119\n14.2.1 基本概念 ..... 119\n14.2.2 Dirichlet 与 Abel 判别法 ..... 120\n14.3 含参反常积分的性质 ..... 121\n14.3.1 含参反常积分的极限与连续性 ..... 121\n14.3.2 含参反常积分的可微性与可积性 ..... 121\n14.3.3 含参反常积分与函数项级数的关系 ..... 122\n\n第 15 章 多重积分 ..... 124\n15.1 二重积分 ..... 124\n15.1.1 二重积分的概念与计算 ..... 124\n15.1.2 二重积分可积性 ..... 125\n15.1.3 二重积分换元法 ..... 125\n15.1.4 极坐标换元 ..... 126\n15.2 三重积分 ..... 127\n15.2.1 投影法 ..... 127\n15.2.2 截面法 ..... 128\n15.2.3 三重积分换元法 ..... 128\n15.2.4 柱面坐标变换与球坐标变换 ..... 128\n\n第 16 章 曲线积分 ..... 130\n16.1 曲线的参数化 ..... 130\n16.2 第一型曲线积分 ..... 130\n16.2.1 第一型曲线积分的基本概念 ..... 130\n16.2.2 使用对称性 ..... 131\n16.2.3 曲线积分中值问题 ..... 132\n16.3 第二型曲线积分 ..... 132\n16.3.1 基本概念与参数计算 ..... 132\n16.3.2 使用对称性 ..... 133\n16.3.3 两类曲线积分的关系 ..... 135\n16.4 Green 公式 ..... 135\n16.4.1 Green 公式及其基本应用 ..... 135\n16.4.2 内部有瑕点 ..... 136\n16.4.3 曲线积分与路径无关性 ..... 137\n\n第 17 章 曲面积分 ..... 138\n17.1 曲面的面积 ..... 138\n17.2 第一型曲面积分 ..... 138\n17.2.1 基本概念与计算公式 ..... 138\n17.2.2 使用对称性 ..... 139\n17.3 第二型曲面积分 ..... 140\n17.3.1 基本概念、投影法、参数方程法 ..... 140\n17.3.2 利用对称性 ..... 141\n17.3.3 两类曲面积分的联系 ..... 141 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":6,"id":"84e46181-5600-4a00-8a58-a7144f0f7cbb","page_size":[595,841],"block_position":"6-1"},{"bbox":[292,803,302,811],"type":"page_number","angle":0,"index":1000001,"text":"vi ","color":{"line":"rgba(164, 164, 164, 1)","fill":"rgba(164, 164, 164, 1)"},"page_idx":6,"is_discarded":true,"id":"2d2fb0eb-943f-4d9b-8efc-8c7f5055aa08","page_size":[595,841],"block_position":"6-2"}],[{"bbox":[518,47,539,58],"type":"header","angle":0,"index":-99999,"text":"目录 ","color":{"line":"rgba(164, 164, 164, 1)","fill":"rgba(164, 164, 164, 1)"},"page_idx":7,"is_discarded":true,"id":"6cd7eb08-5b88-43e2-a57c-39a4fc55ea09","page_size":[595,841],"block_position":"7-0"},{"bbox":[68,71,540,188],"type":"text","angle":0,"index":1,"text":"17.4 Gauss 公式 ..... 142\n17.4.1 Gauss 公式与基本使用 ..... 142\n17.4.2 内部有瑕点 ..... 143\n17.5 Stokes 公式 ..... 143\n17.5.1 Stokes 公式与基本使用 ..... 143\n17.5.2 空间第二型曲线积分与路径无关性 ..... 145\n18 章 实数完备性定理 ..... 146 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":7,"id":"bd7f6769-a824-466e-87ba-5f6c95ecfecb","page_size":[595,841],"block_position":"7-1"},{"bbox":[291,803,302,812],"type":"page_number","angle":0,"index":1000001,"text":"vii ","color":{"line":"rgba(164, 164, 164, 1)","fill":"rgba(164, 164, 164, 1)"},"page_idx":7,"is_discarded":true,"id":"c33b7117-1097-4e01-b56c-1047b1eaf009","page_size":[595,841],"block_position":"7-2"}],[{"bbox":[235,74,369,94],"type":"title","angle":0,"index":0,"level":2,"text":"## 第1章 数学基础","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":8,"id":"aa4c6051-c809-46ae-b6a3-39509500058b","page_size":[595,841],"block_position":"8-0"},{"bbox":[57,122,145,139],"type":"title","angle":0,"index":1,"level":2,"text":"## 1.1 三角函数","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":8,"id":"939436e7-1836-4580-beb9-ad54b86a987d","page_size":[595,841],"block_position":"8-1"},{"bbox":[55,155,319,170],"type":"text","angle":0,"index":2,"merge_prev":false,"text":"1.1.1 和差公式、万能公式、倍角公式、半角公式 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":8,"id":"416dbcf6-f226-4fba-81ab-6b8c5a6e6e20","page_size":[595,841],"block_position":"8-2"},{"bbox":[53,179,166,192],"type":"text","angle":0,"index":3,"merge_prev":false,"text":"一、 $\\sin x, \\cos x$ 和差公式 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":8,"id":"a162b3a2-c874-4ed9-832c-1235b19479e8","page_size":[595,841],"block_position":"8-3"},{"bbox":[215,214,378,228],"type":"interline_equation","angle":0,"index":4,"text":"$$\n\\sin (\\alpha + \\beta) = \\sin \\alpha \\cos \\beta + \\cos \\alpha \\sin \\beta\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":8,"id":"818986cc-aca2-4e11-a82e-dd4bf703bc99","page_size":[595,841],"block_position":"8-4"},{"bbox":[215,232,378,247],"type":"interline_equation","angle":0,"index":5,"text":"$$\n\\sin (\\alpha - \\beta) = \\sin \\alpha \\cos \\beta - \\cos \\alpha \\sin \\beta\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":8,"id":"bbddabd2-2214-422e-8f57-fffc2c6c5ef4","page_size":[595,841],"block_position":"8-5"},{"bbox":[215,251,379,265],"type":"interline_equation","angle":0,"index":6,"text":"$$\n\\cos (\\alpha + \\beta) = \\cos \\alpha \\cos \\beta - \\sin \\alpha \\sin \\beta\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":8,"id":"655d6804-2d8a-43ac-8ef4-744b5a50eff4","page_size":[595,841],"block_position":"8-6"},{"bbox":[215,269,379,284],"type":"interline_equation","angle":0,"index":7,"text":"$$\n\\cos (\\alpha - \\beta) = \\cos \\alpha \\cos \\beta + \\sin \\alpha \\sin \\beta\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":8,"id":"71b2af01-0793-4a3d-a8cd-d134815c78e0","page_size":[595,841],"block_position":"8-7"},{"bbox":[54,307,167,321],"type":"text","angle":0,"index":8,"merge_prev":false,"text":"二、 $\\tan x, \\cot x$ 和差公式 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":8,"id":"2aefdf99-36a8-4b67-97c4-1c3c80155052","page_size":[595,841],"block_position":"8-8"},{"bbox":[231,337,361,362],"type":"interline_equation","angle":0,"index":9,"text":"$$\n\\tan (\\alpha + \\beta) = \\frac {\\tan \\alpha + \\tan \\beta}{1 - \\tan \\alpha \\tan \\beta}\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":8,"id":"f20820ed-1aab-455b-ad17-abbc6e81817e","page_size":[595,841],"block_position":"8-9"},{"bbox":[232,364,361,388],"type":"interline_equation","angle":0,"index":10,"text":"$$\n\\tan (\\alpha - \\beta) = \\frac {\\tan \\alpha - \\tan \\beta}{1 + \\tan \\alpha \\tan \\beta}\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":8,"id":"38e9501e-97e5-4363-95b8-94c4966b6ee0","page_size":[595,841],"block_position":"8-10"},{"bbox":[233,390,359,414],"type":"interline_equation","angle":0,"index":11,"text":"$$\n\\cot (\\alpha + \\beta) = \\frac {\\cot \\alpha \\cot \\beta - 1}{\\cot \\beta + \\cot \\alpha}\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":8,"id":"f70ba7c2-a545-4adf-a86d-069f36c051e0","page_size":[595,841],"block_position":"8-11"},{"bbox":[233,417,359,441],"type":"interline_equation","angle":0,"index":12,"text":"$$\n\\cot (\\alpha - \\beta) = \\frac {\\cot \\alpha \\cot \\beta + 1}{\\cot \\beta - \\cot \\alpha}\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":8,"id":"5f72d955-0a90-48c0-a39a-04789af055bd","page_size":[595,841],"block_position":"8-12"},{"bbox":[52,443,119,456],"type":"text","angle":0,"index":13,"merge_prev":false,"text":"三、万能公式 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":8,"id":"6396c0c7-5c3a-4cca-9192-88599cee8bee","page_size":[595,841],"block_position":"8-13"},{"bbox":[253,472,339,499],"type":"interline_equation","angle":0,"index":14,"text":"$$\n\\sin \\alpha = \\frac {2 \\tan {\\frac {\\alpha}{2}}}{1 + \\tan^ {2} {\\frac {\\alpha}{2}}}\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":8,"id":"6fe5eaae-7fd2-43d2-b045-9ff0688ff6ec","page_size":[595,841],"block_position":"8-14"},{"bbox":[254,501,339,529],"type":"interline_equation","angle":0,"index":15,"text":"$$\n\\cos \\alpha = \\frac {1 - \\tan^ {2} {\\frac {\\alpha}{2}}}{1 + \\tan^ {2} {\\frac {\\alpha}{2}}}\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":8,"id":"ccb8e11d-9674-4473-b58e-a20193181eb5","page_size":[595,841],"block_position":"8-15"},{"bbox":[254,530,340,559],"type":"interline_equation","angle":0,"index":16,"text":"$$\n\\tan \\alpha = \\frac {2 \\tan {\\frac {\\alpha}{2}}}{1 - \\tan^ {2} {\\frac {\\alpha}{2}}}\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":8,"id":"0008947f-58d0-40a7-abd9-c8f2b4383a0d","page_size":[595,841],"block_position":"8-16"},{"bbox":[54,560,119,573],"type":"text","angle":0,"index":17,"merge_prev":false,"text":"四、倍角公式 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":8,"id":"29969906-bc01-446e-9642-e50cf15efb5f","page_size":[595,841],"block_position":"8-17"},{"bbox":[250,597,343,608],"type":"text","angle":0,"index":18,"merge_prev":false,"text":"$\\sin 2\\alpha = 2 \\sin \\alpha \\cos \\alpha$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":8,"id":"262921d2-6a93-4511-9b79-50ed8f7f0866","page_size":[595,841],"block_position":"8-18"},{"bbox":[180,613,413,627],"type":"interline_equation","angle":0,"index":19,"text":"$$\n\\cos 2 \\alpha = \\cos^ {2} \\alpha - \\sin^ {2} \\alpha = 2 \\cos^ {2} \\alpha - 1 = 1 - 2 \\sin^ {2} \\alpha\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":8,"id":"a78c31e0-28da-43a4-b39a-2f9a5ed261c1","page_size":[595,841],"block_position":"8-19"},{"bbox":[250,630,342,655],"type":"interline_equation","angle":0,"index":20,"text":"$$\n\\tan 2 \\alpha = \\frac {2 \\tan \\alpha}{1 - \\tan^ {2} \\alpha}\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":8,"id":"8b510404-0976-40b9-8c93-2ece887fb248","page_size":[595,841],"block_position":"8-20"},{"bbox":[55,656,119,669],"type":"text","angle":0,"index":21,"merge_prev":false,"text":"五、半角公式 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":8,"id":"fa5d0c8a-a92e-4ceb-9364-67a724ac7997","page_size":[595,841],"block_position":"8-21"}],[{"bbox":[480,47,539,58],"type":"header","angle":0,"index":-99999,"text":"1.1 三角函数 ","color":{"line":"rgba(164, 164, 164, 1)","fill":"rgba(164, 164, 164, 1)"},"page_idx":9,"is_discarded":true,"id":"a9594fba-eb56-4278-b4cc-42cfc8c02937","page_size":[595,841],"block_position":"9-0"},{"bbox":[193,86,293,112],"type":"interline_equation","angle":0,"index":1,"text":"$$\n\\sin {\\frac {\\alpha}{2}} = \\pm \\sqrt {\\frac {1 - \\cos \\alpha}{2}}\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":9,"id":"ce6f4337-878b-46de-8cd7-5190dd213b75","page_size":[595,841],"block_position":"9-1"},{"bbox":[193,115,293,141],"type":"interline_equation","angle":0,"index":2,"text":"$$\n\\cos {\\frac {\\alpha}{2}} = \\pm \\sqrt {\\frac {1 + \\cos \\alpha}{2}}\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":9,"id":"8cfa2d32-f83f-41b8-bfab-4c2ff7f9767f","page_size":[595,841],"block_position":"9-2"},{"bbox":[192,142,401,169],"type":"interline_equation","angle":0,"index":3,"text":"$$\n\\tan {\\frac {\\alpha}{2}} = \\frac {\\sin \\alpha}{1 + \\cos \\alpha} = \\frac {1 - \\cos \\alpha}{\\sin \\alpha} = \\pm \\sqrt {\\frac {1 - \\cos \\alpha}{1 + \\cos \\alpha}}\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":9,"id":"02aeeaa9-2c8e-45a1-9d64-5e2a63756ed3","page_size":[595,841],"block_position":"9-3"},{"bbox":[193,171,401,197],"type":"interline_equation","angle":0,"index":4,"text":"$$\n\\cot {\\frac {\\alpha}{2}} = \\frac {1 + \\cos \\alpha}{\\sin \\alpha} = \\frac {\\sin \\alpha}{1 - \\cos \\alpha} = \\pm \\sqrt {\\frac {1 + \\cos \\alpha}{1 - \\cos \\alpha}}\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":9,"id":"561f305c-aed0-48f8-a5fd-65de9014d365","page_size":[595,841],"block_position":"9-4"},{"bbox":[56,212,198,226],"type":"title","angle":0,"index":5,"level":2,"text":"## 1.1.2 和差化积与积化和差","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":9,"id":"bc3474ef-2287-424c-bcc2-49867f4aa5e7","page_size":[595,841],"block_position":"9-5"},{"bbox":[54,236,119,248],"type":"title","angle":0,"index":6,"level":2,"text":"## 一、和差化积","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":9,"id":"dc320a83-7a74-423f-b138-42d1ad0bfde4","page_size":[595,841],"block_position":"9-6"},{"bbox":[212,264,377,289],"type":"interline_equation","angle":0,"index":7,"text":"$$\n\\cos \\alpha + \\cos \\beta = 2 \\cos {\\frac {\\alpha + \\beta}{2}} \\cos {\\frac {\\alpha - \\beta}{2}}\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":9,"id":"189497e8-d109-4b21-a6d2-ae84d4a11925","page_size":[595,841],"block_position":"9-7"},{"bbox":[213,290,382,313],"type":"interline_equation","angle":0,"index":8,"text":"$$\n\\cos \\alpha - \\cos \\beta = - 2 \\sin {\\frac {\\alpha + \\beta}{2}} \\sin {\\frac {\\alpha - \\beta}{2}}\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":9,"id":"98b6d78a-9b3a-4ac0-b9de-0af44fbb3bb9","page_size":[595,841],"block_position":"9-8"},{"bbox":[212,314,374,338],"type":"interline_equation","angle":0,"index":9,"text":"$$\n\\sin \\alpha + \\sin \\beta = 2 \\sin {\\frac {\\alpha + \\beta}{2}} \\cos {\\frac {\\alpha - \\beta}{2}}\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":9,"id":"eb2d45fe-0fb5-4036-b2c0-8d5eec257fb9","page_size":[595,841],"block_position":"9-9"},{"bbox":[213,338,373,362],"type":"interline_equation","angle":0,"index":10,"text":"$$\n\\sin \\alpha - \\sin \\beta = 2 \\cos {\\frac {\\alpha + \\beta}{2}} \\sin {\\frac {\\alpha - \\beta}{2}}\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":9,"id":"1e549995-bc6e-4b6b-ac6c-2d9ef1a5cb12","page_size":[595,841],"block_position":"9-10"},{"bbox":[213,364,333,389],"type":"interline_equation","angle":0,"index":11,"text":"$$\n\\tan \\alpha + \\tan \\beta = \\frac {\\sin (\\alpha + \\beta)}{\\cos \\alpha \\cos \\beta}\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":9,"id":"6b5ce66f-eb40-4a9a-ae1c-32cd0e74b77a","page_size":[595,841],"block_position":"9-11"},{"bbox":[54,391,181,404],"type":"title","angle":0,"index":12,"level":2,"text":"## 二、积化和差：异 $\\sin$ 同cos","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":9,"id":"37d16fdc-d704-4904-acd6-770ceba1a8dc","page_size":[595,841],"block_position":"9-12"},{"bbox":[205,421,384,444],"type":"interline_equation","angle":0,"index":13,"text":"$$\n\\cos \\alpha \\cos \\beta = \\frac {1}{2} [ \\cos (\\alpha + \\beta) + \\cos (\\alpha - \\beta) ]\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":9,"id":"597fb927-44db-4c30-b84b-5cd0dbcc79db","page_size":[595,841],"block_position":"9-13"},{"bbox":[205,446,389,468],"type":"interline_equation","angle":0,"index":14,"text":"$$\n\\sin \\alpha \\sin \\beta = - \\frac {1}{2} [ \\cos (\\alpha + \\beta) - \\cos (\\alpha - \\beta) ]\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":9,"id":"adce76d4-e900-4d53-8ddb-3f3e90e13fef","page_size":[595,841],"block_position":"9-14"},{"bbox":[205,470,380,492],"type":"interline_equation","angle":0,"index":15,"text":"$$\n\\sin \\alpha \\cos \\beta = \\frac {1}{2} [ \\sin (\\alpha + \\beta) + \\sin (\\alpha - \\beta) ]\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":9,"id":"2b4cead4-c685-44b9-a0ec-70118a61b4dd","page_size":[595,841],"block_position":"9-15"},{"bbox":[205,494,380,516],"type":"interline_equation","angle":0,"index":16,"text":"$$\n\\cos \\alpha \\sin \\beta = \\frac {1}{2} [ \\sin (\\alpha + \\beta) - \\sin (\\alpha - \\beta) ]\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":9,"id":"ceaacbc8-454a-4c1b-8d29-7f604a748953","page_size":[595,841],"block_position":"9-16"},{"bbox":[57,532,163,545],"type":"title","angle":0,"index":17,"level":2,"text":"## 1.1.3 交替三角函数","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":9,"id":"27df96d2-bbd3-4d8a-8728-0d68c2087a37","page_size":[595,841],"block_position":"9-17"},{"bbox":[66,555,199,568],"type":"text","angle":0,"index":18,"merge_prev":false,"text":"1. $(-1)^{n}\\sin nx=\\sin(x+\\pi)n$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":9,"id":"1dec425d-986b-4bf6-9c3e-f755dc44cabc","page_size":[595,841],"block_position":"9-18"},{"bbox":[66,571,201,584],"type":"text","angle":0,"index":19,"merge_prev":false,"text":"2. $(-1)^{n}\\cos nx = \\cos (x + \\pi)n$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":9,"id":"824e3a06-567d-475f-87e4-c237a3bfb539","page_size":[595,841],"block_position":"9-19"},{"bbox":[57,602,186,615],"type":"title","angle":0,"index":20,"level":2,"text":"## 1.1.4 重要的三角恒等式","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":9,"id":"8e6865bc-32ae-4601-aef9-a8c8b6f837cf","page_size":[595,841],"block_position":"9-20"},{"bbox":[66,624,246,650],"type":"interline_equation","angle":0,"index":21,"text":"$$\n\\cos \\frac {x}{2 ^ {n}} \\cdot \\cos \\frac {x}{2 ^ {n - 1}} \\dots \\cos \\frac {x}{2} = \\frac {\\sin x}{2 ^ {n} \\sin \\frac {x}{2 ^ {n}}}\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":9,"id":"0cf6a4bc-2e94-49c0-9d51-779c32fc7a59","page_size":[595,841],"block_position":"9-21"},{"bbox":[66,650,301,676],"type":"interline_equation","angle":0,"index":22,"text":"$$\n2. \\sin x + \\sin 2 x + \\dots + \\sin n x = \\frac {\\cos \\frac {x}{2} - \\cos (n + \\frac {1}{2}) x}{2 \\sin \\frac {x}{2}}\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":9,"id":"1cc0fb3d-c3da-441d-87d6-06df6373fa9c","page_size":[595,841],"block_position":"9-22"},{"bbox":[66,675,387,704],"type":"interline_equation","angle":0,"index":23,"text":"$$\n3. \\cos x + \\cos 2 x + \\dots + \\cos n x = \\frac {\\sin (n + \\frac {1}{2}) x - \\sin \\frac {x}{2}}{2 \\sin \\frac {x}{2}} = \\frac {\\sin (n + \\frac {1}{2}) x}{2 \\sin \\frac {x}{2}} - \\frac {1}{2}\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":9,"id":"0509dc87-f292-4dc2-9648-020e75096e4c","page_size":[595,841],"block_position":"9-23"},{"bbox":[66,703,263,726],"type":"text","angle":0,"index":24,"merge_prev":false,"text":"4. $\\arctan \\frac{1}{2k^2} = \\arctan \\frac{k}{k + 1} - \\arctan \\frac{k - 1}{k}$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":9,"id":"3d2cff39-d33f-40ae-aa77-cb051beaf6a0","page_size":[595,841],"block_position":"9-24"},{"bbox":[54,726,271,738],"type":"text","angle":0,"index":25,"merge_prev":false,"text":"证明 (1) 将右侧分母乘到左侧去，用二倍角即可。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":9,"id":"28b1bcb1-2ead-49cb-8259-d74002660c6e","page_size":[595,841],"block_position":"9-25"},{"bbox":[293,802,301,811],"type":"page_number","angle":0,"index":1000025,"text":"2 ","color":{"line":"rgba(164, 164, 164, 1)","fill":"rgba(164, 164, 164, 1)"},"page_idx":9,"is_discarded":true,"id":"a4bb8b7f-3ca5-4a8a-be42-8fd3ccc1f1ca","page_size":[595,841],"block_position":"9-26"}],[{"bbox":[490,47,539,58],"type":"header","angle":0,"index":-99999,"text":"1.2 恒等式 ","color":{"line":"rgba(164, 164, 164, 1)","fill":"rgba(164, 164, 164, 1)"},"page_idx":10,"is_discarded":true,"id":"a9b176d0-4568-4453-b7b3-696f166969b7","page_size":[595,841],"block_position":"10-0"},{"bbox":[73,68,176,90],"type":"text","angle":0,"index":1,"merge_prev":false,"text":"(2) 左边乘 $\\sin \\frac{x}{2}$ , 得到 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":10,"id":"3c14eae5-9523-4e11-a43c-c5864476d9db","page_size":[595,841],"block_position":"10-1"},{"bbox":[98,93,493,123],"type":"interline_equation","angle":0,"index":2,"text":"$$\n\\sin {\\frac {x}{2}} I = \\sum_ {k = 1} ^ {n} \\sin k x \\sin {\\frac {x}{2}} = - \\frac {1}{2} \\sum_ {k = 1} ^ {n} [ \\cos (k + \\frac {1}{2}) x - \\cos (k - \\frac {1}{2}) x ] = \\frac {1}{2} \\left[ \\cos {\\frac {x}{2}} - \\cos (n + \\frac {1}{2}) x \\right]\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":10,"id":"f9c6412e-64e2-4bd1-93dd-15679906891e","page_size":[595,841],"block_position":"10-2"},{"bbox":[74,126,138,139],"type":"text","angle":0,"index":3,"merge_prev":false,"text":"(3) 和 (2) 类似 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":10,"id":"af21e66a-8c4f-4735-a80f-d7eaed860e46","page_size":[595,841],"block_position":"10-3"},{"bbox":[57,158,150,172],"type":"title","angle":0,"index":4,"level":2,"text":"## 1.1.5 反三角函数","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":10,"id":"9e8dcf88-75ab-4083-bced-c2af2dcd4fd4","page_size":[595,841],"block_position":"10-4"},{"bbox":[74,185,210,198],"type":"text","angle":0,"index":5,"merge_prev":false,"text":"定理 1.1 (反三角函数倒数关系) ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":10,"id":"7bdbe923-3c94-4b80-bafb-45762a11e405","page_size":[595,841],"block_position":"10-5"},{"bbox":[242,208,351,232],"type":"interline_equation","angle":0,"index":6,"text":"$$\n\\arctan x + \\arctan {\\frac {1}{x}} = \\frac {\\pi}{2}\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":10,"id":"9a042161-ec94-4dbd-a0bf-fe6e4db2a082","page_size":[595,841],"block_position":"10-6"},{"bbox":[246,232,346,256],"type":"interline_equation","angle":0,"index":7,"text":"$$\n\\operatorname{arccot} x + \\operatorname{arccot} \\frac {1}{x} = \\frac {\\pi}{2}\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":10,"id":"8944b57a-1e10-451a-bac2-fe3264dec7a3","page_size":[595,841],"block_position":"10-7"},{"bbox":[54,266,525,290],"type":"text","angle":0,"index":8,"merge_prev":false,"text":"证明 画一个直角三角形，角 $\\theta_{1}$ 对着边长 $1$ ， $\\theta_{2} = \\frac{\\pi}{2} -\\theta_{1}$ 对着边长 $x$ ， $\\tan \\theta_1 = \\frac{1}{x},\\tan \\theta_2 = x$ ，而两角互补。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":10,"id":"3c43841f-a933-4cad-b66d-54bae43f1819","page_size":[595,841],"block_position":"10-8"},{"bbox":[57,305,151,319],"type":"title","angle":0,"index":9,"level":2,"text":"## 1.1.6 三角不等式","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":10,"id":"03a4bb22-adee-4d0a-ab96-606add375d3b","page_size":[595,841],"block_position":"10-9"},{"bbox":[73,333,173,344],"type":"interline_equation","angle":0,"index":10,"text":"$$\n\\text { 定理 } 1. 2 (\\sin x \\text { 不等式 })\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":10,"id":"ea51e87a-3105-4af5-8884-870e5828548b","page_size":[595,841],"block_position":"10-10"},{"bbox":[68,349,238,372],"type":"interline_equation","angle":0,"index":11,"text":"$$\nx \\in (0, {\\frac {\\pi}{2}}) \\text { 时 }, {\\frac {2 x}{\\pi}} < \\sin x < x < \\tan x\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":10,"id":"68f67ea6-6e15-418f-8a92-7d0795a4efe8","page_size":[595,841],"block_position":"10-11"},{"bbox":[73,392,180,405],"type":"title","angle":0,"index":12,"level":2,"text":"## 定理 1.3 (反三角不等式)","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":10,"id":"6c9703e2-0e86-4c59-93f3-3b79e832ec53","page_size":[595,841],"block_position":"10-12"},{"bbox":[68,410,257,421],"type":"text","angle":0,"index":13,"merge_prev":false,"text":"$\\arcsin x > x > \\arctan x$ ，当 x 为小正数时。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":10,"id":"7fb4c58b-b14d-4c01-afc2-eb5ea1b721e8","page_size":[595,841],"block_position":"10-13"},{"bbox":[57,458,129,474],"type":"title","angle":0,"index":14,"level":2,"text":"## 1.2 恒等式","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":10,"id":"061d0845-0d43-4775-a8cd-2d39f37a2f4c","page_size":[595,841],"block_position":"10-14"},{"bbox":[57,491,186,505],"type":"title","angle":0,"index":15,"level":2,"text":"## 1.2.1 立方和、四次方和","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":10,"id":"62e933ff-b146-48cb-b367-b460b2afb51c","page_size":[595,841],"block_position":"10-15"},{"bbox":[73,518,170,530],"type":"text","angle":0,"index":16,"merge_prev":false,"text":"定理 1.4 (立方和公式) ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":10,"id":"7b5da6b8-a0f1-4080-b94f-18423455d212","page_size":[595,841],"block_position":"10-16"},{"bbox":[68,534,343,550],"type":"interline_equation","angle":0,"index":17,"text":"$$\na ^ {3} + b ^ {3} = (a + b) \\left(a ^ {2} - a b + b ^ {2}\\right), a ^ {3} - b ^ {3} = (a - b) \\left(a ^ {2} + a b + b ^ {2}\\right)\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":10,"id":"bac3bc71-4e3b-41e9-b8d4-4b9cb1ebd014","page_size":[595,841],"block_position":"10-17"},{"bbox":[73,570,190,582],"type":"title","angle":0,"index":18,"level":2,"text":"## 定理 1.5 (四次方和差公式)","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":10,"id":"353c6fdb-8002-4fd6-93a4-5f1141c640f9","page_size":[595,841],"block_position":"10-18"},{"bbox":[68,587,354,601],"type":"interline_equation","angle":0,"index":19,"text":"$$\na ^ {4} + b ^ {4} = (a ^ {2} + b ^ {2}) ^ {2} - 2 a ^ {2} b ^ {2} = (a ^ {2} + b ^ {2} - \\sqrt {2} a b) (a ^ {2} + b ^ {2} - \\sqrt {2} a b)\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":10,"id":"beafaf1c-3838-42db-a82e-9670438f7bcb","page_size":[595,841],"block_position":"10-19"},{"bbox":[57,634,174,647],"type":"title","angle":0,"index":20,"level":2,"text":"## 1.2.2 最大值、最小值","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":10,"id":"77004e56-e397-4aa1-843a-8b4e82be656e","page_size":[595,841],"block_position":"10-20"},{"bbox":[161,667,431,691],"type":"interline_equation","angle":0,"index":21,"text":"$$\n\\min \\{a, b \\} = \\frac {1}{2} (a + b - | a - b |), \\max \\{a, b \\} = \\frac {1}{2} (a + b + | a - b |)\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":10,"id":"fb32baa7-6695-4858-81c6-c07e505251d2","page_size":[595,841],"block_position":"10-21"},{"bbox":[293,803,300,812],"type":"page_number","angle":0,"index":1000021,"text":"3 ","color":{"line":"rgba(164, 164, 164, 1)","fill":"rgba(164, 164, 164, 1)"},"page_idx":10,"is_discarded":true,"id":"c8b1a66d-79d2-4835-b63e-68db7fc4776d","page_size":[595,841],"block_position":"10-22"}],[{"bbox":[490,46,539,58],"type":"header","angle":0,"index":-99999,"text":"1.3 不等式 ","color":{"line":"rgba(164, 164, 164, 1)","fill":"rgba(164, 164, 164, 1)"},"page_idx":11,"is_discarded":true,"id":"88285864-3eed-409c-b14c-940f6b8b9056","page_size":[595,841],"block_position":"11-0"},{"bbox":[57,72,127,86],"type":"title","angle":0,"index":1,"level":2,"text":"## 1.2.3 次方和","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":11,"id":"1331de42-8b2e-41d7-88aa-7820420ccb90","page_size":[595,841],"block_position":"11-1"},{"bbox":[202,105,380,152],"type":"interline_equation","angle":0,"index":2,"text":"$$\n\\left\\{ \\begin{array}{l} 1 + 2 + \\dots + n = \\frac {n (n - 1)}{2} \\\\ 1 ^ {2} + 2 ^ {2} + \\dots + n ^ {2} = \\frac {n (n + 1) (2 n + 1)}{6} \\end{array} \\right.\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":11,"id":"1fbd5cc6-5c32-4b00-945f-72da9b5c6366","page_size":[595,841],"block_position":"11-2"},{"bbox":[57,169,130,185],"type":"title","angle":0,"index":3,"level":2,"text":"## 1.3 不等式","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":11,"id":"01d81115-dfb2-4938-85bc-08ca8e43834d","page_size":[595,841],"block_position":"11-3"},{"bbox":[56,201,151,216],"type":"title","angle":0,"index":4,"level":2,"text":"## 1.3.1 基本不等式","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":11,"id":"e5498f01-d4e0-4da5-9e9e-c67a84069e72","page_size":[595,841],"block_position":"11-4"},{"bbox":[54,225,131,238],"type":"title","angle":0,"index":5,"level":2,"text":"## 一、均值不等式：","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":11,"id":"a9a08298-0bb0-4560-a5cd-fcde3843a7d3","page_size":[595,841],"block_position":"11-5"},{"bbox":[152,251,441,282],"type":"interline_equation","angle":0,"index":6,"text":"$$\n\\frac {n}{\\frac {1}{x _ {1}} + \\cdots + \\frac {1}{x _ {n}}} \\leq \\sqrt [ n ]{x _ {1} x _ {2} \\cdots x _ {n}} \\leq \\frac {x _ {1} + \\cdots + x _ {n}}{n} \\leq \\sqrt {\\frac {x _ {1} ^ {2} + \\cdots + x _ {n} ^ {2}}{n}}\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":11,"id":"c2fb4984-48da-458e-abdf-dd940532d938","page_size":[595,841],"block_position":"11-6"},{"bbox":[54,287,128,301],"type":"title","angle":0,"index":7,"level":2,"text":"## 二、对数不等式","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":11,"id":"cd6430cf-c408-485c-9db7-c00099a9c544","page_size":[595,841],"block_position":"11-7"},{"bbox":[248,316,348,338],"type":"interline_equation","angle":0,"index":8,"text":"$$\n\\frac {x}{1 + x} \\leq \\ln (1 + x) \\leq x\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":11,"id":"401e05fc-ea14-433d-9063-cf321d0bce80","page_size":[595,841],"block_position":"11-8"},{"bbox":[57,353,165,368],"type":"title","angle":0,"index":9,"level":2,"text":"## 1.3.2 Jensen 不等式","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":11,"id":"d5ad20cb-1288-4b8f-9925-c9d59674172b","page_size":[595,841],"block_position":"11-9"},{"bbox":[72,380,182,393],"type":"title","angle":0,"index":10,"level":2,"text":"## 定理 1.6 (Jensen 不等式)","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":11,"id":"5efb361c-51b5-4e9a-ab36-2ce06d08fff0","page_size":[595,841],"block_position":"11-10"},{"bbox":[69,399,225,429],"type":"text","angle":0,"index":11,"merge_prev":false,"text":"若 $\\sum_{i=1}^{n} t_i = 1$ , $f$ 是 (下) 凸函数, 则 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":11,"id":"e5dcc226-f36b-44f9-b834-67eddacaedf4","page_size":[595,841],"block_position":"11-11"},{"bbox":[241,428,351,460],"type":"interline_equation","angle":0,"index":12,"text":"$$\n\\sum_ {i = 1} ^ {n} t _ {i} f (x _ {i}) \\geq f (\\sum_ {i = 1} ^ {n} t _ {i} x _ {i})\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":11,"id":"efab6006-e0b5-439a-b329-354a51b99e75","page_size":[595,841],"block_position":"11-12"},{"bbox":[35,470,538,530],"type":"text","angle":0,"index":13,"merge_prev":false,"text":"练习 1.1Jensen 不等式的应用 (1)ZJU2021: 若 $\\sum_{i=1}^{n} p_{i} = 1, \\quad p_{i} \\in (0,1), \\quad x_{1}, \\cdots, x_{n} \\in R$ ，证明 $\\sum_{i=1}^{n} p_{i}(x_{i} - \\ln p_{i}) \\leq \\ln(\\sum_{i=1}^{n} e^{x_{i}})$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":11,"id":"b244bdfe-2f11-452f-b271-3d8dbdd93c50","page_size":[595,841],"block_position":"11-13"},{"bbox":[52,529,539,592],"type":"text","angle":0,"index":14,"merge_prev":false,"text":"证明 (1) 根据 $x_{i} = \\ln (e^{x_{i}})$ ，因此即证明 $\\sum_{i=1}^{n} p_{i} (\\ln \\frac{e^{x_{i}}}{p_{i}}) \\leq \\ln (\\sum_{i=1}^{n} e^{x_{i}})$ ，设 $t_{i} = \\frac{e^{x_{i}}}{p_{i}}$ ，不等式变为 $\\sum_{i=1}^{n} p_{i} \\ln (t_{i}) \\leq \\ln (\\sum_{i=1}^{n} t_{i} p_{i})$ ，由于 $\\ln x$ 是凹函数，因此得证。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":11,"id":"98d2d4ea-8150-4c74-b93c-3ab5c421ff15","page_size":[595,841],"block_position":"11-14"},{"bbox":[57,608,175,622],"type":"title","angle":0,"index":15,"level":2,"text":"## 1.3.3 调和级数不等式","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":11,"id":"b01cc72f-2c12-4749-a7f8-470c4a3b1e47","page_size":[595,841],"block_position":"11-15"},{"bbox":[73,635,191,648],"type":"title","angle":0,"index":16,"level":2,"text":"## 定理 1.7 (调和级数不等式)","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":11,"id":"9366b722-e783-4595-9491-ac9b2a750253","page_size":[595,841],"block_position":"11-16"},{"bbox":[66,650,473,680],"type":"text","angle":0,"index":17,"merge_prev":false,"text":"$\\ln(n+1)<1+\\frac{1}{2}+\\cdots+\\frac{1}{n}<1+\\ln n,$ 进一步地， $\\lim_{n\\to\\infty}\\frac{1+\\frac{1}{2}+\\cdots+\\frac{1}{n}}{\\ln n}=1$ (即 Euler 常数) ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":11,"id":"a7526b9b-bdd2-402f-8e9f-181e973e9cdc","page_size":[595,841],"block_position":"11-17"},{"bbox":[53,687,509,717],"type":"text","angle":0,"index":18,"merge_prev":false,"text":"证明 由于 $\\frac{x}{1 + x} \\leq \\ln (1 + x) \\leq x$ ，因此 $\\frac{\\frac{1}{n}}{1 + \\frac{1}{n}} < \\ln (1 + \\frac{1}{n}) < \\frac{1}{n}$ ，即 $\\frac{1}{n + 1} < \\ln \\frac{n + 1}{n} < \\frac{1}{n}$ ，求和得到： ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":11,"id":"9fd7c499-2f2c-480b-8a0b-e90b71f045c4","page_size":[595,841],"block_position":"11-18"},{"bbox":[176,719,417,744],"type":"interline_equation","angle":0,"index":20,"text":"$$\n1 + \\frac {1}{2} + \\dots + \\frac {1}{n} < 1 + \\ln \\frac {2}{1} + \\dots + \\ln \\frac {n}{n - 1} = 1 + \\ln n\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":11,"id":"6085e193-57fa-4472-aff4-9e006984366a","page_size":[595,841],"block_position":"11-19"},{"bbox":[54,745,107,757],"type":"text","angle":0,"index":21,"merge_prev":false,"text":"另一侧同理 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":11,"id":"70ff7e88-6f6b-4696-a911-c67f85ec8422","page_size":[595,841],"block_position":"11-20"},{"bbox":[293,803,301,811],"type":"page_number","angle":0,"index":1000018,"text":"4 ","color":{"line":"rgba(164, 164, 164, 1)","fill":"rgba(164, 164, 164, 1)"},"page_idx":11,"is_discarded":true,"id":"69d93be5-5352-4cc8-b7f8-cbcf82a12fa6","page_size":[595,841],"block_position":"11-21"}],[{"bbox":[470,47,539,58],"type":"header","angle":0,"index":-99999,"text":"1.4 几何学基础 ","color":{"line":"rgba(164, 164, 164, 1)","fill":"rgba(164, 164, 164, 1)"},"page_idx":12,"is_discarded":true,"id":"aa1fe2bc-8598-4cd3-adbe-9ed79deb28c7","page_size":[595,841],"block_position":"12-0"},{"bbox":[57,74,158,90],"type":"title","angle":0,"index":1,"level":2,"text":"## 1.4 几何学基础","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":12,"id":"e7e46e72-e63c-4965-942c-4bd2b78d175f","page_size":[595,841],"block_position":"12-1"},{"bbox":[57,107,186,121],"type":"title","angle":0,"index":2,"level":2,"text":"## 1.4.1 常见立体几何体积","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":12,"id":"e0ab2431-d211-4341-9d36-753053ef7014","page_size":[595,841],"block_position":"12-2"},{"bbox":[73,135,161,148],"type":"title","angle":0,"index":3,"level":2,"text":"## 定理 1.8 (椭球面积)","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":12,"id":"3ec5866f-cb65-4f11-b952-43d764dfe0b4","page_size":[595,841],"block_position":"12-3"},{"bbox":[68,151,236,179],"type":"text","angle":0,"index":4,"merge_prev":false,"text":"$ax^{2} + by^{2} + cz^{2}\\leq r^{2}$ 的体积为 $\\frac{4\\pi r^3}{3\\sqrt{abc}}$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":12,"id":"d6ed1d4d-4d02-4dc9-bce0-670fe23b159e","page_size":[595,841],"block_position":"12-4"},{"bbox":[55,190,360,211],"type":"text","angle":0,"index":5,"merge_prev":false,"text":"证明 做仿射变换 $x' = \\sqrt{ax}, y' = \\sqrt{by}, z' = \\sqrt{cz}$ ，而 $S' = \\frac{4}{3}\\pi r^3$ ，因此 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":12,"id":"3c8705e7-aacc-45ba-bd65-194c4d5a33a2","page_size":[595,841],"block_position":"12-5"},{"bbox":[251,215,341,242],"type":"interline_equation","angle":0,"index":6,"text":"$$\nS = \\frac {S ^ {\\prime}}{\\sqrt {a b c}} = \\frac {4 \\pi r ^ {3}}{3 \\sqrt {a b c}}\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":12,"id":"68a20667-26fc-4a97-8aa2-7fdbec337701","page_size":[595,841],"block_position":"12-6"},{"bbox":[294,803,300,811],"type":"page_number","angle":0,"index":1000006,"text":"5 ","color":{"line":"rgba(164, 164, 164, 1)","fill":"rgba(164, 164, 164, 1)"},"page_idx":12,"is_discarded":true,"id":"373f4c82-2d8a-4ded-bccd-797872da30e0","page_size":[595,841],"block_position":"12-7"}],[{"bbox":[236,74,368,94],"type":"title","angle":0,"index":0,"level":1,"text":"# 第 2 章 数列极限","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":13,"id":"32740ac5-4220-4cc1-a690-e987bee9d30a","page_size":[595,841],"block_position":"13-0"},{"bbox":[56,122,229,139],"type":"title","angle":0,"index":1,"level":2,"text":"## 2.1 数列极限的概念与性质","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":13,"id":"94af297c-86cd-426e-8d9d-873096fcd9cc","page_size":[595,841],"block_position":"13-1"},{"bbox":[56,155,174,169],"type":"title","angle":0,"index":2,"level":2,"text":"## 2.1.1 数列极限的概念","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":13,"id":"7ea8ae85-0351-410f-a8fd-d07c9a88577f","page_size":[595,841],"block_position":"13-2"},{"bbox":[73,183,161,195],"type":"title","angle":0,"index":3,"level":2,"text":"## 定义 2.1 (数列极限)","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":13,"id":"16cf078f-e13f-4cc8-a1be-6759b3e179ef","page_size":[595,841],"block_position":"13-3"},{"bbox":[67,198,491,218],"type":"text","angle":0,"index":4,"merge_prev":false,"text":"$a_{n}$ 是实数列， $A\\in \\mathbb{R}$ ，若 $\\forall \\epsilon ,\\exists N,\\forall n > N$ 有 $|a_n - A| < \\epsilon$ ，则称 $a_{n}$ 收敛到 $A$ ，记作 $\\lim_{n\\to \\infty}a_n = A$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":13,"id":"8119b09e-94d0-40ca-b955-1b1e9cf62210","page_size":[595,841],"block_position":"13-4"},{"bbox":[73,238,191,250],"type":"title","angle":0,"index":5,"level":2,"text":"## 定理 2.1 (极限的四则运算)","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":13,"id":"f88b2a59-79fa-4e4e-84b7-8c914184bcc8","page_size":[595,841],"block_position":"13-5"},{"bbox":[68,254,525,284],"type":"text","angle":0,"index":6,"merge_prev":false,"text":"若 $\\lim_{n\\to \\infty}a_n,\\lim_{n\\to \\infty}b_n$ 都存在(不存在时不能拆！)，则 $\\lim_{n\\to \\infty}(a_n\\odot b_n) = \\lim_{n\\to \\infty}a_n\\odot \\lim_{n\\to \\infty}b_n$ ，这里 $\\odot$ 指任意运算(除法额外要求 $\\lim_{n\\to \\infty}b_n\\neq 0)$ 。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":13,"id":"d6483bf3-728c-47f6-be13-3aec2bde3a34","page_size":[595,841],"block_position":"13-6"},{"bbox":[73,309,192,322],"type":"title","angle":0,"index":7,"level":2,"text":"## 定理 2.2 (cauchy 收敛准则)","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":13,"id":"1f5c2c9c-75ac-4a27-a08d-75a58e979ab9","page_size":[595,841],"block_position":"13-7"},{"bbox":[68,326,323,339],"type":"text","angle":0,"index":8,"merge_prev":false,"text":"数列 $a_{n}$ 收敛当且仅当 $\\forall \\epsilon, \\exists N, \\forall m, n > N$ 有 $|a_{n} - a_{m}| < \\epsilon$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":13,"id":"3bcb74c2-dffe-4c3c-a805-7b4ee129f70f","page_size":[595,841],"block_position":"13-8"},{"bbox":[36,356,238,368],"type":"text","angle":0,"index":9,"merge_prev":false,"text":"练习 2.1 几个经典极限 (1) 证明 $\\sin n$ 发散 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":13,"id":"8de7c0e4-325d-44a1-9b0e-c1a1899bde82","page_size":[595,841],"block_position":"13-9"},{"bbox":[73,369,230,389],"type":"text","angle":0,"index":10,"merge_prev":false,"text":"(2) 证明 $a_{n} = 1 + \\frac{1}{2} + \\cdots + \\frac{1}{n}$ 发散 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":13,"id":"8e9d3add-d598-4e6d-a579-941d57ec0f1e","page_size":[595,841],"block_position":"13-10"},{"bbox":[53,388,539,468],"type":"text","angle":0,"index":11,"merge_prev":false,"text":"证明 (1) 显然区间 $\\left[2k\\pi+\\frac{\\pi}{4},2k\\pi+\\frac{3\\pi}{4}\\right]$ 长度大于 1，一定包含一个整数，可以取 $n\\in\\left[2k\\pi+\\frac{\\pi}{4},2k\\pi+\\frac{3\\pi}{4}\\right]$ ，取 $m\\in\\left[2k\\pi+\\pi,2k\\pi+2\\pi\\right]$ ，则 n,m 可取无穷大， $\\left|\\sin n-\\sin m\\right|\\geq\\frac{\\sqrt{2}}{2}$ ，因此根据 Cauchy 收敛准则知道发散。\n(2) $\\forall N$ ，取n=N,m=2N，则 $\\left|a_{n}-a_{m}\\right|=\\frac{1}{N+1}+\\cdots+\\frac{1}{2N}\\geq\\frac{N}{2N}=\\frac{1}{2}$ ，因此根据Cauchy收敛准则逆否命题可知发散 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":13,"id":"756a7e5d-04f8-48c3-8266-f54a69375c66","page_size":[595,841],"block_position":"13-11"},{"bbox":[56,486,174,500],"type":"title","angle":0,"index":12,"level":2,"text":"## 2.1.2 收敛数列的性质","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":13,"id":"a462f863-1bca-4a52-856b-00f2f6b68317","page_size":[595,841],"block_position":"13-12"},{"bbox":[74,513,191,525],"type":"title","angle":0,"index":13,"level":2,"text":"## 定理 2.3 (收敛数列的性质)","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":13,"id":"15c12842-de76-4de0-920e-2e28339be9a7","page_size":[595,841],"block_position":"13-13"},{"bbox":[69,530,158,544],"type":"text","angle":0,"index":14,"merge_prev":false,"text":"若 $\\lim_{n\\to \\infty}a_n = A$ ，则 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":13,"id":"e3839690-4ed3-4cf7-8217-3fa0cc55c4ae","page_size":[595,841],"block_position":"13-14"},{"bbox":[83,545,168,557],"type":"text","angle":0,"index":15,"merge_prev":false,"text":"- 有界性： $a_{n}$ 有界 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":13,"id":"fb028f23-31a3-49cc-b3e0-b2add9690059","page_size":[595,841],"block_position":"13-15"},{"bbox":[83,561,273,573],"type":"text","angle":0,"index":16,"merge_prev":false,"text":"- 保号性：若 $A > a$ ，则 $n \\to \\infty$ 时 $a_{n} > a$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":13,"id":"acae397e-0c87-4191-8283-fac3d33c8e96","page_size":[595,841],"block_position":"13-16"},{"bbox":[83,576,263,588],"type":"text","angle":0,"index":17,"merge_prev":false,"text":"- 收敛当且仅当 $a_{n}$ 所有子列都收敛到 $A$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":13,"id":"457c03ec-fc33-4c35-9132-19f38c26c697","page_size":[595,841],"block_position":"13-17"},{"bbox":[74,609,180,621],"type":"title","angle":0,"index":18,"level":2,"text":"## 定理 2.4 (奇偶子列收敛)","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":13,"id":"16577187-c545-49da-9388-5669568c117f","page_size":[595,841],"block_position":"13-18"},{"bbox":[68,626,255,638],"type":"text","angle":0,"index":19,"merge_prev":false,"text":"$a_{n}$ 收敛到 $A$ 当且仅当 $a_{2k}, a_{2k + 1}$ 收敛到 $A$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":13,"id":"8f4a65ac-8c40-4173-91b4-bde2b84ea3a7","page_size":[595,841],"block_position":"13-19"},{"bbox":[54,655,539,685],"type":"text","angle":0,"index":20,"merge_prev":false,"text":"证明 只右推左：根据偶数列可知 $\\forall \\epsilon, \\exists K, \\forall 2k > K$ 满足 $|a_{2k} - A| < \\epsilon$ ，而对上面的 $\\epsilon, \\exists K', \\forall 2k + 1 > k'$ 满足 $|a_{2k+1} - A| < \\epsilon$ ，因此取 $N = \\max\\{K, K'\\}$ 即可。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":13,"id":"25964f52-b7fb-41d1-b308-99cd63ca6fd4","page_size":[595,841],"block_position":"13-20"}],[{"bbox":[439,46,539,58],"type":"header","angle":0,"index":-99999,"text":"2.2 一些重要极限结论 ","color":{"line":"rgba(164, 164, 164, 1)","fill":"rgba(164, 164, 164, 1)"},"page_idx":14,"is_discarded":true,"id":"75db396a-1480-4e74-9ce0-bd9d657277c7","page_size":[595,841],"block_position":"14-0"},{"bbox":[56,74,201,90],"type":"title","angle":0,"index":1,"level":2,"text":"## 2.2 一些重要极限结论","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":14,"id":"a3497b01-5b05-45a2-be70-d31bad4a814d","page_size":[595,841],"block_position":"14-1"},{"bbox":[56,106,163,121],"type":"title","angle":0,"index":2,"level":2,"text":"## 2.2.1 几个基本极限","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":14,"id":"ce87f59c-d765-488e-9d2e-1cc109128a67","page_size":[595,841],"block_position":"14-2"},{"bbox":[73,135,180,148],"type":"title","angle":0,"index":3,"level":2,"text":"## 定理 2.5 (常用基本极限)","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":14,"id":"2243a8bb-12d4-4e84-a7e7-20b33d5d24dd","page_size":[595,841],"block_position":"14-3"},{"bbox":[69,151,408,163],"type":"text","angle":0,"index":4,"merge_prev":false,"text":"设 $a > 0, b > 1$ ，则基本极限的关系为： $\\log n < n^a < b^n < n! < n^n$ ，更具体地 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":14,"id":"53f69744-3fb9-40ad-ab16-c2d45752defe","page_size":[595,841],"block_position":"14-4"},{"bbox":[83,166,374,184],"type":"text","angle":0,"index":5,"merge_prev":false,"text":"- 根式：(1) $a > 0$ ， $\\lim_{n\\to \\infty}\\sqrt[n]{a} = 1(2)\\lim_{n\\to \\infty}\\sqrt[n]{n} = 1(3)\\lim_{n\\to \\infty}\\sqrt[n]{n!} = +\\infty$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":14,"id":"e1e9eae4-1ddf-4c72-9ec9-af067d827a07","page_size":[595,841],"block_position":"14-5"},{"bbox":[83,185,286,208],"type":"text","angle":0,"index":6,"merge_prev":false,"text":"- 比式：(1) $\\lim_{n\\to \\infty}\\frac{a^n}{n!} = 0(2)a > 1,\\lim_{n\\to \\infty}\\frac{n^k}{a^n} = 0$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":14,"id":"d9d4e0e2-c72d-4483-9f42-4fe74734ac4f","page_size":[595,841],"block_position":"14-6"},{"bbox":[83,209,310,230],"type":"text","angle":0,"index":7,"merge_prev":false,"text":"- 对数：(1) $\\forall k, \\lim_{x \\to \\infty} \\frac{\\ln x}{x^k} = 0$ (2) $\\forall k, \\lim_{x \\to 0^+} x^k \\ln x = 0$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":14,"id":"4a69a392-336f-495b-aa9a-d11f790c2eb4","page_size":[595,841],"block_position":"14-7"},{"bbox":[83,231,203,253],"type":"text","angle":0,"index":8,"merge_prev":false,"text":"- 指数： $\\forall k, \\lim_{x\\to +\\infty}\\frac{x^k}{e^x} = 0$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":14,"id":"f76e4119-8ec3-4783-a40a-e9a1b1edb352","page_size":[595,841],"block_position":"14-8"},{"bbox":[54,269,523,329],"type":"text","angle":0,"index":9,"merge_prev":false,"text":"证明 (1) 根式第一个设 $a_{n} = \\sqrt[n]{a} = 1 + h_{n}$ ，得到 $a = (1 + h_{n})^{n} \\geq 1 + nh_{n}$ ，故 $h_{n} \\leq \\frac{a - 1}{n} \\to 0$ 。第二个设 $a_{n} = \\sqrt[n]{n} = 1 + h_{n}$ ，则 $n = (1 + h_{n})^{n} \\geq \\frac{n(n - 1)}{2} h_{n}^{2}$ ，得到 $0 < h_{n} < \\frac{2}{n - 1}$ ，夹逼准则即可。第三个用Stirling公式，等价于 $\\frac{n}{e}$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":14,"id":"085f363f-8d38-4688-8dc6-2dc79981f939","page_size":[595,841],"block_position":"14-9"},{"bbox":[74,339,161,351],"type":"title","angle":0,"index":10,"level":2,"text":"## 定理 2.6 (重要极限)","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":14,"id":"e76180f9-95cf-41c5-af59-3b7fe623f886","page_size":[595,841],"block_position":"14-10"},{"bbox":[69,355,386,380],"type":"interline_equation","angle":0,"index":11,"text":"$$\n\\lim _ {n \\to \\infty} (1 + \\frac {1}{n}) ^ {n} = e, \\lim _ {n \\to \\infty} (1 - \\frac {1}{n}) ^ {n} = \\frac {1}{e} 。 \\lim _ {n \\to \\infty} (1 + \\frac {1}{n}) ^ {n} = e - \\frac {e}{2 n} + o (\\frac {1}{n})\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":14,"id":"5e6f1a83-d403-4975-8143-f74b26d4837b","page_size":[595,841],"block_position":"14-11"},{"bbox":[54,389,321,414],"type":"text","angle":0,"index":12,"merge_prev":false,"text":"证明 $\\lim_{n\\to \\infty}\\left(1 + \\frac{1}{n}\\right)^n = \\lim_{x\\to 0}(1 + x)^{\\frac{1}{x}} = \\lim_{x\\to 0}e^{\\frac{\\ln(1 + x)}{x}} = e^1 = e$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":14,"id":"7368ff41-92e2-4724-8c8e-8efa5127f3a8","page_size":[595,841],"block_position":"14-12"},{"bbox":[36,419,268,445],"type":"text","angle":0,"index":13,"merge_prev":false,"text":"练习2.2相关练习(1)计算 $\\lim_{n\\to \\infty}\\left(1 + \\frac{1}{n} +\\frac{1}{n^2}\\right)^n$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":14,"id":"a8d32e22-de49-4373-83d2-31daed712c65","page_size":[595,841],"block_position":"14-13"},{"bbox":[53,445,226,463],"type":"text","angle":0,"index":14,"merge_prev":false,"text":"解(1)等价于 $e^{\\frac{\\ln(1 + x + x^2)}{x}} = e^{\\frac{2x + 1}{1 + x + x^2}} = e$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":14,"id":"a68783b3-a168-4317-bceb-5976fa22b80b","page_size":[595,841],"block_position":"14-14"},{"bbox":[56,481,230,496],"type":"title","angle":0,"index":15,"level":2,"text":"## 2.2.2 Stirling 公式与 Wallis 公式","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":14,"id":"89694f23-69ac-4d1b-bf3a-a9204587bef4","page_size":[595,841],"block_position":"14-15"},{"bbox":[73,508,176,521],"type":"title","angle":0,"index":16,"level":2,"text":"## 定理 2.7 (Stirling 公式)","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":14,"id":"454a769c-3bfc-45c5-8b6f-ed0c14f865b5","page_size":[595,841],"block_position":"14-16"},{"bbox":[69,524,299,548],"type":"text","angle":0,"index":17,"merge_prev":false,"text":"当 $n\\to \\infty$ 时， $n!\\approx \\sqrt{2\\pi n} (\\frac{n}{e})^n$ ，进一步地 $\\sqrt[n]{n!}\\sim \\frac{n}{e}$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":14,"id":"c3c28fb4-1476-4db3-9327-49d082bfef6f","page_size":[595,841],"block_position":"14-17"},{"bbox":[73,567,170,579],"type":"title","angle":0,"index":18,"level":2,"text":"## 定理 2.8 (Wallis 公式)","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":14,"id":"b7a53ae0-e9db-418b-b4df-e4d089c4254a","page_size":[595,841],"block_position":"14-18"},{"bbox":[69,583,406,610],"type":"text","angle":0,"index":20,"merge_prev":false,"text":"当 $n\\to \\infty$ 时， $\\frac{(2n)!!}{(2n - 1)!!}\\sim \\sqrt{n\\pi}$ ，更具体地， $\\lim_{n\\to \\infty}\\left[\\frac{(2n)!!}{(2n - 1)!!}\\right]\\frac{1}{2n + 1} = \\frac{\\pi}{2}$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":14,"id":"0e5575ca-b4e7-4c15-a6c2-1c2599cabf46","page_size":[595,841],"block_position":"14-19"},{"bbox":[53,621,539,660],"type":"text","angle":0,"index":21,"merge_prev":false,"text":"证明 根据 $\\sin^{2n+1}x < \\sin^{2n}x < \\sin^{2n-1}x$ 得到 $\\int_{0}^{\\frac{\\pi}{2}}\\sin^{2n+1}xdx < \\int_{0}^{\\frac{\\pi}{2}}\\sin^{2n}xdx < \\int_{0}^{\\frac{\\pi}{2}}\\sin^{2n-1}xdx$ ，根据 Wallis 积分公式下面不等式，并构造 $A_{n}, B_{n}$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":14,"id":"2c370e87-4ea3-4fd7-8ded-6e97221b51de","page_size":[595,841],"block_position":"14-20"},{"bbox":[76,662,517,691],"type":"interline_equation","angle":0,"index":22,"text":"$$\n\\frac {(2 n) ! !}{(2 n + 1) ! !} < \\frac {(2 n - 1) ! !}{(2 n) ! !} \\frac {\\pi}{2} < \\frac {(2 n - 2) ! !}{(2 n - 1) ! !} \\Rightarrow A _ {n} := \\left[ \\frac {(2 n) ! !}{(2 n - 1) ! !} \\right] ^ {2} \\frac {1}{2 n + 1} < \\frac {\\pi}{2} < \\left[ \\frac {(2 n) ! !}{(2 n - 1) ! !} \\right] ^ {2} \\frac {1}{2 n} := B _ {n}\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":14,"id":"d20a1cf3-8390-4c77-9cdb-93e12254c7eb","page_size":[595,841],"block_position":"14-21"},{"bbox":[54,694,482,721],"type":"text","angle":0,"index":23,"merge_prev":false,"text":"此时 $B_{n} - A_{n} = \\left[\\frac{(2n)!!}{(2n - 1)!!}\\right]^{2}\\frac{1}{2n(2n - 1)} = A_{n}\\cdot \\frac{1}{2n} < \\frac{\\pi}{2}\\frac{1}{2n}\\rightarrow 0$ ，因此 $\\lim_{n\\to \\infty}A_n = \\lim_{n\\to \\infty}B_n = \\frac{\\pi}{2}$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":14,"id":"409405bd-e77e-491d-8571-5b0218365bec","page_size":[595,841],"block_position":"14-22"},{"bbox":[293,802,300,811],"type":"page_number","angle":0,"index":1000018,"text":"7 ","color":{"line":"rgba(164, 164, 164, 1)","fill":"rgba(164, 164, 164, 1)"},"page_idx":14,"is_discarded":true,"id":"d96d7d9b-453d-402b-8bcd-3aecc51be79c","page_size":[595,841],"block_position":"14-23"}],[{"bbox":[479,46,539,58],"type":"header","angle":0,"index":-99999,"text":"2.3夹逼准则 ","color":{"line":"rgba(164, 164, 164, 1)","fill":"rgba(164, 164, 164, 1)"},"page_idx":15,"is_discarded":true,"id":"acdecb60-ab5f-43bd-bcac-b14fe6e9c6d5","page_size":[595,841],"block_position":"15-0"},{"bbox":[56,73,146,86],"type":"title","angle":0,"index":1,"level":2,"text":"## 2.2.3 Euler 常数","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":15,"id":"0e6507c3-ffe5-4049-8318-4354b892d31c","page_size":[595,841],"block_position":"15-1"},{"bbox":[73,100,167,112],"type":"title","angle":0,"index":2,"level":2,"text":"## 定理 2.9 (Euler 常数)","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":15,"id":"48ac3e73-b0e3-4758-b2cb-fb683140e66d","page_size":[595,841],"block_position":"15-2"},{"bbox":[68,116,348,149],"type":"text","angle":0,"index":3,"merge_prev":false,"text":"极限 $\\lim_{n\\to \\infty}\\left[\\left(\\sum_{k = 1}^{n}\\frac{1}{k}\\right) - \\ln n\\right]$ 存在且为常数 $\\gamma$ ，记作Euler常数。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":15,"id":"cad8420d-3227-4109-b2db-93120fdc4953","page_size":[595,841],"block_position":"15-3"},{"bbox":[73,170,198,183],"type":"title","angle":0,"index":4,"level":2,"text":"## 推论 2.1 (Euler 常数的应用)","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":15,"id":"58ff9b23-8ccb-4dbc-af11-c6808db76db6","page_size":[595,841],"block_position":"15-4"},{"bbox":[69,188,269,220],"type":"text","angle":0,"index":5,"merge_prev":false,"text":"当 $n\\to \\infty$ 时， $\\sum_{k = 1}^{n}\\frac{1}{k} = \\ln n + \\gamma = \\ln n + o(n)$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":15,"id":"bf8fb267-1cac-4f6f-aef2-f04212ffeb59","page_size":[595,841],"block_position":"15-5"},{"bbox":[56,256,143,272],"type":"title","angle":0,"index":6,"level":2,"text":"## 2.3 夹逼准则","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":15,"id":"bd580796-2305-46ce-b76e-bdfdde132967","page_size":[595,841],"block_position":"15-6"},{"bbox":[73,290,166,303],"type":"title","angle":0,"index":7,"level":2,"text":"## 定理 2.10 (夹逼准则)","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":15,"id":"903d928c-9ae9-499e-842a-6058f46e426c","page_size":[595,841],"block_position":"15-7"},{"bbox":[66,305,536,331],"type":"text","angle":0,"index":8,"merge_prev":false,"text":"$a_{n}, b_{n}, c_{n}$ 为三个序列，若 $\\exists N$ 使得 $n \\geq N$ 时 $a_{n} \\leq b_{n} \\leq c_{n}$ ，且 $\\lim_{n \\to \\infty} a_{n} = \\lim_{n \\to \\infty} c_{n} = A$ ，则 $\\lim_{n \\to \\infty} b_{n} = A$ 。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":15,"id":"b84d082f-77ac-4566-9bd6-0aff81ec9efb","page_size":[595,841],"block_position":"15-8"},{"bbox":[35,354,539,432],"type":"text","angle":0,"index":9,"merge_prev":false,"text":"练习2.3两道经典夹逼准则求极限(1) $\\lim_{n\\to \\infty}\\frac{1}{2}\\cdot \\frac{3}{4}\\dots \\frac{2n - 1}{2n}$ (2) $\\lim_{n\\to \\infty}\\sum_{k = n^2}^{(n + 1)^2}\\frac{1}{\\sqrt{k}}$ 解(1)可用Wallis公式，或者夹逼准则。记 $a_{n} = \\frac{1}{2}\\cdot \\frac{3}{4}\\dots \\frac{2n - 1}{2n},b_{n} = \\frac{2}{3}\\cdot \\frac{4}{5}\\dots \\frac{2n}{2n + 1}$ ，则 $a_{n} < b_{n},a_{n}b_{n} = \\frac{1}{2n + 1}$ 得到 $a_n^2 < a_nb_n = \\frac{1}{2n + 1}\\rightarrow 0$ ，因此 $\\lim_{n\\to \\infty}a_n = 0$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":15,"id":"ac5ededf-c99d-4fc7-a7bf-26dcfe06098f","page_size":[595,841],"block_position":"15-9"},{"bbox":[73,432,465,464],"type":"text","angle":0,"index":10,"merge_prev":false,"text":"(2) 用 $\\frac{1}{n + 1} \\leq \\frac{1}{\\sqrt{k}} \\leq \\frac{1}{n}$ ，从而求和后得到 $\\frac{2n + 2}{n + 1} \\leq \\sum_{k=n^2}^{(n+1)^2} \\frac{1}{\\sqrt{k}} \\leq \\frac{2n + 2}{n}$ ，取极限得到 2。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":15,"id":"1cf2cc79-3403-482c-b0b2-6256ad0c20c9","page_size":[595,841],"block_position":"15-10"},{"bbox":[35,466,405,492],"type":"text","angle":0,"index":11,"merge_prev":false,"text":"练习2.4整体放缩or局部放缩(1) $\\lim_{n\\to \\infty}\\left(\\frac{1}{\\sqrt{n^2 + 1}} +\\frac{1}{\\sqrt{n^2 + 2}} +\\dots +\\frac{1}{\\sqrt{n^2 + n}}\\right)$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":15,"id":"cce1e276-f55a-4ef2-942b-5519297735fd","page_size":[595,841],"block_position":"15-11"},{"bbox":[52,493,538,534],"type":"text","angle":0,"index":12,"merge_prev":false,"text":"解(1)如果是 $\\frac{1}{\\sqrt{n^2 + i^2}}$ 形式则是用定积分，这里看上去就是 $\\sum_{k=1}^{n} \\frac{1}{n} = 1$ ，因此放缩为 $\\frac{n}{\\sqrt{n^2 + n}} \\leq I \\leq \\frac{n}{\\sqrt{n^2}}$ 即可知结果为 1。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":15,"id":"c1e8bc51-44d4-47d8-a471-de5c11685e16","page_size":[595,841],"block_position":"15-12"},{"bbox":[56,557,158,572],"type":"title","angle":0,"index":13,"level":2,"text":"## 2.4 平均值定理","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":15,"id":"6a47ba98-b106-413b-a655-2f26a2d6bac7","page_size":[595,841],"block_position":"15-13"},{"bbox":[56,590,198,603],"type":"title","angle":0,"index":14,"level":2,"text":"## 2.4.1 有限项幂次根号平均","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":15,"id":"56c915ce-d2d3-4da6-8cd4-06092bdcb9a5","page_size":[595,841],"block_position":"15-14"},{"bbox":[73,617,166,629],"type":"title","angle":0,"index":15,"level":2,"text":"## 定理 2.11 (根号平均)","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":15,"id":"ae532b13-d230-4a47-b8bf-ee1137864698","page_size":[595,841],"block_position":"15-15"},{"bbox":[67,633,536,668],"type":"text","angle":0,"index":17,"merge_prev":false,"text":"$a_{1},a_{2},\\cdots,a_{m}$ 是 m 个正数，则 $\\lim_{n\\to\\infty}\\sqrt[n]{a_{1}^{n}+\\cdots+a_{m}^{n}}=\\max\\{a_{1},\\cdots,a_{m}\\}$ (注意这里不是 $\\sqrt[n]{a_{1}^{n}+\\cdots+a_{n}^{n}}$ ，与平均值定理不同) ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":15,"id":"b8ae433f-7d3a-4be6-bf81-6e37792f1374","page_size":[595,841],"block_position":"15-16"},{"bbox":[52,679,537,708],"type":"text","angle":0,"index":18,"merge_prev":false,"text":"证明 设 $A = \\max\\{a_{1}, a_{2}, \\cdots, a_{m}\\}$ ，因此 $\\sqrt[n]{A^{n}} \\leq \\sqrt[n]{a_{1}^{n} + a_{2}^{n} + \\cdots + a_{m}^{n}} \\leq \\sqrt[n]{mA^{n}}$ ，两侧都趋于 A，因此极限为 A。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":15,"id":"3859b996-9e89-48f8-90c7-119370eae896","page_size":[595,841],"block_position":"15-17"},{"bbox":[35,726,417,740],"type":"text","angle":0,"index":19,"merge_prev":false,"text":"练习 2.5 根号平均的两个经典推广 (1) 正数列 $a_{n}$ 收敛到 a > 0，证明 $\\lim_{n\\to\\infty}\\sqrt[n]{a_{n}} = 1$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":15,"id":"089f5bbc-36ba-4b8e-9a4a-b467746942db","page_size":[595,841],"block_position":"15-18"},{"bbox":[73,743,398,764],"type":"text","angle":0,"index":20,"merge_prev":false,"text":"(2) $a_{n}$ 非负且有界 (可能是可数个)，则 $\\lim_{n\\to\\infty}\\sqrt[n]{a_{1}^{n}+a_{2}^{n}+\\cdots+a_{n}^{n}}=\\sup_{n\\geq1}a_{n}$ 证明 (1) 根据保号性， $\\exists N, n > N$ 时 $\\frac{1}{2} a \\leq a_n \\leq \\frac{3}{2} a$ ，因此 $n > N$ 时 $\\sqrt[n]{\\frac{1}{2} a} \\leq \\sqrt[n]{a_n} \\leq \\sqrt[n]{\\frac{3}{2} a}$ ，由夹逼准则得到结果为 1。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":15,"id":"f9fba292-8dc0-417d-a2e8-54ae0ce54070","page_size":[595,841],"block_position":"15-19"},{"bbox":[293,803,300,812],"type":"page_number","angle":0,"index":1000015,"text":"8 ","color":{"line":"rgba(164, 164, 164, 1)","fill":"rgba(164, 164, 164, 1)"},"page_idx":15,"is_discarded":true,"id":"262e824b-42d8-4ee5-9d52-1f5303073d88","page_size":[595,841],"block_position":"15-20"}],[{"bbox":[469,47,539,58],"type":"header","angle":0,"index":-99999,"text":"2.4 平均值定理 ","color":{"line":"rgba(164, 164, 164, 1)","fill":"rgba(164, 164, 164, 1)"},"page_idx":16,"is_discarded":true,"id":"b63723ae-76a0-4f1d-b8c2-f27acfbceb09","page_size":[595,841],"block_position":"16-0"},{"bbox":[53,69,539,106],"type":"text","angle":0,"index":1,"merge_prev":false,"lines_deleted":true,"text":"","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":16,"id":"f9fba292-8dc0-417d-a2e8-54ae0ce54070","page_size":[595,841],"block_position":"16-1"},{"bbox":[54,111,540,159],"type":"text","angle":0,"index":2,"merge_prev":false,"text":"(2) 记 $a = \\sup_{n \\geq 1} a_n$ ，则 $a_n \\leq a$ 且 $\\forall \\epsilon, \\exists k, a_k > a - \\epsilon$ 。当 $n \\geq k$ 时， $a_k^n \\leq a_1^n + a_2^n + \\cdots + a_n^n \\leq na^n$ ，两侧开 n 次根号得到 $a - \\epsilon < \\sqrt[n]{a_1^n + a_2^n + \\cdots + a_n^n} \\leq a\\sqrt[n]{n}$ ，由于 $\\lim_{n \\to \\infty} a\\sqrt[n]{n} = a$ ，对上述 $\\epsilon, \\exists N, \\forall n > N$ 有 $a\\sqrt[n]{n} < a + \\epsilon$ ，根据夹逼准则可证。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":16,"id":"bb693a6b-0db2-48bb-8d32-84d629da401f","page_size":[595,841],"block_position":"16-2"},{"bbox":[55,178,307,192],"type":"title","angle":0,"index":3,"level":2,"text":"## 2.4.2 平均值定理：无穷项算术平均与几何平均","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":16,"id":"a1993657-37b5-4c7c-8bd5-4dd9c89b5f29","page_size":[595,841],"block_position":"16-3"},{"bbox":[72,206,176,217],"type":"title","angle":0,"index":5,"level":2,"text":"## 定理 2.12 (平均值定理)","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":16,"id":"e3813ae4-b5e7-4482-b04b-9d058e1c7c8a","page_size":[595,841],"block_position":"16-4"},{"bbox":[69,222,253,239],"type":"text","angle":0,"index":6,"merge_prev":false,"text":"已知 $\\lim_{n\\to \\infty}a_n = a$ （有限数或正负无穷），则 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":16,"id":"665df5b2-ce1f-4e29-8c95-9ec5a9fda849","page_size":[595,841],"block_position":"16-5"},{"bbox":[80,238,270,257],"type":"text","angle":0,"index":7,"merge_prev":false,"text":"1. 算术平均： $\\lim_{n\\to\\infty}\\frac{a_{1}+a_{2}+\\cdots+a_{n}}{n}=a$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":16,"id":"eea559c6-b0dc-4760-a160-dbedf678700c","page_size":[595,841],"block_position":"16-6"},{"bbox":[80,256,307,275],"type":"text","angle":0,"index":8,"merge_prev":false,"text":"2. 几何平均：若 $a_{n} > 0$ ，则 $\\lim_{n \\to \\infty} \\sqrt[n]{a_{1}a_{2}\\cdots a_{n}} = a$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":16,"id":"9ebaba36-d31a-4a01-a178-7384a4ac815b","page_size":[595,841],"block_position":"16-7"},{"bbox":[54,292,364,306],"type":"text","angle":0,"index":9,"merge_prev":false,"text":"证明 (1) 只证 $a$ 为有限数的情况。 $\\forall \\epsilon, \\exists N, n > N$ 时 $|a_{n} - a| < \\epsilon$ ，因此 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":16,"id":"391c8fc0-5ad2-4b20-a8eb-d7f98950c27b","page_size":[595,841],"block_position":"16-8"},{"bbox":[107,307,487,361],"type":"interline_equation","angle":0,"index":10,"text":"$$\n\\begin{array}{r l} \\left| \\frac {a _ {1} + a _ {2} + \\cdots + a _ {n}}{n} - a \\right| = & \\left| \\frac {(a _ {1} - a) + \\cdots + (a _ {N} - a) + (a _ {N + 1} - a) + \\cdots + (a _ {n} - a)}{n} \\right| \\\\ & \\leq \\frac {| a _ {1} - a | + \\cdots + | a _ {N} - a |}{n} + \\frac {| a _ {N + 1} - a | + \\cdots + | a _ {n} - a |}{n} \\leq 2 \\epsilon \\end{array}\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":16,"id":"48e5e9ce-25b1-4d1f-b9e6-784478df5586","page_size":[595,841],"block_position":"16-9"},{"bbox":[73,362,176,374],"type":"text","angle":0,"index":11,"merge_prev":false,"text":"(2) 取对数则转换为 (1) ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":16,"id":"f4ca4271-3a76-48f3-88fc-a4b806223b4e","page_size":[595,841],"block_position":"16-10"},{"bbox":[35,388,478,420],"type":"text","angle":0,"index":12,"merge_prev":false,"text":"练习 2.6 奇偶子列推广 已知 $\\lim_{n\\to\\infty}a_{2n+1}=a,\\lim_{n\\to\\infty}a_{2n}=b$ ，证明 $\\lim_{n\\to\\infty}\\frac{a_{1}+a_{2}+\\cdots+a_{n}}{n}=\\frac{a+b}{2}$ 证明 分为奇偶数项进行讨论： ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":16,"id":"064019f6-3884-4238-96bc-1252d7320ac4","page_size":[595,841],"block_position":"16-11"},{"bbox":[142,423,452,471],"type":"interline_equation","angle":0,"index":13,"text":"$$\n\\begin{array}{l} \\lim _ {n \\to \\infty} \\frac {a _ {1} + a _ {2} + \\cdots + a _ {2 n}}{2 n} \\\\ = \\frac {1}{2} \\lim _ {n \\to \\infty} \\frac {a _ {1} + a _ {3} + \\cdots + a _ {2 n - 1}}{n} + \\frac {1}{2} \\lim _ {n \\to \\infty} \\frac {a _ {2} + a _ {4} + \\cdots + a _ {2 n}}{n} = \\frac {a + b}{n} \\end{array}\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":16,"id":"72d526e3-d6ad-48e5-83d0-1d6231e7633b","page_size":[595,841],"block_position":"16-12"},{"bbox":[54,471,97,483],"type":"text","angle":0,"index":14,"merge_prev":false,"text":"另一方面 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":16,"id":"583ab37e-664a-4018-aad6-eff5ec9fc053","page_size":[595,841],"block_position":"16-13"},{"bbox":[109,486,485,535],"type":"interline_equation","angle":0,"index":15,"text":"$$\n\\begin{array}{l} \\lim _ {n \\to \\infty} \\frac {a _ {1} + a _ {2} + \\cdots + a _ {2 n + 1}}{2 n + 1} \\\\ = \\lim _ {n \\to \\infty} \\frac {a _ {1} + a _ {2} + \\cdots + a _ {2 n + 1}}{n} \\cdot \\frac {n}{2 n - 1} + \\lim _ {n \\to \\infty} \\frac {a _ {2} + a _ {4} + \\cdots + a _ {2 n - 2}}{n - 1} \\cdot \\frac {n - 1}{2 n - 1} = \\frac {a + b}{2} \\end{array}\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":16,"id":"f1f46fa7-8176-4cdd-af59-b8016065a046","page_size":[595,841],"block_position":"16-14"},{"bbox":[54,537,142,549],"type":"text","angle":0,"index":16,"merge_prev":false,"text":"因此得到结论成立。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":16,"id":"fc422621-ad8c-40ee-a05e-61975bcc949c","page_size":[595,841],"block_position":"16-15"},{"bbox":[35,562,451,598],"type":"text","angle":0,"index":17,"merge_prev":false,"text":"练习2.7倒乘推广 $\\lim_{n\\to \\infty}a_n = a,\\lim_{n\\to \\infty}b_n = b$ ，证明： $\\lim_{n\\to \\infty}\\frac{a_1b_n + a_2b_{n - 1} + \\cdots + a_nb_1}{n} = ab$ 证明由于 $\\lim_{n\\to \\infty}a_n = a$ ，因此 $a_{n}$ 有界，即 $\\exists M$ 使得 $|a_n| < M$ ，故： ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":16,"id":"1074e0d5-9080-4ef3-8465-615b8623806c","page_size":[595,841],"block_position":"16-16"},{"bbox":[116,602,477,617],"type":"interline_equation","angle":0,"index":18,"text":"$$\n\\vert a _ {k} b _ {n - k + 1} - a b \\vert < \\vert a _ {k} \\vert \\cdot \\vert b _ {n - k + 1} - b \\vert + \\vert b \\vert \\cdot \\vert a _ {k} - a \\vert \\leq M \\vert b _ {n - k + 1} - b \\vert + \\vert b \\vert \\cdot \\vert a _ {k} - a \\vert\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":16,"id":"bc33f85b-00cb-4725-9c6e-daf8f3961b73","page_size":[595,841],"block_position":"16-17"},{"bbox":[54,622,101,634],"type":"text","angle":0,"index":19,"merge_prev":false,"text":"因此得到： ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":16,"id":"e3534a2f-79eb-4cb8-b1a6-6b8c9aad9724","page_size":[595,841],"block_position":"16-18"},{"bbox":[167,634,426,711],"type":"interline_equation","angle":0,"index":20,"text":"$$\n\\begin{array}{l} \\left| \\frac {a _ {1} b _ {n} + \\cdots + a _ {n} b _ {1}}{n} - a b \\right| \\\\ \\leq \\frac {| a _ {1} b _ {n} - a b | + \\cdots + | a _ {n} b _ {1} - a b |}{n} \\\\ \\leq M \\cdot \\frac {| b _ {n} - b | + \\cdot + | b _ {1} - b |}{n} + | b | \\cdot \\frac {| a _ {1} - a | + \\cdots + | a _ {n} - a |}{n} \\end{array}\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":16,"id":"03e6de20-feba-4507-ad70-835c50eb8565","page_size":[595,841],"block_position":"16-19"},{"bbox":[54,714,442,740],"type":"text","angle":0,"index":21,"merge_prev":false,"text":"有平均值定理可知右侧极限为0，因此 $\\left|\\frac{a_{1}b_{n}+\\cdots+a_{n}b_{1}}{n}-ab\\right|\\leq0$ ，得到目标极限为0。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":16,"id":"0fe11150-d289-4177-b985-881faa7c38fc","page_size":[595,841],"block_position":"16-20"},{"bbox":[293,802,300,811],"type":"page_number","angle":0,"index":1000003,"text":"9 ","color":{"line":"rgba(164, 164, 164, 1)","fill":"rgba(164, 164, 164, 1)"},"page_idx":16,"is_discarded":true,"id":"324c7003-d24f-4e46-92c2-742364796ca0","page_size":[595,841],"block_position":"16-21"}],[{"bbox":[449,47,539,58],"type":"header","angle":0,"index":-99999,"text":"2.5 使用定积分定义 ","color":{"line":"rgba(164, 164, 164, 1)","fill":"rgba(164, 164, 164, 1)"},"page_idx":17,"is_discarded":true,"id":"a540958c-d174-4040-bcd6-d5d4faedaf2e","page_size":[595,841],"block_position":"17-0"},{"bbox":[55,72,269,86],"type":"title","angle":0,"index":1,"level":2,"text":"## 2.4.3 技巧：一招解决 n 次根号与除以 n","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":17,"id":"4641b351-39c0-45f5-b3f5-38e9e10dbefe","page_size":[595,841],"block_position":"17-1"},{"bbox":[73,100,202,112],"type":"title","angle":0,"index":2,"level":2,"text":"## 定理 2.13 (n 次根号与除以 n)","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":17,"id":"c90b16f6-0f5f-4212-9bac-a7f83d02c7cf","page_size":[595,841],"block_position":"17-2"},{"bbox":[68,116,135,129],"type":"text","angle":0,"index":3,"merge_prev":false,"text":"$a_{n}$ 为数列，则 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":17,"id":"62262fb4-df90-40c7-baa3-c8c5d4f17f10","page_size":[595,841],"block_position":"17-3"},{"bbox":[83,129,277,150],"type":"text","angle":0,"index":4,"merge_prev":false,"text":"若 $\\lim_{n\\to \\infty}(a_n - a_{n - 1}) = a$ ，则 $\\lim_{n\\to \\infty}\\frac{a_n}{n} = a$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":17,"id":"00de4835-c3f4-4276-ae3e-98bd4bb640c9","page_size":[595,841],"block_position":"17-4"},{"bbox":[83,149,258,171],"type":"text","angle":0,"index":5,"merge_prev":false,"text":"若 $\\lim_{n\\to \\infty}\\frac{a_n}{a_{n - 1}} = a$ ，则 $\\lim_{n\\to \\infty}\\sqrt[n]{a_n} = a$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":17,"id":"dcc92ed6-bf72-45d7-9719-479ebf2163b4","page_size":[595,841],"block_position":"17-5"},{"bbox":[54,182,537,215],"type":"text","angle":0,"index":6,"merge_prev":false,"text":"证明 (1) 由于 $a_{n}=a_{1}+(a_{2}-a_{1})+\\cdots+(a_{n}-a_{n-1})$ ，因此 $\\lim_{n\\to\\infty}\\frac{a_{n}}{n}=\\lim_{n\\to\\infty}\\frac{a_{1}+(a_{2}-a_{1})+\\cdots+(a_{n}-a_{n-1})}{n}$ ，根据平均值定理即 a ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":17,"id":"788fc611-7c77-4cb6-95d5-2f1b5cdcb2f2","page_size":[595,841],"block_position":"17-6"},{"bbox":[73,213,471,239],"type":"text","angle":0,"index":7,"merge_prev":false,"text":"(2) 根据 $a_{n} = a_{1} \\cdot \\frac{a_{2}}{a_{1}} \\cdots \\frac{a_{n}}{a_{n-1}}$ , $\\lim_{n \\to \\infty} \\sqrt[n]{a_n} = \\lim_{n \\to \\infty} \\sqrt[n]{a_1 \\cdot \\frac{a_2}{a_1} \\cdots \\frac{a_n}{a_{n-1}}}$ , 根据平均值定理即 $a_{n}$ . ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":17,"id":"1fa17e7f-948a-47d4-99e8-d0c4515e5843","page_size":[595,841],"block_position":"17-7"},{"bbox":[35,248,334,268],"type":"text","angle":0,"index":8,"merge_prev":false,"text":"练习 2.8n 次根号 计算极限 (1) $\\lim_{n\\to\\infty}\\sqrt[n]{a}$ (2) $\\lim_{n\\to\\infty}\\sqrt[n]{n}$ (3) $\\lim_{n\\to\\infty}\\sqrt[n]{n!}$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":17,"id":"6bc70720-4df8-4f4f-acd3-9a41bcdbba92","page_size":[595,841],"block_position":"17-8"},{"bbox":[54,265,195,287],"type":"text","angle":0,"index":9,"merge_prev":false,"text":"解 (1) $\\frac{a_n}{a_{n-1}} = 1$ ，因此极限为 $1$ 。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":17,"id":"5614c1c4-c27f-4058-9194-48f78d04ca83","page_size":[595,841],"block_position":"17-9"},{"bbox":[73,285,218,310],"type":"text","angle":0,"index":10,"merge_prev":false,"text":"(2) $\\frac{a_{n}}{a_{n-1}}=\\frac{n}{n-1}\\rightarrow1$ ，极限为1 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":17,"id":"664af7e0-70c9-41af-a13f-3fe309b30080","page_size":[595,841],"block_position":"17-10"},{"bbox":[73,310,217,331],"type":"text","angle":0,"index":11,"merge_prev":false,"text":"(3) $\\frac{a_n}{a_{n-1}} = n \\to \\infty$ ，因此不收敛 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":17,"id":"0719fa8f-8164-4ca8-b092-24e3848dbe2e","page_size":[595,841],"block_position":"17-11"},{"bbox":[35,335,392,362],"type":"text","angle":0,"index":12,"merge_prev":false,"text":"练习 2.9 除以 n 重点：计算极限 (1) $\\lim_{n\\to\\infty}\\frac{\\sqrt[n]{n!}}{n}$ (2) $\\lim_{n\\to\\infty}\\frac{\\sqrt[n]{n(n+1)\\cdots(2n-1)}}{n}$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":17,"id":"03a78243-da58-41bd-be1a-a049dd5367d7","page_size":[595,841],"block_position":"17-12"},{"bbox":[52,360,459,392],"type":"text","angle":0,"index":13,"merge_prev":false,"text":"解 $(1)a_{n} = \\frac{n!}{n^{n}}$ ， $\\lim_{n\\to \\infty}\\frac{a_n}{a_{n - 1}} = \\frac{\\frac{n!}{n^n}}{\\frac{(n - 1)!}{(n - 1)^{n - 1}}} = \\frac{1}{(1 - \\frac{1}{n})^{n - 1}} = \\frac{1}{\\left[(1 - \\frac{1}{n})^n\\right]^{\\frac{n - 1}{n}}} = \\frac{1}{e}$ 因此极限为 $\\frac{1}{e}$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":17,"id":"46f5a564-d0d5-4acf-9c9b-eda4d66ac322","page_size":[595,841],"block_position":"17-13"},{"bbox":[73,391,233,416],"type":"text","angle":0,"index":14,"merge_prev":false,"text":"(2) $a_{n}=\\frac{n(n+1)\\cdots(2n-1)}{n^{n}}$ ，因此 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":17,"id":"e7dbf214-c09f-4965-9ce0-617c3d97539e","page_size":[595,841],"block_position":"17-14"},{"bbox":[139,419,453,448],"type":"interline_equation","angle":0,"index":15,"text":"$$\n\\lim _ {n \\to \\infty} \\frac {a _ {n}}{a _ {n - 1}} = \\lim _ {n \\to \\infty} \\frac {2 (2 n - 1) (n - 1) ^ {n - 1}}{n ^ {n}} = (4 - \\frac {2}{n}) \\left((1 - \\frac {1}{n}) ^ {n}\\right) ^ {\\frac {n - 1}{n}} = \\frac {4}{e}\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":17,"id":"a4a37024-1a76-47e1-a87b-1307daff4d3a","page_size":[595,841],"block_position":"17-15"},{"bbox":[55,449,117,474],"type":"text","angle":0,"index":16,"merge_prev":false,"text":"因此极限为 $\\frac{4}{e}$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":17,"id":"73a74974-0869-4c93-a6fb-af065454ac04","page_size":[595,841],"block_position":"17-16"},{"bbox":[55,491,186,507],"type":"title","angle":0,"index":17,"level":2,"text":"## 2.5 使用定积分定义","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":17,"id":"59e9029a-85b8-4b17-8dac-dbab0df27103","page_size":[595,841],"block_position":"17-17"},{"bbox":[55,523,198,538],"type":"title","angle":0,"index":18,"level":2,"text":"## 2.5.1 使用定积分计算极限","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":17,"id":"b984f960-6250-4f28-80c2-7396b2466264","page_size":[595,841],"block_position":"17-18"},{"bbox":[72,550,226,564],"type":"title","angle":0,"index":19,"level":2,"text":"## 定理 2.14 (用定积分定义计算极限)","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":17,"id":"e9e9b1d5-a0dd-4110-b028-3d8ba2b3da4f","page_size":[595,841],"block_position":"17-19"},{"bbox":[69,568,180,581],"type":"text","angle":0,"index":20,"merge_prev":false,"text":"若 $f(x)$ 在 $[a,b]$ 连续，则 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":17,"id":"fd1ecd2b-e4aa-4004-bd20-fad6607293e8","page_size":[595,841],"block_position":"17-20"},{"bbox":[102,583,491,613],"type":"interline_equation","angle":0,"index":21,"text":"$$\n\\lim _ {n \\rightarrow \\infty} \\frac {b - a}{n} \\left[ f \\left(a + \\frac {b - a}{n}\\right) + f \\left(a + \\frac {2 (b - a)}{n}\\right) + \\dots + f \\left(a + \\frac {n (b - a)}{n}\\right)\\right] = \\int_ {a} ^ {b} f (x) \\mathrm{d} x\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":17,"id":"12b4c3da-19f2-43cb-8cef-7e9557b8a60b","page_size":[595,841],"block_position":"17-21"},{"bbox":[35,622,489,653],"type":"text","angle":0,"index":23,"merge_prev":false,"text":"笔记 如果是 $\\lim_{n\\to \\infty}\\frac{1}{n + a}\\sum_{k = 1}^{n}f\\left(\\frac{k}{n}\\right)$ ，则可以加一项得到 $\\lim_{n\\to \\infty}\\frac{n}{n + a}\\cdot \\frac{1}{n}\\sum_{k = 1}^{n}f\\left(\\frac{k}{n}\\right)$ 再转换为定积分。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":17,"id":"7989e4f0-c923-472b-8cba-20564802bb49","page_size":[595,841],"block_position":"17-22"},{"bbox":[36,655,363,680],"type":"text","angle":0,"index":24,"merge_prev":false,"text":"练习 2.10 直接判断积分 (1) $\\lim_{n\\to\\infty} n\\left(\\frac{1}{n^{2}+1}+\\frac{1}{n^{2}+2}+\\cdots+\\frac{1}{n^{2}+n^{2}}\\right)$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":17,"id":"4622e52d-329b-46ed-98cb-834c3fe30443","page_size":[595,841],"block_position":"17-23"},{"bbox":[75,681,290,707],"type":"interline_equation","angle":0,"index":25,"text":"$$\n(2) \\lim _ {n \\to \\infty} n \\left[ \\frac {1}{(n + 1) ^ {2}} + \\frac {1}{(n + 2) ^ {2}} + \\dots + \\frac {1}{(n + n) ^ {2}} \\right]\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":17,"id":"3045ac30-96a5-4faa-a24a-63baf3998c71","page_size":[595,841],"block_position":"17-24"},{"bbox":[73,706,302,731],"type":"interline_equation","angle":0,"index":26,"text":"$$\n(3) \\lim _ {n \\rightarrow \\infty} \\left(\\frac {1}{\\sqrt {n ^ {2} + 1 ^ {2}}} + \\frac {1}{\\sqrt {n ^ {2} + 2 ^ {2}}} + \\dots + \\frac {1}{\\sqrt {n ^ {2} + n ^ {2}}}\\right)\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":17,"id":"9b722d5a-67a8-4912-8cfa-a35dacee9c26","page_size":[595,841],"block_position":"17-25"},{"bbox":[74,733,150,761],"type":"text","angle":0,"index":27,"merge_prev":false,"text":"(4) $\\lim_{n\\to \\infty}\\sum_{k = 1}^{n}\\frac{1}{\\sqrt{nk}}$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":17,"id":"906bd667-d0f6-4272-9299-a9ebf68b48b7","page_size":[595,841],"block_position":"17-26"},{"bbox":[292,802,303,812],"type":"page_number","angle":0,"index":1000021,"text":"10 ","color":{"line":"rgba(164, 164, 164, 1)","fill":"rgba(164, 164, 164, 1)"},"page_idx":17,"is_discarded":true,"id":"a17aa51b-2c26-4691-a72c-7118b849bed2","page_size":[595,841],"block_position":"17-27"}],[{"bbox":[449,47,539,58],"type":"header","angle":0,"index":-99999,"text":"2.5 使用定积分定义 ","color":{"line":"rgba(164, 164, 164, 1)","fill":"rgba(164, 164, 164, 1)"},"page_idx":18,"is_discarded":true,"id":"af524c00-a48d-4db2-a8f5-e40a03ffc416","page_size":[595,841],"block_position":"18-0"},{"bbox":[52,69,286,181],"type":"text","angle":0,"index":1,"merge_prev":false,"text":"解(1)等价于 $\\int_0^1\\frac{1}{1 + x^2}\\mathrm{d}x = \\frac{\\pi}{4}$ (2)等价于 $\\int_0^1\\frac{1}{(1 + x)^2}\\mathrm{d}x = \\frac{1}{2}$ (3)等价于 $\\int_0^1\\frac{1}{\\sqrt{1 + x^2}}\\mathrm{d}x = \\ln (1 + \\sqrt{2})$ (4)转换为 $\\frac{1}{n}\\sum_{k = 1}^{n}\\frac{1}{\\sqrt{\\frac{k}{n}}}$ ，即 $\\int_0^1\\frac{1}{\\sqrt{x}}\\mathrm{d}x$ ，极限为2 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":18,"id":"251e5305-2f53-4b9b-af47-fb9ff24e1379","page_size":[595,841],"block_position":"18-1"},{"bbox":[35,185,540,269],"type":"text","angle":0,"index":2,"merge_prev":false,"text":"练习2.11对数 $+$ 积分重点：取对数 $\\lim_{n\\to \\infty}\\frac{\\sqrt[n]{n(n + 1)\\cdots(2n - 1)}}{n}$ 解等价于 $\\exp \\left[\\frac{1}{n}\\ln \\left(\\frac{n(n + 1)\\cdots(2n - 1)}{n^n}\\right)\\right]$ ，指数积分即 $\\int_0^1\\ln (1 + x)\\mathrm{d}x = [(1 + x)\\ln (1 + x) - x]\\bigg|_0^1 = \\ln 4 - 1,$ 因此极限为 $\\frac{4}{e}$ 或者本题也在 $n$ 次根号部分出现了(更推荐 $n$ 次根号的做法) ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":18,"id":"57b589e9-3c65-401e-90d0-0401c45ec3a3","page_size":[595,841],"block_position":"18-2"},{"bbox":[35,278,480,332],"type":"text","angle":0,"index":3,"merge_prev":false,"text":"练习2.12改变有限项(1)重点： $\\lim_{n\\to \\infty}\\left(\\frac{1}{n} +\\frac{1}{n + 1} +\\dots +\\frac{1}{n + (n - 1)}\\right)$ 解(1)可改变有限项不影响极限，变为 $\\lim_{n\\to \\infty}\\frac{1}{n + 1} +\\dots +\\frac{1}{n + n}$ ，因此极限为 $\\int_0^1\\frac{1}{1 + x}\\mathrm{d}x = \\ln 2$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":18,"id":"3c524083-ba67-452a-9c22-833d3755da7b","page_size":[595,841],"block_position":"18-3"},{"bbox":[35,337,405,364],"type":"text","angle":0,"index":4,"merge_prev":false,"text":"练习 2.13 经典夹逼 + 积分 (1) 重点: $I = \\lim_{n \\to \\infty} \\left( \\frac{\\sin \\frac{\\pi}{n}}{n + \\frac{1}{n}} + \\frac{\\sin \\frac{2}{n}\\pi}{n + \\frac{2}{n}} + \\cdots + \\frac{\\sin \\pi}{n + 1} \\right)$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":18,"id":"7b5ff9fc-0968-438f-afa7-95ccb040da75","page_size":[595,841],"block_position":"18-4"},{"bbox":[73,364,352,391],"type":"text","angle":0,"index":5,"merge_prev":false,"text":"(2) 重点: $\\lim_{n \\to \\infty} \\left( \\frac{1}{n^2 + n + 1} + \\frac{2}{n^2 + n + 2} + \\cdots + \\frac{n}{n^2 + n + n} \\right)$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":18,"id":"ba5cecea-64e8-49cd-a0de-26c29f25d6c8","page_size":[595,841],"block_position":"18-5"},{"bbox":[73,391,364,414],"type":"text","angle":0,"index":6,"merge_prev":false,"text":"(3) 重点: $\\lim_{n \\to \\infty} \\left( \\frac{1}{n^2 + n + 1^2} + \\frac{2}{n^2 + n + 2^2} + \\cdots + \\frac{n}{n^2 + n + n^2} \\right)$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":18,"id":"436c336b-5e0d-41d4-bf4a-b51ae30367b1","page_size":[595,841],"block_position":"18-6"},{"bbox":[86,415,320,440],"type":"interline_equation","angle":0,"index":7,"text":"$$\n\\lim _ {n \\rightarrow \\infty} \\left(\\frac {\\sin \\frac {x}{n}}{n + 1} + \\frac {2 \\sin \\frac {2}{n} x}{2 n + 1} + \\dots + \\frac {n \\sin \\frac {n}{n} x}{n ^ {2} + 1}\\right), x \\in (0, \\pi)\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":18,"id":"cb04582b-71ba-4042-8883-94298183c98d","page_size":[595,841],"block_position":"18-7"},{"bbox":[53,440,462,534],"type":"text","angle":0,"index":8,"merge_prev":false,"text":"解 (1) $I = \\lim_{n\\to \\infty}\\frac{1}{n}\\left(\\frac{\\sin\\frac{\\pi}{n}}{1 + \\frac{1}{n^2}} +\\frac{\\sin\\frac{2}{n}\\pi}{1 + \\frac{2}{n^2}} +\\dots +\\frac{\\sin\\pi}{1 + \\frac{n}{n^2}}\\right)$ ，进行 $\\frac{\\sin\\frac{i}{n}\\pi}{n + 1}\\leq \\frac{\\sin\\frac{i}{n}\\pi}{n + \\frac{i}{n}}\\leq \\frac{\\sin\\frac{i}{n}\\pi}{n}$ 放缩： $\\left\\{ \\begin{array}{l}\\lim_{n\\to \\infty}\\frac{1}{n}\\sum_{i = 1}^{n}\\sin \\frac{i}{n}\\pi = \\int_0^1\\sin \\pi x\\mathrm{d}x = \\frac{2}{\\pi}\\\\ \\lim_{n\\to \\infty}\\frac{1}{n + 1}\\sum_{n = 1}^{\\infty}\\sin \\frac{i}{n}\\pi = \\lim_{n\\to \\infty}\\frac{n}{n + 1}\\cdot \\frac{1}{n}\\sum_{i = 1}^{n}\\sin \\frac{i}{n}\\pi = \\frac{2}{\\pi} \\end{array} \\right.$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":18,"id":"05a241f5-f068-4f44-8ccb-6ba22221befe","page_size":[595,841],"block_position":"18-8"},{"bbox":[54,535,177,554],"type":"text","angle":0,"index":9,"merge_prev":false,"text":"因此根据夹逼可知极限为 $\\frac{2}{\\pi}$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":18,"id":"e9639f0d-5441-4270-a702-8b931b28c19a","page_size":[595,841],"block_position":"18-9"},{"bbox":[54,552,539,651],"type":"text","angle":0,"index":10,"merge_prev":false,"text":"(2) 进行放缩: $\\frac{1 + 2 + \\cdots + n}{n^2 + n + n} \\leq I \\leq \\frac{1 + 2 + \\cdots + n}{n^2 + n}$ (或者 $\\frac{i}{n^2 + n + n} \\leq \\frac{i}{n^2 + n + i} \\leq \\frac{i}{n^2}$ 也行)。左右侧分别写为 $\\left\\{\\begin{aligned}&\\sum_{k=1}^{n}\\frac{k}{n^2+2n}=\\frac{1}{n+2}\\sum_{k=1}^{n}\\frac{k}{n}=\\frac{n}{n+2}\\cdot\\frac{1}{n}\\sum_{k=1}^{n}\\frac{k}{n}\\\\&\\sum_{k=1}^{n}\\frac{k}{n^2+n}=\\frac{n}{n+1}\\cdot\\frac{1}{n}\\sum_{k=1}^{n}\\frac{k}{n}\\end{aligned}\\right.$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":18,"id":"f399f939-d6d7-4907-9d4b-9ca6b60a79ac","page_size":[595,841],"block_position":"18-10"},{"bbox":[54,655,540,748],"type":"text","angle":0,"index":11,"merge_prev":false,"text":"因此两侧均等于积分 $\\int_0^1 x\\mathrm{d}x = \\frac{1}{2}$ (3)一方面 $\\frac{i}{n^2 + n + i^2}\\leq \\frac{i}{n^2 + i^2} = \\frac{1}{n}\\cdot \\frac{\\frac{i}{n}}{1 + (\\frac{i}{n})^2}$ ，另一方面 $\\frac{i}{n^2 + n + i}\\geq \\frac{i}{(n + 1)^2 + i^2} = \\frac{1}{n + 1}$ $\\frac{\\frac{i}{n + 1}}{1 + \\left(\\frac{i}{n + 1}\\right)^2}$ ，结果等价于 $\\int_0^1\\frac{x}{1 + x^2}\\mathrm{d}x = \\frac{1}{2}\\ln 2$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":18,"id":"194cbc4c-0a15-43d5-8839-4c3f8f6fbb71","page_size":[595,841],"block_position":"18-11"},{"bbox":[292,803,302,812],"type":"page_number","angle":0,"index":1000011,"text":"11 ","color":{"line":"rgba(164, 164, 164, 1)","fill":"rgba(164, 164, 164, 1)"},"page_idx":18,"is_discarded":true,"id":"482421a3-223c-4135-8cb2-69aa63accdd9","page_size":[595,841],"block_position":"18-12"}],[{"bbox":[459,47,539,58],"type":"header","angle":0,"index":-99999,"text":"2.6 单调有界定理 ","color":{"line":"rgba(164, 164, 164, 1)","fill":"rgba(164, 164, 164, 1)"},"page_idx":19,"is_discarded":true,"id":"3aca3cf3-51e4-4a81-8cce-06f37c1acb8c","page_size":[595,841],"block_position":"19-0"},{"bbox":[52,69,538,128],"type":"text","angle":0,"index":1,"merge_prev":false,"text":"(4) 看上去等价于 $\\lim_{n\\to \\infty}\\sum_{k = 1}^{n}\\frac{k\\sin\\frac{x}{n}}{kn} = \\int_0^1\\sin tx\\mathrm{d}t = \\frac{1 - \\cos x}{x}$ ，放缩为 $\\frac{k\\sin\\frac{k}{n}x}{kn + k}\\leq \\frac{k\\sin\\frac{k}{n}x}{kn + 1}\\leq \\frac{k\\sin\\frac{k}{n}x}{kn}$ 再求和即可。 \n笔记一般不能直接写成积分形式的都需要进行放缩。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":19,"id":"05e2a0d9-6311-46b0-b5f9-aed69d97d573","page_size":[595,841],"block_position":"19-1"},{"bbox":[55,146,258,160],"type":"title","angle":0,"index":2,"level":2,"text":"## 2.5.2 利用定积分定义构造等价无穷小","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":19,"id":"7ddf47a9-9aac-4e20-9e5b-f3104a01dac6","page_size":[595,841],"block_position":"19-2"},{"bbox":[72,173,236,186],"type":"title","angle":0,"index":3,"level":2,"text":"## 定理 2.15 (利用积分构造等价无穷小)","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":19,"id":"5076a634-a779-4d9b-8a76-82a5a7635f79","page_size":[595,841],"block_position":"19-3"},{"bbox":[68,191,277,214],"type":"text","angle":0,"index":4,"merge_prev":false,"text":"对于正整数 k， $1^{k} + 2^{k} + \\cdots + n^{k} \\sim \\frac{1}{k+1} n^{k+1}$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":19,"id":"c1e68c85-3d5c-4767-a3e8-cb09f8dea8a7","page_size":[595,841],"block_position":"19-4"},{"bbox":[54,225,334,251],"type":"text","angle":0,"index":5,"merge_prev":false,"text":"证明 考虑 $\\lim_{n\\to\\infty}\\frac{1}{n^{k+1}}\\left(1^{k}+\\cdots+n^{k}\\right)=\\int_{0}^{1}x^{k}\\mathrm{d}x=\\frac{1}{k+1}$ 即可。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":19,"id":"0cfce2ab-119e-4675-a90a-371aa5de6791","page_size":[595,841],"block_position":"19-5"},{"bbox":[36,259,389,282],"type":"text","angle":0,"index":6,"merge_prev":false,"text":"练习 2.14 利用积分构造等价无穷小 计算 $\\lim_{n\\to\\infty}\\frac{\\ln n}{\\ln(1^{2022}+2^{2022}+\\cdots+n^{2022})}$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":19,"id":"79529512-0358-4dda-8cc0-d1dd83145757","page_size":[595,841],"block_position":"19-6"},{"bbox":[52,282,369,308],"type":"text","angle":0,"index":7,"merge_prev":false,"text":"解 $\\lim_{n\\to\\infty}\\frac{1}{n^{2023}}\\left(1^{2022}+2^{2022}+\\cdots+n^{2022}\\right)=\\int_{0}^{1}x^{2022}\\mathrm{d}x=\\frac{1}{2023}$ ，因此 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":19,"id":"2f55f08b-31c3-4c35-a5bd-41a5421cc1ce","page_size":[595,841],"block_position":"19-7"},{"bbox":[79,311,513,337],"type":"interline_equation","angle":0,"index":8,"text":"$$\n\\lim _ {n \\rightarrow \\infty} \\frac {\\ln n}{\\ln \\left(1 ^ {2 0 2 2} + 2 ^ {2 0 2 2} + \\cdots + n ^ {2 0 2 2}\\right)} = \\lim _ {n \\rightarrow \\infty} \\frac {\\ln n}{2 0 2 3 \\ln n + \\ln \\frac {1}{n ^ {2 0 2 3}} \\left(1 ^ {2 0 2 2} + 2 ^ {2 0 2 2} + \\dots + n ^ {2 0 2 2}\\right)} = \\frac {1}{2 0 2 3}\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":19,"id":"52692db7-1a1c-44d1-8e4b-a76d1c726f40","page_size":[595,841],"block_position":"19-8"},{"bbox":[73,355,226,368],"type":"title","angle":0,"index":9,"level":2,"text":"## 定理 2.16 (已知极限计算求和极限)","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":19,"id":"ec04fd75-8f7d-4adc-b741-574766a42c64","page_size":[595,841],"block_position":"19-9"},{"bbox":[68,372,264,403],"type":"text","angle":0,"index":10,"merge_prev":false,"text":"若 $\\lim_{n\\to \\infty}a_n = A$ ，则 $\\lim_{n\\to \\infty}\\sum_{k = 1}^{n}\\frac{a_{n + k}}{n + k} = A\\ln 2$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":19,"id":"ead8a5ca-3a46-46f8-b034-5967d78c866d","page_size":[595,841],"block_position":"19-10"},{"bbox":[53,415,523,446],"type":"text","angle":0,"index":11,"merge_prev":false,"text":"证明 首先显然 $\\lim_{n\\to \\infty}\\sum_{k = 1}^{n}\\frac{A}{n + k} = A\\ln 2$ ，根据 $\\lim_{n\\to \\infty}a_n = A$ ， $\\forall \\epsilon ,\\exists N,\\forall n > N,|a_n - A| < \\epsilon$ ，因此 $n > N$ 时： ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":19,"id":"760a97f7-a99b-49ef-9c5f-e8e9a9132c5a","page_size":[595,841],"block_position":"19-11"},{"bbox":[177,449,417,482],"type":"interline_equation","angle":0,"index":12,"text":"$$\n\\left| \\sum_ {k = 1} ^ {n} \\frac {a _ {n + k}}{n + k} - \\sum_ {k = 1} ^ {n} \\frac {A}{n + k} \\right| \\leq \\sum_ {k = 1} ^ {n} \\frac {| a _ {n + k} - A |}{n + k} < \\sum_ {k = 1} ^ {n} \\frac {\\epsilon}{n} = \\epsilon\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":19,"id":"b838ff6d-26f8-456e-9d88-9273c5fd3c60","page_size":[595,841],"block_position":"19-12"},{"bbox":[56,504,172,520],"type":"title","angle":0,"index":13,"level":2,"text":"## 2.6 单调有界定理","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":19,"id":"e095c221-7498-46e9-991c-a6d4cd83545d","page_size":[595,841],"block_position":"19-13"},{"bbox":[56,536,210,550],"type":"title","angle":0,"index":14,"level":2,"text":"## 2.6.1 单调有界定理及其推广","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":19,"id":"56ba6b2e-e159-428d-956a-7b90f5322d4e","page_size":[595,841],"block_position":"19-14"},{"bbox":[73,564,186,576],"type":"title","angle":0,"index":15,"level":2,"text":"## 定理 2.17 (单调有界定理)","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":19,"id":"21bef548-53d3-44d5-a1cc-fff61700ab2b","page_size":[595,841],"block_position":"19-15"},{"bbox":[69,581,231,594],"type":"text","angle":0,"index":16,"merge_prev":false,"text":"若数列 $a_{n}$ 单调且有界，则其必收敛。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":19,"id":"74c4fd39-f3f3-4474-ad4b-1c9af04a82f3","page_size":[595,841],"block_position":"19-16"},{"bbox":[36,625,539,702],"type":"text","angle":0,"index":17,"merge_prev":false,"text":"练习 2.15 单调有界定理理论推广 (1) $a_{n}$ 单增， $b_{n}$ 单减， $\\lim_{n\\to\\infty}(b_{n}-a_{n})=0$ ，证明 $\\lim_{n\\to\\infty}a_{n},\\lim_{n\\to\\infty}b_{n}$ 存在且相等\n\n(2) $a_{n}$ 为有界数列，记 $\\overline{a}=\\sup\\{a_{n},a_{n+1},\\cdots\\},\\underline{a}_{n}=\\inf\\{a_{n},a_{n+1},\\cdots\\}$ ，证明 $\\lim_{n\\to\\infty}\\overline{a}_{n},\\lim_{n\\to\\infty}\\underline{a}_{n}$ 存在，且 $\\lim_{n\\to\\infty}a_{n}$ 存在的充要条件为 $\\lim_{n\\to\\infty}\\overline{a}_{n}=\\lim_{n\\to\\infty}\\underline{a}_{n}$ 证明 (1) 由于 $\\lim_{n\\to\\infty}(b_{n}-a_{n})=0,\\exists N,\\forall n>N$ 使得 $|b_{n}-a_{n}|<1$ ，即 $a_{n} N$ 有 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":19,"id":"9372a113-9a51-47f9-84be-1f00a2cc46e0","page_size":[595,841],"block_position":"19-18"},{"bbox":[242,741,351,752],"type":"interline_equation","angle":0,"index":20,"text":"$$\na - \\epsilon < \\underline {{a}} _ {n} \\leq \\overline {{a}} _ {n} < a + \\epsilon\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":19,"id":"5aa67aee-95a3-4a76-a4f7-08330c5d216b","page_size":[595,841],"block_position":"19-19"},{"bbox":[54,758,157,770],"type":"text","angle":0,"index":21,"merge_prev":false,"text":"取极限可知极限均为 a. ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":19,"id":"9bb235e1-cac7-49c1-b85d-b1092365e975","page_size":[595,841],"block_position":"19-20"},{"bbox":[292,803,302,811],"type":"page_number","angle":0,"index":1000018,"text":"12 ","color":{"line":"rgba(164, 164, 164, 1)","fill":"rgba(164, 164, 164, 1)"},"page_idx":19,"is_discarded":true,"id":"4f0ad957-b587-4245-99ac-a6b3ca775c8e","page_size":[595,841],"block_position":"19-21"}],[{"bbox":[459,47,539,58],"type":"header","angle":0,"index":-99999,"text":"2.6 单调有界定理 ","color":{"line":"rgba(164, 164, 164, 1)","fill":"rgba(164, 164, 164, 1)"},"page_idx":20,"is_discarded":true,"id":"ac88b5f5-1bd1-428b-8453-7480023e8a36","page_size":[595,841],"block_position":"20-0"},{"bbox":[35,86,399,104],"type":"text","angle":0,"index":1,"merge_prev":false,"text":"练习2.16两道经典循环根式(1)重点： $a_1 = \\sqrt{c}$ ， $a_{n + 1} = \\sqrt{a_n + c}$ ，求 $\\lim_{n\\to \\infty}a_n$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":20,"id":"49e8338d-bb0c-4a86-b0e0-989f0b16f7dd","page_size":[595,841],"block_position":"20-1"},{"bbox":[73,104,304,129],"type":"text","angle":0,"index":2,"merge_prev":false,"text":"(2) 重点: $a_{n} = \\sqrt{1 + \\sqrt{2 + \\cdots + \\sqrt{n}}}$ , 证明 $a_{n}$ 收敛 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":20,"id":"31680891-9360-439f-8ae7-ff74b4f02a32","page_size":[595,841],"block_position":"20-2"},{"bbox":[53,130,539,190],"type":"text","angle":0,"index":3,"merge_prev":false,"text":"解(1)由于 $a_{n}=\\sqrt{c+\\sqrt{c+\\cdots+\\sqrt{c}}}$ ，根据该通项的形式，每次 $a_{n+1}$ 在 $a_{n}$ 基础上往最里层加入一个 $\\sqrt{c}$ ，因此整体单增。若极限存在，则显然极限为 $a=\\frac{1+\\sqrt{1+4c}}{2}$ ，由于 $a=\\sqrt{a+c}$ ，因此若 $a_{n}\\ln(n+1)-\\ln(n)>0$ (调和级数不等式)，且根据 $\\ln(1+x)\\geq\\frac{x}{1+x}$ 得到 $\\frac{1}{n+1}<\\ln(1+\\frac{1}{n})$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":20,"id":"7e036d23-ed84-4598-82a0-a175e506f11a","page_size":[595,841],"block_position":"20-9"},{"bbox":[55,391,368,417],"type":"text","angle":0,"index":11,"merge_prev":false,"text":"于是 $a_{n + 1} - a_n = \\frac{1}{n + 1} -\\ln (1 + \\frac{1}{n}) < 0$ ，这说明单减有下界，故收敛。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":20,"id":"3fc6aa2f-8946-4f54-b16f-114ceaa77e1d","page_size":[595,841],"block_position":"20-10"},{"bbox":[35,421,354,455],"type":"text","angle":0,"index":12,"merge_prev":false,"text":"练习 2.18 交替级数 $S_{n}=1-\\frac{1}{2}+\\frac{1}{3}-\\cdots+\\frac{(-1)^{n+1}}{n}$ ，计算 $\\lim_{n\\to\\infty}S_{n}$ 解由 Leibniz 定理可知收敛。只需要考虑偶数项 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":20,"id":"4051eb1f-abd0-412c-9574-62ae2b76aa19","page_size":[595,841],"block_position":"20-11"},{"bbox":[52,456,539,497],"type":"text","angle":0,"index":13,"merge_prev":false,"text":"$S_{2n} = (1 + \\frac{1}{2} +\\dots +\\frac{1}{2n}) - 2(\\frac{1}{2} +\\frac{1}{4} +\\dots +\\frac{1}{2n}) = (1 + \\frac{1}{2} +\\dots +\\frac{1}{2n}) - (1 + \\frac{1}{2} +\\dots +\\frac{1}{n}) = (\\frac{1}{n + 1} +\\dots +\\frac{1}{2n})\\to \\ln 2$ 由于收敛数列的奇偶子列收敛到同一极限，因此极限为 $\\ln 2$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":20,"id":"a949cd30-ae36-4d5c-ba4e-7640fb3fe3ea","page_size":[595,841],"block_position":"20-12"},{"bbox":[35,508,466,557],"type":"text","angle":0,"index":14,"merge_prev":false,"text":"练习2.19经典相减法(1)重点： $A > 0, a_{1} > 0$ ， $a_{n + 1} = \\frac{1}{2} (a_n + \\frac{A}{a_n})$ ，证明 $a_{n}$ 收敛，并求极限解(1)首先 $a_{n + 1} = \\frac{1}{2} (a_n + \\frac{A}{a_n}) > \\sqrt{A}$ ，因此 $n \\geq 2$ 时， $a_{n} > \\sqrt{A}$ ，而前后项相减得到 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":20,"id":"b8901284-f42c-407e-9bd6-2180aac75e0c","page_size":[595,841],"block_position":"20-13"},{"bbox":[210,559,382,585],"type":"interline_equation","angle":0,"index":15,"text":"$$\na _ {n + 1} - a _ {n} = \\frac {1}{2} (\\frac {A}{a _ {n}} - a _ {n}) = \\frac {1}{2} \\cdot \\frac {A - a _ {n} ^ {2}}{a _ {n}}\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":20,"id":"4b553064-2b50-42ae-9cb4-d7e0c5478c66","page_size":[595,841],"block_position":"20-14"},{"bbox":[52,587,539,610],"type":"text","angle":0,"index":16,"merge_prev":false,"text":"结合 $n \\geq 2$ 时 $a_{n} > \\sqrt{A}$ 得到 $a_{n + 1} - a_{n} < 0$ ，因此单减，显然 $a_{n} > 0$ ，因此收敛。两侧取极限得到 $\\lim_{n\\to \\infty}a_n = \\sqrt{A}$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":20,"id":"708e5086-375b-4eff-9c04-356cbfd755c0","page_size":[595,841],"block_position":"20-15"},{"bbox":[55,623,127,637],"type":"title","angle":0,"index":17,"level":2,"text":"## 2.6.3 相除法","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":20,"id":"66f5bb2d-fe67-4616-aa43-563dbbb58d1d","page_size":[595,841],"block_position":"20-16"},{"bbox":[35,646,364,662],"type":"text","angle":0,"index":18,"merge_prev":false,"text":"练习2.20两道经典相除法(1)重点： $a_1 = 1, a_{n + 1} = \\sqrt{2a_n}$ ，求 $\\lim_{n\\to \\infty}a_n$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":20,"id":"9c1b50a9-d4e3-4b0b-b842-e1739b3c0fba","page_size":[595,841],"block_position":"20-17"},{"bbox":[71,661,417,685],"type":"text","angle":0,"index":19,"merge_prev":false,"text":"(2) 重点: $c > 0, 0 < x_{1} < \\frac{1}{c}, x_{n+1} = x_{n}(2 - cx_{n})$ , 证明: $x_{n}$ 收敛并求其极限 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":20,"id":"1907caa7-19dd-471d-9a97-399bc75afda6","page_size":[595,841],"block_position":"20-18"},{"bbox":[53,687,539,719],"type":"text","angle":0,"index":20,"merge_prev":false,"text":"解 (1) $\\frac{a_{n+1}}{a_n} = \\sqrt{\\frac{2}{a_n}}$ ，当 $a_n < 2$ 时， $a_{n+1} > a_n$ 。根据 $a_{n+1} = \\sqrt{2a_n}$ ，当 $a_n < 2$ 时， $a_{n+1} < 2$ ，单调有界故收敛，两侧同取极限得到极限为 $A = 2$ 。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":20,"id":"cb944afe-ac23-4975-bd6d-2f3473b7b0d3","page_size":[595,841],"block_position":"20-19"},{"bbox":[55,720,539,766],"type":"text","angle":0,"index":21,"merge_prev":false,"text":"(2) $\\frac{x_{n + 1}}{x_n} = 2 - cx_n$ ，当 $x_{n} < \\frac{1}{c}$ 时， $x_{n + 1} > x_n$ 。设 $f(x) = 2x - cx^2, f'(x) = 2 - 2cx$ ， $f(x)$ 在 $x = \\frac{1}{c}$ 取最大，因此 $x_{n} < \\frac{1}{c}$ 时 $x_{n + 1} < \\frac{1}{c}$ 有界，根据单调有界收敛，两侧取极限得到极限为 $\\frac{1}{c}$ 。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":20,"id":"84a50ff2-aad4-42e7-bf5c-5725abe04b95","page_size":[595,841],"block_position":"20-20"},{"bbox":[292,803,302,812],"type":"page_number","angle":0,"index":1000009,"text":"13 ","color":{"line":"rgba(164, 164, 164, 1)","fill":"rgba(164, 164, 164, 1)"},"page_idx":20,"is_discarded":true,"id":"16e96574-8467-424d-b6f1-fd5de4d4b201","page_size":[595,841],"block_position":"20-21"}],[{"bbox":[478,47,539,58],"type":"header","angle":0,"index":-99999,"text":"2.7 递推问题 ","color":{"line":"rgba(164, 164, 164, 1)","fill":"rgba(164, 164, 164, 1)"},"page_idx":21,"is_discarded":true,"id":"e88823db-1c3f-4c21-9556-f70565306255","page_size":[595,841],"block_position":"21-0"},{"bbox":[55,72,151,86],"type":"title","angle":0,"index":1,"level":2,"text":"## 2.6.4 不等式放缩","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":21,"id":"da24d607-17ea-4d4a-b734-c6094c5d7a03","page_size":[595,841],"block_position":"21-1"},{"bbox":[36,97,539,132],"type":"text","angle":0,"index":2,"merge_prev":false,"text":"练习 2.21 两道经典双数列 + 不等式放缩 (1) 重点: $a_1 > b_1 > 0$ , $a_{n+1} = \\frac{a_n + b_n}{2}, b_{n+1} = \\sqrt{a_nb_n}$ , 证明 $\\lim_{n \\to \\infty} a_n, \\lim_{n \\to \\infty} b_n$ 存在且相等。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":21,"id":"246113ee-aabb-47ba-af0a-4751a9bdf748","page_size":[595,841],"block_position":"21-2"},{"bbox":[62,132,498,158],"type":"text","angle":0,"index":3,"merge_prev":false,"text":"(2) 重点: $a_1 > b_1 > 0$ , $a_{n+1} = \\frac{a_n + b_n}{2}$ , $b_{n+1} = \\frac{2a_nb_n}{a_n + b_n}$ , 证明 $\\lim_{n \\to \\infty} a_n, \\lim_{n \\to \\infty} b_n$ 存在, 并求其值 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":21,"id":"f7523ae2-6630-4389-8e45-a18b4ac09d36","page_size":[595,841],"block_position":"21-3"},{"bbox":[53,156,379,169],"type":"text","angle":0,"index":4,"merge_prev":false,"text":"解(1)根据均值不等式 $a_{n + 1}\\geq b_{n + 1}$ ，根据 $a_1 > b_1$ 可知 $\\forall n,a_n > b_n$ ，从而 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":21,"id":"8f082dda-7497-4f29-8b05-c11c5adc8fea","page_size":[595,841],"block_position":"21-4"},{"bbox":[155,172,436,193],"type":"interline_equation","angle":0,"index":5,"text":"$$\na _ {n + 1} = \\frac {a _ {n} + b _ {n}}{2} \\leq \\frac {a _ {n} + a _ {n}}{2} = a _ {n}, b _ {n + 1} = \\sqrt {a _ {n} b _ {n}} \\geq \\sqrt {b _ {n} b _ {n}} = b _ {n}\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":21,"id":"5538a3d4-9d19-43ac-bfdc-e88f6182f5e0","page_size":[595,841],"block_position":"21-5"},{"bbox":[53,195,540,227],"type":"text","angle":0,"index":6,"merge_prev":false,"text":"因此 $a_{n}$ 单减， $b_{n}$ 单增，即 $a_{1} \\geq a_{2} \\geq \\cdots \\geq a_{n} \\geq b_{n} \\geq b_{n-1} \\cdots \\geq b_{1}$ ，这说明 $a_{n}, b_{n}$ 均单调有界，故均收敛。 $a_{n+1} = \\frac{a_{n} + b_{n}}{2}$ 两侧取极限可知极限相等。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":21,"id":"c87273df-9820-44a4-81bc-ed1836b76cba","page_size":[595,841],"block_position":"21-6"},{"bbox":[72,228,394,241],"type":"text","angle":0,"index":7,"merge_prev":false,"text":"(2) 根据均值不等式可知 $a_{n} \\geq b_{n}$ , 由于 $a_{1} > b_{1}$ , 因此 $\\forall n, a_{n} \\geq b_{n}$ , 因此 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":21,"id":"f09f243b-4a2e-41f7-9591-4cc882b96519","page_size":[595,841],"block_position":"21-7"},{"bbox":[149,243,442,272],"type":"interline_equation","angle":0,"index":8,"text":"$$\na _ {n + 1} = \\frac {a _ {n} + b _ {n}}{2} \\leq \\frac {a _ {n} + a _ {n}}{2} = a _ {n}, b _ {n + 1} = \\frac {2}{\\frac {1}{a _ {n}} + \\frac {1}{b _ {n}}} \\geq \\frac {2}{\\frac {1}{b _ {n}} + \\frac {1}{b _ {n}}} = b _ {n}\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":21,"id":"f42de9df-4bfd-496e-b6d0-348fdf617985","page_size":[595,841],"block_position":"21-8"},{"bbox":[53,275,539,317],"type":"text","angle":0,"index":9,"merge_prev":false,"text":"$a_{n}$ 单减， $b_{n}$ 单增，且 $a_{1} \\geq a_{2} \\geq \\cdots \\geq a_{n} \\geq b_{n} \\geq \\cdots \\geq b_{1}$ ，单调有界故均收敛。 $a_{n+1} = \\frac{a_{n} + b_{n}}{2}$ 两侧取极限可知极限满足 $a = \\frac{a + b}{2}$ ，故相等。而 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":21,"id":"8f301522-dcb3-46a0-b34a-2c6c53e265b6","page_size":[595,841],"block_position":"21-9"},{"bbox":[211,320,381,345],"type":"interline_equation","angle":0,"index":10,"text":"$$\na _ {n + 1} b _ {n + 1} = \\frac {a _ {n} + b _ {n}}{2} \\cdot \\frac {2 a _ {n} b _ {n}}{a _ {n} + b _ {n}} = a _ {n} b _ {n}\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":21,"id":"cb698b16-69a8-44f9-aea6-f8353a902d9b","page_size":[595,841],"block_position":"21-10"},{"bbox":[54,349,238,362],"type":"text","angle":0,"index":11,"merge_prev":false,"text":"因此 $ab = a_{1}b_{1}$ ，即极限为 $a = b = \\sqrt{a_1b_1}$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":21,"id":"d24edb15-f27a-463c-b5b1-b67c8db45e6c","page_size":[595,841],"block_position":"21-11"},{"bbox":[35,380,484,393],"type":"text","angle":0,"index":12,"merge_prev":false,"text":"练习 2.22 前后项不等式放缩 (1) 重点: $0 < a_{n} < 2$ , $(2 - a_{n})a_{n+1} \\geq 1$ , 证明 $a_{n}$ 收敛, 并求其极限 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":21,"id":"7e005df5-331a-4cd3-be69-624f7b28fa60","page_size":[595,841],"block_position":"21-12"},{"bbox":[73,393,348,415],"type":"text","angle":0,"index":13,"merge_prev":false,"text":"(2) 重点: $x_{n} > 0$ , $x_{n+1} + \\frac{4}{x_{n}} < 4$ , 证明 $x_{n}$ 收敛并求 $\\lim_{n \\to \\infty} x_{n}$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":21,"id":"f55bf617-b544-46d7-95a3-1857a2afd19e","page_size":[595,841],"block_position":"21-13"},{"bbox":[53,415,539,452],"type":"text","angle":0,"index":14,"merge_prev":false,"text":"解(1)显然 $a_{n}(2 - a_{n})\\leq \\left[\\frac{a_{n} + (2 - a_{n})}{2}\\right]^{2} = 1\\leq a_{n + 1}(2 - a_n)$ ，因此 $a_{n}\\leq a_{n + 1}$ ，根据单调有界定理可知收敛，两侧取极限得到极限为1 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":21,"id":"0d5dd684-2c72-4073-babb-71f7c0acf395","page_size":[595,841],"block_position":"21-14"},{"bbox":[54,453,539,491],"type":"text","angle":0,"index":15,"merge_prev":false,"text":"(2) 显然 $x_{n} + \\frac{4}{x_{n}} \\geq 2\\sqrt{x_{n} \\cdot \\frac{4}{x_{n}}} = 4 > x_{n+1} + \\frac{4}{x_{n}}$ ，因此 $x_{n} > x_{n+1}$ ，单减。另外有 $0 < x_{n+1} < x_{n+1} + \\frac{4}{x_{n}} < 4$ 可知 $x_{n}$ 有界，故收敛，两侧取极限得到极限为 2。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":21,"id":"196df9a1-2ad9-4a37-baea-03323df5269a","page_size":[595,841],"block_position":"21-15"},{"bbox":[52,491,309,504],"type":"text","angle":0,"index":16,"merge_prev":false,"text":"笔记 前后项不等式先用同项进行放缩，可得出单调性结论 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":21,"id":"d6378b99-a284-4192-a010-de1cc8382af0","page_size":[595,841],"block_position":"21-16"},{"bbox":[55,528,143,543],"type":"title","angle":0,"index":17,"level":2,"text":"## 2.7 递推问题","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":21,"id":"f8816f73-fe4d-460c-a1e1-d92cde46d139","page_size":[595,841],"block_position":"21-17"},{"bbox":[56,560,163,575],"type":"title","angle":0,"index":18,"level":2,"text":"## 2.7.1 极限压缩定理","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":21,"id":"31530235-1ee2-4a4a-a1d3-955746f8144e","page_size":[595,841],"block_position":"21-18"},{"bbox":[72,587,186,600],"type":"title","angle":0,"index":19,"level":2,"text":"## 定理 2.18 (极限压缩定理)","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":21,"id":"a3910c99-d81f-4093-86f6-8fbd370df971","page_size":[595,841],"block_position":"21-19"},{"bbox":[68,604,390,618],"type":"text","angle":0,"index":20,"merge_prev":false,"text":"$A \\in \\mathbb{R}$ , 若 $\\exists r \\in (0,1)$ 使得 $|a_n - A| < r |a_{n-1} - A|$ , 则数列 $a_n$ 收敛到 $A$ 。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":21,"id":"8072f381-7c51-47b7-8447-e7ee599e9318","page_size":[595,841],"block_position":"21-20"},{"bbox":[35,649,537,678],"type":"text","angle":0,"index":21,"merge_prev":false,"text":"练习2.23经典极限压缩定理题(1)设 $x_{n + 1} = \\cos x_n, x_1 \\in [0, \\frac{\\pi}{3}]$ ，证明 $\\lim_{n \\to \\infty} x_n$ 存在且极限为 $\\cos x - x = 0$ 的根解(1)考虑 $f(x) = \\cos x - x$ ， $f'(x) = \\sin x - 1 \\leq 0$ 单减，因此有唯一解，设 $\\cos a = a$ 。由于 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":21,"id":"2bf81a2c-5f05-434e-9af2-940ab020e56c","page_size":[595,841],"block_position":"21-21"},{"bbox":[161,685,432,698],"type":"interline_equation","angle":0,"index":23,"text":"$$\n\\left| x _ {n + 1} - a \\right| = \\left| \\cos x _ {n} - a \\right| = \\left| \\cos x _ {n} - \\cos a \\right| = \\left| \\sin \\xi \\right| \\cdot \\left| x _ {n} - a \\right|\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":21,"id":"6fc25e30-e303-4072-8cb5-1804b13a9a71","page_size":[595,841],"block_position":"21-22"},{"bbox":[54,703,283,726],"type":"text","angle":0,"index":24,"merge_prev":false,"text":"而由于 $\\xi \\in [0, \\frac{\\pi}{3}]$ , $|\\sin \\xi| < \\frac{\\sqrt{3}}{2}$ , 因此 $|x_{n} - a| \\to 0$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":21,"id":"114a20ba-862e-419a-b78a-752ae4cd0443","page_size":[595,841],"block_position":"21-23"},{"bbox":[292,803,303,811],"type":"page_number","angle":0,"index":1000021,"text":"14 ","color":{"line":"rgba(164, 164, 164, 1)","fill":"rgba(164, 164, 164, 1)"},"page_idx":21,"is_discarded":true,"id":"9240f0b5-37eb-44f2-916f-ee91918f8030","page_size":[595,841],"block_position":"21-24"}],[{"bbox":[477,47,539,58],"type":"header","angle":0,"index":-99999,"text":"2.8 Stolz 定理 ","color":{"line":"rgba(164, 164, 164, 1)","fill":"rgba(164, 164, 164, 1)"},"page_idx":22,"is_discarded":true,"id":"d1fa4e09-a281-42a7-a853-ce2011cc2fd7","page_size":[595,841],"block_position":"22-0"},{"bbox":[56,72,163,86],"type":"title","angle":0,"index":1,"level":2,"text":"## 2.7.2 压缩映射定理","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":22,"id":"dd9fbb85-fe28-418d-836a-497bb7a17b2a","page_size":[595,841],"block_position":"22-1"},{"bbox":[73,100,186,112],"type":"title","angle":0,"index":2,"level":2,"text":"## 定理 2.19 (压缩映射定理)","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":22,"id":"1f896253-8c09-4081-831d-c744337a63f1","page_size":[595,841],"block_position":"22-2"},{"bbox":[69,116,365,131],"type":"text","angle":0,"index":3,"merge_prev":false,"text":"若存在 $r \\in (0,1)$ 使得 $|a_{n} - a_{n-1}| \\leq r |a_{n} - a_{n-1}|$ ，则数列 $a_{n}$ 收敛。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":22,"id":"8c6fe7da-11bc-476a-8ee6-053642ce3961","page_size":[595,841],"block_position":"22-3"},{"bbox":[73,151,201,163],"type":"title","angle":0,"index":4,"level":2,"text":"## 推论 2.2 (导数判断压缩映射)","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":22,"id":"19f05b47-8679-418f-afb1-120ce3b0ee7c","page_size":[595,841],"block_position":"22-4"},{"bbox":[69,167,398,181],"type":"text","angle":0,"index":5,"merge_prev":false,"text":"若 $a_{n + 1} = f(a_n)$ ， $f$ 可导，且 $\\exists r\\in (0,1)$ 使得 $|f'(x)|\\leq r < 1$ ，则 $a_{n}$ 收敛。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":22,"id":"18d29ef2-eba1-4b7c-80cd-7ec961136fd5","page_size":[595,841],"block_position":"22-5"},{"bbox":[35,213,257,225],"type":"text","angle":0,"index":6,"merge_prev":false,"text":"练习 2.24 两个经典循环式 证明收敛并求极限: ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":22,"id":"f44fb42d-a16d-4adf-81a5-51bca75d39de","page_size":[595,841],"block_position":"22-6"},{"bbox":[73,226,257,247],"type":"interline_equation","angle":0,"index":7,"text":"$$\n(1) x _ {1} = 2, x _ {2} = 2 + \\frac {1}{2}, \\dots , x _ {n + 1} = 2 + \\frac {1}{a _ {n}}\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":22,"id":"f6abd44a-2800-4341-9135-49582ec5cdf5","page_size":[595,841],"block_position":"22-7"},{"bbox":[73,249,252,264],"type":"interline_equation","angle":0,"index":8,"text":"$$\n(2) x _ {1} = 1, x _ {2} = \\sqrt {2}, \\dots , x _ {n + 1} = \\sqrt {2 \\sqrt {x _ {n}}}\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":22,"id":"06f59b68-d4ee-4fcf-beee-9bd67e990e2a","page_size":[595,841],"block_position":"22-8"},{"bbox":[52,265,509,291],"type":"text","angle":0,"index":9,"merge_prev":false,"text":"解 (1) $|x_{n+1}-x_{n}|=\\left|\\frac{1}{x_{n}}-\\frac{1}{x_{n-1}}\\right|=\\frac{|x_{n-1}-x_{n}|}{|x_{n}x_{n-1}|}\\leq\\frac{1}{4}|x_{n}-x_{n-1}|$ ，因此收敛，取极限得到极限为 $1+\\sqrt{2}$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":22,"id":"b9922e45-5a9b-4859-83a3-c3d2ab15b4b8","page_size":[595,841],"block_position":"22-9"},{"bbox":[73,292,280,317],"type":"text","angle":0,"index":10,"merge_prev":false,"text":"(2) 设 $f(x) = \\sqrt{2x}$ , $|f'(x)| = \\frac{\\sqrt{2}}{2\\sqrt{x}} \\leq \\frac{\\sqrt{2}}{2} < 1$ 。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":22,"id":"dc16724d-865e-4953-a621-be63f0ef58cf","page_size":[595,841],"block_position":"22-10"},{"bbox":[56,337,148,352],"type":"title","angle":0,"index":11,"level":2,"text":"## 2.8 Stolz 定理","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":22,"id":"937ef83a-1f46-4997-9f58-afe204b7f555","page_size":[595,841],"block_position":"22-11"},{"bbox":[73,370,169,382],"type":"title","angle":0,"index":12,"level":2,"text":"## 定理 2.20 (Stolz 定理)","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":22,"id":"c1974be0-8ae8-4cf8-9435-8a94485f6bd1","page_size":[595,841],"block_position":"22-12"},{"bbox":[68,387,281,398],"type":"text","angle":0,"index":13,"merge_prev":false,"text":"Stolz 定理分为以下两种，下面的 $A$ 可以为无穷。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":22,"id":"50693b42-e39c-4653-88f0-f35cbea44eba","page_size":[595,841],"block_position":"22-13"},{"bbox":[83,399,406,423],"type":"text","angle":0,"index":14,"merge_prev":false,"text":"- $\\frac{\\infty}{\\infty}$ 型： $b_{n}$ 严格单增趋于 $\\infty$ ，且 $\\lim_{n\\to \\infty}\\frac{a_n - a_{n - 1}}{b_n - b_{n - 1}} = A$ ，则 $\\lim_{n\\to \\infty}\\frac{a_n}{b_n} = A$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":22,"id":"ba4c82a0-c869-41fb-b7a1-33f052f3c159","page_size":[595,841],"block_position":"22-14"},{"bbox":[83,423,468,446],"type":"text","angle":0,"index":15,"merge_prev":false,"text":"- $\\frac{0}{0}$ 型： $b_{n}$ 严格单减趋于0， $a_{n}$ 也收敛到0，且 $\\lim_{n\\to \\infty}\\frac{a_n - a_{n - 1}}{b_n - b_{n - 1}} = A$ ，则 $\\lim_{n\\to \\infty}\\frac{a_n}{b_n} = A$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":22,"id":"46d2aa4f-ab67-4f37-8fda-f02afe0102c3","page_size":[595,841],"block_position":"22-15"},{"bbox":[35,478,269,490],"type":"text","angle":0,"index":16,"merge_prev":false,"text":"练习 2.25 直接使用 用 Stolz 定理证明平均值定理 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":22,"id":"689ac35f-c450-4150-9964-e9da71b4e775","page_size":[595,841],"block_position":"22-16"},{"bbox":[37,489,520,513],"type":"text","angle":0,"index":17,"merge_prev":false,"text":"证明 设 $a_{n}=x_{1}+x_{2}+\\cdots+x_{n}, b_{n}=n$ ，则 $\\lim_{n\\to\\infty}\\frac{a_{n}-a_{n-1}}{b_{n}-b_{n-1}}=x_{n}\\to a$ ，则 $\\frac{a_{n}}{b_{n}}=\\frac{x_{1}+x_{2}+\\cdots+x_{n}}{n}\\to a$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":22,"id":"4cf017e1-48d3-4f6b-94c7-f1af1705fbf7","page_size":[595,841],"block_position":"22-17"},{"bbox":[33,515,438,528],"type":"text","angle":0,"index":18,"merge_prev":false,"text":"笔记 由于 Stolz 定理可以证出平均值定理，在使用平均值定理时就不需要使用放缩法了。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":22,"id":"8e0ac6a5-d6fd-4c41-9654-41eac10a262e","page_size":[595,841],"block_position":"22-18"},{"bbox":[35,533,372,545],"type":"text","angle":0,"index":19,"merge_prev":false,"text":"练习 2.26 几道经典题目 (1) ZJU2022.2: $x_{0} > 0, x_{n} = \\arctan x_{n-1}$ ，证明： ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":22,"id":"282c5f15-4bbb-4411-b68e-902d6843ef07","page_size":[595,841],"block_position":"22-19"},{"bbox":[74,549,141,562],"type":"text","angle":0,"index":20,"merge_prev":false,"text":"(a) $\\lim_{n\\to \\infty}x_n = 0$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":22,"id":"3efe2a5a-54c4-46e7-80ea-192c44520675","page_size":[595,841],"block_position":"22-20"},{"bbox":[73,563,186,576],"type":"text","angle":0,"index":21,"merge_prev":false,"text":"(b) $\\sqrt[n]{nx_{n}}$ 收敛，并求极限 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":22,"id":"c19a0089-27d3-4a3e-8d51-b5b83882849c","page_size":[595,841],"block_position":"22-21"},{"bbox":[52,579,508,592],"type":"text","angle":0,"index":22,"merge_prev":false,"text":"解(1)(a)由于 $x_{n} = \\arctan x_{n - 1} < x_{n - 1}$ 因此单减，根据单调有界定理则极限存在，且两侧取极限得到零 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":22,"id":"f268cf43-7514-4bfb-b653-9a43aec1b3ed","page_size":[595,841],"block_position":"22-22"},{"bbox":[73,592,315,618],"type":"text","angle":0,"index":23,"merge_prev":false,"text":"(b) 考虑 $\\lim_{n\\to \\infty}nx_n^2$ ，由于 $nx_{n}^{2} = \\frac{n}{\\frac{1}{x_{n}^{2}}}$ ，根据Stolz定理即 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":22,"id":"3873b298-45d5-4faa-b346-f9e2366285d3","page_size":[595,841],"block_position":"22-23"},{"bbox":[78,619,514,650],"type":"interline_equation","angle":0,"index":24,"text":"$$\n\\lim _ {n \\to \\infty} \\frac {1}{\\frac {1}{x _ {n} ^ {2}} - \\frac {1}{x _ {n - 1} ^ {2}}} = \\lim _ {n \\to \\infty} \\frac {x _ {n - 1} ^ {2} x _ {n} ^ {2}}{x _ {n - 1} ^ {2} - x _ {n} ^ {2}} = \\lim _ {n \\to \\infty} \\frac {x _ {n - 1} ^ {2} \\arctan^ {2} x _ {n - 1}}{x _ {n - 1} ^ {2} - \\arctan^ {2} x _ {n - 1}} = \\frac {x _ {n - 1} ^ {2} (x _ {n - 1} ^ {2} + \\frac {2}{3} x _ {n - 1} ^ {4} + \\frac {1}{9} x _ {n - 1} ^ {6})}{- \\frac {2}{3} x _ {n - 1} ^ {4} - \\frac {1}{9} x _ {n - 1} ^ {6}} \\frac {3}{2}\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":22,"id":"b5033f7f-f6e6-40d2-9828-62dbf827f4c2","page_size":[595,841],"block_position":"22-24"},{"bbox":[55,656,157,681],"type":"text","angle":0,"index":25,"merge_prev":false,"text":"因此 $\\lim_{n\\to \\infty}\\sqrt{n} x_n = \\frac{\\sqrt{6}}{2}$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":22,"id":"6f012c35-17e4-49c1-86a7-b72e5339f3f0","page_size":[595,841],"block_position":"22-25"},{"bbox":[292,803,302,811],"type":"page_number","angle":0,"index":1000025,"text":"15 ","color":{"line":"rgba(164, 164, 164, 1)","fill":"rgba(164, 164, 164, 1)"},"page_idx":22,"is_discarded":true,"id":"c832f75d-6be3-46db-a3c4-5c85ba8667cb","page_size":[595,841],"block_position":"22-26"}],[{"bbox":[419,47,539,58],"type":"header","angle":0,"index":-99999,"text":"2.9 积分形式极限：分段法 ","color":{"line":"rgba(164, 164, 164, 1)","fill":"rgba(164, 164, 164, 1)"},"page_idx":23,"is_discarded":true,"id":"60233f37-4692-42c5-9566-409c85e66305","page_size":[595,841],"block_position":"23-0"},{"bbox":[55,74,230,90],"type":"title","angle":0,"index":1,"level":2,"text":"## 2.9 积分形式极限：分段法","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":23,"id":"6fe4dd31-620d-4ca0-8e96-d523b63a454d","page_size":[595,841],"block_position":"23-1"},{"bbox":[55,106,163,121],"type":"title","angle":0,"index":2,"level":2,"text":"## 2.9.1 定积分形极限","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":23,"id":"0ccbbe92-c654-4778-886d-65292c6f4770","page_size":[595,841],"block_position":"23-2"},{"bbox":[35,131,208,142],"type":"text","angle":0,"index":3,"merge_prev":false,"text":"练习 2.27 基础拆分法 求下列极限: ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":23,"id":"45d3e7c3-e993-48bf-88c4-ba8c266ea29d","page_size":[595,841],"block_position":"23-3"},{"bbox":[88,143,505,174],"type":"text","angle":0,"index":4,"merge_prev":false,"text":"(1) $\\lim_{n\\to \\infty}\\int_0^{\\frac{\\pi}{2}}\\sin^n x\\mathrm{d}x$ (2) $\\lim_{n\\to \\infty}\\int_0^{\\frac{\\pi}{2}}(1 - \\sin x)^n\\mathrm{d}x$ (3) $\\lim_{n\\to \\infty}\\int_0^1\\frac{x^n}{1 + \\sqrt{x}}\\mathrm{d}x$ (4) $\\lim_{n\\to \\infty}\\int_0^1 e^{x^n}\\mathrm{d}x$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":23,"id":"6d8651a9-1a9f-4c8d-9c89-8d112fef5541","page_size":[595,841],"block_position":"23-4"},{"bbox":[52,175,432,196],"type":"text","angle":0,"index":5,"merge_prev":false,"text":"解 (1) 显然 $x \\in [0, \\frac{\\pi}{2})$ 时， $\\lim_{n \\to \\infty} \\sin^n x = 0$ ，但 $x = \\frac{\\pi}{2}$ 时， $\\sin x = 1$ ，因此做以下拆分： ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":23,"id":"ca407177-1df1-479b-a38d-c7da1a8dc686","page_size":[595,841],"block_position":"23-5"},{"bbox":[190,197,402,227],"type":"interline_equation","angle":0,"index":6,"text":"$$\n\\int_ {0} ^ {\\frac {\\pi}{2}} \\sin^ {n} x \\mathrm{d} x = \\int_ {0} ^ {\\frac {\\pi}{2} - \\delta} \\sin^ {n} x \\mathrm{d} x + \\int_ {\\frac {\\pi}{2} - \\delta} ^ {\\frac {\\pi}{2}} \\sin^ {n} x \\mathrm{d} x\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":23,"id":"027ef39c-a2e8-412f-8acf-e5f5e86c61e8","page_size":[595,841],"block_position":"23-6"},{"bbox":[53,230,506,260],"type":"text","angle":0,"index":7,"merge_prev":false,"text":"前者 $\\int_0^{\\frac{\\pi}{2} -\\delta}\\sin^n x\\mathrm{d}x\\leq \\frac{\\pi}{2}\\sin^n (\\frac{\\pi}{2} -\\delta)\\to 0$ ，后者 $\\int_{\\frac{\\pi}{2} -\\delta}^{\\frac{\\pi}{2}}\\sin^n x\\mathrm{d}x\\leq \\int_{\\frac{\\pi}{2} -\\delta}^{\\frac{\\pi}{2}}1\\mathrm{d}x = \\delta$ ，因此积分极限等于0。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":23,"id":"ad18d6b3-01cc-42ab-84aa-8dc8be604966","page_size":[595,841],"block_position":"23-7"},{"bbox":[73,259,151,270],"type":"text","angle":0,"index":8,"merge_prev":false,"text":"(2)(3) 同理都是 0 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":23,"id":"785a8642-63fd-4513-a908-7bde6beaa08e","page_size":[595,841],"block_position":"23-8"},{"bbox":[73,270,243,297],"type":"text","angle":0,"index":9,"merge_prev":false,"text":"(4) 可以考虑 $\\int_{0}^{1} e^{x^{n}} - 1 dx$ ，结果为 1. ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":23,"id":"918f2e4b-4670-4605-8529-09b1c8e6bd80","page_size":[595,841],"block_position":"23-9"},{"bbox":[35,301,415,329],"type":"text","angle":0,"index":10,"merge_prev":false,"text":"练习2.28用分段法证明命题(1) $f(x)$ 在 $[0,1]$ 连续，证明 $\\lim_{n\\to \\infty}\\int_0^1 f(\\sqrt[n]{x})\\mathrm{d}x = f(1)$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":23,"id":"693a75dc-2895-4c75-af53-066c098a6a89","page_size":[595,841],"block_position":"23-10"},{"bbox":[55,344,175,359],"type":"title","angle":0,"index":11,"level":2,"text":"## 2.9.2 反常积分形极限","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":23,"id":"707d6ece-553f-4fb3-8cf8-0e308581dace","page_size":[595,841],"block_position":"23-11"},{"bbox":[72,371,217,385],"type":"title","angle":0,"index":12,"level":2,"text":"## 定理 2.21 (函数形式平均值定理)","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":23,"id":"7ca9b1c1-3083-4b5e-8a10-5ccdc8783b91","page_size":[595,841],"block_position":"23-12"},{"bbox":[67,388,371,405],"type":"text","angle":0,"index":13,"merge_prev":false,"text":"$f(x)$ 在 $[a, +\\infty)$ 有定义且在任意有限区间可积， $\\lim_{x\\to +\\infty}f(x) = A$ ，则 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":23,"id":"a2b58c56-2668-4258-9545-656238b56893","page_size":[595,841],"block_position":"23-13"},{"bbox":[243,410,350,436],"type":"interline_equation","angle":0,"index":14,"text":"$$\n\\lim _ {x \\to + \\infty} \\frac {1}{x} \\int_ {0} ^ {x} f (t) \\mathrm{d} t = A\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":23,"id":"ee73894f-27bb-49cc-b970-db6964fbcf9e","page_size":[595,841],"block_position":"23-14"},{"bbox":[53,445,308,471],"type":"text","angle":0,"index":15,"merge_prev":false,"text":"证明 先斩后奏，已知 $A = \\lim_{x\\to +\\infty}\\frac{1}{x}\\int_{a}^{x}Adt$ ，因此只需证 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":23,"id":"889f7805-f3bb-4031-af75-72968910c397","page_size":[595,841],"block_position":"23-15"},{"bbox":[147,473,446,501],"type":"interline_equation","angle":0,"index":16,"text":"$$\n\\lim _ {x \\rightarrow + \\infty} \\left(\\frac {1}{x} \\int_ {a} ^ {x} f (t) \\mathrm{d} t - \\frac {1}{x} \\int_ {a} ^ {x} A \\mathrm{d} t\\right) = \\lim _ {x \\rightarrow + \\infty} \\frac {1}{x} \\int_ {a} ^ {x} (f (t) - A) \\mathrm{d} t = 0\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":23,"id":"42432caa-32f8-446c-9b8a-2ef264f6535a","page_size":[595,841],"block_position":"23-16"},{"bbox":[52,503,436,530],"type":"text","angle":0,"index":17,"merge_prev":false,"text":"$\\forall \\epsilon, \\exists M, x \\geq M$ 时 $|f(x) - A| < \\epsilon$ ，因此可有 $N, x > N$ 时 $\\frac{1}{x} \\int_{a}^{M} |f(t) - A| \\mathrm{d}t < \\epsilon$ ，得到 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":23,"id":"959e4143-af17-445b-ad06-2581cdcfd16b","page_size":[595,841],"block_position":"23-17"},{"bbox":[111,533,481,561],"type":"interline_equation","angle":0,"index":18,"text":"$$\n\\left| \\frac {1}{x} \\int_ {a} ^ {x} (f (t) - A) \\mathrm{d} t \\right| \\leq \\frac {1}{x} \\int_ {a} ^ {M} | f (t) - A | \\mathrm{d} t + \\frac {1}{x} \\int_ {M} ^ {x} | f (t) - A | \\mathrm{d} t < \\epsilon + \\frac {x - M}{x} \\cdot \\epsilon < 2 \\epsilon\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":23,"id":"6d34c561-5b17-490e-8cd9-13fbbad20441","page_size":[595,841],"block_position":"23-18"},{"bbox":[54,563,117,575],"type":"text","angle":0,"index":19,"merge_prev":false,"text":"因此结论成立 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":23,"id":"25c83d75-c021-42c3-8e5a-f58c74493438","page_size":[595,841],"block_position":"23-19"},{"bbox":[33,568,365,600],"type":"text","angle":0,"index":20,"merge_prev":false,"text":"因此结论成立笔记如果有说明 $f(x)$ 连续，则 $\\int_{a}^{x}f(t)\\mathrm{d}t$ 可导，可用洛必达法则证明。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":23,"id":"dc0192d0-b7c3-4468-a165-26d7e0da3ba1","page_size":[595,841],"block_position":"23-20"},{"bbox":[35,602,540,653],"type":"text","angle":0,"index":21,"merge_prev":false,"text":"练习2.29相关练习已知 $f(x)$ 在 $[0, +\\infty)$ 非负，对 $\\forall A > 0$ ， $xf(x)$ 在 $[0, A]$ 可积，且 $\\int_0^{+\\infty} f(x) \\mathrm{d}x$ 可积，证明： $\\lim_{A \\to +\\infty} \\frac{1}{A} \\int_0^A xf(x) \\mathrm{d}x = 0$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":23,"id":"5934224b-2ddc-49df-aacc-d9219e7a1615","page_size":[595,841],"block_position":"23-21"},{"bbox":[52,652,540,707],"type":"text","angle":0,"index":22,"merge_prev":false,"text":"证明 由于 $\\int_0^{+\\infty}f(x)\\mathrm{d}x$ 收敛，因此 $\\forall \\epsilon ,\\exists M,\\forall A^{\\prime \\prime}\\geq A^{\\prime}\\geq M$ 都有 $0\\leq \\int_{A^{\\prime}}^{A^{\\prime \\prime}}f(x)\\mathrm{d}x < \\epsilon$ ，固定上面的 $M$ $\\exists N,\\forall A > N$ 有 $0 < \\frac{1}{A}\\int_0^M xf(x)\\mathrm{d}x < \\epsilon$ ，进而 $A > N$ 时 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":23,"id":"cd08add5-4afe-489d-89d1-3e1a2a412172","page_size":[595,841],"block_position":"23-22"},{"bbox":[111,709,481,739],"type":"interline_equation","angle":0,"index":23,"text":"$$\n0 < \\frac {1}{A} \\int_ {0} ^ {A} x f (x) \\mathrm{d} x = \\frac {1}{A} \\int_ {0} ^ {M} x f (x) \\mathrm{d} x + \\frac {1}{A} \\int_ {M} ^ {A} x f (x) \\mathrm{d} x < \\epsilon + \\frac {1}{A} \\int_ {M} ^ {A} A f (x) \\mathrm{d} x < 2 \\epsilon\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":23,"id":"0a1f6d97-6f92-4120-ae39-8a0de995cee8","page_size":[595,841],"block_position":"23-23"},{"bbox":[292,803,303,812],"type":"page_number","angle":0,"index":1000023,"text":"16 ","color":{"line":"rgba(164, 164, 164, 1)","fill":"rgba(164, 164, 164, 1)"},"page_idx":23,"is_discarded":true,"id":"eae197c5-d1e5-4c61-a79b-9ef46d7a7ee5","page_size":[595,841],"block_position":"23-24"}],[{"bbox":[419,47,539,58],"type":"header","angle":0,"index":-99999,"text":"2.9 积分形式极限：分段法 ","color":{"line":"rgba(164, 164, 164, 1)","fill":"rgba(164, 164, 164, 1)"},"page_idx":24,"is_discarded":true,"id":"2a4c0ad0-0cd4-4641-a237-bfa3c6deffc7","page_size":[595,841],"block_position":"24-0"},{"bbox":[55,72,188,86],"type":"title","angle":0,"index":1,"level":2,"text":"## 2.9.3 周期函数积分极限","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":24,"id":"d94daef5-b3ac-479f-bc1d-dd56e56dfa4c","page_size":[595,841],"block_position":"24-1"},{"bbox":[35,95,442,122],"type":"text","angle":0,"index":2,"merge_prev":false,"text":"练习2.30 $f(x)$ 是 $\\mathbb{R}$ 上周期为 $T$ 的可积函数，证明： $\\forall a\\in \\mathbb{R},\\int_0^T f(x)\\mathrm{d}x = \\int_a^{a + T}f(x)\\mathrm{d}x$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":24,"id":"d818fca6-957d-41fc-8930-7e55c4ccc41e","page_size":[595,841],"block_position":"24-2"},{"bbox":[52,121,418,149],"type":"text","angle":0,"index":3,"merge_prev":false,"text":"证明 令 $x = T + u$ ，则 $\\int_{T}^{a + T} f(x) \\mathrm{d}x = \\int_{0}^{a} f(T + u) \\mathrm{d}(T + u) = \\int_{0}^{a} f(x) \\mathrm{d}x$ ，因此 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":24,"id":"464258b9-ebdf-4cd4-9894-4b7c7d485ab3","page_size":[595,841],"block_position":"24-3"},{"bbox":[138,151,455,179],"type":"interline_equation","angle":0,"index":4,"text":"$$\n\\int_ {a} ^ {a + T} f (x) \\mathrm{d} x = \\int_ {a} ^ {0} f (x) \\mathrm{d} x + \\int_ {0} ^ {T} f (x) \\mathrm{d} x + \\int_ {T} ^ {T + a} f (x) \\mathrm{d} x = \\int_ {0} ^ {T} f (x) \\mathrm{d} x\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":24,"id":"ddc4a131-bc81-449a-9598-acbe6e5fd720","page_size":[595,841],"block_position":"24-4"},{"bbox":[35,188,449,236],"type":"text","angle":0,"index":5,"merge_prev":false,"text":"练习2.31 $f(x)$ 是周期为 $T$ 的连续函数，证明： $\\lim_{x\\to +\\infty}\\frac{1}{x}\\int_0^x f(t)\\mathrm{d}t = \\frac{1}{T}\\int_0^T f(t)\\mathrm{d}t$ 证明对 $\\forall x > 0$ ，都 $\\exists n_x\\in \\mathbb{N}$ 使得 $x - n_{x}T\\in [0,T)$ ，此时 $\\lim_{x\\to \\infty}\\frac{n_x}{x} = \\frac{1}{T}$ 。根据周期性可知 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":24,"id":"88d3b91f-0eb1-4d20-a0ac-0765fccb1a63","page_size":[595,841],"block_position":"24-5"},{"bbox":[177,238,415,266],"type":"interline_equation","angle":0,"index":6,"text":"$$\n\\lim _ {x \\rightarrow + \\infty} \\frac {1}{x} \\int_ {0} ^ {n _ {x} T} f (t) \\mathrm{d} t = \\frac {n _ {x}}{x} \\int_ {0} ^ {T} f (t) \\mathrm{d} t = \\frac {1}{T} \\int_ {0} ^ {T} f (t) \\mathrm{d} t\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":24,"id":"89d0d330-e7dd-4d4c-8677-8c042cea7ec2","page_size":[595,841],"block_position":"24-6"},{"bbox":[53,268,539,305],"type":"text","angle":0,"index":7,"merge_prev":false,"text":"由于 $||f(x)| - |f(y)|| \\leq |f(x) - f(y)|$ ，从振幅角度来看，若 $f(x)$ 可积，则 $|f(x)|$ 也可积，且显然 $|f(x)|$ 也是周期函数。因此 $\\int_0^T |f(x)|\\mathrm{d}x$ 有界，此时： ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":24,"id":"9182bb1c-4a07-46a6-ad0a-9fd60c6e3238","page_size":[595,841],"block_position":"24-7"},{"bbox":[179,309,413,339],"type":"interline_equation","angle":0,"index":8,"text":"$$\n\\int_ {n _ {x} T} ^ {x} f (t) \\mathrm{d} t \\leq \\int_ {n _ {x} T} ^ {(n _ {x} + 1) T} f (x) \\mathrm{d} x \\leq \\int_ {0} ^ {T} | f (x) | \\mathrm{d} x \\leq M\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":24,"id":"12fc1760-0b25-4414-9fb6-b7a141067bf4","page_size":[595,841],"block_position":"24-8"},{"bbox":[54,340,188,367],"type":"text","angle":0,"index":9,"merge_prev":false,"text":"因此 $\\lim_{x\\to \\infty}\\frac{1}{x}\\int_{n_xT}^x f(x)\\mathrm{d}x = 0$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":24,"id":"0a672f9e-3741-40f9-a0b4-eeedda6beeef","page_size":[595,841],"block_position":"24-9"},{"bbox":[35,378,305,390],"type":"text","angle":0,"index":10,"merge_prev":false,"text":"练习 2.32 进阶难度 $f(x) \\geq 0$ ，周期为 T，且连续，证明： ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":24,"id":"04e786db-d23a-44e0-805b-370e26ada756","page_size":[595,841],"block_position":"24-10"},{"bbox":[54,423,78,435],"type":"text","angle":0,"index":11,"merge_prev":false,"text":"证明 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":24,"id":"f68034c1-54ed-4246-ab35-09663f1678ce","page_size":[595,841],"block_position":"24-11"},{"bbox":[213,393,381,421],"type":"interline_equation","angle":0,"index":12,"text":"$$\n\\lim _ {n \\rightarrow \\infty} n \\int_ {n} ^ {+ \\infty} \\frac {f (x)}{x ^ {2}} \\mathrm{d} x = \\frac {1}{T} \\int_ {0} ^ {T} f (x) \\mathrm{d} x\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":24,"id":"af60b1ad-9b4b-413c-abe2-13b0423117a3","page_size":[595,841],"block_position":"24-12"},{"bbox":[292,802,302,811],"type":"page_number","angle":0,"index":1000012,"text":"17 ","color":{"line":"rgba(164, 164, 164, 1)","fill":"rgba(164, 164, 164, 1)"},"page_idx":24,"is_discarded":true,"id":"a1f05b05-8d0a-4015-9d0d-f0611af5ca3d","page_size":[595,841],"block_position":"24-13"}],[{"bbox":[35,391,47,407],"type":"image","angle":0,"index":0,"img_path":"/8eb8fb666bbaa9a816146c4a4a4c8ebfda7fa464a434b62b0448bbe8a2aa3b16.jpg","color":{"line":"rgba(89, 92, 220, 1)","fill":"rgba(89, 92, 220, 1)"},"img_ratio":0.2,"page_idx":25,"id":"e3f2e2a0-acd2-4057-9962-24c3e301b0c7","page_size":[595,841],"block_position":"25-0"},{"bbox":[201,74,402,94],"type":"title","angle":0,"index":1,"level":1,"text":"# 第 3 章 函数极限与连续性","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":25,"id":"dbc8ed0a-b6af-48af-933e-ae6276f26d5f","page_size":[595,841],"block_position":"25-1"},{"bbox":[55,122,158,139],"type":"title","angle":0,"index":2,"level":2,"text":"## 3.1 函数的极限","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":25,"id":"c819ea6d-7c0a-4a19-955b-7c3edeca235c","page_size":[595,841],"block_position":"25-2"},{"bbox":[55,155,163,169],"type":"title","angle":0,"index":3,"level":2,"text":"## 3.1.1 函数极限定义","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":25,"id":"7759eb33-95c8-4c13-a872-8681ff6b616b","page_size":[595,841],"block_position":"25-3"},{"bbox":[71,183,161,195],"type":"title","angle":0,"index":4,"level":2,"text":"## 定义 3.1 (函数极限)","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":25,"id":"039cc13b-683c-44b2-98db-4fe0bfba7ad6","page_size":[595,841],"block_position":"25-4"},{"bbox":[68,198,534,233],"type":"text","angle":0,"index":5,"merge_prev":false,"text":"$x_0 \\in \\mathbb{R}$ , $f$ 在 $x_0$ 邻域有定义 ( $x_0$ 处可以没有), $\\forall \\epsilon, \\exists \\delta$ , s.t. $\\forall x \\in [x_0 - \\delta, x_0 + \\delta] - \\{x_0\\}$ 有 $|f(x) - A| < \\epsilon$ , 则称 $A$ 为 $f(x)$ 在 $x_0$ 处极限。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":25,"id":"44890504-3ca3-45cd-9242-9baca5a21f65","page_size":[595,841],"block_position":"25-5"},{"bbox":[35,260,431,277],"type":"text","angle":0,"index":6,"merge_prev":false,"text":"练习3.1Riemann函数Riemann函数定义如下，证明对 $\\forall x_0\\in [0,1]$ ，有 $\\lim_{x\\to x_0}R(x) = 0$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":25,"id":"8b9c2f15-0aa6-4300-86a3-1e2910390a09","page_size":[595,841],"block_position":"25-6"},{"bbox":[188,282,403,325],"type":"text","angle":0,"index":7,"merge_prev":false,"text":"$R(x) = \\left\\{ \\begin{array}{ll}\\frac{1}{q}, & x = \\frac{p}{q} (\\frac{p}{q}\\text{是既约真分数})\\\\ 0, & x\\text{是}(0,1)\\text{中的无理数或者}0,1\\text{本身} \\end{array} \\right.$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":25,"id":"1ba5bd20-fb4c-4e7c-8635-a725507de833","page_size":[595,841],"block_position":"25-7"},{"bbox":[52,327,439,354],"type":"text","angle":0,"index":8,"merge_prev":false,"text":"证明 $\\forall\\epsilon$ ，满足 $\\frac{1}{q}\\geq\\epsilon$ 的正整数 q 只有有限个，从而 $\\frac{p}{q}$ 有限，令这些数为 $x_{1},\\cdots,x_{k}$ ，取 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":25,"id":"16b17031-6738-46f7-9795-e04a70ba6f19","page_size":[595,841],"block_position":"25-8"},{"bbox":[218,354,373,368],"type":"interline_equation","angle":0,"index":9,"text":"$$\n\\delta = \\min \\left\\{\\left| x _ {i} - x _ {0} \\right|: i = 1, 2, \\dots , k \\right\\}\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":25,"id":"eedc0945-a63b-4e1c-9b41-24421d97c3c0","page_size":[595,841],"block_position":"25-9"},{"bbox":[54,374,329,389],"type":"text","angle":0,"index":10,"merge_prev":false,"text":"当 $x$ 属于 $x_0$ 的 $\\delta$ 邻域时，显然 $R(x) < \\epsilon$ ，因此 $\\lim_{x\\to x_0}R(x) = 0$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":25,"id":"61c59138-8cc2-425f-b691-da830a49c6cd","page_size":[595,841],"block_position":"25-10"},{"bbox":[52,395,539,424],"type":"text","angle":0,"index":11,"merge_prev":false,"text":"笔记 上述练习说明 Riemann 函数在每个点极限为 0，在 $(0,1)$ 的无理点和 0,1 连续，在 $(0,1)$ 有理点间断。这说明 Riemann 函数是几乎处处连续的，根据实变可知其 Riemann 可积。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":25,"id":"bceab60e-d3d8-4e8b-9586-299e2c768f27","page_size":[595,841],"block_position":"25-11"},{"bbox":[72,432,182,444],"type":"title","angle":0,"index":12,"level":2,"text":"## 定理 3.1 (极限的唯一性)","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":25,"id":"4500f181-dc5b-4416-ab14-a67d82653cbe","page_size":[595,841],"block_position":"25-12"},{"bbox":[69,449,226,461],"type":"text","angle":0,"index":13,"merge_prev":false,"text":"若函数的极限存在，则它是唯一的。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":25,"id":"617940bb-3425-487c-a5aa-e88dd260f5d0","page_size":[595,841],"block_position":"25-13"},{"bbox":[54,477,315,491],"type":"text","angle":0,"index":14,"merge_prev":false,"text":"证明 反设有两个极限，取 $\\epsilon$ 为差距的一半，根据定义可证。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":25,"id":"64cc1ac3-263c-4a3a-b945-15a96fb91227","page_size":[595,841],"block_position":"25-14"},{"bbox":[55,509,174,523],"type":"title","angle":0,"index":15,"level":2,"text":"## 3.1.2 函数极限的性质","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":25,"id":"ba6bc487-bc33-4bd1-a17f-1c80020c34ba","page_size":[595,841],"block_position":"25-15"},{"bbox":[73,536,201,549],"type":"title","angle":0,"index":16,"level":2,"text":"## 定理 3.2 (函数极限四则运算)","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":25,"id":"7e9199c2-45d3-46b5-8152-a9aeab04008c","page_size":[595,841],"block_position":"25-16"},{"bbox":[67,553,526,589],"type":"text","angle":0,"index":17,"merge_prev":false,"text":"若 $f(x), g(x)$ 在 $x_0$ 处极限存在， $f + g, f - g, f \\times g, \\frac{f}{g}$ 极限存在且极限为对应极限的运算 (除法要求分母极限非 0)。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":25,"id":"cb3ea788-a20f-486c-a946-d088a3dd7835","page_size":[595,841],"block_position":"25-17"},{"bbox":[35,605,434,622],"type":"text","angle":0,"index":18,"merge_prev":false,"text":"练习3.2函数绝对值的极限设 $\\lim_{x\\to x_0}f(x) = A$ ，则 $\\lim_{x\\to x_0}|f(x)| = |A|$ ，反之不一定成立。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":25,"id":"2c948dde-1ef8-48cd-809d-af217337a1a8","page_size":[595,841],"block_position":"25-18"},{"bbox":[45,619,539,654],"type":"text","angle":0,"index":19,"merge_prev":false,"text":"证明 由于 $\\lim_{x\\to x_0}f(x) = A$ ， $\\forall \\epsilon ,\\exists \\delta ,\\forall x\\in U(x_0,\\delta)$ 有 $|f(x) - A| < \\epsilon$ ，而 $||f(x)| - |A||\\leq |f(x) - A|$ ，因此极限存在。反之有反例 $f(x) = \\mathrm{sgn}(x)$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":25,"id":"5421299c-2b75-4f1e-8a7f-156061f1cb63","page_size":[595,841],"block_position":"25-19"},{"bbox":[73,662,161,674],"type":"title","angle":0,"index":20,"level":2,"text":"## 定理 3.3 (归结原则)","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":25,"id":"cd088d6d-61bd-4a0e-8db2-5b0b25221fa2","page_size":[595,841],"block_position":"25-20"},{"bbox":[67,679,380,698],"type":"text","angle":0,"index":21,"merge_prev":false,"text":"$\\lim_{x\\to x_{0}}f(x)=A$ 当且仅当任意收敛到 $x_{0}$ 的数列 $x_{n}$ ，有 $\\lim_{n\\to\\infty}f(x_{n})=A$ 。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":25,"id":"acc4afac-2cc6-4cfc-a086-8fbbccf77bb3","page_size":[595,841],"block_position":"25-21"},{"bbox":[54,714,539,744],"type":"text","angle":0,"index":22,"merge_prev":false,"text":"证明 (1) 左推右: $\\forall \\epsilon, \\exists \\delta, |x - x_0| < \\delta$ 时 $|f(x) - A| < \\epsilon$ , 对于 $x_n$ , 由于 $\\lim_{n \\to \\infty} x_n = x_0$ , $\\exists N, \\forall n > N, |x_n - x_0| < \\delta$ , 因此 $|f(x_n) - A| < \\epsilon$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":25,"id":"87c9e386-64c0-48ee-9d90-cab0ed01a203","page_size":[595,841],"block_position":"25-22"}],[{"bbox":[449,47,539,58],"type":"header","angle":0,"index":-99998,"text":"3.2 函数极限的计算 ","color":{"line":"rgba(164, 164, 164, 1)","fill":"rgba(164, 164, 164, 1)"},"page_idx":26,"is_discarded":true,"id":"00d5402e-f826-4075-8724-968907106a6d","page_size":[595,841],"block_position":"26-0"},{"bbox":[54,69,539,109],"type":"text","angle":0,"index":2,"merge_prev":false,"text":"(2) 右推左：反设 $\\lim_{x\\to x_{0}}f(x)\\neq A$ ，则 $\\exists\\epsilon_{0},\\forall\\delta,\\exists x\\in U(x_{0},\\delta)$ 满足 $|f(x)-A|>\\epsilon_{0}$ ，每次取 $\\delta'=\\delta,\\frac{\\delta}{2},\\cdots,\\frac{\\delta}{n}$ ，每个 $\\delta'$ 邻域对应一个 $x_{n}$ ，但是总有 $|f(x_{n})-A|>\\epsilon_{0}$ ，这与 $\\lim_{n\\to\\infty}f(x_{n})=A$ 矛盾。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":26,"id":"3b73b074-7dbc-4b8a-9a6d-34223194288a","page_size":[595,841],"block_position":"26-1"},{"bbox":[35,118,307,139],"type":"text","angle":0,"index":3,"merge_prev":false,"text":"练习3.3用归结原则证明不收敛 证明： $\\lim_{x\\to 0}\\sin \\frac{1}{x}$ 不存在。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":26,"id":"01447ed6-ad7b-45c6-874c-94fbd990b62a","page_size":[595,841],"block_position":"26-2"},{"bbox":[54,140,487,166],"type":"text","angle":0,"index":4,"merge_prev":false,"text":"证明 令 $x_{n} = \\frac{1}{(2n + \\frac{1}{2})\\pi}, x_{n}^{\\prime} = \\frac{1}{2n\\pi}$ ，两者均收敛至0，但是 $f(x_{n}) \\to 1$ ， $f(x_{n}^{\\prime}) \\to 0$ ，从而无极限。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":26,"id":"0c6b8b16-25a6-4e91-b4a5-aa57862ef56a","page_size":[595,841],"block_position":"26-3"},{"bbox":[55,182,174,196],"type":"title","angle":0,"index":5,"level":2,"text":"## 3.1.3 无穷大与无穷小","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":26,"id":"12667f82-4518-4ac7-aaa1-152a4e9cfa67","page_size":[595,841],"block_position":"26-4"},{"bbox":[73,210,149,222],"type":"title","angle":0,"index":6,"level":2,"text":"## 定义 3.2 (O 与 o)","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":26,"id":"e6014ef7-e328-4988-bbde-d3cc0e5c30dd","page_size":[595,841],"block_position":"26-5"},{"bbox":[67,225,526,264],"type":"text","angle":0,"index":7,"merge_prev":false,"text":"$f, g$ 在 $x_0$ 附近有定义， $g(x) \\neq 0$ ，若 $\\lim_{x \\to x_0} \\frac{|f(x)|}{|g(x)|} \\leq M$ ，则记 $f(x) = O(g(x))$ ；若 $\\lim_{x \\to x_0} \\frac{|f(x)|}{|g(x)|} = 0$ ，则记 $f(x) = o(g(x))$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":26,"id":"c4764e38-7012-4e83-a580-c079612e4b33","page_size":[595,841],"block_position":"26-6"},{"bbox":[73,285,129,296],"type":"title","angle":0,"index":8,"level":2,"text":"## 定义 3.3 (Ω)","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":26,"id":"cdf47692-274a-499e-b8eb-5a333e61d03f","page_size":[595,841],"block_position":"26-7"},{"bbox":[67,301,448,315],"type":"text","angle":0,"index":9,"merge_prev":false,"text":"$f, g$ 是两个算法，若 $\\exists C \\in \\mathbb{R}^{+}$ 使得 $n > n_{0}$ 时 $0 \\leq Cg(n) \\leq f(n)$ ，则记 $f(n) = \\Omega(g(n))$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":26,"id":"590cdb76-968b-485b-bdcd-f66f7850148f","page_size":[595,841],"block_position":"26-8"},{"bbox":[73,335,129,348],"type":"title","angle":0,"index":10,"level":2,"text":"## 定义 3.4 ( $\\Theta$ )","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":26,"id":"e1a24fc1-2833-418b-bfb4-3875e91b29af","page_size":[595,841],"block_position":"26-9"},{"bbox":[68,352,314,365],"type":"text","angle":0,"index":11,"merge_prev":false,"text":"若 $f(n)$ 是 $\\Omega (g(n))$ 且 $O(g(n))$ 的，则记 $f(n) = \\Theta (g(n))$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":26,"id":"e5374bbb-ef81-4185-a745-e76872b1a288","page_size":[595,841],"block_position":"26-10"},{"bbox":[53,381,539,409],"type":"text","angle":0,"index":12,"merge_prev":false,"text":"笔记 记忆方法：全以多项式为例，O 表示阶数 (次数) 大于等于，o 表示阶数大于， $\\Omega$ 表示阶数小于等于， $\\Theta$ 表示阶数等于。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":26,"id":"f1545091-7e3a-48fa-be15-1c3be2bc587c","page_size":[595,841],"block_position":"26-11"},{"bbox":[55,428,187,442],"type":"title","angle":0,"index":13,"level":2,"text":"## 3.1.4 使用定义直接计算","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":26,"id":"04943fa0-4034-40ae-86e7-4416a3dee377","page_size":[595,841],"block_position":"26-12"},{"bbox":[35,453,307,468],"type":"text","angle":0,"index":14,"merge_prev":false,"text":"练习 3.4 几个经典函数极限 (1) 计算 $\\lim_{n\\to\\infty}\\sin^{2}(\\pi\\sqrt{n^{2}+n})$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":26,"id":"667e2029-381b-401a-b646-8a13bac7f7ad","page_size":[595,841],"block_position":"26-13"},{"bbox":[73,469,249,494],"type":"text","angle":0,"index":15,"merge_prev":false,"text":"(2) $\\lim_{x\\to +\\infty}\\frac{[xf(x)]}{x}$ ，其中 $\\lim_{x\\to +\\infty}f(x) = a$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":26,"id":"a37b6fb0-9e60-45c6-8153-cb22f6d63685","page_size":[595,841],"block_position":"26-14"},{"bbox":[53,492,446,514],"type":"text","angle":0,"index":16,"merge_prev":false,"text":"解 (1) $\\sin^2 (\\pi \\sqrt{n^2 + n}) = \\sin^2 (\\pi \\sqrt{n^2 + n} -\\pi n) = \\sin^2 (\\frac{n\\pi}{\\sqrt{n^2 + n} + n})$ ，极限为 $\\sin^2 (\\frac{\\pi}{2}) = 1$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":26,"id":"9f9bb230-1556-4622-b79c-aa3e371984ce","page_size":[595,841],"block_position":"26-15"},{"bbox":[54,514,539,556],"type":"text","angle":0,"index":17,"merge_prev":false,"text":"(2) 由于 $xf(x) - 1 < [xf(x)] \\leq xf(x)$ ，因此 $f(x) - \\frac{1}{x} < \\frac{[xf(x)]}{x} \\leq f(x)$ ，显然 $\\lim_{x \\to +\\infty} \\left( f(x) - \\frac{1}{x} \\right) = \\lim_{x \\to +\\infty} f(x) = a$ ，因此由夹逼准则，极限为 $a$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":26,"id":"915664d8-4057-4cea-a9fb-6a3927a4b58c","page_size":[595,841],"block_position":"26-16"},{"bbox":[35,563,525,601],"type":"text","angle":0,"index":18,"merge_prev":false,"text":"练习3.5几个经典递推问题 $(1)f(x)$ 在 $(0, +\\infty)$ 单调增， $\\lim_{x\\to +\\infty}\\frac{f(2x)}{f(x)} = 1$ ，证明 $\\forall a > 0$ 有 $\\lim_{x\\to +\\infty}\\frac{f(ax)}{f(x)} = 1$ 证明 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":26,"id":"91e07ee4-3c72-4d8a-aa68-e7f8c8d0b03b","page_size":[595,841],"block_position":"26-17"},{"bbox":[55,624,187,640],"type":"title","angle":0,"index":19,"level":2,"text":"## 3.2 函数极限的计算","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":26,"id":"53e868db-4525-427b-81b9-7d0df908fb91","page_size":[595,841],"block_position":"26-18"},{"bbox":[55,656,175,671],"type":"text","angle":0,"index":20,"merge_prev":false,"text":"3.2.1 等价无穷小公式 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":26,"id":"5da983d8-eb8f-48d2-960a-3cb024c34055","page_size":[595,841],"block_position":"26-19"},{"bbox":[303,689,427,712],"type":"interline_equation","angle":0,"index":21,"text":"$$\n1 - \\cos x \\sim \\frac {1}{2} x ^ {2} \\tan x \\sim x\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":26,"id":"49ad9c40-7136-45c8-bc18-a0f2cd3798ef","page_size":[595,841],"block_position":"26-20"},{"bbox":[145,713,271,725],"type":"text","angle":0,"index":22,"merge_prev":false,"text":"反三角： $\\arcsin x\\sim x$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":26,"id":"53c16781-9f80-4859-821b-e076c87f3767","page_size":[595,841],"block_position":"26-21"},{"bbox":[303,715,361,724],"type":"text","angle":0,"index":23,"merge_prev":false,"text":"$\\arctan x \\sim x$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":26,"id":"c2dca423-e392-4239-8a05-df25ee0ce3b8","page_size":[595,841],"block_position":"26-22"},{"bbox":[145,728,277,740],"type":"text","angle":0,"index":24,"merge_prev":false,"text":"指数、对数： $\\ln (1 + x)\\sim x$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":26,"id":"dfe019ac-9727-47a6-9803-f3b299d397b4","page_size":[595,841],"block_position":"26-23"},{"bbox":[303,729,353,740],"type":"interline_equation","angle":0,"index":25,"text":"$$\ne ^ {x} - 1 \\sim x\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":26,"id":"a43e32df-fe8b-421a-b9c3-4b0918fa58bb","page_size":[595,841],"block_position":"26-24"},{"bbox":[381,729,449,740],"type":"interline_equation","angle":0,"index":26,"text":"$$\na ^ {x} - 1 \\sim x \\ln a\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":26,"id":"791675c3-e2d3-4596-87c3-c3ce132d7f2c","page_size":[595,841],"block_position":"26-25"},{"bbox":[145,744,298,756],"type":"text","angle":0,"index":27,"merge_prev":false,"text":"其他： $(1 + x)^{\\alpha} - 1\\sim \\alpha x$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":26,"id":"421a683b-f6fd-446c-b467-525d4c3983de","page_size":[595,841],"block_position":"26-26"},{"bbox":[292,803,302,811],"type":"page_number","angle":0,"index":999999,"text":"19 ","color":{"line":"rgba(164, 164, 164, 1)","fill":"rgba(164, 164, 164, 1)"},"page_idx":26,"is_discarded":true,"id":"9c9c88aa-24d0-452c-a258-92d524a931ba","page_size":[595,841],"block_position":"26-27"}],[{"bbox":[459,47,539,58],"type":"header","angle":0,"index":-99999,"text":"3.3 函数的连续性 ","color":{"line":"rgba(164, 164, 164, 1)","fill":"rgba(164, 164, 164, 1)"},"page_idx":27,"is_discarded":true,"id":"5e72ee06-b156-4769-a4f5-35610b47334a","page_size":[595,841],"block_position":"27-0"},{"bbox":[55,73,176,87],"type":"title","angle":0,"index":1,"level":2,"text":"## 3.2.2 Taylor 展开公式","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":27,"id":"d711ac9b-adb8-4366-839b-c801236e6596","page_size":[595,841],"block_position":"27-1"},{"bbox":[58,116,116,131],"type":"text","angle":0,"index":2,"merge_prev":false,"text":"指数、对数: ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":27,"id":"faffa035-98f6-42d9-b525-c0007e001a8b","page_size":[595,841],"block_position":"27-2"},{"bbox":[128,110,184,141],"type":"interline_equation","angle":0,"index":3,"text":"$$\ne ^ {x} = \\sum_ {n = 0} ^ {\\infty} \\frac {x ^ {n}}{n !}\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":27,"id":"d716f7c7-c0d9-4a75-9729-040f5fd04566","page_size":[595,841],"block_position":"27-3"},{"bbox":[368,110,489,139],"type":"interline_equation","angle":0,"index":4,"text":"$$\n\\ln (1 + x) = \\sum_ {n = 1} ^ {\\infty} (- 1) ^ {n - 1} \\frac {x ^ {n}}{n}\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":27,"id":"3d8747fc-bc60-4952-b55c-7416c07a404e","page_size":[595,841],"block_position":"27-4"},{"bbox":[58,147,86,159],"type":"text","angle":0,"index":5,"merge_prev":false,"text":"三角： ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":27,"id":"46bbd4f4-9bf7-4819-98c9-ff23ad4c4e33","page_size":[595,841],"block_position":"27-5"},{"bbox":[127,139,246,169],"type":"interline_equation","angle":0,"index":6,"text":"$$\n\\sin x = \\sum_ {n = 0} ^ {\\infty} (- 1) ^ {n} \\frac {x ^ {2 n + 1}}{(2 n + 1) !}\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":27,"id":"2ba8e54c-f7c5-42a8-83f8-a5050b9eefd0","page_size":[595,841],"block_position":"27-6"},{"bbox":[370,140,472,169],"type":"interline_equation","angle":0,"index":7,"text":"$$\n\\cos x = \\sum_ {n = 0} ^ {\\infty} (- 1) ^ {n} \\frac {x ^ {2 n}}{(2 n) !}\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":27,"id":"4dce0112-6284-48e3-8460-1764d1d60caa","page_size":[595,841],"block_position":"27-7"},{"bbox":[59,175,96,189],"type":"text","angle":0,"index":8,"merge_prev":false,"text":"反三角： ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":27,"id":"6111b1b6-b2d2-4932-b481-f83ac69aba3c","page_size":[595,841],"block_position":"27-8"},{"bbox":[127,168,362,198],"type":"interline_equation","angle":0,"index":9,"text":"$$\n\\arcsin x = \\sum_ {n = 0} ^ {\\infty} \\left(\\frac {(2 n) !}{2 ^ {2 n} (n !) ^ {2}}\\right) \\frac {x ^ {2 n + 1}}{2 n + 1} = x + \\frac {1}{6} x ^ {3} + o \\left(x ^ {3}\\right)\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":27,"id":"136629b3-a3cb-4b23-b605-92eb698fd6b6","page_size":[595,841],"block_position":"27-9"},{"bbox":[370,168,575,198],"type":"interline_equation","angle":0,"index":10,"text":"$$\n\\arctan x = \\sum_ {n = 0} ^ {\\infty} \\frac {(- 1) ^ {n} x ^ {2 n + 1}}{2 n + 1} = x - \\frac {1}{3} x ^ {3} + o (x ^ {3})\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":27,"id":"f91d0a6a-6ff9-4056-8f3e-bad1330c1463","page_size":[595,841],"block_position":"27-10"},{"bbox":[58,204,86,218],"type":"text","angle":0,"index":11,"merge_prev":false,"text":"分式: ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":27,"id":"5f5882be-369f-4633-aac2-16278db6feb0","page_size":[595,841],"block_position":"27-11"},{"bbox":[128,199,223,227],"type":"interline_equation","angle":0,"index":12,"text":"$$\n{\\frac {1}{1 + x}} = \\sum_ {n = 0} ^ {\\infty} (- 1) ^ {n} x ^ {n}\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":27,"id":"8cf25ed5-1a08-4825-9c6a-54f09f304b41","page_size":[595,841],"block_position":"27-12"},{"bbox":[370,199,440,227],"type":"interline_equation","angle":0,"index":13,"text":"$$\n{\\frac {1}{1 - x}} = \\sum_ {n = 0} ^ {\\infty} x ^ {n}\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":27,"id":"e9bbd560-fc7c-4731-80c1-0c8b79c48cc7","page_size":[595,841],"block_position":"27-13"},{"bbox":[58,233,86,247],"type":"text","angle":0,"index":14,"merge_prev":false,"text":"其他： ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":27,"id":"e55d1399-1d2f-4f17-a718-8b531e8ed6df","page_size":[595,841],"block_position":"27-14"},{"bbox":[128,227,308,253],"type":"interline_equation","angle":0,"index":15,"text":"$$\n(1 + x) ^ {\\alpha} = 1 + \\alpha x + \\frac {\\alpha (\\alpha - 1)}{2} x ^ {2} + o (x ^ {2})\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":27,"id":"5f798c5a-3fa2-4374-a9f1-1a4e950594cc","page_size":[595,841],"block_position":"27-15"},{"bbox":[370,227,471,258],"type":"interline_equation","angle":0,"index":16,"text":"$$\n(1 + x) ^ {\\alpha} = \\sum_ {n = 0} ^ {\\infty} {\\binom {\\alpha} {n}} x ^ {n}\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":27,"id":"5a71d526-866c-4555-a36a-12c36c528859","page_size":[595,841],"block_position":"27-16"},{"bbox":[55,275,172,292],"type":"title","angle":0,"index":17,"level":2,"text":"## 3.3 函数的连续性","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":27,"id":"034ae229-df8e-40bf-8828-02d14e5f06d6","page_size":[595,841],"block_position":"27-17"},{"bbox":[55,308,174,323],"type":"title","angle":0,"index":18,"level":2,"text":"## 3.3.1 函数连续的定义","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":27,"id":"720f578d-9990-4216-8df2-238bd2360214","page_size":[595,841],"block_position":"27-18"},{"bbox":[72,335,161,349],"type":"title","angle":0,"index":19,"level":2,"text":"## 定义 3.5 (单点连续)","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":27,"id":"9667eca0-b9cb-41b8-bc03-b0884e2bfefd","page_size":[595,841],"block_position":"27-19"},{"bbox":[67,352,389,370],"type":"text","angle":0,"index":20,"merge_prev":false,"text":"$f:[a,b]\\to \\mathbb{R},x_0\\in [a,b]$ 若 $\\lim_{x\\to x_0}f(x) = f(x_0)$ ，则称 $f(x)$ 在 $x_0$ 处连续。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":27,"id":"a2b81d1f-3324-4937-a5f8-6264bc342bbb","page_size":[595,841],"block_position":"27-20"},{"bbox":[52,387,539,416],"type":"text","angle":0,"index":21,"merge_prev":false,"text":"笔记 有时也可以展开用 $\\forall \\epsilon, \\exists \\delta$ 使得 $\\forall x \\in (x_0 - \\delta, x_0 + \\delta)$ 满足 $|f(x) - f(x_0)| < \\epsilon$ 证明。例如证明 Riemann 函数在无理点连续。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":27,"id":"f78b2ec9-6d78-43b9-bc3a-e7ecb58cd8df","page_size":[595,841],"block_position":"27-21"},{"bbox":[72,423,181,437],"type":"title","angle":0,"index":22,"level":2,"text":"## 定义 3.6 (单点左右连续)","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":27,"id":"3f5738e7-d6c7-4e68-bd48-1bbae3c1df32","page_size":[595,841],"block_position":"27-22"},{"bbox":[68,441,289,455],"type":"text","angle":0,"index":23,"merge_prev":false,"text":"若 $f(x_0^+) = f(x_0)$ 则称 $f(x)$ 右连续。左连续同理。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":27,"id":"8a741e15-15a2-4c7c-b799-9679106533e7","page_size":[595,841],"block_position":"27-23"},{"bbox":[52,470,463,484],"type":"text","angle":0,"index":24,"merge_prev":false,"text":"笔记 注意右连续是指右极限等于当前点的函数值，而非左侧极限靠过来等于当前点的函数值！ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":27,"id":"2986ec7e-cfa6-418b-b369-22ce25a2b08d","page_size":[595,841],"block_position":"27-24"},{"bbox":[35,502,417,515],"type":"text","angle":0,"index":25,"merge_prev":false,"text":"练习 3.6Dirichlet 函数的连续性 判断下面 Dirichlet 函数的连续性、单调性、周期性 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":27,"id":"87a1eebc-f9c3-4ea3-8fa5-6b74c2a9ed57","page_size":[595,841],"block_position":"27-25"},{"bbox":[239,518,352,555],"type":"interline_equation","angle":0,"index":26,"text":"$$\nD (x) = {\\left\\{ \\begin{array}{l l} {1,} & {x \\text {为有理数}} \\\\ {0} & {x \\text {为无理数}} \\end{array} \\right.}\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":27,"id":"8dc3485c-793b-44f1-9532-2d5563006fd1","page_size":[595,841],"block_position":"27-26"},{"bbox":[52,559,193,571],"type":"text","angle":0,"index":27,"merge_prev":false,"text":"解(1)周期: 任意有理数(显然) ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":27,"id":"1eb5292d-5b8c-4826-84f1-c03809f1ae15","page_size":[595,841],"block_position":"27-27"},{"bbox":[53,575,539,602],"type":"text","angle":0,"index":28,"merge_prev":false,"text":"(2) 连续性：处处不连续，因为根据有理数和无理数的稠密性，总能在小邻域中找到相差 1 的点，因此不连续。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":27,"id":"c3a2152f-975a-4679-85d7-2fbedee72601","page_size":[595,841],"block_position":"27-28"},{"bbox":[35,621,412,634],"type":"text","angle":0,"index":29,"merge_prev":false,"text":"练习 3.7Riemann 函数的连续性 证明 Riemann 函数在无理点连续，在有理点间断。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":27,"id":"9b4b1849-63ff-4dab-9dec-c6d7d269d57e","page_size":[595,841],"block_position":"27-29"},{"bbox":[223,636,368,678],"type":"interline_equation","angle":0,"index":30,"text":"$$\nR (x) = {\\left\\{ \\begin{array}{l l} { {\\frac {1}{q}},} & {x = {\\frac {p}{q}} \\text {为既约分数}} \\\\ {0,} & {x \\text {为无理数}} \\end{array} \\right.}\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":27,"id":"2573ee12-12e4-4d10-a566-11727a634f7e","page_size":[595,841],"block_position":"27-30"},{"bbox":[52,682,495,706],"type":"text","angle":0,"index":31,"merge_prev":false,"text":"证明 (1) 有理点: 设 $x_0 = \\frac{p}{q}$ 为有理点, $R(x_0) = \\frac{1}{q}$ , 无理点列趋于 $x_0$ 时, 极限为 0 , 与函数值不同。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":27,"id":"9fbe8b90-719b-4091-892b-a81e6bef64be","page_size":[595,841],"block_position":"27-31"},{"bbox":[52,708,539,741],"type":"text","angle":0,"index":32,"merge_prev":false,"text":"(2) 无理点： $x_{0}$ 是无理点， $\\forall \\epsilon, \\frac{1}{q} > \\epsilon$ 的个数有限，设这些点为 $x_{1}, \\cdots, x_{n}$ ，取 $\\delta < \\min\\{|x_{0} - x_{i}| : i = 1 \\sim n\\}$ ， $x \\in (x_{0} - \\delta, x_{0} + \\delta) - \\{x_{0}\\}$ 时 $|f(x_{0}) - f(x)| < \\epsilon$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":27,"id":"2468c093-9d0f-4e7e-a801-fe8450baa82b","page_size":[595,841],"block_position":"27-32"},{"bbox":[35,758,463,772],"type":"text","angle":0,"index":34,"merge_prev":false,"text":"练习 3.8 初等函数的连续性 f, g, h 连续，证明： $\\max\\{f(x), g(x)\\}, \\min\\{f(x), g(x)\\}, |f(x)|$ 连续 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":27,"id":"2ed30d71-8c6f-4eeb-81ab-a9ca6dd4ac24","page_size":[595,841],"block_position":"27-33"},{"bbox":[291,803,303,812],"type":"page_number","angle":0,"index":1000032,"text":"20 ","color":{"line":"rgba(164, 164, 164, 1)","fill":"rgba(164, 164, 164, 1)"},"page_idx":27,"is_discarded":true,"id":"ac703749-fe02-436f-b4f9-5f0f0bddef10","page_size":[595,841],"block_position":"27-34"}],[{"bbox":[459,47,538,58],"type":"header","angle":0,"index":-99998,"text":"3.3 函数的连续性 ","color":{"line":"rgba(164, 164, 164, 1)","fill":"rgba(164, 164, 164, 1)"},"page_idx":28,"is_discarded":true,"id":"c9601d10-cc0d-4455-be9b-6015898d6c49","page_size":[595,841],"block_position":"28-0"},{"bbox":[54,69,394,92],"type":"text","angle":0,"index":2,"merge_prev":false,"text":"证明 (1) 最大、最小：根据 $\\max \\{f(x), g(x)\\} = \\frac{1}{2}[f(x) + g(x) + |f(x) - g(x)|]$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":28,"id":"c5f8731f-f263-4236-812c-197dc541a989","page_size":[595,841],"block_position":"28-1"},{"bbox":[73,92,232,105],"type":"text","angle":0,"index":3,"merge_prev":false,"text":"(2) 绝对值: $|f(x)| = \\sqrt{|f(x)|^2}$ 连续 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":28,"id":"f022908c-7283-40ae-a09e-8d97271f1e94","page_size":[595,841],"block_position":"28-2"},{"bbox":[55,123,163,137],"type":"title","angle":0,"index":4,"level":2,"text":"## 3.3.2 间断点的概念","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":28,"id":"dfb43063-8c8e-41d4-9d8b-4e0135f50844","page_size":[595,841],"block_position":"28-3"},{"bbox":[72,151,181,163],"type":"title","angle":0,"index":5,"level":2,"text":"## 定义 3.7 (第一类间断点)","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":28,"id":"62e535ae-8cbf-4161-9686-d513d23277cd","page_size":[595,841],"block_position":"28-4"},{"bbox":[67,168,286,179],"type":"text","angle":0,"index":6,"merge_prev":false,"text":"跳跃点与可去间断点统称为 $f(x)$ 的第一类间断点 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":28,"id":"06c7262f-8bae-4b14-bf7e-de3929d3ac6a","page_size":[595,841],"block_position":"28-5"},{"bbox":[83,183,371,195],"type":"text","angle":0,"index":7,"merge_prev":false,"text":"- 跳跃点：若 $f(x)$ 在 $x_0$ 处左右极限 $f(x_0^-), f(x_0^+)$ 存在但不相等， ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":28,"id":"7d1fa707-eb80-4d72-a176-8656e41c9e17","page_size":[595,841],"block_position":"28-6"},{"bbox":[83,198,404,212],"type":"text","angle":0,"index":8,"merge_prev":false,"text":"- 可去间断点：若 $f(x)$ 在 $x_0$ 处左右极限存在且相等，即 $f(x_0^-) = f(x_0^+)$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":28,"id":"1e47ad07-2d29-4c02-8bfb-809b22ec1248","page_size":[595,841],"block_position":"28-7"},{"bbox":[35,244,539,307],"type":"text","angle":0,"index":9,"merge_prev":false,"text":"练习3.9第一类间断点的性质(1)若 $f(x)$ 在 $[a,b]$ 至多有第一类间断点，证明： $f(x)$ 在 $[a,b]$ 有界证明(1)若没有间断点，则连续，显然有界。否则反设无界，则对 $M = 1$ ，存在 $x_{1} \\in [a,b]$ 使得 $|f(x_1)| > 1$ ， $M = 2$ 存在 $x_{2} \\in [a,b]$ 使得 $|f(x_2)| > 2$ ，以此类推构造 $x_{n}$ 。由于 $x_{n} \\in [a,b]$ ，根据致密性定理，存在收敛子列 $x_{n_k}$ ，而由于 $f(x)$ 只有第一类间断点，因此 $\\lim_{k \\to +\\infty} f(x_{n_k})$ 存在，但其趋于无穷，矛盾。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":28,"id":"be20b6d3-2500-4eae-9787-b38fee4d93cd","page_size":[595,841],"block_position":"28-8"},{"bbox":[72,326,181,338],"type":"title","angle":0,"index":10,"level":2,"text":"## 定义 3.8 (第二类间断点)","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":28,"id":"a3e2ecdc-34ae-4294-932a-d05553d73c5a","page_size":[595,841],"block_position":"28-9"},{"bbox":[68,343,456,357],"type":"text","angle":0,"index":11,"merge_prev":false,"text":"若 $f(x_0^+)$ 和 $f(x_0^-)$ 至少一个不存在或者不是有限的数，则称 $x_0$ 是 $f(x)$ 的第二类间断点。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":28,"id":"a9fa0442-c415-48f6-aee2-a257ec690cb3","page_size":[595,841],"block_position":"28-10"},{"bbox":[35,389,538,418],"type":"text","angle":0,"index":12,"merge_prev":false,"text":"练习 3.10 判断间断点类型 (1) 已知 $f(x) = \\text{sgn}(x)$ , $g(x) = x - [x]$ ，讨论 $f(g(x))$ 和 $g(f(x))$ 的间断点类型\n\n解 (1) $f(g(x))$ 在 x 为整数时等于 0，在 x 非整数时大于 0，因此在整数处取可去间断点。 $g(f(x)) = 0$ ，因此连续 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":28,"id":"ec92a950-6e19-473f-93da-65b8e438f542","page_size":[595,841],"block_position":"28-11"},{"bbox":[35,435,539,463],"type":"text","angle":0,"index":13,"merge_prev":false,"text":"练习 3.11 单调函数的间断点 $f(x)$ 在 $x_{0}$ 邻域单增， $f(x_{0}^{-}), f(x_{0}^{+})$ 均存在，证明如下性质，即单调函数的间断点都是跳跃间断点 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":28,"id":"fa6172e7-21df-47c0-8625-233364b5a248","page_size":[595,841],"block_position":"28-12"},{"bbox":[196,466,396,487],"type":"interline_equation","angle":0,"index":14,"text":"$$\nf (x _ {0} ^ {-}) = \\sup _ {x \\in U _ {-} ^ {o} (x _ {0})} f (x), f (x _ {0} ^ {+}) = \\inf _ {x \\in U _ {+} ^ {o} (x _ {0})} f (x)\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":28,"id":"17bd5a6f-eccb-493a-939b-d0c1aa69ae9d","page_size":[595,841],"block_position":"28-13"},{"bbox":[45,496,537,540],"type":"text","angle":0,"index":15,"merge_prev":false,"text":"证明 取 $x_{1} \\in U_{+}^{o}(x_{0})$ ，则 $\\forall x \\in U_{-}^{o}(x_{0})$ 有 $f(x) \\leq f(x_{1})$ ，从而根据确界原理有上确界 $A$ ，从而 $f(x) \\leq A$ 且 $\\forall \\epsilon, \\exists x'$ 满足 $f(x') > A - \\epsilon$ ，从而 $\\forall x \\in (x_{0} - \\delta, x_{0})$ 有 $|f(x) - A| < \\epsilon$ 笔记 单调函数左右极限均存在是一个会在考题中使用的结论。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":28,"id":"8d861741-30ed-48e6-93fd-662c5be90c27","page_size":[595,841],"block_position":"28-14"},{"bbox":[35,558,504,587],"type":"text","angle":0,"index":16,"merge_prev":false,"text":"练习3.12单调函数右连续 $f(x)$ 是 $\\mathbb{R}$ 上单调函数，定义 $g(x) = f(x + 0)$ ，证明 $g(x)$ 在 $\\mathbb{R}$ 每一点右连续证明只需要证明 $\\forall \\epsilon, \\exists \\delta$ ，当 $|x - x_0| < \\delta$ 时，满足 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":28,"id":"ee909f76-bf62-45fd-8c89-cc183c1d20ad","page_size":[595,841],"block_position":"28-15"},{"bbox":[257,594,337,608],"type":"interline_equation","angle":0,"index":17,"text":"$$\n| g (x) - g (x _ {0}) | < \\epsilon\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":28,"id":"86895456-9a40-4303-b2aa-0f5c096bf2f0","page_size":[595,841],"block_position":"28-16"},{"bbox":[53,613,451,634],"type":"text","angle":0,"index":18,"merge_prev":false,"text":"根据 $g(x_0) = \\lim_{y\\to x_0^+}f(y)$ 得到 $y\\in (x_0,x_0 + \\delta)$ 时有 $|f(y) - g(x_0)| < \\epsilon$ ，令 $y\\to x^{+}$ 即可得到 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":28,"id":"67263475-56da-40a4-9778-1757775549c8","page_size":[595,841],"block_position":"28-17"},{"bbox":[257,637,337,650],"type":"interline_equation","angle":0,"index":19,"text":"$$\n| g (x) - g (x _ {0}) | < \\epsilon\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":28,"id":"7c2004a2-6d82-4dee-9f80-bbeb7948fc9e","page_size":[595,841],"block_position":"28-18"},{"bbox":[35,671,438,707],"type":"text","angle":0,"index":20,"merge_prev":false,"text":"练习3.13极限的连续性 $f(x)$ 只有可去间断点， $g(x) = \\lim_{y\\to x}f(y)$ ，证明 $g(x)$ 是连续函数证明可去间断点说明极限存在， $g(x_0) = \\lim_{y\\to x_0}f(y)$ ，即对 $\\forall \\epsilon, \\exists \\delta, y \\in U^o(x_0, \\delta)$ 时： ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":28,"id":"365dccb8-f4bf-40d6-b9c1-909910055f3d","page_size":[595,841],"block_position":"28-19"},{"bbox":[232,711,361,724],"type":"interline_equation","angle":0,"index":21,"text":"$$\ng (x _ {0}) - \\epsilon < f (y) < g (x _ {0}) + \\epsilon\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":28,"id":"e642672c-358a-4f86-8224-18e78f07ce6c","page_size":[595,841],"block_position":"28-20"},{"bbox":[54,729,231,747],"type":"text","angle":0,"index":22,"merge_prev":false,"text":"根据 $g(x) = \\lim_{y\\to x}f(y)$ 和保不等式性得到 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":28,"id":"6507db03-ea6d-4bfd-8b8d-9ec56b27a362","page_size":[595,841],"block_position":"28-21"},{"bbox":[232,751,360,763],"type":"interline_equation","angle":0,"index":23,"text":"$$\ng (x _ {0}) - \\epsilon \\leq g (x) \\leq g (x _ {0}) + \\epsilon\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":28,"id":"4a042881-e90c-4dfb-a10b-cfe93e66f782","page_size":[595,841],"block_position":"28-22"},{"bbox":[291,803,302,812],"type":"page_number","angle":0,"index":999999,"text":"21 ","color":{"line":"rgba(164, 164, 164, 1)","fill":"rgba(164, 164, 164, 1)"},"page_idx":28,"is_discarded":true,"id":"b614bd47-86ae-4d36-9bb8-2d2fd75c9864","page_size":[595,841],"block_position":"28-23"}],[{"bbox":[459,47,538,58],"type":"header","angle":0,"index":-99999,"text":"3.3 函数的连续性 ","color":{"line":"rgba(164, 164, 164, 1)","fill":"rgba(164, 164, 164, 1)"},"page_idx":29,"is_discarded":true,"id":"34bd5114-f1e2-4ef3-9571-82506f8d8867","page_size":[595,841],"block_position":"29-0"},{"bbox":[54,72,275,84],"type":"text","angle":0,"index":1,"merge_prev":false,"text":"因此 $g(x)$ 在 $x_{0}$ 连续，由 $x_{0}$ 任意性可知结论成立。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":29,"id":"ebad83a6-2a69-43ba-8ad4-9ee6a9634658","page_size":[595,841],"block_position":"29-1"},{"bbox":[35,98,539,140],"type":"text","angle":0,"index":2,"merge_prev":false,"text":"练习 3.14 第一类间断点的经典题目 $f(x)$ 在 $(a,b)$ 至多只有第一类间断点，且对 $\\forall x,y\\in(a,b)$ 有 $f\\left(\\frac{x+y}{2}\\right)\\leq\\frac{f(x)+f(y)}{2}$ ，证明： $f(x)$ 在 $(a,b)$ 连续 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":29,"id":"145d8d10-6c10-4acd-85ea-9ce35497ada8","page_size":[595,841],"block_position":"29-2"},{"bbox":[52,141,321,155],"type":"text","angle":0,"index":3,"merge_prev":false,"text":"证明 $\\forall x_{0}\\in(a,b),\\forall h$ 满足 $x_{0}\\pm h\\in(a,b)$ 时，根据条件可知： ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":29,"id":"c49a9135-1331-4e47-8d83-a56396565de0","page_size":[595,841],"block_position":"29-3"},{"bbox":[152,158,439,182],"type":"interline_equation","angle":0,"index":4,"text":"$$\nf (x _ {0}) \\leq \\frac {f (x _ {0} + h) + f (x _ {0} - h)}{2} \\quad f (x _ {0} + \\frac {h}{2}) \\leq \\frac {f (x _ {0}) + f (x _ {0} + h)}{2}\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":29,"id":"69a94e80-7590-4ff1-be42-37698759ad95","page_size":[595,841],"block_position":"29-4"},{"bbox":[54,184,268,195],"type":"text","angle":0,"index":5,"merge_prev":false,"text":"前者取 $h \\rightarrow 0^{+}$ ，后者分别取 $h \\rightarrow 0^{+}, h \\rightarrow 0^{-}$ 则 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":29,"id":"8d95e7f1-e827-4ab6-b8e9-e304f1499be0","page_size":[595,841],"block_position":"29-5"},{"bbox":[219,198,364,264],"type":"interline_equation","angle":0,"index":6,"text":"$$\n\\left\\{ \\begin{array}{l} f (x _ {0}) \\leq \\frac {f (x _ {0} + 0) + f (x _ {0} - 0)}{2} \\\\ f (x _ {0} + 0) \\leq \\frac {f (x _ {0}) + f (x _ {0} + 0)}{2} \\\\ f (x _ {0} - 0) \\leq \\frac {f (x _ {0}) + f (x _ {0} - 0)}{2} \\end{array} \\right.\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":29,"id":"fa84ca71-bcaf-47c4-b7be-f5d5eb346a2a","page_size":[595,841],"block_position":"29-6"},{"bbox":[54,266,327,280],"type":"text","angle":0,"index":7,"merge_prev":false,"text":"综合得到 $f(x_0 - 0) = f(x_0 + 0) = f(x_0)$ ，因此在任意点连续。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":29,"id":"df26e8af-807b-46e4-8074-5dff841e3965","page_size":[595,841],"block_position":"29-7"},{"bbox":[56,297,151,312],"type":"title","angle":0,"index":8,"level":2,"text":"## 3.3.3 介值性定理","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":29,"id":"6bacfb8c-f8a9-4f40-be46-98fcbe453eb1","page_size":[595,841],"block_position":"29-8"},{"bbox":[72,324,181,337],"type":"title","angle":0,"index":9,"level":2,"text":"## 定理 3.4 (零点存在定理)","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":29,"id":"2bd3e597-cee6-433d-b762-57642645d6d9","page_size":[595,841],"block_position":"29-9"},{"bbox":[186,343,243,354],"type":"interline_equation","angle":0,"index":10,"text":"$$\nf (a) f (b) < 0\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":29,"id":"0b340444-f4c6-448b-8e9c-17185d61ef17","page_size":[595,841],"block_position":"29-10"},{"bbox":[263,343,308,354],"type":"interline_equation","angle":0,"index":11,"text":"$$\n\\exists \\xi \\in (a, b)\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":29,"id":"5970bde6-08be-44fc-ac9e-7f3bc1c7a7f4","page_size":[595,841],"block_position":"29-11"},{"bbox":[333,343,371,354],"type":"interline_equation","angle":0,"index":12,"text":"$$\nf (\\xi) = 0\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":29,"id":"f296c45d-69d9-411a-aa10-6e1a4aaf06d5","page_size":[595,841],"block_position":"29-12"},{"bbox":[54,369,539,401],"type":"text","angle":0,"index":13,"merge_prev":false,"text":"证明 (1) 使用闭区间套定理进行证明：取 $a_1 = a, b_1 = b, c_1 = \\frac{a_1 + b_1}{2}$ ，根据 $f(c_1)$ 正负性选取 $a_2, b_2$ ，以此类推，由于 $\\lim_{n\\to \\infty}|a_n - b_n| = 0$ ， $f(a_n)f(b_n) < 0$ ，根据闭区间套 $\\exists \\xi$ 使得 $f^2 (\\xi)\\leq 0$ ，因此得出 $f(\\xi) = 0$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":29,"id":"f08ba1ce-1d46-4958-922d-6444364e9a23","page_size":[595,841],"block_position":"29-13"},{"bbox":[54,403,539,446],"type":"text","angle":0,"index":14,"merge_prev":false,"text":"(2) 使用有限覆盖定理证明：根据连续函数局部保号性， $\\forall x \\in [a, b]$ ， $\\exists \\delta_x$ 使得 $f(x)$ 在 $(x - \\delta_x, x + \\delta_x) \\cap [a, b]$ 中保号，取遍 $[a, b]$ 显然是一族开覆盖，根据有限覆盖定理可找到有限覆盖，不妨设这有限覆盖相互有重叠，则全部等号，这与 $f(a)f(b) < 0$ 矛盾。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":29,"id":"6929a33b-96d6-4cb9-aee2-d0c4e6b2b0d3","page_size":[595,841],"block_position":"29-14"},{"bbox":[72,456,171,469],"type":"title","angle":0,"index":16,"level":2,"text":"## 定理 3.5 (介值性定理)","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":29,"id":"15b25429-22bb-4557-9d19-3ad59ee92c36","page_size":[595,841],"block_position":"29-15"},{"bbox":[67,472,532,502],"type":"text","angle":0,"index":17,"merge_prev":false,"text":"$f(x)$ 是 $[a,b]$ 中的连续函数， $a\\leq x_1 < x_2\\leq b$ ， $f(x_{1})\\neq f(x_{2})$ ，则 $f(x)$ 在 $[a,b]$ 可取遍 $f(x_{1}),f(x_{2})$ 间任意值。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":29,"id":"7116e42d-3c52-445b-916c-42f1ff52e9ed","page_size":[595,841],"block_position":"29-16"},{"bbox":[35,532,539,560],"type":"text","angle":0,"index":18,"merge_prev":false,"text":"练习 3.15 构造辅助函数 (1) $f(x), g(x)$ 是 [0,1] 上的连续函数，且在 [0,1] 有相同最大值，证明 $\\exists \\xi \\in [a,b]$ 使得 $f(\\xi) = g(\\xi)$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":29,"id":"208579e3-bbb2-4ef8-9f85-27915aba8422","page_size":[595,841],"block_position":"29-17"},{"bbox":[72,563,470,576],"type":"text","angle":0,"index":19,"merge_prev":false,"text":"(2) 重点: $f(x)$ 是 $[0, 2a]$ 连续函数, 且 $f(0) = f(2a)$ , 证明 $\\exists \\xi \\in [0, a]$ 使得 $f(\\xi) = f(\\xi + a)$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":29,"id":"7754e069-b832-4710-b6f0-bcd7a25209cc","page_size":[595,841],"block_position":"29-18"},{"bbox":[72,578,518,594],"type":"text","angle":0,"index":20,"merge_prev":false,"text":"(3) 重点： $f(x)$ 是 $[0,1]$ 上连续函数， $f(0) = f(1) = 0$ ，证明 $\\forall n \\in \\mathbb{N}^{+}, \\exists \\xi \\in [0,1]$ 使得 $f(\\xi + \\frac{1}{n}) = f(\\xi)$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":29,"id":"2598a11a-98b1-4398-9752-29a6a2e3e806","page_size":[595,841],"block_position":"29-19"},{"bbox":[52,595,538,640],"type":"text","angle":0,"index":21,"merge_prev":false,"text":"(4) 重点: $f(x)$ 是 $[0,1]$ 上非负连续函数, $f(0) = f(1) = 0$ , 证明: 对 $\\forall a \\in (0,1), \\exists \\xi \\in [0,1]$ , 使得 $f(\\xi) = f(\\xi + a)$ 证明 (1) 问题在于最大值取值点不一定一样: 若最大值对应 $f(x_1), g(x_2)$ , 若 $x_1 = x_2$ , 则显然成立。若 $x_1 \\neq x_2$ , 则取 $F(x) = f(x) - g(x)$ , $F(x_1) \\geq 0, F(x_2) \\leq 0$ , 从而由零点存在定理成立。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":29,"id":"897650b7-4c84-4efa-805b-dd6021d02928","page_size":[595,841],"block_position":"29-20"},{"bbox":[54,642,537,669],"type":"text","angle":0,"index":22,"merge_prev":false,"text":"(2) 取 $F(x) = f(x) - f(x + a)$ , $F(0) = f(0) - f(a)$ , $F(a) = f(a) - f(2a) = f(a) - f(0)$ ，因此 $F(0) = -F(a)$ ，显然成立。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":29,"id":"674bc287-20bc-48b7-9f44-9e4d9331a01d","page_size":[595,841],"block_position":"29-21"},{"bbox":[101,667,535,691],"type":"interline_equation","angle":0,"index":23,"text":"$$\nF (x) = f \\left(x + \\frac {1}{n}\\right) - f (x), F (0) = f \\left(\\frac {1}{n}\\right) - f (0), F \\left(\\frac {1}{n}\\right) = f \\left(\\frac {2}{n}\\right) - f \\left(\\frac {1}{n}\\right), \\dots , F \\left(\\frac {n - 1}{n}\\right) = f (1) - f \\left(\\frac {n - 1}{n}\\right)\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":29,"id":"57150992-3694-4659-8217-327c294d6e97","page_size":[595,841],"block_position":"29-22"},{"bbox":[54,690,530,708],"type":"text","angle":0,"index":24,"merge_prev":false,"text":"得到 $F(0)+F(\\frac{1}{n})+\\cdots+F(\\frac{n-1}{n})=f(1)-f(0)=0$ ，要么这些全部为零，要么有正有负，根据定理可知。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":29,"id":"a6a527c9-c718-4138-bfd6-d771cc6d4755","page_size":[595,841],"block_position":"29-23"},{"bbox":[53,709,539,739],"type":"text","angle":0,"index":25,"merge_prev":false,"text":"(4) 取 $F(x) = f(x) - f(x + a)$ ， $F(0) = f(0) - f(a) = -f(a)$ , $F(1 - a) = f(1 - a) - f(1) = f(1 - a)$ ，由于 $f$ 非负，因此 $F(0)F(1 - a) \\leq 0$ 。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":29,"id":"9cc436c5-8d76-4f14-926f-1b87df2beec5","page_size":[595,841],"block_position":"29-24"},{"bbox":[291,802,303,811],"type":"page_number","angle":0,"index":1000014,"text":"22 ","color":{"line":"rgba(164, 164, 164, 1)","fill":"rgba(164, 164, 164, 1)"},"page_idx":29,"is_discarded":true,"id":"4fa3a4a5-8d7e-4e0e-b018-347800b60f2b","page_size":[595,841],"block_position":"29-25"}],[{"bbox":[459,47,539,58],"type":"header","angle":0,"index":-99999,"text":"3.4 连续性的应用 ","color":{"line":"rgba(164, 164, 164, 1)","fill":"rgba(164, 164, 164, 1)"},"page_idx":30,"is_discarded":true,"id":"9e818cd6-eed5-42ef-89c0-bf5f6140ab76","page_size":[595,841],"block_position":"30-0"},{"bbox":[72,74,211,85],"type":"title","angle":0,"index":1,"level":2,"text":"## 定理 3.6 (连续函数平均值定理)","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":30,"id":"709f5a4f-7aec-42a4-befa-0377fa0ca3dd","page_size":[595,841],"block_position":"30-1"},{"bbox":[68,89,348,104],"type":"text","angle":0,"index":2,"merge_prev":false,"text":"$f(x)$ 在 $[a,b]$ 连续， $x_{1},\\dots ,x_{n}\\in [a,b]$ ，证明： $\\exists \\xi \\in [a,b]$ ，使得 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":30,"id":"448ce27a-4d53-43a1-8af1-6d71d2db49dd","page_size":[595,841],"block_position":"30-2"},{"bbox":[229,106,363,129],"type":"interline_equation","angle":0,"index":3,"text":"$$\nf (\\xi) = \\frac {1}{n} [ f (x _ {1}) + \\dots + f (x _ {n}) ]\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":30,"id":"212f0d32-97ed-46f5-be14-5fe2aaf66c9d","page_size":[595,841],"block_position":"30-3"},{"bbox":[53,144,364,158],"type":"text","angle":0,"index":4,"merge_prev":false,"text":"证明 不妨设 $f(x_{1}),\\cdots,f(x_{n})$ 中最大值为 $f(x_{n})$ ，最小值为 $f(x_{1})$ ，则 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":30,"id":"5bd2901e-553e-465b-b759-82583db920cb","page_size":[595,841],"block_position":"30-4"},{"bbox":[207,160,385,184],"type":"interline_equation","angle":0,"index":5,"text":"$$\nf (x _ {1}) \\leq \\frac {1}{n} [ f (x _ {1}) + \\dots + f (x _ {n}) ] \\leq f (x _ {n})\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":30,"id":"5f1b495f-d40d-4c92-8687-db14570b09bd","page_size":[595,841],"block_position":"30-5"},{"bbox":[53,185,487,217],"type":"text","angle":0,"index":6,"merge_prev":false,"text":"若 $f(x_{1}) = \\frac{1}{n}\\sum_{i = 1}^{n}f(x_{i})$ 或 $f(x_{n}) = \\frac{1}{n}\\sum_{i = 1}^{n}f(x_{i})$ ，则取 $\\xi = x_{1},x_{n}$ 。否则由连续函数介值性定理可证。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":30,"id":"46e623bc-ed2a-42f1-8cf1-35256e0cdd14","page_size":[595,841],"block_position":"30-6"},{"bbox":[55,232,223,247],"type":"title","angle":0,"index":7,"level":2,"text":"## 3.3.4 介值性定理的无连续推广","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":30,"id":"dd78a212-c69a-4da2-8ad6-52f2c558b8af","page_size":[595,841],"block_position":"30-7"},{"bbox":[73,257,381,269],"type":"text","angle":0,"index":8,"merge_prev":false,"text":"有时没有连续的条件却能推出介值性定理的结论，常见的条件如单调。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":30,"id":"dd85b933-be35-4bec-920b-af68e101a2d6","page_size":[595,841],"block_position":"30-8"},{"bbox":[35,287,540,368],"type":"text","angle":0,"index":10,"merge_prev":false,"text":"练习3.16单调介值定理 核心： $f(x)$ 在[0,1]上单增， $f(0) > 0,f(1) < 1$ ， $k\\in \\mathbb{N}^{+}$ ，证明 $\\exists x_0\\in (0,1)$ 使得 $f(x_0) = x_0^k$ 证明取 $F(x) = f(x) - x^k$ ，显然 $F(0) = f(0) > 0,F(1) = f(1) - 1 < 0$ ，不停地二分取闭区间套，有 $\\lim_{n\\to \\infty}|a_n-$ $b_{n}| = 0$ ， $\\lim_{n\\to \\infty}a_n = \\lim_{n\\to \\infty}b_n = x_0$ ， $F(a_{n}) < 0,F(b_{n}) > 0\\Rightarrow f(a_{n}) > a_{n}^{k},f(b_{n}) < b_{n}^{k}$ ，由于 $f(x)$ 单增， $a_{n}^{k} < f(a_{n})\\leq f(x_{0})\\leq f(b_{n}) < b_{n}^{k}$ ，因此取极限可知 $x_0^k\\leq f(x_0)\\leq x_0^k$ 得证。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":30,"id":"c4084f04-2e73-4300-9b4b-ff58395717fa","page_size":[595,841],"block_position":"30-9"},{"bbox":[33,366,363,380],"type":"text","angle":0,"index":11,"merge_prev":false,"text":"笔记 单调介值性定理不能将极限取进函数，但是用了两侧夹逼的方式。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":30,"id":"19efbe44-6e40-4c25-a9d1-6722a61b0f09","page_size":[595,841],"block_position":"30-10"},{"bbox":[35,399,240,412],"type":"text","angle":0,"index":12,"merge_prev":false,"text":"练习 3.17 证明：闭区间连续函数一定有界 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":30,"id":"dbd4c461-0dae-424d-b94f-0b670316307d","page_size":[595,841],"block_position":"30-11"},{"bbox":[53,414,246,427],"type":"text","angle":0,"index":13,"merge_prev":false,"text":"证明 根据局部有界性以及有限开覆盖可证。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":30,"id":"1d387f4d-9df1-4ad1-ae00-d75d403d48d5","page_size":[595,841],"block_position":"30-12"},{"bbox":[55,450,171,466],"type":"title","angle":0,"index":14,"level":2,"text":"## 3.4 连续性的应用","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":30,"id":"aec81aec-8c46-46c2-865e-1c60237aab87","page_size":[595,841],"block_position":"30-13"},{"bbox":[35,479,428,492],"type":"text","angle":0,"index":15,"merge_prev":false,"text":"练习3.18几道经典题目(1) $f(x)$ 在 $x = 0$ 连续， $f(2x) = f(x)$ ，证明 $f(x)$ 是常值函数 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":30,"id":"6014bcae-b47d-4487-9af4-00d0ad091875","page_size":[595,841],"block_position":"30-14"},{"bbox":[72,495,351,508],"type":"text","angle":0,"index":16,"merge_prev":false,"text":"(2) $f(x)$ 在 x=0,1 连续， $f(x^{2})=f(x)$ ，证明 $f(x)$ 是常值函数 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":30,"id":"71c2cf44-33a2-4e80-8510-8355645fb7b3","page_size":[595,841],"block_position":"30-15"},{"bbox":[73,511,435,523],"type":"text","angle":0,"index":17,"merge_prev":false,"text":"(3) $f(x), g(x)$ 在 R 上连续，且在每个有理点有 $f(x) = g(x)$ ，证明 $f(x)$ 恒等于 $g(x)$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":30,"id":"de5166b3-9e09-4a56-baab-adf5513f7542","page_size":[595,841],"block_position":"30-16"},{"bbox":[52,525,539,573],"type":"text","angle":0,"index":18,"merge_prev":false,"text":"(4) 容易被吓到： $f(x) \\in C[a, b]$ ，且对 $\\forall x \\in [a, b]$ 都有 $y \\in [a, b]$ 使得 $|f(y)| \\leq \\frac{1}{2} |f(x)|$ ，证明 $f(x)$ 在 $[a, b]$ 存在零点\n\n证明 (1) $f(x) = f\\left(\\frac{x}{2}\\right) = \\cdots = f\\left(\\frac{x}{2n}\\right) = f(0)$ ，因此为常值 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":30,"id":"c8081398-64b2-426d-bac0-1a28d90ec33e","page_size":[595,841],"block_position":"30-17"},{"bbox":[53,576,539,603],"type":"text","angle":0,"index":19,"merge_prev":false,"text":"(2)x 为正直接 $f(x)=f(x^{\\frac{1}{2}})=\\cdots=f(x^{\\frac{1}{2^{n}}})=f(1)$ ，x 为负平方一下转换为正，x=0 根据连续性可知等于附近，因此整体常数 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":30,"id":"1e286a34-235b-4630-9d80-798e6247dd67","page_size":[595,841],"block_position":"30-18"},{"bbox":[73,608,211,619],"type":"text","angle":0,"index":20,"merge_prev":false,"text":"(3) 根据有理点稠密性 + 连续性 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":30,"id":"9b10d96e-a299-4deb-933a-fe4d5cfb1ada","page_size":[595,841],"block_position":"30-19"},{"bbox":[35,622,539,668],"type":"text","angle":0,"index":21,"merge_prev":false,"text":"(4) 假设无零点，根据闭区间连续函数有最小值，这与 $|f(y)| \\leq \\frac{1}{2} |f(x)|$ 矛盾\n\n练习 3.19 几道经典最值问题 (1) 重点：设 $f(x)$ 在 $(a, b)$ 连续，且 $\\lim_{x \\to a^{+}} f(x) = \\lim_{x \\to b^{-}} f(x) = A$ ，这里 $A$ 可以有限，也可以为 $\\pm \\infty$ ，证明 $f(x)$ 在 $(a, b)$ 上有最大或最小值 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":30,"id":"a2f13295-b1ac-43ac-95ac-a6e087dd3cd1","page_size":[595,841],"block_position":"30-20"},{"bbox":[52,671,539,700],"type":"text","angle":0,"index":22,"merge_prev":false,"text":"(2) 重点： $f(x)$ 在 $\\mathbb{R}$ 上连续， $\\lim_{x\\to+\\infty}f(x)=\\lim_{x\\to-\\infty}f(x)=A$ ，这里 A 可以有限，也可以为 $\\pm\\infty$ ，证明 $f(x)$ 在 $\\mathbb{R}$ 上有最大或最小值 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":30,"id":"836ab503-f2c5-4c4f-9155-54db37e375ee","page_size":[595,841],"block_position":"30-21"},{"bbox":[53,703,539,734],"type":"text","angle":0,"index":23,"merge_prev":false,"text":"(3) 重点： $f(x)$ 在 $[a, +\\infty)$ 连续， $\\lim_{x \\to +\\infty} f(x) = A$ ，这里 $A$ 为有限数，证明 $f(x)$ 在 $[a, +\\infty)$ 存在最大值或最小值 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":30,"id":"dbff57c8-cc4d-42cd-80b8-4873d201e9f8","page_size":[595,841],"block_position":"30-22"},{"bbox":[53,735,184,748],"type":"text","angle":0,"index":24,"merge_prev":false,"text":"证明 (1) 对 $A$ 进行分类讨论: ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":30,"id":"a84ba612-d8c8-45ec-a26e-66aa4c5491e3","page_size":[595,841],"block_position":"30-23"},{"bbox":[291,802,303,812],"type":"page_number","angle":0,"index":1000008,"text":"23 ","color":{"line":"rgba(164, 164, 164, 1)","fill":"rgba(164, 164, 164, 1)"},"page_idx":30,"is_discarded":true,"id":"c36a9679-8820-4e8b-a8de-b6fed668ca97","page_size":[595,841],"block_position":"30-24"}],[{"bbox":[480,47,539,58],"type":"header","angle":0,"index":-99999,"text":"3.5 一致连续 ","color":{"line":"rgba(164, 164, 164, 1)","fill":"rgba(164, 164, 164, 1)"},"page_idx":31,"is_discarded":true,"id":"c424473b-68b5-4ec9-b9da-36b18b6e6b15","page_size":[595,841],"block_position":"31-0"},{"bbox":[54,71,539,116],"type":"text","angle":0,"index":1,"merge_prev":false,"text":"(a) 先考虑 $A$ 为有限值。如果 $f(x) \\equiv A$ ，则显然。若 $\\exists x_0, f(x_0) > A$ ，则 $\\exists \\delta, \\forall x \\in (a, a + \\delta) \\cup (b - \\delta, b)$ 满足 $f(x) < f(x_0)$ ，由于 $f(x)$ 在 $[a + \\delta, b - \\delta]$ 上连续，因此 $\\exists \\xi$ 使得 $f(\\xi)$ 在 $[a + \\delta, b - \\delta]$ 最大，而 $\\forall x \\in (a, a + \\delta] \\cup [b - \\delta, b)$ ，自然也有 $f(\\xi) \\geq f(x_0) \\geq f(x)$ ，因此 $f(\\xi)$ 在 $(a, b)$ 最大。若 $f(x_0) < A$ 也同理有最小值。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":31,"id":"e535096e-15a0-429e-a1ea-0d73154e9c5c","page_size":[595,841],"block_position":"31-1"},{"bbox":[53,118,539,147],"type":"text","angle":0,"index":2,"merge_prev":false,"text":"(b) 若 $A = +\\infty$ ，则对 $\\forall x_0 \\in (a, b)$ ， $f(x_0)$ 为有限数， $\\exists \\delta, \\forall x \\in (a, a + \\delta) \\cup (b - \\delta, b)$ 有 $f(x) > f(x_0)$ ，由于 $f(x)$ 在 $[a + \\delta, b - \\delta]$ 连续，因此有最小值，且最小值比 $f(x_0)$ 还小，因此在 $(a, b)$ 上取最小值。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":31,"id":"fe447e03-de91-4065-af9c-afab06e90f31","page_size":[595,841],"block_position":"31-2"},{"bbox":[73,149,231,161],"type":"text","angle":0,"index":3,"merge_prev":false,"text":"(2) 和 (1) 同理，只是把 $\\exists \\delta$ 换成 $\\exists N$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":31,"id":"dc42108a-610b-465f-afd5-f4c2d4f08671","page_size":[595,841],"block_position":"31-3"},{"bbox":[53,164,539,194],"type":"text","angle":0,"index":4,"merge_prev":false,"text":"(3) 若 $f(x) \\equiv A$ ，则显然。若 $\\exists x_0, f(x_0) > A$ ，则 $\\exists M, \\forall x > M$ 有 $f(x_0) \\geq f(x)$ ，由于在 $[a, M]$ 连续，有最大值 $f(\\xi)$ ，且 $\\forall x > M, f(\\xi) > f(x_0) > f(x)$ ，因此有最大值。最小值同理 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":31,"id":"7f4ca819-9f66-447c-89c4-41a47e69fc55","page_size":[595,841],"block_position":"31-4"},{"bbox":[55,216,143,231],"type":"title","angle":0,"index":5,"level":2,"text":"## 3.5 一致连续","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":31,"id":"730598bd-1759-49f7-865c-21ba74b1df71","page_size":[595,841],"block_position":"31-5"},{"bbox":[55,248,174,263],"type":"title","angle":0,"index":6,"level":2,"text":"## 3.5.1 一致连续的概念","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":31,"id":"335fa620-f5bc-497f-ab5d-d1540e48b3c8","page_size":[595,841],"block_position":"31-6"},{"bbox":[73,275,161,288],"type":"title","angle":0,"index":7,"level":2,"text":"## 定义 3.9 (一致连续)","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":31,"id":"e94e2bf8-fed0-477b-9151-9ef59ecb89c7","page_size":[595,841],"block_position":"31-7"},{"bbox":[67,290,534,325],"type":"text","angle":0,"index":8,"merge_prev":false,"text":"$I$ 是任意区间 (任意开闭，有界无界)， $f: I \\to \\mathbb{R}$ ， $\\forall \\epsilon, \\exists \\delta$ 对 $\\forall x_1, x_2 \\in I$ 满足 $|x_1 - x_2| < \\delta$ 时有 $|f(x_1) - f(x_2)| < \\epsilon$ ，则称 $f$ 在 $I$ 一致连续 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":31,"id":"91885636-4f68-4d17-b976-f43a15412bea","page_size":[595,841],"block_position":"31-8"},{"bbox":[73,342,181,354],"type":"title","angle":0,"index":9,"level":2,"text":"## 定理 3.7 (常用充要条件)","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":31,"id":"b9eee5b6-f1b6-41c9-b5d3-764e376ed852","page_size":[595,841],"block_position":"31-9"},{"bbox":[67,357,527,394],"type":"text","angle":0,"index":10,"merge_prev":false,"text":"$f(x)$ 在区间 $I$ 上有定义，则 $f(x)$ 在 $I$ 一致连续当且仅当任意数列 $x_{n}^{\\prime}, x_{n}^{\\prime \\prime}$ ，若 $\\lim_{n\\to \\infty}x_n' - x_n'' = 0$ ，则 $\\lim_{n\\to \\infty}[f(x_n') - f(x_n')] = 0$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":31,"id":"d3b5640a-8bfa-4c53-a486-eea10f1d6118","page_size":[595,841],"block_position":"31-10"},{"bbox":[54,408,157,421],"type":"text","angle":0,"index":12,"merge_prev":false,"text":"证明 (1) 左推右: 显然 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":31,"id":"f4f8a303-969c-48cc-ad63-732a79f19f0f","page_size":[595,841],"block_position":"31-11"},{"bbox":[53,423,539,459],"type":"text","angle":0,"index":13,"merge_prev":false,"text":"(2) 右推左：反设不一致连续， $\\exists \\epsilon_0, \\forall \\delta, \\exists x', x''$ ，虽然 $|x' - x''| < \\delta$ ，但是 $|f(x') - f(x'')| \\geq \\epsilon_0$ 。取 $\\delta = 1, \\frac{1}{2}, \\cdots, \\frac{1}{n}, \\cdots, \\exists x_n', x_n''$ , $|x_n' - x_n''| < \\frac{1}{n}$ , $|f(x_n') - f(x_n'')| \\geq \\epsilon_0$ ，这与条件矛盾。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":31,"id":"9e2a714a-3146-42f7-b61d-50024de4cd9b","page_size":[595,841],"block_position":"31-12"},{"bbox":[52,461,274,473],"type":"text","angle":0,"index":14,"merge_prev":false,"text":"笔记 证明不一致连续绝大多数情况都用上述定理。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":31,"id":"3a200264-bfea-4558-a712-e7fceb56b6f1","page_size":[595,841],"block_position":"31-13"},{"bbox":[35,487,512,528],"type":"text","angle":0,"index":15,"merge_prev":false,"text":"练习3.20连续但不一致连续的例子： $(1)x^{2},(0, + \\infty)(2)\\sin x^{2},[0, + \\infty)(3)\\frac{x + 2}{x + 1}\\sin \\frac{1}{x},(0,1)(4)\\frac{1}{x},x\\in (0,1)$ 证明 $(1)x_{n} = \\sqrt{n},y_{n} = \\sqrt{n + 1},\\lim_{n\\to \\infty}(x_n - y_n) = 0,\\lim_{n\\to \\infty}[f(x_n) - f(y_n)] = -1$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":31,"id":"e3da78fb-daf6-416e-96f8-e8fd72bc3b3f","page_size":[595,841],"block_position":"31-14"},{"bbox":[73,528,229,552],"type":"text","angle":0,"index":16,"merge_prev":false,"text":"(2) 取 $x_{n} = \\sqrt{2n\\pi}, y_{n} = \\sqrt{2n\\pi + \\frac{\\pi}{2}}$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":31,"id":"232a7d4c-0db1-4eaf-bd51-fe6c9b144153","page_size":[595,841],"block_position":"31-15"},{"bbox":[73,553,214,576],"type":"text","angle":0,"index":17,"merge_prev":false,"text":"(3) 取 $x_{n} = \\frac{1}{2n\\pi}, y_{n} = \\frac{1}{2n\\pi + \\frac{\\pi}{2}}$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":31,"id":"de8e640e-d555-4fca-bb5a-5d778423cbb4","page_size":[595,841],"block_position":"31-16"},{"bbox":[73,576,190,600],"type":"text","angle":0,"index":18,"merge_prev":false,"text":"(4) 取 $x_{n} = \\frac{1}{n}, y_{n} = \\frac{1}{n + 1}$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":31,"id":"84e519f5-df38-449a-9197-543af8129342","page_size":[595,841],"block_position":"31-17"},{"bbox":[35,612,364,624],"type":"text","angle":0,"index":19,"merge_prev":false,"text":"练习3.21一致连续的证明 (1) 证明 $f(x) = \\cos \\sqrt{x}$ 在 $[0, +\\infty)$ 一致连续 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":31,"id":"fce9104e-2d96-4118-a2ce-1059fefdea4b","page_size":[595,841],"block_position":"31-18"},{"bbox":[73,627,252,640],"type":"text","angle":0,"index":20,"merge_prev":false,"text":"(2) 证明 $f(x) = \\sqrt{x}$ 在 $[0, +\\infty)$ 一致连续 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":31,"id":"ba2973a3-f723-4c45-9ac2-e7d5000730e7","page_size":[595,841],"block_position":"31-19"},{"bbox":[73,643,257,655],"type":"text","angle":0,"index":21,"merge_prev":false,"text":"(3) 证明 $f(x) = \\sin x$ 在 $[0, +\\infty)$ 一致连续 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":31,"id":"eea6215e-2a67-461c-895a-0148b4b9304f","page_size":[595,841],"block_position":"31-20"},{"bbox":[53,659,220,671],"type":"text","angle":0,"index":22,"merge_prev":false,"text":"证明 (1) 方法 1: 不妨设 $x_{1} > x_{2}$ , 则 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":31,"id":"2a680e9c-c1e1-4407-bfef-9f50cfebd87a","page_size":[595,841],"block_position":"31-21"},{"bbox":[132,673,459,755],"type":"interline_equation","angle":0,"index":23,"text":"$$\n\\begin{array}{l} | f (x _ {1}) - f (x _ {2}) | = | \\cos \\sqrt {x _ {1}} - \\cos \\sqrt {x _ {2}} | = 2 \\left| \\sin \\frac {\\sqrt {x _ {1}} + \\sqrt {x _ {2}}}{2} \\sin \\frac {\\sqrt {x _ {1}} - \\sqrt {x _ {2}}}{2} \\right| \\\\ \\leq 2 \\left| \\sin \\frac {\\sqrt {x _ {1}} - \\sqrt {x _ {2}}}{2} \\right| \\leq | \\sqrt {x _ {1}} - \\sqrt {x _ {2}} | = \\frac {| x _ {1} - x _ {2} |}{\\sqrt {x _ {1}} + \\sqrt {x _ {2}}} \\\\ = \\frac {x _ {1} - x _ {2}}{\\sqrt {(x _ {1} - x _ {2}) + x _ {2}} + \\sqrt {x _ {2}}} \\leq \\frac {x _ {1} - x _ {2}}{\\sqrt {x _ {1} - x _ {2}}} = \\sqrt {x _ {1} - x _ {2}} \\end{array}\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":31,"id":"fea35a2b-6240-4600-b780-adc87a424ab2","page_size":[595,841],"block_position":"31-22"},{"bbox":[54,758,343,772],"type":"text","angle":0,"index":24,"merge_prev":false,"text":"当 $\\delta = \\epsilon^2$ 时， $|x_1 - x_2| < \\delta$ 有 $|f(x_1) - f(x_2)| < \\epsilon$ ，因此一致连续 方法 2: $(\\cos \\sqrt{x})' = \\sin \\sqrt{x} \\frac{1}{2 \\sqrt{x}}$ 在 $[1, +\\infty)$ 有界，而 $[0, 1]$ 用 Cantor 定理，因此 Lip 连续推出一致连续 (2) 分段 +Lip 连续 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":31,"id":"71f48c5d-664a-460a-869a-2f89e68c31e2","page_size":[595,841],"block_position":"31-23"},{"bbox":[291,803,303,811],"type":"page_number","angle":0,"index":1000010,"text":"24 ","color":{"line":"rgba(164, 164, 164, 1)","fill":"rgba(164, 164, 164, 1)"},"page_idx":31,"is_discarded":true,"id":"3658b281-9892-4f09-8190-6ddaddc0e9a0","page_size":[595,841],"block_position":"31-24"}],[{"bbox":[479,47,539,58],"type":"header","angle":0,"index":-99998,"text":"3.5 一致连续 ","color":{"line":"rgba(164, 164, 164, 1)","fill":"rgba(164, 164, 164, 1)"},"page_idx":32,"is_discarded":true,"id":"e6117087-776c-4b8e-9771-52a9c3fc9cb3","page_size":[595,841],"block_position":"32-0"},{"bbox":[73,70,523,106],"type":"text","angle":0,"index":2,"merge_prev":false,"lines_deleted":true,"text":"","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":32,"id":"71f48c5d-664a-460a-869a-2f89e68c31e2","page_size":[595,841],"block_position":"32-1"},{"bbox":[55,124,202,138],"type":"title","angle":0,"index":3,"level":2,"text":"## 3.5.2 Cantor 定理及其推广","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":32,"id":"638984ee-a7c4-435d-bfd0-82e12c13158f","page_size":[595,841],"block_position":"32-2"},{"bbox":[73,152,174,163],"type":"title","angle":0,"index":4,"level":2,"text":"## 定理 3.8 (Cantor 定理)","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":32,"id":"fac413e9-c406-44eb-84b4-a2c40a10d76a","page_size":[595,841],"block_position":"32-3"},{"bbox":[66,157,537,187],"type":"table_body","angle":0,"index":5,"img_path":"/","id":"6f416002-53e7-4916-aaab-a28484efa772","color":{"line":"rgba(103, 194, 63, 1)","fill":"rgba(103, 194, 63, 1)"},"table_body":"
定理 3.8 (Cantor 足理)
若 f(x) 在闭区间 [a,b] 连续,则其在 [a,b] 上一致连续。
","page_idx":32,"page_size":[595,841],"block_position":"32-4"},{"bbox":[53,198,539,247],"type":"text","angle":0,"index":6,"merge_prev":false,"text":"证明 反设在 $[a, b]$ 不一致连续，根据一致连续充要条件， $\\exists \\epsilon_0, \\exists x_n, y_n$ 满足 $\\lim_{n \\to \\infty} (x_n - y_n) = 0$ ，但 $|f(x_n) - f(y_n)| \\geq \\epsilon_0$ 。由于 $x_n$ 有界，根据致密性定理可知 $x_n$ 有收敛子列，设 $\\lim_{k \\to \\infty} x_{n_k} = x_0$ ，从而 $\\lim_{k \\to \\infty} y_{n_k} = \\lim_{k \\to \\infty} [(y_{n_k} - x_{n_k}) + x_{n_k}] = x_0$ ，因此推出 $\\epsilon_0 \\leq \\lim_{k \\to \\infty} |f(x_{n_k}) - f(y_{n_k})| = |f(x_0) - f(x_0)| = 0$ （这一步用了连续条件），矛盾。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":32,"id":"1df77d8a-9322-4036-b57a-d8959e8d69b7","page_size":[595,841],"block_position":"32-5"},{"bbox":[35,250,203,264],"type":"title","angle":0,"index":7,"level":2,"text":"## 笔记 Cantor 定理的其他证明方式:","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":32,"id":"179fd7be-8a46-484d-81e5-2366370266d7","page_size":[595,841],"block_position":"32-6"},{"bbox":[68,268,539,296],"type":"text","angle":0,"index":8,"merge_prev":false,"text":"- 有限覆盖定理： $\\forall \\epsilon, \\exists \\delta_{x_0}, \\forall x \\in B(x_0, \\delta_{x_0})$ ，有 $|f(x) - f(x_0)| < \\epsilon$ ，取遍闭区间，有有限覆盖，上述 $\\epsilon$ 不变，当取 $\\delta = \\min\\{\\delta_x\\}$ 时，即有 $\\forall a, b$ 满足 $|a - b| < \\delta$ 都有 $|f(a) - f(b)| < \\epsilon$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":32,"id":"7b2794ad-c72f-44a8-9802-43496277345c","page_size":[595,841],"block_position":"32-7"},{"bbox":[68,297,539,349],"type":"text","angle":0,"index":9,"merge_prev":false,"text":"- 闭区间套： $f(x)$ 在 $[a, b]$ 不一致连续，则一定在 $[a, \\frac{a + b}{2}]$ ， $[\\frac{a + b}{2}, b]$ 中有一个区间中不一致连续，不断取区间套，最后套住 $\\xi$ ，由于 $f$ 在 $\\xi$ 连续， $\\forall \\epsilon, \\exists \\delta, \\forall x, |x - \\xi| < \\delta$ 有 $|f(\\xi) - f(x)| < \\frac{\\epsilon}{2}$ ，同理取 $y$ ，得到 $|f(x) - f(y)| < \\epsilon$ ，而存在 $a_n, b_n$ 满足 $[a_n, b_n] \\subseteq [\\xi - \\delta, \\xi + \\delta]$ ，这与不一致连续矛盾 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":32,"id":"b05f694c-b24d-44b0-87c0-94ef1f8bc53d","page_size":[595,841],"block_position":"32-8"},{"bbox":[69,349,261,362],"type":"text","angle":0,"index":10,"merge_prev":false,"text":"- 单调有界定理：和闭区间套一种构造方式 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":32,"id":"0357b99f-7dbc-4168-8b5c-0c6ef7fffa78","page_size":[595,841],"block_position":"32-9"},{"bbox":[73,371,231,383],"type":"title","angle":0,"index":11,"level":2,"text":"## 定理 3.9 (开区间一致连续充要条件)","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":32,"id":"67ccabe5-a557-43c3-a3f3-2342d876bd79","page_size":[595,841],"block_position":"32-10"},{"bbox":[64,380,537,414],"type":"table_body","angle":0,"index":12,"img_path":"/","id":"8b365f95-b60a-4b24-924d-0c19b19e7881","color":{"line":"rgba(103, 194, 63, 1)","fill":"rgba(103, 194, 63, 1)"},"table_body":"
f(x) 在 (a,b) 连续,则 f(x) 在 (a,b) 一致连续的充要条件是 \\lim_{x\\to a^{+}}f(x),\\lim_{x\\to b^{-}}f(x) 存在(有限)
","page_idx":32,"page_size":[595,841],"block_position":"32-11"},{"bbox":[54,423,539,451],"type":"text","angle":0,"index":13,"merge_prev":false,"text":"证明 (1) 左推右：根据 $f(x)$ 在 $(a, b)$ 一致连续， $\\forall \\epsilon, \\exists \\delta, \\forall x', x'', |x' - x''| < \\delta$ ，有 $|f(x') - f(x'')| < \\epsilon$ 。取 $x', x'' \\in (a, a + \\delta)$ ，显然 $|f(x') - f(x'')| < \\epsilon$ ，即右极限的 Cauchy 准则，另一侧同理。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":32,"id":"1e42a1e8-25ad-4aa9-be6a-ee51c951068e","page_size":[595,841],"block_position":"32-12"},{"bbox":[73,454,317,466],"type":"text","angle":0,"index":14,"merge_prev":false,"text":"(2) 右推左：补充定义两个端点，根据 Cantor 定理可知。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":32,"id":"7a9b3a7f-b5c8-4ba3-ab4c-099c7973ddca","page_size":[595,841],"block_position":"32-13"},{"bbox":[35,485,540,530],"type":"text","angle":0,"index":15,"merge_prev":false,"text":"练习 3.22 开区间一致连续应用 (1)(东北师大) 设 $f(x)$ 在 $(a,b)$ 上一致连续，证明 $f(x)$ 在 $(a,b)$ 上有界\n\n证明 (1) 由于一致连续，因此 $f(x)$ 在 a,b 的分别存在右、左极限，补充定义为 $F(x)$ ， $F(x)$ 在 $[a,b]$ 上连续有界，且 $f(x)$ 和 $F(x)$ 在 $(a,b)$ 相等，因此 $f(x)$ 在 $(a,b)$ 有界。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":32,"id":"4c689c07-2329-44f3-9c5c-868cbb0366fb","page_size":[595,841],"block_position":"32-14"},{"bbox":[73,551,206,564],"type":"title","angle":0,"index":16,"level":2,"text":"## 定理 3.10 (无穷区间一致连续)","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":32,"id":"3b6843d9-56ae-4ad1-8577-563891984d84","page_size":[595,841],"block_position":"32-15"},{"bbox":[63,559,537,595],"type":"table_body","angle":0,"index":17,"img_path":"/","id":"9743e3c0-5913-40c8-923c-515f75cd903d","color":{"line":"rgba(103, 194, 63, 1)","fill":"rgba(103, 194, 63, 1)"},"table_body":"
定理3.10(元为区间一致连续)
f(x)在[a,+∞)连续,limx→+∞f(x)=A(有限),则f(x)在[a,+∞)一致连续
","page_idx":32,"page_size":[595,841],"block_position":"32-16"},{"bbox":[53,602,539,649],"type":"text","angle":0,"index":18,"merge_prev":false,"text":"证明 由于 $\\lim_{x\\to +\\infty}f(x) = A$ ，根据Cauchy收敛准则， $\\forall \\epsilon ,\\exists M,\\forall x^{\\prime},x^{\\prime \\prime} > M$ 有 $|f(x^{\\prime}) - f(x^{\\prime \\prime})| < \\epsilon$ 。根据Cantor定理， $f(x)$ 在 $[a,M + 1]$ 一致连续，对 $\\forall \\epsilon ,\\exists \\delta \\in (0,1)$ ，只要 $|x^{\\prime} - x^{\\prime \\prime}| < \\delta$ 时，两点要么同时落在 $[a,M + 1]$ ，要么同时落在 $[M, + \\infty)$ ，根据上面的命题可知一致连续。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":32,"id":"ba5cb84a-59bd-45bd-82d1-23cf7ef6acd4","page_size":[595,841],"block_position":"32-17"},{"bbox":[52,651,539,679],"type":"text","angle":0,"index":19,"merge_prev":false,"text":"笔记 $[a, +\\infty)$ 一致连续并不代表无穷处 $f(x)$ 极限一定存在，只是一个充分条件，例如 $f(x) = x$ 极限不存在，但一致连续。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":32,"id":"1864d6a6-0935-4127-a9de-9df4b5417d96","page_size":[595,841],"block_position":"32-18"},{"bbox":[35,692,449,737],"type":"text","angle":0,"index":20,"merge_prev":false,"text":"练习 3.23 开区间一致连续 证明：(1) $x \\sin \\frac{1}{x}$ 在 (0,1) 一致连续 (2) $\\sin \\frac{1}{x}$ 在 (0,1) 不一致连续证明 (1) 因为 $\\lim_{x \\to 0} x \\sin \\frac{1}{x} = 1$ ，因此一致连续 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":32,"id":"4343c560-a6ff-49d4-b8c9-f1805e642dd1","page_size":[595,841],"block_position":"32-19"},{"bbox":[73,738,261,759],"type":"text","angle":0,"index":21,"merge_prev":false,"text":"(2) 因为 $\\lim_{x\\to 0}\\sin \\frac{1}{x}$ 不存在，故不一致连续。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":32,"id":"88da2915-8bdb-43bb-b682-be6ef1a9e4eb","page_size":[595,841],"block_position":"32-20"},{"bbox":[291,803,302,812],"type":"page_number","angle":0,"index":999999,"text":"25 ","color":{"line":"rgba(164, 164, 164, 1)","fill":"rgba(164, 164, 164, 1)"},"page_idx":32,"is_discarded":true,"id":"20450c00-f698-4389-834b-7df7cc03454d","page_size":[595,841],"block_position":"32-21"}],[{"bbox":[479,46,539,58],"type":"header","angle":0,"index":-99999,"text":"3.5 一致连续 ","color":{"line":"rgba(164, 164, 164, 1)","fill":"rgba(164, 164, 164, 1)"},"page_idx":33,"is_discarded":true,"id":"4eac4ed7-a5c9-4e68-8414-d1c2873a13b5","page_size":[595,841],"block_position":"33-0"},{"bbox":[55,72,196,87],"type":"title","angle":0,"index":1,"level":2,"text":"## 3.5.3 Lip 连续与一致连续","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":33,"id":"5099ddaf-7d9b-4b76-8079-4046dfb8767c","page_size":[595,841],"block_position":"33-1"},{"bbox":[72,100,214,112],"type":"title","angle":0,"index":2,"level":2,"text":"## 定理 3.11 (Lip 连续与一致连续)","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":33,"id":"2eaf3575-6073-4c6b-8a12-c265e7f62c6c","page_size":[595,841],"block_position":"33-2"},{"bbox":[68,116,317,130],"type":"text","angle":0,"index":3,"merge_prev":false,"text":"$f(x)$ 在区间 $I$ 上满足Lip条件，则 $f(x)$ 在 $I$ 上一致连续 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":33,"id":"6fbc023b-ba7d-4e2f-8696-28ddb43fa541","page_size":[595,841],"block_position":"33-3"},{"bbox":[53,143,387,164],"type":"text","angle":0,"index":4,"merge_prev":false,"text":"证明 Lip 条件即 $\\forall x', x''$ 有 $|f(x') - f(x'')| \\leq L|x' - x''|$ ，取 $\\forall \\epsilon, \\delta = \\frac{\\epsilon}{L}$ 即可。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":33,"id":"881e990f-3405-43b6-b6b9-d880d658b2dc","page_size":[595,841],"block_position":"33-4"},{"bbox":[71,173,221,186],"type":"text","angle":0,"index":5,"merge_prev":false,"text":"推论 3.1 (导数有界一定一致连续) ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":33,"id":"57d06934-951d-4cd7-87e2-99db62c51344","page_size":[595,841],"block_position":"33-5"},{"bbox":[68,190,410,203],"type":"text","angle":0,"index":6,"merge_prev":false,"text":"若 $f'(x)$ 在 $I$ 上有界，则根据Lagrange中值定理可知Lip连续，从而一致连续 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":33,"id":"40107262-b6b2-46e0-b144-a21b13c305c9","page_size":[595,841],"block_position":"33-6"},{"bbox":[73,224,281,237],"type":"title","angle":0,"index":7,"level":2,"text":"## 推论 3.2 (导数在无穷处趋于无穷则不一致连续)","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":33,"id":"bdeceff0-cbe2-47be-a3d6-0124e904a4f2","page_size":[595,841],"block_position":"33-7"},{"bbox":[69,241,301,259],"type":"text","angle":0,"index":8,"merge_prev":false,"text":"若 $\\lim_{x\\to +\\infty}f'(x) = \\infty$ ，则 $f(x)$ 在 $[a, + \\infty)$ 不一致连续 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":33,"id":"6a21de11-47b7-4cbd-be47-505bef497b5c","page_size":[595,841],"block_position":"33-8"},{"bbox":[53,275,343,293],"type":"text","angle":0,"index":9,"merge_prev":false,"text":"证明 根据 $\\lim_{x\\to +\\infty}|f'(x)| = \\infty$ 可知 $|x' - x''|$ 控制不住导数的增长。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":33,"id":"e744ab69-e6c1-44f9-b04e-9d1cdd588385","page_size":[595,841],"block_position":"33-9"},{"bbox":[52,295,539,329],"type":"text","angle":0,"index":10,"merge_prev":false,"text":"笔记 注意上述两个推论都是充分条件。前者例如 $\\sqrt{x}$ 在 $[0,1]$ 一致连续，其导数 $\\frac{1}{2\\sqrt{x}}$ 却不有界。后者例如 $x^2$ 在无穷处导数趋于无穷，则一定不一致连续。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":33,"id":"6654133a-b4bc-4e74-b1eb-89863185933e","page_size":[595,841],"block_position":"33-10"},{"bbox":[35,347,539,375],"type":"text","angle":0,"index":11,"merge_prev":false,"text":"练习 3.24 基础练习 判断一致连续性：(1) $\\sin x, x \\in R$ (2) $x^{k}, x \\in (1, +\\infty)$ (3) $\\cos \\sqrt{x}, x \\in (1, +\\infty)$ (4) $x \\ln x, x \\in (0, +\\infty)$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":33,"id":"8e899274-3bdf-4b34-a23d-4197817a234a","page_size":[595,841],"block_position":"33-11"},{"bbox":[53,379,193,391],"type":"text","angle":0,"index":12,"merge_prev":false,"text":"解(1)一致连续，因为导数有界 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":33,"id":"6d73fbf3-486d-4be6-8ba1-88665c12aa25","page_size":[595,841],"block_position":"33-12"},{"bbox":[53,394,538,423],"type":"text","angle":0,"index":13,"merge_prev":false,"text":"(2) 若 $k \\leq 0$ ，则 $\\lim_{x \\to \\infty} f(x)$ 存在，补充定义 $x = 1$ ，根据 Cantor 定理的无穷推广可知一致连续。若 $0 < k \\leq 1$ ，则导数有界，因此一致连续。若 $k > 1$ ，因为导数在无穷处趋于无穷，则不一致连续 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":33,"id":"2cc2bf9b-6cfc-4183-b590-a3da075b5b0c","page_size":[595,841],"block_position":"33-13"},{"bbox":[73,426,182,438],"type":"text","angle":0,"index":14,"merge_prev":false,"text":"(3) 导数有界，一致连续 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":33,"id":"3013aaad-a09d-4cb5-a2a5-e8afc2b92fa2","page_size":[595,841],"block_position":"33-14"},{"bbox":[73,441,265,454],"type":"text","angle":0,"index":15,"merge_prev":false,"text":"(4) 导数在无穷处趋于无穷，则不一致连续。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":33,"id":"8cc43a80-f2ef-4d09-ab95-3d2bbc642b31","page_size":[595,841],"block_position":"33-15"},{"bbox":[56,472,163,486],"type":"title","angle":0,"index":16,"level":2,"text":"## 3.5.4 区间合并问题","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":33,"id":"39b14552-3ae3-40e4-8116-6bc1d0cc9955","page_size":[595,841],"block_position":"33-16"},{"bbox":[72,499,226,512],"type":"title","angle":0,"index":17,"level":2,"text":"## 定理 3.12 (一致连续区间合并定理)","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":33,"id":"82a58598-1d28-4afb-8253-0ddb52234c47","page_size":[595,841],"block_position":"33-17"},{"bbox":[68,516,533,546],"type":"text","angle":0,"index":18,"merge_prev":false,"text":"设区间 $I_{1}$ 右端点为 $c \\in I_{1}(\\text{闭})$ ， $I_{2}$ 左端点为 $c \\in I_{2}(\\text{闭})$ ， $f(x)$ 在 $I_{1}, I_{2}$ 均一致连续，则其在 $I_{1} \\cup I_{2}$ 也一致连续 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":33,"id":"6adcff9e-9eb2-48d4-8b4d-739f1b56de6e","page_size":[595,841],"block_position":"33-18"},{"bbox":[54,558,540,607],"type":"text","angle":0,"index":19,"merge_prev":false,"text":"证明 由于 $f(x)$ 在 $I_1, I_2$ ，对 $\\forall \\epsilon$ ，存在公共 $\\delta_1$ ，当 $x', x''$ 同时落于 $I_1$ 或 $I_2$ 时，若 $|x' - x''| < \\delta_1$ ，则 $|f(x') - f(x'')| < \\epsilon$ 。若 $x', x''$ 分别落于 $I_1, I_2$ 中，则 $\\exists \\delta_2, |x' - c| < \\delta_2$ 时， $|f(x') - f(c)| < \\frac{\\epsilon}{2}$ 且 $|x'' - c| < \\delta_2$ 时 $|f(x'') - f(c)| < \\frac{\\epsilon}{2}$ ，取 $\\delta = \\min\\{\\delta_1, \\delta_2\\}$ ，得出 $\\forall x', x'' \\in I_1 \\cup I_2$ ，当 $|x' - x''| < \\delta$ 时 $|f(x') - f(x'')| < \\epsilon$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":33,"id":"3510e745-72b9-4e76-ad91-daaa0ab6771c","page_size":[595,841],"block_position":"33-19"},{"bbox":[52,607,264,618],"type":"text","angle":0,"index":20,"merge_prev":false,"text":"笔记 如果边界点为开区间点，则可以补上定义。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":33,"id":"d6c887c7-30b5-4a7e-a09e-c83aba800227","page_size":[595,841],"block_position":"33-20"},{"bbox":[35,633,539,700],"type":"text","angle":0,"index":21,"merge_prev":false,"text":"练习3.25区间合并练习(1)考虑 $f(x) = \\frac{|\\sin x|}{x}$ 在 $(-1,0)\\cup (0,1)$ 上是否一致连续证明(1)由于 $\\lim_{x\\to 0^{+}}f(x) = 1,\\lim_{x\\to 0^{-}}f(x) = -1,\\lim_{x\\to 1^{-}}f(x),\\lim_{x\\to -1^{+}}f(x)$ 显然存在，因此 $f(x)$ 在 $(-1,0),(0,1)$ 显然一致连续。但是由于0处左右极限不相同，取 $x_{n}$ 单减趋于 $0,y_{n}$ 单增趋于0，则 $|x_{n} - y_{n}|\\rightarrow 0$ ，而 $|f(x_n) - f(y_n)|\\to 2$ ，因此不一致连续。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":33,"id":"d88a00db-62ee-495c-9df1-8cb4bb4b6273","page_size":[595,841],"block_position":"33-21"},{"bbox":[291,803,303,812],"type":"page_number","angle":0,"index":1000021,"text":"26 ","color":{"line":"rgba(164, 164, 164, 1)","fill":"rgba(164, 164, 164, 1)"},"page_idx":33,"is_discarded":true,"id":"92d336d6-88d6-4872-b5b5-674866d795c2","page_size":[595,841],"block_position":"33-22"}],[{"bbox":[429,47,539,58],"type":"header","angle":0,"index":-99999,"text":"3.6 一致连续的经典问题 ","color":{"line":"rgba(164, 164, 164, 1)","fill":"rgba(164, 164, 164, 1)"},"page_idx":34,"is_discarded":true,"id":"8f591387-0d6d-489c-a7d2-a45e2320d172","page_size":[595,841],"block_position":"34-0"},{"bbox":[55,72,246,86],"type":"title","angle":0,"index":1,"level":2,"text":"## 3.5.5 四则运算、复合的一致连续性","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":34,"id":"13ac4f9b-a132-4197-a0f9-1bc88c86616b","page_size":[595,841],"block_position":"34-1"},{"bbox":[73,100,226,112],"type":"title","angle":0,"index":2,"level":2,"text":"## 定理 3.13 (四则运算的一致连续性)","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":34,"id":"78eb0ee0-4a76-4c36-a639-03acdba9770f","page_size":[595,841],"block_position":"34-2"},{"bbox":[69,116,239,129],"type":"text","angle":0,"index":3,"merge_prev":false,"text":"设 $f(x), g(x)$ 在区间 $I$ 上一致连续，则 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":34,"id":"8b8b656f-93fd-4f7b-b9cd-389cd889360c","page_size":[595,841],"block_position":"34-3"},{"bbox":[83,132,339,144],"type":"text","angle":0,"index":4,"merge_prev":false,"text":"- 数乘、加减： $\\forall k \\in \\mathbb{R}$ ， $kf(x), f(x) \\pm g(x)$ 在 $I$ 一致连续 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":34,"id":"ae354a33-b8e8-4558-89b5-e3090625c983","page_size":[595,841],"block_position":"34-4"},{"bbox":[83,148,521,160],"type":"text","angle":0,"index":5,"merge_prev":false,"text":"- 乘法： $I$ 为有限区间时， $f(x)g(x)$ 在 $I$ 一致连续。 $I$ 为无穷区间时， $f(x)g(x)$ 在 $I$ 不一定一致连续 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":34,"id":"cb2aafbc-20dc-44c5-ac96-b195e43d7435","page_size":[595,841],"block_position":"34-5"},{"bbox":[83,161,295,185],"type":"text","angle":0,"index":6,"merge_prev":false,"text":"- 除法： $f(x) \\neq 0$ ， $\\frac{1}{f(x)}$ ， $\\frac{g(x)}{f(x)}$ 不一定一致连续 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":34,"id":"9cb6437e-f78d-497f-b747-6885f2cf1072","page_size":[595,841],"block_position":"34-6"},{"bbox":[54,202,186,215],"type":"title","angle":0,"index":7,"level":2,"text":"## 证明 (1) 直接三角不等式放缩","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":34,"id":"ce521a34-bf23-46a8-a132-9cd454221979","page_size":[595,841],"block_position":"34-7"},{"bbox":[53,217,538,247],"type":"text","angle":0,"index":8,"merge_prev":false,"text":"(2) $|f(x')g(x')-f(x'')g(x'')|\\leq|f(x')g'(x)-f(x')g(x'')|+|f(x')g(x'')-f(x'')g(x'')|\\leq|f(x')||g(x')-g(x'')|+|g(x'')||f(x')-f(x'')|<2M\\epsilon$ ，因此若M有限显然成立，M不有限时不一定。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":34,"id":"d7a0350c-2fc0-4fb0-8c48-e3e4c05acba7","page_size":[595,841],"block_position":"34-8"},{"bbox":[53,264,539,300],"type":"text","angle":0,"index":9,"merge_prev":false,"text":"例题3.1四则运算反例(1)乘法： $x$ 在 $(1, + \\infty)$ 一致连续，但是 $x^{2}$ 在 $(1, + \\infty)$ 不一致连续(2)倒数： $x$ 在 $(0,1)$ 一致连续，但是 $\\frac{1}{x}$ 在 $(0,1)$ 不一致连续 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":34,"id":"f5d0a427-1c50-4a30-8d32-d766ea6089c4","page_size":[595,841],"block_position":"34-9"},{"bbox":[36,312,539,340],"type":"text","angle":0,"index":10,"merge_prev":false,"text":"练习3.26四则运算练习 $f(x)$ 在 $[a, +\\infty)$ 连续， $g(x)$ 在 $[a, +\\infty)$ 一致连续， $\\lim_{x \\to \\infty} [f(x) - g(x)] = 0$ ，证明： $f(x)$ 在 $[a, +\\infty)$ 上一致连续 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":34,"id":"b643816b-dfba-456a-9da8-e4777431caeb","page_size":[595,841],"block_position":"34-10"},{"bbox":[52,343,539,373],"type":"text","angle":0,"index":11,"merge_prev":false,"text":"证明 由于 $f(x) - g(x)$ 在无穷远处极限存在，其一致连续。 $g(x)$ 一致连续，根据四则运算可知 $f(x) = [f(x) - g(x)] + g(x)$ 一致连续。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":34,"id":"db75d253-4c6d-4593-8482-90a6edd87e52","page_size":[595,841],"block_position":"34-11"},{"bbox":[34,373,233,386],"type":"text","angle":0,"index":12,"merge_prev":false,"text":"笔记 上述练习本身也是一个重要的结论。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":34,"id":"00665615-0530-4fc7-8f7f-ec222769a850","page_size":[595,841],"block_position":"34-12"},{"bbox":[72,409,205,422],"type":"title","angle":0,"index":13,"level":2,"text":"## 定理 3.14 (复合的一致连续性)","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":34,"id":"82052843-c2c2-4ac5-886f-2357b63bfd77","page_size":[595,841],"block_position":"34-13"},{"bbox":[68,426,458,439],"type":"text","angle":0,"index":14,"merge_prev":false,"text":"$f(x)$ 在 $I_{1}$ 一致连续， $g(x)$ 在 $I_{2}$ 一致连续， $f(x)$ 值域在 $I_{2}$ 中，则 $g(f(x))$ 在 $I_{1}$ 一致连续 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":34,"id":"86c9bc90-1666-415f-ae38-d04e0dce170b","page_size":[595,841],"block_position":"34-14"},{"bbox":[53,456,269,469],"type":"text","angle":0,"index":15,"merge_prev":false,"text":"证明 根据定义套即可。外层取 $\\epsilon, \\delta$ ，内层取 $\\delta, \\delta'$ 。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":34,"id":"cd63fe02-8f8a-45fe-acab-bf1c51618cde","page_size":[595,841],"block_position":"34-15"},{"bbox":[35,487,522,516],"type":"text","angle":0,"index":16,"merge_prev":false,"text":"练习 3.27 复合一致连续性 设 $f(x)$ 在 $[0, +\\infty)$ 满足 Lip 条件， $\\alpha \\in (0, 1)$ ，证明： $f(x^{\\alpha})$ 在 $[0, +\\infty)$ 一致连续证明 $\\alpha \\in (0, 1)$ 时， $x^{\\alpha}$ 在 $[0, +\\infty)$ 一致连续，而 f 满足 Lip 条件也一致连续，根据复合一致连续即可。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":34,"id":"513f3781-aa2b-4c9a-a4db-328192df8e5d","page_size":[595,841],"block_position":"34-16"},{"bbox":[55,539,215,554],"type":"title","angle":0,"index":17,"level":2,"text":"## 3.6 一致连续的经典问题","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":34,"id":"f0af148c-0313-4faa-bb8a-92af531d8d60","page_size":[595,841],"block_position":"34-17"},{"bbox":[35,567,373,580],"type":"text","angle":0,"index":18,"merge_prev":false,"text":"练习3.28设 $f(x)$ 在 $\\mathbb{R}$ 上一致连续，证明 $\\exists a, b > 0$ 使得 $|f(x)| \\leq a|x| + b$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":34,"id":"34c53208-8d53-49f2-a16c-4890d0a4e465","page_size":[595,841],"block_position":"34-18"},{"bbox":[52,582,441,596],"type":"text","angle":0,"index":19,"merge_prev":false,"text":"证明 由于 $f(x)$ 一致连续，对 $\\epsilon = 1, \\exists \\delta$ ，当 $|x' - x''| < \\delta$ 时 $|f(x') - f(x'')| \\leq 1$ ，不妨设 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":34,"id":"c9d8cea7-b000-4795-9403-6cdac3951a6e","page_size":[595,841],"block_position":"34-19"},{"bbox":[246,602,346,616],"type":"interline_equation","angle":0,"index":20,"text":"$$\n| f (x) | \\leq M, x \\in [ - \\delta , \\delta ]\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":34,"id":"93622259-a457-4421-9ace-b99bd431e562","page_size":[595,841],"block_position":"34-20"},{"bbox":[54,619,379,641],"type":"text","angle":0,"index":21,"merge_prev":false,"text":"对 $\\forall x > \\delta, \\exists n_x$ 使得 $x - n_x\\delta = x_0 \\in [0, \\delta)$ ，此时 $n_x = \\frac{x - x_0}{\\delta} \\leq \\frac{|x|}{\\delta}$ ，因此 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":34,"id":"0d398029-977b-4bbd-93e0-e568ebbfbbda","page_size":[595,841],"block_position":"34-21"},{"bbox":[122,645,470,676],"type":"interline_equation","angle":0,"index":23,"text":"$$\n| f (x) | \\leq \\sum_ {k = 0} ^ {n - 1} | f (x - k \\delta) - f (x - (k + 1) \\delta) | + | f (x - n _ {x} \\delta) | \\leq n _ {x} + M \\leq \\frac {| x |}{\\delta} + M\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":34,"id":"346178da-e3aa-44b6-895e-b2b8424e6127","page_size":[595,841],"block_position":"34-22"},{"bbox":[54,679,355,704],"type":"text","angle":0,"index":24,"merge_prev":false,"text":"同理 $\\forall x < -\\delta$ 也有 $|f(x)| < \\frac{|x|}{\\delta} + M$ ，综上，取 $a = \\frac{1}{\\delta}, b = M$ 即可。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":34,"id":"c17279c7-f535-4bd8-8dd5-18f21964190f","page_size":[595,841],"block_position":"34-23"},{"bbox":[291,802,302,811],"type":"page_number","angle":0,"index":1000021,"text":"27 ","color":{"line":"rgba(164, 164, 164, 1)","fill":"rgba(164, 164, 164, 1)"},"page_idx":34,"is_discarded":true,"id":"0f7b53e3-9123-4ffa-9008-fa927e5fc273","page_size":[595,841],"block_position":"34-24"}],[{"bbox":[34,240,48,256],"type":"image","angle":0,"index":0,"img_path":"/c122ebe756b85cc934a9cc595d279b73472d6fd550c0d6f9bccc6b54fa5a2a43.jpg","color":{"line":"rgba(89, 92, 220, 1)","fill":"rgba(89, 92, 220, 1)"},"img_ratio":0.2,"page_idx":35,"id":"deecc4d6-7247-4377-9ad1-5e200d95a213","page_size":[595,841],"block_position":"35-0"},{"bbox":[35,656,48,673],"type":"image","angle":0,"index":1,"img_path":"/17fd49c281bf8c586d120b9c43414c3c152dbfa67f7288b38082014cd6080b4a.jpg","color":{"line":"rgba(89, 92, 220, 1)","fill":"rgba(89, 92, 220, 1)"},"img_ratio":0.2,"page_idx":35,"id":"68f58a33-d6e7-4a7b-99e1-b007fd1f58b5","page_size":[595,841],"block_position":"35-1"},{"bbox":[226,74,377,94],"type":"title","angle":0,"index":2,"level":2,"text":"## 第 4 章 一元微分学","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":35,"id":"d46ca838-2e74-4100-ba3a-cea7694e0981","page_size":[595,841],"block_position":"35-2"},{"bbox":[55,122,158,139],"type":"title","angle":0,"index":3,"level":2,"text":"## 4.1 导数与微分","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":35,"id":"8f19848a-c6cf-4c49-8414-01a2262658aa","page_size":[595,841],"block_position":"35-3"},{"bbox":[55,155,187,170],"type":"title","angle":0,"index":4,"level":2,"text":"## 4.1.1 导数概念与可导性","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":35,"id":"f6273c1c-8396-4e50-b6dc-a60c37dd5cc1","page_size":[595,841],"block_position":"35-4"},{"bbox":[60,184,539,234],"type":"table_body","angle":0,"index":5,"img_path":"/","id":"b7270ce9-70c3-4a04-bcfd-fe4d606efd57","color":{"line":"rgba(103, 194, 63, 1)","fill":"rgba(103, 194, 63, 1)"},"table_body":"
定义4.1(导数、左右导数)
导数定义为\\lim_{x\\to x_0}\\frac{f(x)-f(x_0)}{x-x_0},左右导数定义为\\lim_{x\\to x_0^-}\\frac{f(x)-f(x_0)}{x-x_0},\\lim_{x\\to x_0^+}\\frac{f(x)-f(x_0)}{x-x_0}
","page_idx":35,"page_size":[595,841],"block_position":"35-5"},{"bbox":[33,243,337,258],"type":"text","angle":0,"index":6,"merge_prev":false,"text":"笔记 $f(x)$ 在 $x_0$ 处可导当且仅当其在 $x_0$ 处左右导数存在且相等。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":35,"id":"70deb120-ff2f-4fb8-b73f-29fcd7c9a0c7","page_size":[595,841],"block_position":"35-6"},{"bbox":[35,275,368,289],"type":"text","angle":0,"index":7,"merge_prev":false,"text":"练习 4.1 用定义求导函数 设 $g(0)=g'(0)=0$ ， $f(x)$ 定义如下，求 $f'(0)$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":35,"id":"c51cb7cf-5b02-4f1d-9310-8680d6af997a","page_size":[595,841],"block_position":"35-7"},{"bbox":[233,292,360,330],"type":"interline_equation","angle":0,"index":8,"text":"$$\nf (x) = \\left\\{ \\begin{array}{l l} g (x) \\sin {\\frac {1}{x}}, & x \\neq 0 \\\\ 0, & x = 0 \\end{array} \\right.\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":35,"id":"be66b939-9360-428b-9f83-7302eca6f82e","page_size":[595,841],"block_position":"35-8"},{"bbox":[52,334,497,362],"type":"text","angle":0,"index":9,"merge_prev":false,"text":"解 根据定义可知 $f'(0) = \\lim_{\\Delta x \\to 0} \\frac{f(\\Delta x) - f(0)}{\\Delta x} = \\lim_{\\Delta x \\to 0} \\frac{g(\\Delta x) \\sin \\frac{1}{\\Delta x}}{\\Delta x} = \\lim_{\\Delta x \\to 0} \\frac{g(\\Delta x) - g(0)}{\\Delta x - 0} \\sin \\frac{1}{\\Delta x} = 0$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":35,"id":"56b0e8bb-1931-4fd3-b0b8-2c327f8cc744","page_size":[595,841],"block_position":"35-9"},{"bbox":[35,367,523,391],"type":"text","angle":0,"index":10,"merge_prev":false,"text":"练习4.2可导概念推广(1)重点： $f(x)$ 在 $x = 0$ 连续， $\\lim_{x\\to 0}\\frac{f(2x) - f(x)}{x} = A$ ，证明 $f^{\\prime}(0)$ 存在且 $f^{\\prime}(0) = A$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":35,"id":"c53b7380-4d2c-432e-959e-610235b1e924","page_size":[595,841],"block_position":"35-10"},{"bbox":[72,389,512,414],"type":"text","angle":0,"index":11,"merge_prev":false,"text":"(2) 重点：若 $\\lim_{\\Delta x \\to 0} \\frac{f(x_0 + \\Delta x) - f(x_0 - \\Delta x)}{\\Delta x}$ ，并不能推出 $x_0$ 处导数存在，例如 $|x|$ 。但反命题成立 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":35,"id":"188c9659-dc2d-4eb6-a66e-2275055cc59e","page_size":[595,841],"block_position":"35-11"},{"bbox":[52,412,465,436],"type":"text","angle":0,"index":12,"merge_prev":false,"text":"证明 (1) 根据极限存在可知 $\\forall \\epsilon, \\exists \\delta, |x| < \\delta$ 时 $A - \\frac{\\epsilon}{2} < \\frac{f(2x) - f(x)}{x} < A + \\frac{\\epsilon}{2}$ ，同理可以得到： ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":35,"id":"e5f57c2f-ad00-471c-a2e8-34c3cfaf418a","page_size":[595,841],"block_position":"35-12"},{"bbox":[114,438,480,464],"type":"interline_equation","angle":0,"index":13,"text":"$$\nA - \\frac {\\epsilon}{2} < \\frac {f (\\frac {x}{2 ^ {k - 1}}) - f (\\frac {x}{2 ^ {k}})}{x / 2 ^ {k}} < A + \\frac {\\epsilon}{2} \\Rightarrow \\frac {1}{2 ^ {k}} (A - \\frac {\\epsilon}{2}) < \\frac {f (\\frac {x}{2 ^ {k - 1}}) - f (\\frac {x}{2 ^ {k}})}{x} < \\frac {1}{2 ^ {k}} (A + \\frac {\\epsilon}{2})\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":35,"id":"c5230bcb-137f-4bfc-9414-725228fed8ac","page_size":[595,841],"block_position":"35-13"},{"bbox":[52,467,539,513],"type":"text","angle":0,"index":14,"merge_prev":false,"text":"取 $x_{k}=\\frac{x}{2^{k}}$ ，将 $k=1,2,\\cdots,n$ 全部相加得到 $(1-\\frac{1}{2^{n}})(A-\\frac{\\epsilon}{2})<\\frac{f(x)-f(x_{n})}{x}<(1-\\frac{1}{2^{n}})(A+\\frac{\\epsilon}{2})$ ，令 $n\\to\\infty$ 得到 $A-\\frac{\\epsilon}{2}<\\frac{f(x)-f(0)}{x} 0$ , 证明 $\\exists \\xi \\in (a, b)$ 使得 $\\frac{af(b) - bf(a)}{a - b} = f(\\xi) - \\xi f'(\\xi)$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":40,"id":"648ca7c7-d9ca-4c3c-ba87-2b0b862f8319","page_size":[595,841],"block_position":"40-9"},{"bbox":[73,371,409,384],"type":"text","angle":0,"index":10,"merge_prev":false,"text":"(2) 实例： $0 < a < b < \\infty$ ，证明 $\\exists\\theta \\in (a, b)$ 使得 $ae^{b} - be^{a} = (1 - \\theta)e^{\\theta}(a - b)$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":40,"id":"31747626-8466-4109-86be-c4ca8162786c","page_size":[595,841],"block_position":"40-10"},{"bbox":[52,385,539,423],"type":"text","angle":0,"index":11,"merge_prev":false,"text":"证明 (1) 由于左侧有交叉项，考虑分子分母同除以 $ab$ ，凑 Cauchy 中值定理形式即 $\\frac{\\left(\\frac{f(x)}{x}\\right)'}{\\left(\\frac{1}{x}\\right)'}$ 在 $\\xi$ 处取值，因此即证明： ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":40,"id":"54b11dec-ca28-4c3c-bd5d-37b408d05054","page_size":[595,841],"block_position":"40-11"},{"bbox":[249,421,344,453],"type":"interline_equation","angle":0,"index":13,"text":"$$\n\\frac {\\frac {f (b)}{b} - \\frac {f (a)}{a}}{\\frac {1}{b} - \\frac {1}{a}} = \\frac {\\left(\\frac {f (x)}{x}\\right) ^ {\\prime}}{\\left(\\frac {1}{x}\\right) ^ {\\prime}}\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":40,"id":"e9ca4db1-f929-4513-91b8-99103d1e6b84","page_size":[595,841],"block_position":"40-12"},{"bbox":[54,459,196,470],"type":"text","angle":0,"index":14,"merge_prev":false,"text":"根据 Cauchy 中值定理即可证明。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":40,"id":"b333f36e-209f-4813-94ba-accbfad9b381","page_size":[595,841],"block_position":"40-13"},{"bbox":[73,471,208,502],"type":"text","angle":0,"index":15,"merge_prev":false,"text":"(2) 等价于 $\\frac{\\frac{e^{b}}{b}-\\frac{e^{a}}{a}}{\\frac{1}{b}-\\frac{1}{a}}=e^{\\theta}-\\theta e^{\\theta}$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":40,"id":"ce2ad1b8-adbd-42b8-9fb8-9d253c96eb96","page_size":[595,841],"block_position":"40-14"},{"bbox":[56,517,222,531],"type":"title","angle":0,"index":16,"level":2,"text":"## 4.2.2 中值问题：构造辅助函数","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":40,"id":"dfbab0bf-3142-4125-a517-e0fc6717593f","page_size":[595,841],"block_position":"40-15"},{"bbox":[73,544,171,557],"type":"title","angle":0,"index":17,"level":2,"text":"## 定理 4.9 (直接积分法)","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":40,"id":"261d8b51-6bbe-49f0-bbc5-cc141d59cf96","page_size":[595,841],"block_position":"40-16"},{"bbox":[68,560,330,587],"type":"text","angle":0,"index":18,"merge_prev":false,"text":"要构造 $f^{\\prime}(\\xi) = g(\\xi)$ ，则构造 $F(x) = f(x) - \\int_0^x g(t)\\mathrm{d}t$ 即可 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":40,"id":"0849089f-bda8-4b16-bb82-4f049fde1ddc","page_size":[595,841],"block_position":"40-17"},{"bbox":[73,608,226,620],"type":"title","angle":0,"index":19,"level":2,"text":"## 定理 4.10 (常用中值定理辅助函数)","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":40,"id":"5d7dc4e1-63bc-4c01-ac12-3d53e1564e96","page_size":[595,841],"block_position":"40-18"},{"bbox":[67,624,524,654],"type":"text","angle":0,"index":20,"merge_prev":false,"text":"看到 $f'(\\xi) + f(\\xi)g(\\xi)$ ，需要构造 $f(x)e^{\\int g(x)dx}$ ，因为 $[f(x)e^{\\int g(x)dx}]' = [f'(x) + f(x)g(x)]e^{\\int g(x)dx}$ ，具体有下面这些技巧： ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":40,"id":"d064f8de-4561-4244-9ac4-a687c2e743ac","page_size":[595,841],"block_position":"40-19"},{"bbox":[83,657,210,671],"type":"interline_equation","angle":0,"index":21,"text":"$$\nm f (\\xi) + n f ^ {\\prime} (\\xi): f (x) e ^ {\\frac {m}{n} x}\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":40,"id":"2b68286a-484b-4657-801b-880ecce9a300","page_size":[595,841],"block_position":"40-20"},{"bbox":[83,672,310,686],"type":"text","angle":0,"index":22,"merge_prev":false,"text":"- $mf(\\xi) + n\\xi f'(\\xi)$ : $x^{m}f^{n}(x)$ (注意不是 $x^{\\frac{m}{n}}f(x)!$ ) ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":40,"id":"bac1e59d-0789-4269-ad73-d1c9d108646b","page_size":[595,841],"block_position":"40-21"},{"bbox":[83,685,524,710],"type":"text","angle":0,"index":23,"merge_prev":false,"text":"- $mf(\\xi) - n\\xi f'(\\xi)$ : $\\frac{f^n(x)}{x^m}$ 或者 $\\frac{x^m}{f^n(x)}$ ，但此时一般要配上 $g(x) = \\frac{1}{x^m}$ ，用 Cauchy 中值定理消去分母 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":40,"id":"62371b86-5f18-41b0-85da-055357ec2d6e","page_size":[595,841],"block_position":"40-22"},{"bbox":[131,713,487,741],"type":"interline_equation","angle":0,"index":24,"text":"$$\n\\left[ \\frac {f ^ {n} (x)}{x ^ {m}} \\right] = \\frac {f ^ {n - 1} (x)}{x ^ {m + 1}} [ n x f ^ {\\prime} (x) - m f (x) ], \\quad \\left[ \\frac {x ^ {m}}{f ^ {n} (x)} \\right] ^ {\\prime} = \\frac {x ^ {m - 1}}{f ^ {n + 1} (x)} [ m f (x) - n x f ^ {\\prime} (x) ]\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":40,"id":"a663bcd2-d3c9-4920-b2a8-534565851861","page_size":[595,841],"block_position":"40-23"},{"bbox":[84,743,317,756],"type":"interline_equation","angle":0,"index":25,"text":"$$\nn f ^ {\\prime} (\\xi) f (1 - \\xi) - m f (\\xi) f ^ {\\prime} (1 - \\xi): f ^ {n} (x) f ^ {m} (1 - x)\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":40,"id":"c93966f1-8a56-499f-a1a8-f8285975c1ab","page_size":[595,841],"block_position":"40-24"},{"bbox":[291,802,302,812],"type":"page_number","angle":0,"index":1000011,"text":"33 ","color":{"line":"rgba(164, 164, 164, 1)","fill":"rgba(164, 164, 164, 1)"},"page_idx":40,"is_discarded":true,"id":"f610cca2-3a9a-458c-86cd-ad451705aefe","page_size":[595,841],"block_position":"40-25"}],[{"bbox":[458,47,539,58],"type":"header","angle":0,"index":-99999,"text":"4.2 微分中值定理 ","color":{"line":"rgba(164, 164, 164, 1)","fill":"rgba(164, 164, 164, 1)"},"page_idx":41,"is_discarded":true,"id":"6a1a6ac7-3478-41ae-b421-4d428547a675","page_size":[595,841],"block_position":"41-0"},{"bbox":[82,79,417,166],"type":"interline_equation","angle":0,"index":1,"text":"$$\n\\begin{array}{l} \\bullet m f ^ {\\prime} (x) g (x) + n f (x) g ^ {\\prime} (x): f ^ {m} (x) g ^ {n} (x) \\\\ \\bullet m f ^ {\\prime} (x) g (x) - n f (x) g ^ {\\prime} (x): \\frac {f ^ {m} (x)}{g ^ {n} (x)} \\\\ \\left[ \\frac {f ^ {m} (x)}{g ^ {n} (x)} \\right] ^ {\\prime} = \\frac {f ^ {m - 1} (x)}{g ^ {n + 1} (x)} [ m f ^ {\\prime} (x) g (x) - n f (x) g ^ {\\prime} (x) ] \\\\ \\bullet f (\\xi) g ^ {\\prime \\prime} (\\xi) - g (\\xi) f ^ {\\prime \\prime} (\\xi): f ^ {\\prime} (x) g (x) - f (x) g ^ {\\prime} (x) \\end{array}\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":41,"id":"4ac0d0c1-d9d9-4c4c-80a2-00b2d301f6b1","page_size":[595,841],"block_position":"41-1"},{"bbox":[33,177,537,211],"type":"text","angle":0,"index":2,"merge_prev":false,"text":"笔记 主要看高阶导数项的系数是啥，是常数的话肯定是 $e^{ax}f(x)$ ，是 $\\xi$ 的话是 $x^{a}f(x)$ ，是 $-\\xi$ 的话是 $\\frac{f(x)}{x^{a}}$ ，是 $(1-\\xi)$ 的话是 $(1-x)^{a}f(x)$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":41,"id":"e6b5b5fe-1116-437e-9598-384543fbfd6c","page_size":[595,841],"block_position":"41-2"},{"bbox":[35,227,538,256],"type":"text","angle":0,"index":3,"merge_prev":false,"text":"练习4.18行列式求导 $f(x), g(x), h(x)$ 在 $[a, b]$ 连续， $(a, b)$ 可导，证明： $\\exists \\xi \\in (a, b)$ 下面等式成立。若 $h(x) = 1$ 则为Cauchy中值定理，若 $h(x) = 1, g(x) = x$ 则为Lagrange中值定理。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":41,"id":"9cc6f6fb-263e-449d-87a7-1d29b319ad89","page_size":[595,841],"block_position":"41-3"},{"bbox":[238,260,358,307],"type":"interline_equation","angle":0,"index":4,"text":"$$\n\\left| \\begin{array}{c c c} f (a) & g (a) & h (a) \\\\ f (b) & g (b) & h (b) \\\\ f ^ {\\prime} (\\xi) & g ^ {\\prime} (\\xi) & h ^ {\\prime} (\\xi) \\end{array} \\right| = 0\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":41,"id":"867f2188-f474-4278-b8aa-78028e7908d3","page_size":[595,841],"block_position":"41-4"},{"bbox":[52,311,427,324],"type":"text","angle":0,"index":5,"merge_prev":false,"text":"证明 构造下面的 $F(x)$ ，注意到 $F(a) = F(b) = 0$ 以及行列式求导即对一行求导即可。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":41,"id":"2af2b8bc-2b75-4e02-9ade-6db5c0d650a2","page_size":[595,841],"block_position":"41-5"},{"bbox":[230,328,358,376],"type":"interline_equation","angle":0,"index":6,"text":"$$\nF (x) = \\left| \\begin{array}{c c c} f (a) & g (a) & h (a) \\\\ f (b) & g (b) & h (b) \\\\ f (x) & g (x) & h (x) \\end{array} \\right.\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":41,"id":"3f8e1cf6-1f46-4e8c-bc66-c73bc0b5396f","page_size":[595,841],"block_position":"41-6"},{"bbox":[35,384,531,416],"type":"text","angle":0,"index":7,"merge_prev":false,"text":"练习4.19直接积分法(1)重点： $f(x)$ 在[1,2]连续， $(1,2)$ 可微，证明 $\\exists \\xi \\in (1,2)$ 使得 $f(2) - f(1) = \\frac{1}{2}\\xi^2 f'\\left(\\xi\\right)$ (2)重点： $f(x)$ 在 $[a,b]$ 连续， $(a,b)$ 可导， $\\exists \\xi \\in (a,b)$ 使得 $2\\xi [f(b) - f(a)] = (b^{2} - a^{2})f^{\\prime}(\\xi)$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":41,"id":"aaa4591e-cdb8-4feb-aba6-3561582be93c","page_size":[595,841],"block_position":"41-7"},{"bbox":[50,401,532,451],"type":"text","angle":0,"index":8,"merge_prev":false,"text":"(2) 重点： $f(x)$ 在 $[a, b]$ 连续， $(a, b)$ 可导， $\\exists \\xi \\in (a, b)$ 使得 $2\\xi[f(b) - f(a)] = (b^2 - a^2)f'(\\xi)$ 证明 (1) 由于只有单项，只需要直接用积分法：所求等价于 $f'(\\xi) = \\frac{2}{\\xi^2}[f(2) - f(1)]$ ，构造 $F(x) = f(x) + \\frac{2}{x}[f(2) - f(1)]$ ，此时 $F(1) = F(2) = 2f(2) - f(1)$ ，因此根据 Rolle 中值定理可知 $\\exists \\xi \\in (1, 2), F'(\\xi) = 0$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":41,"id":"c6df3b56-10af-410b-9e9b-340109f13ae0","page_size":[595,841],"block_position":"41-8"},{"bbox":[72,452,530,466],"type":"text","angle":0,"index":9,"merge_prev":false,"text":"(2) 直接用积分法： $F(x) = (b^{2} - a^{2})f(x) - x^{2}[f(b) - f(a)]$ ，根据 $F(b) = F(a)$ 以及 Rolle 中值定理可知。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":41,"id":"26abdfa1-c43f-4178-94b1-99c38bc6dd4b","page_size":[595,841],"block_position":"41-9"},{"bbox":[35,483,539,512],"type":"text","angle":0,"index":10,"merge_prev":false,"text":"练习 4.20 简单的辅助函数法 (1) $f(x)$ 在 $[a, b]$ 连续， $(a, b)$ 可导， $f(a) < 0, f(b) < 0$ ， $\\exists c \\in (a, b)$ 使得 $f(c) > 0$ ，证明： $\\xi \\in (a, b)$ 使得 $f(\\xi) + f'(\\xi) = 0$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":41,"id":"cf0c2951-3884-4911-a258-a25cc3859449","page_size":[595,841],"block_position":"41-10"},{"bbox":[54,513,540,576],"type":"text","angle":0,"index":11,"merge_prev":false,"text":"(2) 重点： $f(x)$ 在 $[a, b]$ 上 $n$ 阶连续可微，在 $(a, b)$ 上存在 $n + 1$ 阶导数， $f^{(k)}(a) = f^{(k)}(b), k = 0, 1, \\cdots, n$ ，证明： $\\exists \\xi \\in (a, b)$ 使得 $f(\\xi) = f^{(n+1)}(\\xi)$ (3) 重点： $f(x)$ 在 $[0, 1]$ 连续， $(0, 1)$ 可导， $f(0) = f(1) = 0, f(\\frac{1}{2}) = 1$ ，证明： $\\forall \\lambda > 0, \\exists \\xi \\in (0, 1)$ 使得 $f'(\\xi) - \\lambda[f(\\xi) - \\xi] = 1$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":41,"id":"96e175a9-8aa7-4714-bdda-041cd222b133","page_size":[595,841],"block_position":"41-11"},{"bbox":[52,576,539,626],"type":"text","angle":0,"index":12,"merge_prev":false,"text":"(4) 重点: $f(x)$ 在 $[0,1]$ 可微, $f(0) = 0$ , 对 $\\forall x \\in (0,1), f(x) \\neq 0$ , 证明: $\\forall n, \\exists \\xi \\in (0,1)$ 使得 $n\\frac{f'(\\xi)}{f(\\xi)} = \\frac{f'(1 - \\xi)}{f(1 - \\xi)}$ 证明 (1) 构造 $F(x) = e^x f(x)$ , 则 $F(a) < 0, F(b) < 0, F(c) > 0$ , 根据连续函数零点存在定理可知有两个零点, 再根据 Rolle 中值定理可知。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":41,"id":"7fd5934c-4216-46e1-8a99-500b2992ad46","page_size":[595,841],"block_position":"41-12"},{"bbox":[73,629,509,642],"type":"text","angle":0,"index":13,"merge_prev":false,"text":"(2) 构造 $F(x) = [f(x) + f'(x) + \\cdots + f^{(n)}(x)]e^{-x}$ ， $F(a) = F(b) = 0$ ，根据 $F'(\\xi) = 0$ 可以得到结论 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":41,"id":"c309e460-946c-4d11-8a84-b2e1c600fb79","page_size":[595,841],"block_position":"41-13"},{"bbox":[53,645,539,683],"type":"text","angle":0,"index":14,"merge_prev":false,"text":"(3) 考虑配为 $[f'(\\xi) - 1] - \\lambda[f(\\xi) - \\xi] = 0$ ，故构造 $F(x) = [f(x) - x]e^{-\\lambda x}$ ，此时 $F(0) = 0, F(\\frac{1}{2}) > 0, F(1) < 0$ ，在 $[\\frac{1}{2}, 1]$ 中 $F(x)$ 有一个零点，根据 Rolle 可知。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":41,"id":"1c65a170-9524-4e5e-8e16-ae131089a934","page_size":[595,841],"block_position":"41-14"},{"bbox":[71,682,402,696],"type":"text","angle":0,"index":15,"merge_prev":false,"text":"(4) 即 $nf'(\\xi)f(1-\\xi)-f(\\xi)f'(1-\\xi)=0$ ，因此构造 $F(x)=f^n(x)f(1-x)$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":41,"id":"796e8e77-9f0e-4e99-a77c-f6d853e3957b","page_size":[595,841],"block_position":"41-15"},{"bbox":[35,713,539,743],"type":"text","angle":0,"index":16,"merge_prev":false,"text":"练习 4.21 进阶难度的辅助函数法 (1)(上交 2021) $f(x)$ 在 $[a, c]$ 可导， $f_{-}^{\\prime}(c) = 0$ ，证明： $\\exists \\xi \\in (a, c)$ 使得 $f^{\\prime}(\\xi) = 2(f(\\xi) - f(a))$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":41,"id":"4e4ca872-be8c-4b10-b6ff-571ae5db6f15","page_size":[595,841],"block_position":"41-16"},{"bbox":[72,741,428,766],"type":"text","angle":0,"index":17,"merge_prev":false,"text":"(2) $f(x)$ 在 $[0,1]$ 二阶可导， $f(0)=f(1)$ ，证明： $\\exists\\xi\\in(0,1)$ ，使得 $f''(\\xi)=\\frac{3f'(\\xi)}{1-\\xi}$ 证明 (1) 首先显然最终等式可写为 $[f(x) - f(a)]' - 2(f(x) - f(a)) = 0$ ，因此构造 $F(x) = (f(x) - f(a))e^{-2x}$ ， $F'(x) = [f'(x) - 2(f(x) - f(a))]e^{-2x}$ ，显然 $F(a) = 0$ ，但另一侧信息不足。根据条件可知 $F_{-}'(c) = -2[f(c) - f(a)]e^{-2c}$ ，下面对 $F(c)$ 情况进行讨论。(a) 若 $F(c) = 0$ 则显然成立 (b) 若 $F(c) \\neq 0$ ，不妨设 $F(c) > 0$ ，根据 $F(c) = -2F_{-}'(c)$ （根据表达式）可看出 $F_{-}'(c) < 0$ ，根据 Lagrange 中值定理： ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":41,"id":"0a32a123-57b9-4990-a402-3685e17fe28c","page_size":[595,841],"block_position":"41-17"},{"bbox":[291,802,303,811],"type":"page_number","angle":0,"index":1000017,"text":"34 ","color":{"line":"rgba(164, 164, 164, 1)","fill":"rgba(164, 164, 164, 1)"},"page_idx":41,"is_discarded":true,"id":"86ecbc23-a432-4301-9634-ea3fa47e2b64","page_size":[595,841],"block_position":"41-18"}],[{"bbox":[459,47,539,58],"type":"header","angle":0,"index":-99998,"text":"4.2 微分中值定理 ","color":{"line":"rgba(164, 164, 164, 1)","fill":"rgba(164, 164, 164, 1)"},"page_idx":42,"is_discarded":true,"id":"06b45426-2bc3-40b6-b576-eb52b1072902","page_size":[595,841],"block_position":"42-0"},{"bbox":[53,71,539,132],"type":"text","angle":0,"index":2,"merge_prev":false,"lines_deleted":true,"text":"","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":42,"id":"0a32a123-57b9-4990-a402-3685e17fe28c","page_size":[595,841],"block_position":"42-1"},{"bbox":[179,137,413,151],"type":"interline_equation","angle":0,"index":3,"text":"$$\n0 < F (c) = F (c) - F (a) = F ^ {\\prime} (\\eta) (c - a) \\Rightarrow F ^ {\\prime} (\\eta) > 0\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":42,"id":"c1b8a392-aed4-471c-88dc-8375072c360e","page_size":[595,841],"block_position":"42-2"},{"bbox":[54,157,212,169],"type":"text","angle":0,"index":4,"merge_prev":false,"text":"根据导数介值性定理可知结论成立。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":42,"id":"af5d455b-d5fa-4b43-9538-bd8ece70ef75","page_size":[595,841],"block_position":"42-3"},{"bbox":[53,172,539,202],"type":"text","angle":0,"index":5,"merge_prev":false,"text":"(2) 等价于 $(1 - \\xi)f''(\\xi) - 3f'(\\xi) = 0$ ，因此构造 $F(x) = (1 - x)^3 f'(x)$ ，根据 $f(0) = f(1) = 0$ ，根据 Rolle 可知 $\\exists \\eta$ 使得 $f'(\\eta) = 0$ ，因此 $F(\\eta) = 0$ ，再根据 $F(1) = 0$ 可知 $\\exists \\xi$ 使得 $F'(\\xi) = 0$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":42,"id":"9e8a0808-9c40-4917-816b-fb3caeee46b7","page_size":[595,841],"block_position":"42-4"},{"bbox":[55,218,211,232],"type":"title","angle":0,"index":6,"level":2,"text":"## 4.2.3 带积分的微分中值问题","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":42,"id":"3ef420f3-8e82-4e2b-9b31-6209b2479a07","page_size":[595,841],"block_position":"42-5"},{"bbox":[73,240,411,268],"type":"text","angle":0,"index":7,"merge_prev":false,"text":"由于变限积分 $\\int_{a}^{x}f(t)\\mathrm{d}t$ 的导数为 $f(x)$ ，该性质常常会应用于微分中值定理。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":42,"id":"6e8876f0-bb8b-495f-8a8d-6f349053eda1","page_size":[595,841],"block_position":"42-6"},{"bbox":[35,268,539,315],"type":"text","angle":0,"index":8,"merge_prev":false,"text":"练习4.22基础问题 (1)重点： $f(x)$ 在[0,1]连续，(0,1)可导， $\\int_0^1 xf(x)\\mathrm{d}x = f(1)$ ，证明： $\\exists \\xi \\in (0,1)$ 使得 $f^{\\prime}(\\xi) = -\\frac{f(\\xi)}{\\xi}$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":42,"id":"ede72cfc-b92b-421a-ac28-40aafd05c227","page_size":[595,841],"block_position":"42-7"},{"bbox":[72,314,411,341],"type":"text","angle":0,"index":9,"merge_prev":false,"text":"(2)ZJU2021: $f(x)$ 在 $\\mathbb{R}$ 连续， $g(x) = f(x)\\int_0^x f(t)\\mathrm{d}t$ 单减，证明 $f(x)$ 恒为0 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":42,"id":"385a303a-8a0f-45b0-abdd-d9cb1d9330d0","page_size":[595,841],"block_position":"42-8"},{"bbox":[52,343,446,376],"type":"text","angle":0,"index":10,"merge_prev":false,"text":"(3) 重点: $f(x)$ 是 $(0,1)$ 上连续函数, 证明: $\\exists c \\in (0,1)$ 使得 $\\int_{0}^{c} f(x) \\mathrm{d}x = (1 - c)f(c)$ 证明 (1) 所证形式即 $\\xi f'(\\xi) + f(\\xi) = 0$ , 因此 $F(x) = xf(x)$ , 根据积分第一中值定理可知 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":42,"id":"eb0e0864-c734-4856-a8ef-33096bbedf74","page_size":[595,841],"block_position":"42-9"},{"bbox":[210,380,382,407],"type":"interline_equation","angle":0,"index":11,"text":"$$\n\\int_ {0} ^ {1} x f (x) d x = \\eta f (\\eta) \\cdot 1 = F (\\eta) = f (1)\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":42,"id":"fa7d681c-db2f-40a0-a5c1-7757c26c0b4b","page_size":[595,841],"block_position":"42-10"},{"bbox":[54,409,262,421],"type":"text","angle":0,"index":12,"merge_prev":false,"text":"而 $F(1)=f(1)$ ，因此根据 Rolle 中值定理可知。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":42,"id":"a69ba716-977c-450d-aff8-5e486c8b2605","page_size":[595,841],"block_position":"42-11"},{"bbox":[54,423,538,457],"type":"text","angle":0,"index":13,"merge_prev":false,"text":"(2) $F(x)=\\frac{1}{2}(\\int_{0}^{x}f(t)\\mathrm{d}t)^{2}$ ，根据条件 $F'(x)$ 单减， $F'(0)=0$ ，根据 $F(0)=0$ 得出 $F(x)\\leq0$ ，但 $F(x)\\geq0$ ，因此 $F(x)$ 恒为0，推出 $f(x)$ 恒为0。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":42,"id":"8c7b097d-e652-45d6-a914-5a0734434870","page_size":[595,841],"block_position":"42-12"},{"bbox":[54,458,539,492],"type":"text","angle":0,"index":14,"merge_prev":false,"text":"(3) 所求即 $\\int_0^c f(x)\\mathrm{d}x - (1 - c)f(c) = 0$ ，构造 $F(x) = (1 - x)\\int_0^x f(t)\\mathrm{d}t$ ，显然 $F(0) = F(1) = 0$ ，根据 Rolle 可知。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":42,"id":"c45347c6-ea57-4722-b8a0-b6560f246288","page_size":[595,841],"block_position":"42-13"},{"bbox":[35,505,541,598],"type":"text","angle":0,"index":15,"merge_prev":false,"text":"练习4.23进阶问题 $f(x)$ 在[0,2]可导，(0,2)上三阶可导， $f(0) = f'(0) = 0, \\int_0^2 f(x) \\mathrm{d}x = 8 \\int_0^1 f(x) \\mathrm{d}x$ ，证明： $\\exists \\xi \\in (0,2)$ 使得 $f'''(\\xi) = 0$ 。证明构造 $F(x) = \\int_0^{2x} f(t) dt - 8 \\int_0^x f(t) dt$ ，得到 $F(0) = F(1) = 0$ ，故 $\\exists \\xi_1, F'(\\xi_1) = 0$ ，根据 $F'(x) = 2f(2x) - 8f(x)$ ，因此 $F'(0) = 0$ ，得到 $\\exists \\xi_2$ 使得 $F''(\\xi_2) = 0$ 。再根据 $F''(x) = 4f'(2x) - 8f'(x)$ ， $F''(0) = 0$ ，得到 $\\exists \\xi_3$ 使得 $F'''(\\xi_3) = 0$ 。而 $F'''(x) = 8f''(2x) - 8f''(x)$ ，代入 $\\xi_3$ 得到 $f''(2\\xi_3) = f''(\\xi_3)$ ，因此 $\\exists \\xi$ 使得 $f'''(\\xi) = 0$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":42,"id":"9e2dcd4e-54dc-4cad-a9d9-fbad150bfa5b","page_size":[595,841],"block_position":"42-14"},{"bbox":[55,614,211,628],"type":"title","angle":0,"index":16,"level":2,"text":"## 4.2.4 多个中值点的中值问题","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":42,"id":"85903ca5-6ceb-4170-a21f-f3940f382a77","page_size":[595,841],"block_position":"42-15"},{"bbox":[35,638,470,650],"type":"text","angle":0,"index":17,"merge_prev":false,"text":"练习4.24经典两个中值点的问题设 $f(x)$ 在[0,1]连续，(0,1)可导， $f(0) = 0,f(1) = 1$ ，证明： ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":42,"id":"dc8a8216-2c4c-4769-a184-ac617cf5ab6b","page_size":[595,841],"block_position":"42-16"},{"bbox":[73,654,213,666],"type":"text","angle":0,"index":18,"merge_prev":false,"text":"(1) $\\exists\\xi\\in(0,1)$ 使得 $f(\\xi)=1-\\xi$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":42,"id":"c555fbff-19e8-46ef-90df-63a5bfbc9537","page_size":[595,841],"block_position":"42-17"},{"bbox":[73,669,296,682],"type":"text","angle":0,"index":19,"merge_prev":false,"text":"(2) 存在互异的 $\\eta_1, \\eta_2 \\in (0,1)$ 使得 $f'(\\eta_1)f'(\\eta_2) = 1$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":42,"id":"029d68ce-da59-4221-b4f5-0f2c62552775","page_size":[595,841],"block_position":"42-18"},{"bbox":[73,684,323,697],"type":"text","angle":0,"index":20,"merge_prev":false,"text":"(3) 存在互异的 $\\eta_1, \\eta_2 \\in (0,1)$ 使得 $f'(\\eta_1)(f'(\\eta_2) + 1) = 2$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":42,"id":"d4360665-0b65-4849-b71f-17c7a78dd933","page_size":[595,841],"block_position":"42-19"},{"bbox":[53,700,449,713],"type":"text","angle":0,"index":21,"merge_prev":false,"text":"证明 (1) 构造 $F(x) = f(x) - 1 + x$ ，则 $F(0) = -1, F(1) = 1$ ，根据连续零点存在定理可得 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":42,"id":"05350b2c-242c-47b5-88be-50ee61b29378","page_size":[595,841],"block_position":"42-20"},{"bbox":[72,716,395,729],"type":"text","angle":0,"index":22,"merge_prev":false,"text":"(2) 对 (1) 中的 $\\xi$ ，在 $[0, \\xi]$ , $[\\xi, 1]$ 上用 Lagrange 中值定理，得到 $\\eta_{1}, \\eta_{2}$ 满足 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":42,"id":"ccdf7f61-3366-4e57-9395-58a0863f3a80","page_size":[595,841],"block_position":"42-21"},{"bbox":[165,730,427,757],"type":"interline_equation","angle":0,"index":23,"text":"$$\nf ^ {\\prime} (\\eta_ {1}) = \\frac {f (\\xi) - 0}{\\xi} = \\frac {1 - \\xi}{\\xi}, f ^ {\\prime} (\\eta_ {2}) = \\frac {f (1) - f (\\xi)}{1 - \\xi} = \\frac {\\xi}{1 - \\xi}\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":42,"id":"993e9fd6-354a-4583-a454-3d5565c7ec6d","page_size":[595,841],"block_position":"42-22"},{"bbox":[54,759,150,772],"type":"text","angle":0,"index":24,"merge_prev":false,"text":"因此 $f^{\\prime}(\\eta_{1})f^{\\prime}(\\eta_{2}) = 1$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":42,"id":"0d93b28b-a67c-4b42-8a57-4f20abcffdb7","page_size":[595,841],"block_position":"42-23"},{"bbox":[291,803,302,812],"type":"page_number","angle":0,"index":999999,"text":"35 ","color":{"line":"rgba(164, 164, 164, 1)","fill":"rgba(164, 164, 164, 1)"},"page_idx":42,"is_discarded":true,"id":"efd0a0f3-a336-4139-a606-e20e6677e975","page_size":[595,841],"block_position":"42-24"}],[{"bbox":[458,47,539,58],"type":"header","angle":0,"index":-99998,"text":"4.2 微分中值定理 ","color":{"line":"rgba(164, 164, 164, 1)","fill":"rgba(164, 164, 164, 1)"},"page_idx":43,"is_discarded":true,"id":"e8f21653-1ad6-4d67-9bcb-d304e610a53e","page_size":[595,841],"block_position":"43-0"},{"bbox":[72,72,377,84],"type":"text","angle":0,"index":2,"merge_prev":false,"text":"(3) 提示：先找 $\\xi$ 满足 $f(\\xi) = 2(1 - \\xi)$ ，再用两次 Lagrange 中值定理。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":43,"id":"0bfce9e8-7fcc-4e13-8e78-1562e77e641c","page_size":[595,841],"block_position":"43-1"},{"bbox":[55,103,164,116],"type":"title","angle":0,"index":3,"level":2,"text":"## 4.2.5 中值等式问题","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":43,"id":"e930ac2c-4161-4dd6-83ca-74696c2660e2","page_size":[595,841],"block_position":"43-2"},{"bbox":[52,126,539,155],"type":"text","angle":0,"index":4,"merge_prev":false,"text":"中值等式问题本质很像多项式插值中 Cauchy 常数的构造方法，根据目标等式的形式，将某个常数换成 x，以此得出辅助函数在某些位置相等或等于 0。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":43,"id":"21292a1d-fc0b-48b5-9c8d-a0c09f52864d","page_size":[595,841],"block_position":"43-3"},{"bbox":[35,173,538,205],"type":"text","angle":0,"index":5,"merge_prev":false,"text":"练习 4.25 利用等式构造辅助函数 (1) 重点: $f(x)$ 在 $[a, b]$ 二阶可导, $f(a) = f(b) = 0$ , 证明对 $\\forall x \\in (a, b)$ , 都存在 $\\xi \\in (a, b)$ , 使得 $f(x) = \\frac{f''(\\xi)}{2}(x - a)(x - b)$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":43,"id":"59f1e11f-5dad-4c6a-b8dc-efe288df3227","page_size":[595,841],"block_position":"43-4"},{"bbox":[72,206,368,219],"type":"text","angle":0,"index":6,"merge_prev":false,"text":"(2) 变形： $f(x)$ 在 $[a, b]$ 二阶可导，a < c < b，证明 $\\exists \\xi \\in (a, b)$ 使得 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":43,"id":"61c62254-5a6e-4949-a163-c44b5501cc3a","page_size":[595,841],"block_position":"43-5"},{"bbox":[201,221,393,247],"type":"interline_equation","angle":0,"index":7,"text":"$$\n\\frac {f (b) - f (a)}{b - a} - \\frac {f (c) - f (a)}{c - a} = \\frac {1}{2} f ^ {\\prime \\prime} (\\xi) (b - c)\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":43,"id":"4268edf2-3fc5-4768-ac2c-ad544a7a097e","page_size":[595,841],"block_position":"43-6"},{"bbox":[52,250,540,301],"type":"text","angle":0,"index":8,"merge_prev":false,"text":"证明 (1) 固定 $x$ ，记常数 $k = \\frac{2f(x)}{(x - a)(x - b)}$ ，构造 $F(t) = f(t) - \\frac{k}{2}(t - a)(t - b)$ ，满足 $F(x) = F(a) = F(b) = 0$ ， $\\exists \\xi_1 \\in (a, x), \\xi_2 \\in (x, b)$ 使得 $F'(\\xi_1) = F'(\\xi_2) = 0$ ，根据 Rolle 定理可知 $\\exists \\xi \\in (\\xi_1, \\xi_2)$ 满足 $F''(\\xi) = 0$ ，因此 $f''(\\xi) = k$ ，结论成立。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":43,"id":"494c188c-ed3e-4838-a372-3ee125aa0648","page_size":[595,841],"block_position":"43-7"},{"bbox":[72,300,446,327],"type":"text","angle":0,"index":9,"merge_prev":false,"text":"(2) 构造常数 $k = \\frac{(c - a)[f(b) - f(a)] - (b - a)[f(c) - f(a)]}{\\frac{1}{2}(b - c)(b - a)(c - a)}$ ，辅助函数将 $c$ 用 $x$ 代替 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":43,"id":"608ec79c-2e1c-4e83-bc33-a2368c60aeab","page_size":[595,841],"block_position":"43-8"},{"bbox":[130,330,462,353],"type":"interline_equation","angle":0,"index":10,"text":"$$\nF (x) = (x - a) [ f (b) - f (a) ] - (b - a) [ f (x) - f (a) ] - \\frac {k}{2} (b - x) (b - a) (x - a)\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":43,"id":"67bcb2cb-eae4-4699-9be7-afd2c91fe6fd","page_size":[595,841],"block_position":"43-9"},{"bbox":[52,354,422,368],"type":"text","angle":0,"index":11,"merge_prev":false,"text":"得到 $F(a)=F(b)=F(c)=0$ ，因此根据两次 Rolle 得到 $F''(\\xi)=0$ ，从而得到结论。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":43,"id":"5d80330a-9837-4b16-9e1b-72d24a2c6447","page_size":[595,841],"block_position":"43-10"},{"bbox":[35,386,382,399],"type":"text","angle":0,"index":12,"merge_prev":false,"text":"练习4.26另外几道经典题(1) $f(x)$ 在 $[a,b]$ 三阶可导，证明 $\\exists \\xi \\in (a,b)$ 使得 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":43,"id":"f2f757b1-3816-478d-bf76-a3c7d4171e66","page_size":[595,841],"block_position":"43-11"},{"bbox":[171,401,421,423],"type":"interline_equation","angle":0,"index":13,"text":"$$\nf (b) = f (a) + \\frac {1}{2} (b - a) [ f ^ {\\prime} (a) + f ^ {\\prime} (b) ] - \\frac {1}{1 2} (b - a) ^ {3} f ^ {\\prime \\prime \\prime} (\\xi)\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":43,"id":"afc097c6-be2d-4b9a-aa2c-82a7a2175128","page_size":[595,841],"block_position":"43-12"},{"bbox":[72,426,506,439],"type":"text","angle":0,"index":14,"merge_prev":false,"text":"(2) 配合拉格朗日插值或拉格朗日余项的 Taylor 展开： $f(x)$ 在 $[a, b]$ 二阶可导，证明 $\\exists \\xi \\in (a, b)$ 使得 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":43,"id":"5db25cbd-bcf4-476c-baae-6034fd20e489","page_size":[595,841],"block_position":"43-13"},{"bbox":[202,442,390,465],"type":"interline_equation","angle":0,"index":15,"text":"$$\nf (b) - 2 f (\\frac {a + b}{2}) + f (a) = \\frac {1}{4} (b - a) ^ {2} f ^ {\\prime \\prime} (\\xi)\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":43,"id":"4b87e79c-5c02-4d6d-98bf-5827dd9e6c6d","page_size":[595,841],"block_position":"43-14"},{"bbox":[73,466,395,480],"type":"text","angle":0,"index":16,"merge_prev":false,"text":"(3) 需配合 Taylor 展开: $f(x)$ 在 $[a, b]$ 上二阶可导, 证明 $\\exists \\xi \\in (a, b)$ , 使得 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":43,"id":"1e9f2f69-5162-4c1b-8614-8c54ce1cdf63","page_size":[595,841],"block_position":"43-15"},{"bbox":[185,482,408,509],"type":"interline_equation","angle":0,"index":17,"text":"$$\n\\int_ {a} ^ {b} f (x) \\mathrm{d} x = (b - a) f \\left(\\frac {a + b}{2}\\right) + \\frac {1}{2 4} (b - a) ^ {3} f ^ {\\prime \\prime} (\\xi)\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":43,"id":"8b36451f-a029-436a-add5-a3a37be16f2d","page_size":[595,841],"block_position":"43-16"},{"bbox":[52,513,279,543],"type":"text","angle":0,"index":18,"merge_prev":false,"text":"证明 (3) 设 $k = \\frac{\\int_{a}^{b} f(x) \\, \\mathrm{d}x - (b - a)f\\left(\\frac{a + b}{2}\\right)}{\\frac{1}{24}(b - a)^{3}}$ ，构造 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":43,"id":"78924b0d-ea9f-4394-bf0a-0ba26548c517","page_size":[595,841],"block_position":"43-17"},{"bbox":[177,545,416,571],"type":"interline_equation","angle":0,"index":19,"text":"$$\nF (x) = \\int_ {a} ^ {x} f (t) \\mathrm{d} t - (x - a) f \\left(\\frac {x + a}{2}\\right) + \\frac {1}{2 4} (x - a) ^ {3} k\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":43,"id":"a03b201a-8f74-43d5-83db-6e8c9fd08304","page_size":[595,841],"block_position":"43-18"},{"bbox":[54,574,340,587],"type":"text","angle":0,"index":20,"merge_prev":false,"text":"$F(a)=F(b)=0$ ，根据 Rolle 得到 $\\exists\\eta\\in(a,b)$ 使得 $F'(\\eta)=0$ ，即 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":43,"id":"dc55a6f6-977b-406a-8009-e1d169b3643c","page_size":[595,841],"block_position":"43-19"},{"bbox":[174,590,420,615],"type":"interline_equation","angle":0,"index":21,"text":"$$\nf (\\eta) - f \\left(\\frac {a + \\eta}{2}\\right) - f ^ {\\prime} \\left(\\frac {a + \\eta}{2}\\right) \\frac {\\eta - a}{2} - \\frac {k}{8} (\\eta - a) ^ {2} = 0\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":43,"id":"8a9e2f4d-87ba-4554-9967-fc815b16fc8c","page_size":[595,841],"block_position":"43-20"},{"bbox":[54,619,279,644],"type":"text","angle":0,"index":22,"merge_prev":false,"text":"而根据 $\\eta$ 在 $\\frac{a+\\eta}{2}$ 的 Taylor 展开， $\\exists\\xi\\in\\left(\\frac{a+\\eta}{2},\\eta\\right)$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":43,"id":"2703cb8b-c7c3-4aaf-b289-00ec3816f0e1","page_size":[595,841],"block_position":"43-21"},{"bbox":[167,648,426,675],"type":"interline_equation","angle":0,"index":23,"text":"$$\nf (\\eta) = f \\left(\\frac {a + \\eta}{2}\\right) + f ^ {\\prime} \\left(\\frac {a + \\eta}{2}\\right) \\frac {\\eta - a}{2} + \\frac {f ^ {\\prime \\prime} (\\xi)}{2} \\left(\\frac {\\eta - a}{2}\\right) ^ {2}\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":43,"id":"406d8075-55f7-4fce-b18f-e1fb6c93f042","page_size":[595,841],"block_position":"43-22"},{"bbox":[54,678,162,691],"type":"text","angle":0,"index":24,"merge_prev":false,"text":"因此对比得到 $k = f''(\\xi)$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":43,"id":"3d964a46-a47c-48a1-bd67-8548e54426d4","page_size":[595,841],"block_position":"43-23"},{"bbox":[55,709,151,723],"type":"title","angle":0,"index":25,"level":2,"text":"## 4.2.6 中值不等式","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":43,"id":"31bd61d2-bc77-4c70-8359-ff99489f614c","page_size":[595,841],"block_position":"43-24"},{"bbox":[35,731,539,756],"type":"text","angle":0,"index":26,"merge_prev":false,"text":"练习4.27基于Taylor展开(1) $f(x)$ 在 $[a,b]$ 二阶可导， $|f(x)| \\leq A, |f''(x)| \\leq B$ ，证明 $|f'(x)| \\leq \\frac{2A}{b - a} + \\frac{B}{2}(b - a)$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":43,"id":"19cca192-6b2e-4322-870d-618783175c2d","page_size":[595,841],"block_position":"43-25"},{"bbox":[72,754,349,768],"type":"text","angle":0,"index":27,"merge_prev":false,"text":"(3) 特例 1: 区间 $[0,2]$ , $|f(x)| \\leq 1$ , $|f''(x)| \\leq 2$ 证明 $|f'(x)| \\leq 3$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":43,"id":"e3d964cf-b729-4a92-9976-d7f94ac0558e","page_size":[595,841],"block_position":"43-26"},{"bbox":[291,803,303,812],"type":"page_number","angle":0,"index":999999,"text":"36 ","color":{"line":"rgba(164, 164, 164, 1)","fill":"rgba(164, 164, 164, 1)"},"page_idx":43,"is_discarded":true,"id":"857c4c60-6948-4271-bb65-c4b177b73ba0","page_size":[595,841],"block_position":"43-27"}],[{"bbox":[409,47,539,58],"type":"header","angle":0,"index":-99999,"text":"4.3 洛必达法则与 Taylor 公式 ","color":{"line":"rgba(164, 164, 164, 1)","fill":"rgba(164, 164, 164, 1)"},"page_idx":44,"is_discarded":true,"id":"ed1efb61-913c-4623-8a3c-c82e177242a4","page_size":[595,841],"block_position":"44-0"},{"bbox":[73,70,397,91],"type":"text","angle":0,"index":1,"merge_prev":false,"text":"(4) 特例 2：区间 $[0,1]$ ， $|f(x)| \\leq A, |f''(x)| \\leq B$ ，证明 $|f'(x)| \\leq 2A + \\frac{1}{2}B$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":44,"id":"293051b9-bc1f-4b35-8525-848c290f4d09","page_size":[595,841],"block_position":"44-1"},{"bbox":[73,93,408,112],"type":"text","angle":0,"index":2,"merge_prev":false,"text":"(5) 推论：将 $|f(x)| \\leq A$ 的条件改为 $f(a) = f(b)$ ，则得到 $|f'(x)| \\leq \\frac{B}{2}(b - a)$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":44,"id":"2ad5067d-898f-465d-888b-3bb24d38cae5","page_size":[595,841],"block_position":"44-2"},{"bbox":[54,112,508,146],"type":"text","angle":0,"index":3,"merge_prev":false,"text":"(6) 推论特例 1: $f(x)$ 在 $[0,1]$ 二阶可导, $f(0) = f(1) = f(\\frac{1}{2}) = 0$ , $|f''(x)| \\leq M$ , 证明 $|f'(x)| \\leq \\frac{M}{2}$ 证明 注意这里是已知点在 $x$ 处展开, 并使用 Lagrange 余项! 不要写成 $x$ 在 $a, b$ 展开, 或用 Peano 余项。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":44,"id":"e864979a-cbfe-4c78-b431-fdcc35889edf","page_size":[595,841],"block_position":"44-3"},{"bbox":[73,148,297,161],"type":"text","angle":0,"index":4,"merge_prev":false,"text":"(1) 根据 Taylor 展开得到 $\\exists \\xi \\in (a, x), \\eta \\in (x, b)$ 使得 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":44,"id":"1490258b-9aa2-4882-baad-aa97646d688b","page_size":[595,841],"block_position":"44-4"},{"bbox":[189,163,393,208],"type":"interline_equation","angle":0,"index":5,"text":"$$\n\\left\\{ \\begin{array}{l} f (a) = f (x) + f ^ {\\prime} (x) (a - x) + \\frac {f ^ {\\prime \\prime} (\\xi)}{2} (a - x) ^ {2} \\\\ f (b) = f (x) + f ^ {\\prime} (x) (b - x) + \\frac {f ^ {\\prime \\prime} (\\eta)}{2} (b - x) ^ {2} \\end{array} \\right.\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":44,"id":"0feff7ef-0e59-4a67-82fb-6e3c432d70c0","page_size":[595,841],"block_position":"44-5"},{"bbox":[54,211,455,235],"type":"text","angle":0,"index":6,"merge_prev":false,"text":"两式相减得到 $f(b)-f(a)=f'(x)(b-a)+\\frac{f''(\\eta)}{2}(b-x)^{2}-\\frac{f''(\\xi)}{2}(a-x)^{2}$ ，结合 A, B 得到 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":44,"id":"21c1442f-8308-4eef-93a6-e7b29622d8ad","page_size":[595,841],"block_position":"44-6"},{"bbox":[193,237,399,261],"type":"interline_equation","angle":0,"index":7,"text":"$$\n\\left| f ^ {\\prime} (x) \\right| \\leq \\frac {2 A}{b - a} + \\frac {B}{2 (b - a)} \\left[ (a - x) ^ {2} + (b - x) ^ {2} \\right]\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":44,"id":"f0607768-48c7-458d-96f8-7c2ab8f5ff67","page_size":[595,841],"block_position":"44-7"},{"bbox":[53,265,445,287],"type":"text","angle":0,"index":8,"merge_prev":false,"text":"而 $(a - x)^2 +(b - x)^2 < (b - a)^2$ （用二次函数最值)，因此得到 $|f'(x)| < \\frac{2A}{b - a} +\\frac{B}{2} (b - a)$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":44,"id":"6289f4c9-240a-42a7-9b69-8018f4e62c29","page_size":[595,841],"block_position":"44-8"},{"bbox":[72,286,540,312],"type":"text","angle":0,"index":9,"merge_prev":false,"text":"(5) 由于 $f(b) - f(a) = f'(x)(b - a) + \\frac{f''(\\eta)}{2} (b - x)^2 - \\frac{f''(\\xi)}{2} (a - x)^2 = 0$ ，因此 $f(x)$ 项被消除，后面同理。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":44,"id":"534b0c98-c33e-49e5-bb78-acec7ad46f4b","page_size":[595,841],"block_position":"44-9"},{"bbox":[36,323,539,357],"type":"text","angle":0,"index":10,"merge_prev":false,"text":"练习 4.28 几道经典配凑放缩 (1) $f(x)$ 在 $[a, b]$ 连续， $(a, b)$ 可导， $f(a) = f(b), |f'(x)| \\leq 1$ ，证明： $\\forall x_1, x_2$ 满足 $|f(x_1) - f(x_2)| \\leq \\frac{b - a}{2}$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":44,"id":"8eb21d7e-ede6-46fb-893e-0cf11acec42c","page_size":[595,841],"block_position":"44-10"},{"bbox":[52,357,393,370],"type":"text","angle":0,"index":11,"merge_prev":false,"text":"证明 (1) 首先根据 Lagrange 中值定理得到 $\\left|f(x_{1})-f(x_{2})\\right|\\leq x_{2}-x_{1}$ ，再根据 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":44,"id":"d4d78fc9-edef-4d3f-96d5-3e5c5e790aad","page_size":[595,841],"block_position":"44-11"},{"bbox":[143,375,449,390],"type":"interline_equation","angle":0,"index":12,"text":"$$\n\\left| f \\left(x _ {1}\\right) - f \\left(x _ {2}\\right) \\right| = \\left| f \\left(x _ {1}\\right) - f (a) + f (b) - f \\left(x _ {2}\\right) \\right| \\leq \\left(x _ {1} - a\\right) + \\left(b - x _ {2}\\right)\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":44,"id":"c1e87bbe-9a9f-4a52-9636-9aecc73eae8c","page_size":[595,841],"block_position":"44-12"},{"bbox":[54,392,223,415],"type":"text","angle":0,"index":13,"merge_prev":false,"text":"因此相加得到 $\\left|f(x_{1})-f(x_{2})\\right|\\leq\\frac{b-a}{2}$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":44,"id":"69c6d651-72fb-465c-88a7-1bcf6ab405a3","page_size":[595,841],"block_position":"44-13"},{"bbox":[55,430,246,444],"type":"title","angle":0,"index":14,"level":2,"text":"## 4.2.7 导数极限定理与导数介值定理","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":44,"id":"67f8ebf3-93a4-4fb9-973e-5ef942bd3305","page_size":[595,841],"block_position":"44-14"},{"bbox":[72,457,236,470],"type":"title","angle":0,"index":15,"level":2,"text":"## 定理 4.11 (导数介值定理：达布定理)","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":44,"id":"b09349f8-8849-4a8c-ab99-8b3559c5b2e9","page_size":[595,841],"block_position":"44-15"},{"bbox":[63,465,537,496],"type":"table_body","angle":0,"index":16,"img_path":"/","id":"63ecc95d-c2f0-4670-8600-e393cf9b4ea4","color":{"line":"rgba(103, 194, 63, 1)","fill":"rgba(103, 194, 63, 1)"},"table_body":"
定理4.11(等数升值定理:达布定理)
f(x)在[a,b]可导,f′+(a)≠f′-b),k介于f′+(a),f′-b之间,则∃ξ∈(a,b)使得f′(ξ)=k
","page_idx":44,"page_size":[595,841],"block_position":"44-16"},{"bbox":[53,505,539,534],"type":"text","angle":0,"index":17,"merge_prev":false,"text":"证明 构造 $F(x) = f(x) - kx$ ，则 $F'(x) = f'(x) - k$ ，由于 $k$ 介于 $f_{+}'(a), f_{-}'(b)$ 之间，因此 $F_{+}'(a)F_{-}'(b) < 0$ ，不妨设 $F_{+}'(a) > 0, F_{-}'(b) < 0$ ，则 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":44,"id":"b2997d37-6516-48df-805b-c0e906084e17","page_size":[595,841],"block_position":"44-17"},{"bbox":[193,537,401,562],"type":"interline_equation","angle":0,"index":18,"text":"$$\n\\lim _ {x \\to a ^ {+}} \\frac {F (x) - F (a)}{x - a} > 0, \\lim _ {x \\to a ^ {-}} \\frac {F (x) - F (b)}{x - b} < 0\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":44,"id":"cc7a104c-53ab-41e4-8cbf-9f155c2dbfd1","page_size":[595,841],"block_position":"44-18"},{"bbox":[52,565,552,616],"type":"text","angle":0,"index":19,"merge_prev":false,"text":"根据极限保号性得到 $\\exists x_{1},x_{2}\\in (a,b),x_{1} < x_{2}$ 使得 $\\frac{F(x_1) - F(a)}{x_1 - a} >0,\\frac{F(x_2) - F(b)}{x_2 - b} < 0$ ，即 $F(x_{1}) > F(a),F(x_{2})>$ $F(b)$ ，由于 $F(x)$ 连续，存在 $\\xi \\in [a,b]$ 使得 $F(\\xi)$ 最大值，根据费马定理得到 $F^{\\prime}(\\xi) = 0$ ，即 $f^{\\prime}(\\xi) = k$ 笔记导数介值定理在ZJU2020中出现过 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":44,"id":"1a1c3ebd-ef27-4505-8dc7-bdbfb8e7e68a","page_size":[595,841],"block_position":"44-19"},{"bbox":[291,802,302,812],"type":"page_number","angle":0,"index":1000019,"text":"37 ","color":{"line":"rgba(164, 164, 164, 1)","fill":"rgba(164, 164, 164, 1)"},"page_idx":44,"is_discarded":true,"id":"4c1bce05-4613-44bd-b617-897ae897ad08","page_size":[595,841],"block_position":"44-20"}],[{"bbox":[409,47,539,58],"type":"header","angle":0,"index":-99999,"text":"4.3 洛必达法则与 Taylor 公式 ","color":{"line":"rgba(164, 164, 164, 1)","fill":"rgba(164, 164, 164, 1)"},"page_idx":45,"is_discarded":true,"id":"db654561-be54-40fa-af4c-5ae3af94620d","page_size":[595,841],"block_position":"45-0"},{"bbox":[55,74,248,91],"type":"title","angle":0,"index":1,"level":2,"text":"## 4.3 洛必达法则与 Taylor 公式","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":45,"id":"8e047de8-7d91-4764-80eb-f3853682973a","page_size":[595,841],"block_position":"45-1"},{"bbox":[55,106,151,121],"type":"title","angle":0,"index":2,"level":2,"text":"## 4.3.1 洛必达法则","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":45,"id":"e888a86d-b187-4f54-ae26-3d6853c16113","page_size":[595,841],"block_position":"45-2"},{"bbox":[72,134,176,148],"type":"title","angle":0,"index":3,"level":2,"text":"## 定理 4.12 (洛必达法则)","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":45,"id":"947d63f2-aa6e-4472-8c95-ae20e479490e","page_size":[595,841],"block_position":"45-3"},{"bbox":[62,143,537,212],"type":"table_body","angle":0,"index":4,"img_path":"/","id":"5bacc2ec-6e30-4d40-bec1-0407e82a99d4","color":{"line":"rgba(103, 194, 63, 1)","fill":"rgba(103, 194, 63, 1)"},"table_body":"
f,g满足(1) \\lim_{x\\to x_0}f(x)=\\lim_{x\\to x_0}g(x)=0 或者 \\lim_{x\\to x_0}g(x)=\\infty(2)x_0 空心邻域中 f'(x),g'(x) 有定义(非常重要的条件!)则\\lim_{x\\to x_0}\\frac{f(x)}{g(x)}=\\lim_{x\\to x_0}\\frac{f'(x)}{g'(x)}
","page_idx":45,"page_size":[595,841],"block_position":"45-4"},{"bbox":[52,221,315,234],"type":"text","angle":0,"index":5,"merge_prev":false,"text":"笔记 $x_0$ 空心邻域中 $f(x), g(x)$ 有定义且可导条件非常重要！ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":45,"id":"09f745a8-7a38-4cf8-9816-8434fbf9cc07","page_size":[595,841],"block_position":"45-5"},{"bbox":[35,252,426,265],"type":"text","angle":0,"index":6,"merge_prev":false,"text":"练习 4.29 空心邻域中没定义则只能用定义 已知 $g(0)=g'(0)=0, g''(0)=3$ ，求 $f'(0)$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":45,"id":"9adba0ad-8402-4579-999f-b40dc0b49574","page_size":[595,841],"block_position":"45-6"},{"bbox":[243,269,349,307],"type":"interline_equation","angle":0,"index":7,"text":"$$\nf (x) = \\left\\{ \\begin{array}{l l} \\frac {g (x)}{x}, & x \\neq 0 \\\\ 0, & x = 0 \\end{array} \\right.\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":45,"id":"82a29ba1-0162-4698-b798-11b6eb01bc1d","page_size":[595,841],"block_position":"45-7"},{"bbox":[52,312,465,325],"type":"text","angle":0,"index":8,"merge_prev":false,"text":"证明 可以直接用 Peano 余项的 Taylor 展开。但这里展现其证明过程，先代入并用一次洛必达， ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":45,"id":"f455d540-499c-456a-918c-cb10e4967011","page_size":[595,841],"block_position":"45-8"},{"bbox":[205,327,387,352],"type":"interline_equation","angle":0,"index":9,"text":"$$\n\\lim _ {x \\rightarrow 0} \\frac {f (x) - f (0)}{x - 0} = \\lim _ {x \\rightarrow 0} \\frac {g (x)}{x ^ {2}} = \\lim _ {x \\rightarrow 0} \\frac {g ^ {\\prime} (x)}{2 x}\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":45,"id":"94207df5-f0e9-4621-8f5e-a0b94e5dbd50","page_size":[595,841],"block_position":"45-9"},{"bbox":[52,354,539,396],"type":"text","angle":0,"index":10,"merge_prev":false,"text":"注意后面不能继续用洛必达了，原因在于 $g'(x)$ 在 $x_0$ 空心邻域中不一定有定义，极限 $\\lim_{x\\to 0}g''(x)$ 没有意义。因此需要用导数定义： $\\lim_{x\\to 0}\\frac{g'(x) - g'(0)}{2x} = \\frac{1}{2}\\lim_{x\\to 0}g''(x) = \\frac{3}{2}$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":45,"id":"a721e79f-5945-4c7d-bb9f-93fbe6ae055e","page_size":[595,841],"block_position":"45-10"},{"bbox":[52,397,502,411],"type":"text","angle":0,"index":11,"merge_prev":false,"text":"笔记 该证明过程是 Peano 余项的 Taylor 公式的证明思路，极度重要，保证最后一次不能用洛必达法则。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":45,"id":"1f2b5920-4e02-41fa-808a-1eff77a24041","page_size":[595,841],"block_position":"45-11"},{"bbox":[56,428,151,443],"type":"title","angle":0,"index":12,"level":2,"text":"## 4.3.2 Taylor 公式","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":45,"id":"04ddeb49-1729-4f22-82b6-a0af33634e9a","page_size":[595,841],"block_position":"45-12"},{"bbox":[71,455,227,468],"type":"title","angle":0,"index":13,"level":2,"text":"## 定理 4.13 (Peano 余项 Taylor 公式)","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":45,"id":"ba61d1f7-3947-49e7-acd8-e63b5f973494","page_size":[595,841],"block_position":"45-13"},{"bbox":[67,473,410,486],"type":"text","angle":0,"index":14,"merge_prev":false,"text":"若 $f$ 在 $x_0$ 存在直至 $n$ 阶导数 (只需要单点导数，比Lagrange要求低很多)，则 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":45,"id":"ed95a75d-e6ee-4dc9-ad02-1dc783058b4b","page_size":[595,841],"block_position":"45-14"},{"bbox":[138,488,455,514],"type":"interline_equation","angle":0,"index":15,"text":"$$\nf (x) = f (x _ {0}) + f ^ {\\prime} (x _ {0}) (x - x _ {0}) + \\dots + \\frac {f ^ {(n)} (x _ {0})}{n !} (x - x _ {0}) ^ {n} + o ((x - x _ {0}) ^ {n})\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":45,"id":"2396fd23-17a4-4111-a9d5-e7da4aa382dd","page_size":[595,841],"block_position":"45-15"},{"bbox":[52,524,539,558],"type":"text","angle":0,"index":16,"merge_prev":false,"text":"证明 设 $T_{n}(x)=f(x_{0})+f'(x_{0})(x-x_{0})+\\cdots+\\frac{f^{(n)}(x_{0})}{n!}(x-x_{0})^{n}$ ， $R_{n}(x)=f(x)-T_{n}(x)$ ， $Q_{n}(x)=(x-x_{0})^{n}$ ，先做 n-1 次洛必达，再用一次导数定义可以得到： ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":45,"id":"6e41ff40-f187-4b07-b98a-d678a01559b9","page_size":[595,841],"block_position":"45-16"},{"bbox":[114,560,479,590],"type":"interline_equation","angle":0,"index":17,"text":"$$\n\\lim _ {x \\to x _ {0}} \\frac {R _ {n} (x)}{Q _ {n} (x)} = \\lim _ {x \\to x _ {0}} \\frac {R _ {n} ^ {(n - 1)} (x)}{Q _ {n} ^ {(n - 1)} (x)} = \\frac {1}{n !} \\lim _ {x \\to x _ {0}} \\left[ \\frac {f ^ {(n - 1)} (x) - f ^ {(n - 1)} (x _ {0})}{x - x _ {0}} - f ^ {(n)} (x _ {0}) \\right] = 0\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":45,"id":"e81e3d38-5961-4f9e-8f06-65c847b7625a","page_size":[595,841],"block_position":"45-17"},{"bbox":[73,604,242,617],"type":"title","angle":0,"index":18,"level":2,"text":"## 定理 4.14 (Lagrange 余项 Taylor 公式)","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":45,"id":"5a8afe06-3f42-4c50-9fe7-e30344b2b31f","page_size":[595,841],"block_position":"45-18"},{"bbox":[68,620,480,634],"type":"text","angle":0,"index":19,"merge_prev":false,"text":"若 $f$ 在 $[a,b]$ 存在直至 $n + 1$ 阶导数，且在 $(a,b)$ 满足 $n$ 阶连续可导(条件比Peano强得多)，则 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":45,"id":"e65acdae-76dd-4ac8-8b0c-5e1b25d12bb2","page_size":[595,841],"block_position":"45-19"},{"bbox":[117,636,475,663],"type":"interline_equation","angle":0,"index":20,"text":"$$\nf (x) = f \\left(x _ {0}\\right) + f ^ {\\prime} \\left(x _ {0}\\right) \\left(x - x _ {0}\\right) + \\dots + \\frac {f ^ {(n)} \\left(x _ {0}\\right)}{n !} \\left(x - x _ {0}\\right) ^ {n} + \\frac {f ^ {(n + 1)} (\\xi)}{(n + 1) !} \\left(x - x _ {0}\\right) ^ {n + 1}\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":45,"id":"e38eba4b-b97e-443b-b623-28e1026f6cab","page_size":[595,841],"block_position":"45-20"},{"bbox":[52,676,539,735],"type":"text","angle":0,"index":21,"merge_prev":false,"text":"证明 设 $T_{n}(x)=f(x_{0})+f'(x_{0})(x-x_{0})+\\cdots+\\frac{f^{(n)}(x_{0})}{n!}(x-x_{0})^{n}$ ，构造 $F(t)=f(t)-T_{n}(t), G(t)=(x-t)^{n+1}$ ，即证明 $\\frac{F(x_{0})}{G(x_{0})}=\\frac{f^{(n+1)}(\\xi)}{(n+1)!}$ ，因为 $F(x)=G(x)=0, F'(t)=-\\frac{f^{(n+1)}(t)}{n!}(x-t)^{n}, G'(t)=-(n+1)(x-t)^{n}$ ，因此根据 Cauchy 中值定理可知 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":45,"id":"4d7ec240-6242-4c9d-8a1f-e137b8afd928","page_size":[595,841],"block_position":"45-21"},{"bbox":[195,738,397,766],"type":"interline_equation","angle":0,"index":22,"text":"$$\n\\frac {F (x _ {0})}{G (x _ {0})} = \\frac {F (x _ {0}) - F (x)}{G (x _ {0}) - G (x)} = \\frac {F ^ {\\prime} (\\xi)}{G ^ {\\prime} (\\xi)} = \\frac {f ^ {(n + 1)} (\\xi)}{(n + 1) !}\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":45,"id":"a7fe74a5-bb59-49bf-881d-a4715882d269","page_size":[595,841],"block_position":"45-22"},{"bbox":[291,803,302,812],"type":"page_number","angle":0,"index":1000022,"text":"38 ","color":{"line":"rgba(164, 164, 164, 1)","fill":"rgba(164, 164, 164, 1)"},"page_idx":45,"is_discarded":true,"id":"471fabb0-3057-458c-aee8-0146223b9559","page_size":[595,841],"block_position":"45-23"}],[{"bbox":[489,47,539,58],"type":"header","angle":0,"index":-99999,"text":"4.4 凸函数 ","color":{"line":"rgba(164, 164, 164, 1)","fill":"rgba(164, 164, 164, 1)"},"page_idx":46,"is_discarded":true,"id":"02052452-cc56-4d31-a6a3-2ac0caf9fb7b","page_size":[595,841],"block_position":"46-0"},{"bbox":[55,74,129,90],"type":"title","angle":0,"index":1,"level":2,"text":"## 4.4 凸函数","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":46,"id":"98840aa0-6688-4495-b988-21824bfbf88e","page_size":[595,841],"block_position":"46-1"},{"bbox":[55,106,163,121],"type":"title","angle":0,"index":2,"level":2,"text":"## 4.4.1 函数的凹凸性","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":46,"id":"02af6cdf-efb1-46c6-9eb4-69ad33f577ad","page_size":[595,841],"block_position":"46-2"},{"bbox":[73,134,151,147],"type":"title","angle":0,"index":3,"level":2,"text":"## 定义 4.3 (凸函数)","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":46,"id":"bd19041b-135a-4dd1-a5db-c3b618f46ab1","page_size":[595,841],"block_position":"46-3"},{"bbox":[67,151,456,163],"type":"text","angle":0,"index":4,"merge_prev":false,"text":"$f$ 是区间 $I$ 上的函数，若对 $\\forall x_{1},x_{2}\\in I$ 和 $\\forall \\lambda \\in (0,1)$ 满足以下条件，则称 $f(x)$ 为凸函数 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":46,"id":"621271cc-0e70-4921-97af-d40095cbc189","page_size":[595,841],"block_position":"46-4"},{"bbox":[198,171,395,185],"type":"interline_equation","angle":0,"index":5,"text":"$$\nf (\\lambda x _ {1} + (1 - \\lambda) x _ {2}) \\leq \\lambda f (x _ {1}) + (1 - \\lambda) f (x _ {2})\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":46,"id":"e8877641-a7eb-4cac-b082-666e1652695b","page_size":[595,841],"block_position":"46-5"},{"bbox":[74,205,181,217],"type":"title","angle":0,"index":6,"level":2,"text":"## 引理 4.1 (定比分点公式)","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":46,"id":"427a2ebc-777d-48b0-93ec-968d5387239b","page_size":[595,841],"block_position":"46-6"},{"bbox":[68,222,337,234],"type":"text","angle":0,"index":7,"merge_prev":false,"text":"对于一点 $x \\in [x_1, x_2]$ ，则 $x$ 可由其与 $x_1, x_2$ 的距离唯一表出： ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":46,"id":"2b522e5b-35f2-44e0-a659-268500e23d1b","page_size":[595,841],"block_position":"46-7"},{"bbox":[233,238,359,260],"type":"interline_equation","angle":0,"index":8,"text":"$$\nx = \\frac {x _ {2} - x}{x _ {2} - x _ {1}} x _ {1} + \\frac {x - x _ {1}}{x _ {2} - x _ {1}} x _ {2}\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":46,"id":"44db57d7-bda4-4cea-a663-6af85259d3cf","page_size":[595,841],"block_position":"46-8"},{"bbox":[68,263,391,275],"type":"text","angle":0,"index":9,"merge_prev":false,"text":"记忆方式：首先保证分子都是正的，而且 $x_{1}$ 对应的权重是 x 到 $x_{2}$ 的距离 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":46,"id":"ccd1cb18-b9ce-4656-b4de-70c38b292406","page_size":[595,841],"block_position":"46-9"},{"bbox":[73,296,205,308],"type":"title","angle":0,"index":10,"level":2,"text":"## 定理 4.15 (凸函数的等价条件)","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":46,"id":"d05ad776-3682-49c4-a27b-1ff2f33ec6ab","page_size":[595,841],"block_position":"46-10"},{"bbox":[67,312,224,324],"type":"text","angle":0,"index":11,"merge_prev":false,"text":"$f$ 是 $I$ 上的凸函数与下面条件等价： ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":46,"id":"1d8b0f8a-b6c5-4824-aa19-c500aa369df4","page_size":[595,841],"block_position":"46-11"},{"bbox":[82,328,367,340],"type":"text","angle":0,"index":12,"merge_prev":false,"text":"- 一阶可导最常用： $\\forall x_0 \\in I$ ，有 $f(x) \\geq f(x_0) + f'(x_0)(x - x_0)$ 。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":46,"id":"5cfd5c28-1a7b-4433-8742-de320cc1cb80","page_size":[595,841],"block_position":"46-12"},{"bbox":[83,343,337,356],"type":"text","angle":0,"index":13,"merge_prev":false,"text":"- 若不可导，上述命题等价于： $f(x) \\geq f(x_0) + k(x - x_0)$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":46,"id":"e91a3b6c-fa2f-473a-b2d3-05a636d57e12","page_size":[595,841],"block_position":"46-13"},{"bbox":[83,359,248,371],"type":"text","angle":0,"index":14,"merge_prev":false,"text":"若 $f(x)$ 可导，则等价于 $f'(x)$ 单增 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":46,"id":"666ac87b-8f05-4269-b131-f7b7b8e1f27f","page_size":[595,841],"block_position":"46-14"},{"bbox":[83,375,317,387],"type":"text","angle":0,"index":15,"merge_prev":false,"text":"- 二阶可导：若 $f(x)$ 二阶可导，则等价于 $f''(x) \\geq 0$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":46,"id":"7496cf47-204f-44f8-aad7-397003c8e0c1","page_size":[595,841],"block_position":"46-15"},{"bbox":[35,416,539,449],"type":"text","angle":0,"index":16,"merge_prev":false,"text":"练习 4.30 不二阶可导凸性质的应用 (1) 重点：函数 $f(x)$ 满足 $f(0)=0$ ， $f'(x)$ 单增，证明： $F(x)=\\frac{f(x)}{x}$ 在 $(0,+\\infty)$ 单增 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":46,"id":"706e4c69-efc0-457d-8e20-58a5027f35d3","page_size":[595,841],"block_position":"46-16"},{"bbox":[52,450,539,464],"type":"text","angle":0,"index":17,"merge_prev":false,"text":"证明 (1) $f'(x)$ 单增等价于可导，而没说二阶可导，因此取 y > x > 0，只能用 $f(y) \\geq f(x) + f'(x)(y - x)$ 。此时 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":46,"id":"8e2cfdd0-f5ba-4687-a743-285505ab28be","page_size":[595,841],"block_position":"46-17"},{"bbox":[123,466,468,549],"type":"interline_equation","angle":0,"index":18,"text":"$$\n\\begin{array}{l} F (y) - F (x) = \\frac {x f (y) - y f (x)}{x y} \\geq \\frac {x f (x) + x f ^ {\\prime} (x) (y - x) - (y - x) f (x) - x f (x)}{x y} \\\\ \\quad = \\frac {(y - x) (x f ^ {\\prime} (x) - f (x))}{x y} = \\frac {(y - x) (f ^ {\\prime} (x) - \\frac {f (x) - f (0)}{x - 0})}{y} \\\\ \\quad = \\frac {(y - x) (f ^ {\\prime} (x) - f ^ {\\prime} (\\xi))}{y} \\geq 0 \\end{array}\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":46,"id":"b8c08ce3-06d4-4e27-928d-7098db56833c","page_size":[595,841],"block_position":"46-18"},{"bbox":[35,565,539,594],"type":"text","angle":0,"index":19,"merge_prev":false,"text":"练习 4.31 凸函数等价条件应用 (1) $f(x)$ 在 R 二阶连续可导，证明 $f''(x) \\geq 0$ 当且仅当 $\\forall x \\in \\mathbb{R}, \\forall h, f(x + h) + f(x - h) - 2f(x) \\geq 0$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":46,"id":"802f2f9c-86ad-4f7f-b262-2549e3becaff","page_size":[595,841],"block_position":"46-19"},{"bbox":[71,597,489,609],"type":"text","angle":0,"index":20,"merge_prev":false,"text":"(2) $f(x)$ 为 I 上凸函数当且仅当 $\\forall x_{1}, x_{2} \\in I$ 满足 $\\varphi(\\lambda) = f(\\lambda x_{1} + (1 - \\lambda)x_{2})$ 为 [0,1] 上的凸函数 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":46,"id":"b1e7e911-5e10-4a3a-a26b-692922a754cf","page_size":[595,841],"block_position":"46-20"},{"bbox":[52,609,461,627],"type":"text","angle":0,"index":21,"merge_prev":false,"text":"证明 (1) $f''(x) \\geq 0$ 等价于凸函数，等价于 $\\frac{f(x+h)-f(x)}{h} \\geq \\frac{f(x)-f(x-h)}{h}$ ，即得到结论。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":46,"id":"da92f682-ba48-46c6-a3eb-6558c11bf09b","page_size":[595,841],"block_position":"46-21"},{"bbox":[53,629,537,659],"type":"text","angle":0,"index":22,"merge_prev":false,"text":"(2.1) 右推左：要证 $f(\\lambda x_1 + (1 - \\lambda)x_2) \\leq \\lambda f(x_1) + (1 - \\lambda)f(x_2)$ ，等价于 $\\varphi(\\lambda) \\leq \\lambda \\varphi(1) + (1 - \\lambda)\\varphi(0)$ ，而 $\\varphi(\\lambda) = \\varphi(\\lambda \\cdot 1 + (1 - \\lambda) \\cdot 0)$ ，根据 $\\varphi$ 的凸性可直接得出。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":46,"id":"f85f27f3-d947-4634-8f64-72eead912ede","page_size":[595,841],"block_position":"46-22"},{"bbox":[72,661,389,674],"type":"text","angle":0,"index":23,"merge_prev":false,"text":"(2.2) 左推右：要证 $\\varphi(\\alpha t_1 + (1 - \\alpha)t_2) \\leq \\alpha \\varphi(t_1) + (1 - \\alpha)\\varphi(t_2)$ ，等价于 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":46,"id":"36d14ab3-3320-477d-be11-fe1c1f182251","page_size":[595,841],"block_position":"46-23"},{"bbox":[68,681,523,693],"type":"interline_equation","angle":0,"index":24,"text":"$$\nf \\left(\\left(\\alpha t _ {1} + (1 - \\alpha) t _ {2}\\right) x _ {1} + \\left(1 - \\alpha t _ {1} - (1 - \\alpha) t _ {2}\\right) x _ {2}\\right) \\leq \\alpha f \\left(t _ {1} x _ {1} + \\left(1 - t _ {1}\\right) x _ {2}\\right) + (1 - \\alpha) f \\left(t _ {2} x _ {1} + \\left(1 - t _ {2}\\right) x _ {2}\\right)\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":46,"id":"96c66238-9341-4caa-a73a-5d77369c09ff","page_size":[595,841],"block_position":"46-24"},{"bbox":[54,700,295,712],"type":"text","angle":0,"index":26,"merge_prev":false,"text":"根据 $f(x)$ 的凸性，右侧大于等于左侧 (可能有点麻烦)。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":46,"id":"a0c11b2c-1c62-429c-800d-29062448e429","page_size":[595,841],"block_position":"46-25"},{"bbox":[291,802,302,811],"type":"page_number","angle":0,"index":1000024,"text":"39 ","color":{"line":"rgba(164, 164, 164, 1)","fill":"rgba(164, 164, 164, 1)"},"page_idx":46,"is_discarded":true,"id":"c82f67d2-92d2-42ae-8e02-c7adf37314de","page_size":[595,841],"block_position":"46-26"}],[{"bbox":[489,47,539,58],"type":"header","angle":0,"index":-99999,"text":"4.4 凸函数 ","color":{"line":"rgba(164, 164, 164, 1)","fill":"rgba(164, 164, 164, 1)"},"page_idx":47,"is_discarded":true,"id":"f94a9240-afef-4b1c-9d78-4360cd537a3e","page_size":[595,841],"block_position":"47-0"},{"bbox":[73,73,185,86],"type":"title","angle":0,"index":1,"level":2,"text":"## 定理 4.16 (凸性质的保持)","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":47,"id":"ee6c12e5-b5ec-4862-966b-b3d2175568e5","page_size":[595,841],"block_position":"47-1"},{"bbox":[69,89,236,101],"type":"text","angle":0,"index":2,"merge_prev":false,"text":"考虑凸函数的四则运算、复合、最值： ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":47,"id":"88a5b725-aed4-4207-880c-da5f95c67e08","page_size":[595,841],"block_position":"47-2"},{"bbox":[83,105,326,117],"type":"text","angle":0,"index":3,"merge_prev":false,"text":"- 凸函数的数乘和加法保持凸性质，差、积、商不一定 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":47,"id":"cbfdf138-9832-485c-86eb-bbf1c8888b5a","page_size":[595,841],"block_position":"47-3"},{"bbox":[83,121,478,133],"type":"text","angle":0,"index":4,"merge_prev":false,"text":"- 复合：若 $f(x)$ 为凸函数， $g(x)$ 为凸增函数，则 $g(f(x))$ 为凸函数，去掉增性质则不一定 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":47,"id":"61c92a7c-4582-420b-848e-33741dcb4b17","page_size":[595,841],"block_position":"47-4"},{"bbox":[83,137,390,149],"type":"text","angle":0,"index":5,"merge_prev":false,"text":"- 最值：若 $f(x), g(x)$ 均为 $I$ 上凸函数，则 $\\max \\{f(x), g(x)\\}$ 为凸函数 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":47,"id":"5af5ad3a-2742-47fd-b375-24781b7a04b7","page_size":[595,841],"block_position":"47-5"},{"bbox":[53,166,538,195],"type":"text","angle":0,"index":6,"merge_prev":false,"text":"证明 (3) 最值： $F(\\lambda x_{1} + (1 - \\lambda)x_{2}) = \\max \\{f(\\lambda x_{1} + (1 - \\lambda)x_{2}), g(\\lambda x_{1} + (1 - \\lambda)x_{2})\\} \\leq \\max \\{\\lambda f(x_{1}) + (1 - \\lambda)f(x_{2}), \\lambda g(x_{1}) + (1 - \\lambda)g(x_{2})\\}$ ，进行放缩 $F \\geq f, g$ 得到 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":47,"id":"ee6ce762-721e-4219-9961-dbc0fbe038c6","page_size":[595,841],"block_position":"47-6"},{"bbox":[195,201,398,215],"type":"interline_equation","angle":0,"index":7,"text":"$$\nF (\\lambda x _ {1} + (1 - \\lambda) x _ {2}) \\leq \\lambda F (x _ {1}) + (1 - \\lambda) F (x _ {2})\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":47,"id":"63b4e26e-7096-4291-b255-99771e56b5da","page_size":[595,841],"block_position":"47-7"},{"bbox":[74,240,141,253],"type":"title","angle":0,"index":8,"level":2,"text":"## 定义 4.4 (拐点)","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":47,"id":"c8db653d-aa10-4a8a-9e42-621dc9c90a99","page_size":[595,841],"block_position":"47-8"},{"bbox":[67,257,524,285],"type":"text","angle":0,"index":9,"merge_prev":false,"text":"$y = f(x)$ 在 $x_0$ 两侧分别为严格凸和严格凹的，则称 $x_0$ 为 $f(x)$ 的拐点。若 $f$ 在 $x_0$ 二阶可导，则 $x_0$ 是拐点的必要条件为 $f''(x_0) = 0$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":47,"id":"7c17969d-fd7e-42a2-80f3-e563ab979a57","page_size":[595,841],"block_position":"47-9"},{"bbox":[57,318,186,333],"type":"title","angle":0,"index":10,"level":2,"text":"## 4.4.2 凸函数的基本性质","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":47,"id":"23e98f02-4def-4a93-91b1-ce914daf8fda","page_size":[595,841],"block_position":"47-10"},{"bbox":[73,346,196,358],"type":"title","angle":0,"index":11,"level":2,"text":"## 定理 4.17 (凸函数基本性质)","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":47,"id":"b4a6ed8f-52a9-4cf7-a802-868fa7532e12","page_size":[595,841],"block_position":"47-11"},{"bbox":[69,363,216,375],"type":"text","angle":0,"index":12,"merge_prev":false,"text":"设 $f(x)$ 在区间 $I$ 上为凸函数，则 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":47,"id":"c4484788-a058-4964-8f68-6473dd2cfc3a","page_size":[595,841],"block_position":"47-12"},{"bbox":[83,378,248,391],"type":"text","angle":0,"index":13,"merge_prev":false,"text":"- $f(x)$ 在 $I$ 中任意一点有左右导函数 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":47,"id":"6c477ccd-7b5b-4bdb-9200-80cf9a3b271d","page_size":[595,841],"block_position":"47-13"},{"bbox":[83,393,398,407],"type":"text","angle":0,"index":14,"merge_prev":false,"text":"- $f(x)$ 在 $I$ 区间内部连续，端点未知，若是开区间，则在整个区间连续 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":47,"id":"ff261482-2316-4833-8c86-c34fb2e725c3","page_size":[595,841],"block_position":"47-14"},{"bbox":[54,418,538,451],"type":"text","angle":0,"index":15,"merge_prev":false,"text":"证明 (1) 只证右导数: $F(h) = \\frac{f(x_0 + h) - f(x_0)}{h}$ 为增函数, 根据有界的单调函数有左右极限可知左右导数存在。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":47,"id":"cae597a5-b73a-4cd1-9f93-580b46a081d8","page_size":[595,841],"block_position":"47-15"},{"bbox":[54,454,538,483],"type":"text","angle":0,"index":16,"merge_prev":false,"text":"(2) 根据左右导数的存在性可知左右连续 (否则导数定义中分母趋于 0，分子不趋于 0，显然无极限)，因此在内部点连续，端点单边连续。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":47,"id":"9e363a0e-3145-4e35-b795-748ae4a25ac9","page_size":[595,841],"block_position":"47-16"},{"bbox":[73,505,186,518],"type":"title","angle":0,"index":17,"level":2,"text":"## 定理 4.18 (凸函数的最值)","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":47,"id":"ffa6ae73-b4d6-4386-a0b7-b9d6624fb266","page_size":[595,841],"block_position":"47-17"},{"bbox":[69,521,459,535],"type":"text","angle":0,"index":18,"merge_prev":false,"text":"设 $f(x)$ 是 $[a, b]$ 上的凸函数，则 $f(x) \\leq \\max \\{f(a), f(b)\\}$ ，即凸函数的最值只在端点取到。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":47,"id":"62aa849a-d36f-4945-bad7-12881085b483","page_size":[595,841],"block_position":"47-18"},{"bbox":[54,552,343,565],"type":"text","angle":0,"index":19,"merge_prev":false,"text":"证明 $\\forall x\\in [a,b]$ ，设 $x = \\lambda a + (1 - \\lambda)b$ ，根据 $f(x)$ 为凸函数可知： ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":47,"id":"17e1fdd5-f78d-4318-abcf-95241bbfb6ea","page_size":[595,841],"block_position":"47-19"},{"bbox":[193,571,398,585],"type":"interline_equation","angle":0,"index":20,"text":"$$\nf (x) \\leq \\lambda f (a) + (1 - \\lambda) f (b) \\leq \\max \\{f (a), f (b) \\}\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":47,"id":"d7ff8b02-3d84-4b23-b030-9931add58b0a","page_size":[595,841],"block_position":"47-20"},{"bbox":[73,610,187,624],"type":"title","angle":0,"index":21,"level":2,"text":"## 定理 4.19 (Jensen 不等式)","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":47,"id":"6110311e-ffd4-43dc-bba6-c8f52b97109b","page_size":[595,841],"block_position":"47-21"},{"bbox":[68,629,361,658],"type":"text","angle":0,"index":22,"merge_prev":false,"text":"$f(x)$ 是 $[a,b]$ 上的凸函数，则对 $\\forall x_{i}\\in [a,b],\\lambda_{i} > 0,\\sum_{i = 1}^{n}\\lambda_{i} = 1$ ，有 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":47,"id":"7b24dc5e-c099-4069-8799-96151c7e8417","page_size":[595,841],"block_position":"47-22"},{"bbox":[235,661,358,694],"type":"interline_equation","angle":0,"index":23,"text":"$$\nf \\left(\\sum_ {i = 1} ^ {n} \\lambda_ {i} x _ {i}\\right) \\leq \\sum_ {i = 1} ^ {n} \\lambda_ {i} f (x _ {i})\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":47,"id":"3111e4b5-ec3f-404d-8671-7e9082e9b7e9","page_size":[595,841],"block_position":"47-23"},{"bbox":[291,803,303,812],"type":"page_number","angle":0,"index":1000023,"text":"40 ","color":{"line":"rgba(164, 164, 164, 1)","fill":"rgba(164, 164, 164, 1)"},"page_idx":47,"is_discarded":true,"id":"68dcada5-f1cc-4dfb-91f0-a0274046ef59","page_size":[595,841],"block_position":"47-24"}],[{"bbox":[409,47,539,58],"type":"header","angle":0,"index":-99998,"text":"4.5 一元极值与导数零点问题 ","color":{"line":"rgba(164, 164, 164, 1)","fill":"rgba(164, 164, 164, 1)"},"page_idx":48,"is_discarded":true,"id":"812b6c54-8a1b-4649-9eb7-e094cc128fdf","page_size":[595,841],"block_position":"48-0"},{"bbox":[35,68,453,101],"type":"text","angle":0,"index":2,"merge_prev":false,"text":"练习 4.32Jensen 不等式经典题 正数 $p_{1}, \\cdots, p_{n}$ 满足 $\\sum_{i=1}^{n} p_{i} = 1$ ，对 $\\forall x_{1}, \\cdots, x_{n} \\in R$ ，证明： ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":48,"id":"267c5523-3e22-42de-b4d9-c12243599aef","page_size":[595,841],"block_position":"48-1"},{"bbox":[223,104,367,137],"type":"interline_equation","angle":0,"index":3,"text":"$$\n\\sum_ {i = 1} ^ {n} p _ {i} \\left(x _ {i} - \\ln p _ {i}\\right) \\leq \\ln \\left(\\sum_ {i = 1} ^ {n} e ^ {x _ {i}}\\right)\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":48,"id":"aaef12cc-0911-4b2c-b21e-ccbb60bb9a6c","page_size":[595,841],"block_position":"48-2"},{"bbox":[52,138,432,171],"type":"text","angle":0,"index":4,"merge_prev":false,"text":"证明 即证明 $\\sum_{i=1}^{n}p_{i}\\ln\\frac{e^{x_{i}}}{p_{i}}\\leq\\ln\\left(\\sum_{i=1}^{n}e^{x_{i}}\\right)$ ，视 $e^{x_{i}}=t_{i}$ ，根据 $\\ln t$ 的凹性即可得到结论。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":48,"id":"c418282c-11d6-46e2-a378-6eb5b6dcc435","page_size":[595,841],"block_position":"48-3"},{"bbox":[55,187,211,201],"type":"title","angle":0,"index":5,"level":2,"text":"## 4.4.3 利用凹凸性证明不等式","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":48,"id":"a27e4b03-efc4-48da-aab8-72dd7be16013","page_size":[595,841],"block_position":"48-4"},{"bbox":[35,211,537,237],"type":"text","angle":0,"index":6,"merge_prev":false,"text":"练习4.33对中点位置的估计(1)经典：设 $\\varphi (x)$ 是[0,1]上连续凸函数，证明 $\\varphi \\left(\\frac{1}{2}\\right)\\leq \\int_0^1\\varphi (x)\\mathrm{d}x\\leq \\frac{\\varphi(0) + \\varphi(1)}{2}$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":48,"id":"b21ffe96-06e0-4833-99b0-3f6aa0638365","page_size":[595,841],"block_position":"48-5"},{"bbox":[52,238,476,275],"type":"text","angle":0,"index":7,"merge_prev":false,"text":"(2) 太经典： $f(x)$ 是 $[a, b]$ 上连续凸函数，证明 $f\\left(\\frac{a + b}{2}\\right) \\leq \\frac{1}{b - a} \\int_{a}^{b} f(x) \\, \\mathrm{d}x \\leq \\frac{f(a) + f(b)}{2}$ 证明 (1) 由于 $x = x \\cdot 1 + (1 - x) \\cdot 0$ ，因此 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":48,"id":"2a838cf5-e2a1-4369-af92-8ae0ff684f1e","page_size":[595,841],"block_position":"48-6"},{"bbox":[176,278,415,306],"type":"interline_equation","angle":0,"index":8,"text":"$$\n\\int_ {0} ^ {1} \\varphi (x) \\mathrm{d} x \\leq \\int_ {0} ^ {1} [ (1 - x) \\varphi (0) + x \\varphi (1) ] = \\frac {\\varphi (0) + \\varphi (1)}{2}\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":48,"id":"7b53112d-9d6e-412a-bb47-0833aa73511c","page_size":[595,841],"block_position":"48-7"},{"bbox":[54,307,414,335],"type":"text","angle":0,"index":9,"merge_prev":false,"text":"另一方面做变量替换 $x = 1 - t$ ，则 $\\int_0^1\\varphi (x)\\mathrm{d}x = \\int_0^1\\varphi (1 - x)\\mathrm{d}x$ （区间再现），得到 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":48,"id":"61209424-d3c1-4fc4-b5de-cb0b57976397","page_size":[595,841],"block_position":"48-8"},{"bbox":[131,337,460,365],"type":"interline_equation","angle":0,"index":10,"text":"$$\n\\int_ {0} ^ {1} \\varphi (x) \\mathrm{d} x = \\int_ {0} ^ {1} \\frac {1}{2} \\varphi (x) + \\frac {1}{2} \\varphi (1 - x) \\mathrm{d} x \\geq \\int_ {0} ^ {1} \\varphi \\left(\\frac {x}{2} + \\frac {1 - x}{2}\\right) \\mathrm{d} x = \\varphi \\left(\\frac {1}{2}\\right)\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":48,"id":"6f2075c4-2e20-4b4f-bd22-80c0309061a0","page_size":[595,841],"block_position":"48-9"},{"bbox":[72,368,476,396],"type":"text","angle":0,"index":11,"merge_prev":false,"text":"(2) 作替换 $x = \\lambda a + (1 - \\lambda)b = b - (b - a)\\lambda$ ，则 $\\frac{1}{b - a}\\int_{a}^{b}f(x)\\mathrm{d}x = \\int_{0}^{1}\\varphi (\\lambda)\\mathrm{d}\\lambda$ ，转换为 (1) ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":48,"id":"20eac30a-94ae-4b2b-bc20-2247058e65ec","page_size":[595,841],"block_position":"48-10"},{"bbox":[55,413,244,430],"type":"title","angle":0,"index":12,"level":2,"text":"## 4.5 一元极值与导数零点问题","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":48,"id":"be42e190-0307-4409-aaab-7479b8a94997","page_size":[595,841],"block_position":"48-11"},{"bbox":[55,446,210,460],"type":"title","angle":0,"index":13,"level":2,"text":"## 4.5.1 一元极值的概念与判定","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":48,"id":"8c159e77-c6f2-4fda-b593-633cb932c786","page_size":[595,841],"block_position":"48-12"},{"bbox":[73,473,226,486],"type":"title","angle":0,"index":14,"level":2,"text":"## 定理 4.20 (一元极值三大充分条件)","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":48,"id":"956a6794-9395-4064-9240-5699e3f3767c","page_size":[595,841],"block_position":"48-13"},{"bbox":[67,491,336,503],"type":"text","angle":0,"index":15,"merge_prev":false,"text":"$f(x)$ 可导，且满足必要条件 $f'(x_{0})=0$ ，则有三大充分条件： ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":48,"id":"63b2d8dd-cb70-4204-bbd8-1b6a8ee9f00a","page_size":[595,841],"block_position":"48-14"},{"bbox":[80,506,524,533],"type":"text","angle":0,"index":16,"merge_prev":false,"text":"1. $f(x)$ 在 $x_0$ 连续， $U^o (x_0,\\delta)$ 可导，若左邻域 $f^{\\prime}(x)\\leq 0$ ，右邻域 $f^{\\prime}(x)\\geq 0$ ，则为极小值。反之为极大值 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":48,"id":"045ab99d-a68a-45a9-8a9f-dd0f9c5f1311","page_size":[595,841],"block_position":"48-15"},{"bbox":[79,537,478,550],"type":"text","angle":0,"index":17,"merge_prev":false,"text":"2. 若在 $U(x_{0},\\delta)$ 二阶可导 $f''(x_{0})\\neq0$ ，则 $f''(x_{0})<0$ 时取极大值， $f''(x_{0})>0$ 时取极小值 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":48,"id":"572f1782-0007-40d2-923c-14d2ca4bbb3b","page_size":[595,841],"block_position":"48-16"},{"bbox":[80,551,528,582],"type":"text","angle":0,"index":18,"merge_prev":false,"text":"3. 若在 $U(x_{0}, \\delta)$ 存在 n 阶导数，且 $f'(x_{0}), \\cdots, f^{(n-1)}(x_{0}) = 0, f^{(n)}(x_{0}) \\neq 0$ 。则 n 为偶时， $x_{0}$ 为极值点， $f^{(n)}(x_{0}) < 0$ 极大， $f^{(n)}(x_{0}) > 0$ 时极小，n 为奇时不取极值。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":48,"id":"f32f29c6-1e9d-40f7-a88a-9570a860e6a4","page_size":[595,841],"block_position":"48-17"},{"bbox":[52,592,557,646],"type":"text","angle":0,"index":19,"merge_prev":false,"text":"证明 只证明第三充分条件。用 Peano 余项的 Taylor 展开，根据前面导数为 0 得到 $f(x)-f(x_{0})=\\left[\\frac{f^{(n)}(x_{0})}{n!}+o(1)\\right](x-x_{0})^{n}$ 。若 n 为偶且导数大于 0，则显然为极小值，极大值同理。若 n 为奇数，则 $(x-x_{0})^{n}$ 符号不定，因此无法确定。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":48,"id":"d424863b-ec81-4a49-a5ff-a496e56a708d","page_size":[595,841],"block_position":"48-18"},{"bbox":[55,664,164,678],"type":"title","angle":0,"index":20,"level":2,"text":"## 4.5.2 导数零点问题","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":48,"id":"31efbfb5-1615-4639-946d-52fefd2f00b4","page_size":[595,841],"block_position":"48-19"},{"bbox":[54,687,539,717],"type":"text","angle":0,"index":21,"merge_prev":false,"text":"要善于取出区间中的极值点或最值点，利用 $f'(x_0) = 0$ 的性质加上中值定理或 Taylor 展开研究更高阶的导数。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":48,"id":"c06903fa-127a-4457-9cb1-08ca91dd263d","page_size":[595,841],"block_position":"48-20"},{"bbox":[35,735,539,764],"type":"text","angle":0,"index":22,"merge_prev":false,"text":"练习 4.34 几道经典导数零点问题 (1) 经典: $f(x)$ 在 $[a, b]$ 连续, 在 $(a, b)$ 二阶可导, $f(a) = f(b) = 0$ , $\\exists c \\in (a, b)$ 满足 $f(c) > 0$ , 证明 $\\exists \\xi \\in (a, b)$ 使得 $f''(\\xi) < 0$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":48,"id":"bf1d032c-e958-4d03-910e-14cd4b599c3b","page_size":[595,841],"block_position":"48-21"},{"bbox":[291,803,302,811],"type":"page_number","angle":0,"index":999999,"text":"41 ","color":{"line":"rgba(164, 164, 164, 1)","fill":"rgba(164, 164, 164, 1)"},"page_idx":48,"is_discarded":true,"id":"1fb11c31-a1c0-4ebb-a5f4-b62c668c9503","page_size":[595,841],"block_position":"48-22"}],[{"bbox":[409,47,539,58],"type":"header","angle":0,"index":-99999,"text":"4.5 一元极值与导数零点问题 ","color":{"line":"rgba(164, 164, 164, 1)","fill":"rgba(164, 164, 164, 1)"},"page_idx":49,"is_discarded":true,"id":"c983690d-9941-4407-bc21-6e784be754dc","page_size":[595,841],"block_position":"49-0"},{"bbox":[52,71,537,100],"type":"text","angle":0,"index":1,"merge_prev":false,"text":"证明 由于 $f(a) = f(b) = 0$ ， $f(c) > 0$ ，由于 $f(x)$ 连续，总能找到最大值 $f(x_0)$ ，由于是最值点， $f'(x_0) = 0$ ，考虑 $f(a)$ 在 $x_0$ 处 Lagrange 余项的 Taylor 展开： ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":49,"id":"8bcd7412-6ed0-49cc-ba48-6189ed34b878","page_size":[595,841],"block_position":"49-1"},{"bbox":[224,101,368,126],"type":"interline_equation","angle":0,"index":2,"text":"$$\nf (a) = f (x _ {0}) + \\frac {1}{2} f ^ {\\prime \\prime} (\\xi) (a - x _ {0}) ^ {2}\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":49,"id":"17b90a84-f6de-4d74-b44c-2f7793ad2668","page_size":[595,841],"block_position":"49-2"},{"bbox":[55,127,192,140],"type":"text","angle":0,"index":3,"merge_prev":false,"text":"由于 $f(a)=0$ ，得到 $f''(\\xi)<0$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":49,"id":"960e8fbd-e393-4778-a377-f40369a7995a","page_size":[595,841],"block_position":"49-3"},{"bbox":[291,803,302,811],"type":"page_number","angle":0,"index":1000003,"text":"42 ","color":{"line":"rgba(164, 164, 164, 1)","fill":"rgba(164, 164, 164, 1)"},"page_idx":49,"is_discarded":true,"id":"38f391d3-34e3-41e6-84dd-eb05e134e5ae","page_size":[595,841],"block_position":"49-4"}],[{"bbox":[236,74,368,94],"type":"title","angle":0,"index":0,"level":2,"text":"## 第5章 不定积分","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":50,"id":"dcf0d2ef-4f24-4b45-83cc-cf27ec9dfdbe","page_size":[595,841],"block_position":"50-0"},{"bbox":[55,122,201,139],"type":"title","angle":0,"index":1,"level":2,"text":"## 5.1 常用不定积分公式","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":50,"id":"6d0808b6-aedf-4c52-818b-7182445aa430","page_size":[595,841],"block_position":"50-1"},{"bbox":[73,152,121,164],"type":"text","angle":0,"index":2,"merge_prev":false,"text":"必背部分: ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":50,"id":"c5719d7a-605b-4a74-8e52-2e288a795b01","page_size":[595,841],"block_position":"50-2"},{"bbox":[71,185,119,199],"type":"text","angle":0,"index":3,"merge_prev":false,"text":"普通三角： ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":50,"id":"78ad7a5f-7f9c-4d9c-aeb0-81cdc17717fc","page_size":[595,841],"block_position":"50-3"},{"bbox":[132,180,239,205],"type":"interline_equation","angle":0,"index":4,"text":"$$\n\\int \\sin x \\mathrm{d} x = - \\cos x + c\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":50,"id":"817c8b46-39e6-4fca-ab23-cf6168202aa3","page_size":[595,841],"block_position":"50-4"},{"bbox":[71,209,115,221],"type":"text","angle":0,"index":5,"merge_prev":false,"text":"sec、csc: ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":50,"id":"cec080f8-216b-40ee-a587-2638e015102e","page_size":[595,841],"block_position":"50-5"},{"bbox":[138,204,238,226],"type":"interline_equation","angle":0,"index":6,"text":"$$\n\\int \\sec^ {2} x \\mathrm{d} x = \\tan x + c\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":50,"id":"da31e482-019f-4b6e-971a-b617fa4b93b3","page_size":[595,841],"block_position":"50-6"},{"bbox":[312,180,411,205],"type":"interline_equation","angle":0,"index":7,"text":"$$\n\\int \\cos x \\mathrm{d} x = \\sin x + c\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":50,"id":"5eb69b9e-e571-4797-9e54-213db157189c","page_size":[595,841],"block_position":"50-7"},{"bbox":[321,205,425,222],"type":"interline_equation","angle":0,"index":8,"text":"$$\n\\csc^ {2} x \\mathrm{d} x = - \\cot x + c\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":50,"id":"d729bc57-3e22-4cf9-9cc0-08fc0354efaa","page_size":[595,841],"block_position":"50-8"},{"bbox":[134,225,263,250],"type":"interline_equation","angle":0,"index":9,"text":"$$\n\\int_ {a} ^ {b} \\sec x \\mathrm{d} x = \\ln | \\sec x + \\tan x |\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":50,"id":"9066d951-a9d8-4cdc-b71f-06787ab1e121","page_size":[595,841],"block_position":"50-9"},{"bbox":[317,225,521,248],"type":"interline_equation","angle":0,"index":10,"text":"$$\n\\int \\sec^ {3} x \\mathrm{d} x = \\frac {1}{2} \\tan x \\sec x + \\frac {1}{2} \\ln | \\tan x + \\sec x |\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":50,"id":"1084e3c7-86f9-4ce2-99f2-f23536ce30d5","page_size":[595,841],"block_position":"50-10"},{"bbox":[71,252,119,265],"type":"text","angle":0,"index":11,"merge_prev":false,"text":"乘积三角： ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":50,"id":"5da6aedd-580b-4752-8a43-c77ba42af457","page_size":[595,841],"block_position":"50-11"},{"bbox":[141,250,253,267],"type":"interline_equation","angle":0,"index":12,"text":"$$\n\\sec x \\tan x \\mathrm{d} x = \\sec x + c\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":50,"id":"03873201-c3a5-482a-bab7-cb2098e1ff51","page_size":[595,841],"block_position":"50-12"},{"bbox":[322,251,442,267],"type":"interline_equation","angle":0,"index":13,"text":"$$\n\\csc x \\cot x \\mathrm{d} x = - \\csc x + c\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":50,"id":"ecfc9454-68f0-4f6e-8124-9c22cbf0601d","page_size":[595,841],"block_position":"50-13"},{"bbox":[71,274,99,287],"type":"text","angle":0,"index":14,"merge_prev":false,"text":"分式: ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":50,"id":"4fd513cc-7b57-4e0c-a5ad-711c1aeeafaf","page_size":[595,841],"block_position":"50-14"},{"bbox":[134,269,270,293],"type":"interline_equation","angle":0,"index":15,"text":"$$\n\\int {\\frac {1}{x ^ {2} + a ^ {2}}} \\mathrm{d} x = \\frac {1}{a} \\arctan {\\frac {x}{a}} + c\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":50,"id":"a229ba82-e0b8-4275-82cf-51b37d7d4437","page_size":[595,841],"block_position":"50-15"},{"bbox":[318,269,460,294],"type":"interline_equation","angle":0,"index":16,"text":"$$\n\\int_ {a} \\frac {1}{x ^ {2} - a ^ {2}} \\mathrm{d} x = \\frac {1}{2 a} \\ln | \\frac {x - a}{x + a} | + c\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":50,"id":"e5246a48-c16d-4d13-b01f-bdbfb8bb1d7c","page_size":[595,841],"block_position":"50-16"},{"bbox":[133,291,307,317],"type":"interline_equation","angle":0,"index":17,"text":"$$\n\\int_ {c} \\frac {1}{\\sqrt {x ^ {2} + a ^ {2}}} \\mathrm{d} x = \\ln | x + \\sqrt {x ^ {2} + a ^ {2}} | + c\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":50,"id":"e238770c-2f5f-4300-af51-e100983222bb","page_size":[595,841],"block_position":"50-17"},{"bbox":[313,292,488,317],"type":"interline_equation","angle":0,"index":18,"text":"$$\n\\int {\\frac {1}{\\sqrt {x ^ {2} - a ^ {2}}}} \\mathrm{d} x = \\ln | x + \\sqrt {x ^ {2} - a ^ {2}} | + c\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":50,"id":"49fda126-ef06-473d-9b43-530bc6ad94c4","page_size":[595,841],"block_position":"50-18"},{"bbox":[133,317,267,339],"type":"interline_equation","angle":0,"index":19,"text":"$$\n\\int_ {a} \\frac {1}{\\sqrt {a ^ {2} - x ^ {2}}} \\mathrm{d} x = \\arcsin {\\frac {x}{a}} + c\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":50,"id":"61655342-a62e-4ea6-9f42-50afe8f91a9e","page_size":[595,841],"block_position":"50-19"},{"bbox":[71,342,99,354],"type":"text","angle":0,"index":20,"merge_prev":false,"text":"对数: ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":50,"id":"7ab2e0bd-09f5-4cf8-b603-7d8e9cce8f99","page_size":[595,841],"block_position":"50-20"},{"bbox":[136,339,246,360],"type":"interline_equation","angle":0,"index":21,"text":"$$\n\\int \\ln x \\mathrm{d} x = x \\ln x - x + c\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":50,"id":"a9149230-63be-4850-b4e4-32eb7f5e25b4","page_size":[595,841],"block_position":"50-21"},{"bbox":[312,335,498,362],"type":"interline_equation","angle":0,"index":22,"text":"$$\n\\int \\ln (1 + x) \\mathrm{d} x = (1 + x) \\ln (1 + x) - x + c\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":50,"id":"8ef3a373-a8a5-4772-ae7b-c5ef8f77c5e6","page_size":[595,841],"block_position":"50-22"},{"bbox":[73,363,121,376],"type":"text","angle":0,"index":23,"merge_prev":false,"text":"会算部分: ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":50,"id":"138535d7-6f2b-4c36-8e27-c6c86b6dbf2d","page_size":[595,841],"block_position":"50-23"},{"bbox":[73,397,95,411],"type":"text","angle":0,"index":24,"merge_prev":false,"text":"三角： ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":50,"id":"3e1c7dc4-6ae3-43f0-ae52-a9f074376690","page_size":[595,841],"block_position":"50-24"},{"bbox":[127,392,254,416],"type":"interline_equation","angle":0,"index":25,"text":"$$\n\\int \\tan x \\mathrm{d} x = - \\ln | \\cos x | + c\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":50,"id":"22c3e241-4260-4991-8011-d24df319a250","page_size":[595,841],"block_position":"50-25"},{"bbox":[398,391,512,416],"type":"interline_equation","angle":0,"index":26,"text":"$$\n\\int \\cot x \\mathrm{d} x = \\ln | \\sin x | + c\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":50,"id":"4d154df3-4267-48aa-abd9-c7f1eb7a41de","page_size":[595,841],"block_position":"50-26"},{"bbox":[67,419,114,433],"type":"text","angle":0,"index":27,"merge_prev":false,"text":"平方三角： ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":50,"id":"2e5c366a-b282-4d79-8b01-329d51fd667d","page_size":[595,841],"block_position":"50-27"},{"bbox":[127,415,255,439],"type":"interline_equation","angle":0,"index":28,"text":"$$\n\\int_ {a} ^ {b} \\sin^ {2} x \\mathrm{d} x = \\frac {x}{2} - \\frac {\\sin 2 x}{4} + c\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":50,"id":"db5a2ee2-4929-4ef4-9df6-36e115c49cab","page_size":[595,841],"block_position":"50-28"},{"bbox":[398,415,526,439],"type":"interline_equation","angle":0,"index":29,"text":"$$\n\\int \\cos^ {2} x \\mathrm{d} x = \\frac {x}{2} + \\frac {\\sin 2 x}{4} + c\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":50,"id":"ea263a7b-eb3b-44f1-ba3e-2b6886d2d05a","page_size":[595,841],"block_position":"50-29"},{"bbox":[127,438,251,459],"type":"interline_equation","angle":0,"index":30,"text":"$$\n\\int \\tan^ {2} x \\mathrm{d} x = \\tan x - x + c\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":50,"id":"445762c7-b680-4501-8c2d-e43a87fbe5e4","page_size":[595,841],"block_position":"50-30"},{"bbox":[396,438,527,461],"type":"interline_equation","angle":0,"index":31,"text":"$$\n\\int \\cot^ {2} x \\mathrm{d} x = - \\cot x - x + c\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":50,"id":"70db9f1e-ec50-45b3-9e4c-5f2371f1703c","page_size":[595,841],"block_position":"50-31"},{"bbox":[67,464,114,477],"type":"text","angle":0,"index":32,"merge_prev":false,"text":"复合三角： ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":50,"id":"c11f3a03-5dce-4cce-8a3d-70914576fd56","page_size":[595,841],"block_position":"50-32"},{"bbox":[127,459,340,483],"type":"interline_equation","angle":0,"index":33,"text":"$$\n\\int_ {a} ^ {b} e ^ {a x} \\cos b x \\mathrm{d} x = \\frac {e ^ {a x}}{a ^ {2} + b ^ {2}} (a \\cos b x + b \\sin b x) + c\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":50,"id":"46622df9-bd10-47f4-a4c4-8a2feae96e1b","page_size":[595,841],"block_position":"50-33"},{"bbox":[128,481,339,505],"type":"interline_equation","angle":0,"index":34,"text":"$$\n\\int e ^ {a x} \\sin b x \\mathrm{d} x = \\frac {e ^ {a x}}{a ^ {2} + b ^ {2}} (a \\sin b x - b \\cos b x) + c\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":50,"id":"5a53f4c3-476e-4e31-bfd4-e279021a2d70","page_size":[595,841],"block_position":"50-34"},{"bbox":[127,503,308,528],"type":"interline_equation","angle":0,"index":35,"text":"$$\n\\int_ {a} ^ {b} x \\cos n x \\mathrm{d} x = \\frac {1}{n ^ {2}} \\cos n x + \\frac {x}{n} \\sin n x + c\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":50,"id":"b4b862e6-b5ad-476a-9179-8f6667c75e51","page_size":[595,841],"block_position":"50-35"},{"bbox":[127,528,307,549],"type":"interline_equation","angle":0,"index":36,"text":"$$\n\\int x \\sin n x \\mathrm{d} x = \\frac {1}{n ^ {2}} \\sin n x - \\frac {x}{n} \\cos n x + c\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":50,"id":"a36cc226-1614-4813-a9b0-d135c8ca5347","page_size":[595,841],"block_position":"50-36"},{"bbox":[67,552,95,565],"type":"text","angle":0,"index":37,"merge_prev":false,"text":"根式: ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":50,"id":"7df57237-e3ed-4bf6-8a20-7b93965eed79","page_size":[595,841],"block_position":"50-37"},{"bbox":[127,548,392,571],"type":"interline_equation","angle":0,"index":38,"text":"$$\n\\int_ {0} ^ {1} \\sqrt {x ^ {2} \\pm a ^ {2}} \\mathrm{d} x = \\frac {1}{2} \\left[ x \\sqrt {x ^ {2} \\pm a ^ {2}} \\pm a ^ {2} \\ln | x + \\sqrt {x ^ {2} \\pm a ^ {2}} | \\right] + c\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":50,"id":"b4dbe91c-8c48-4830-9e09-8f195a7e13db","page_size":[595,841],"block_position":"50-38"},{"bbox":[127,570,352,594],"type":"interline_equation","angle":0,"index":39,"text":"$$\n\\int {\\sqrt {a ^ {2} - x ^ {2}}} \\mathrm{d} x = \\frac {1}{2} \\left[ x \\sqrt {a ^ {2} - x ^ {2}} + a ^ {2} \\arcsin {\\frac {x}{a}} \\right] + c\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":50,"id":"e5ff4c31-2299-44b6-b782-8f9710839b98","page_size":[595,841],"block_position":"50-39"},{"bbox":[56,613,201,629],"type":"title","angle":0,"index":40,"level":2,"text":"## 5.2 不定积分基本方法","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":50,"id":"2928a686-cfed-44f2-83c9-fa82b084bc01","page_size":[595,841],"block_position":"50-40"},{"bbox":[73,646,181,660],"type":"title","angle":0,"index":41,"level":2,"text":"## 定理 5.1 (分部积分公式)","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":50,"id":"b21eb621-9847-4498-967e-9821f1b33b5a","page_size":[595,841],"block_position":"50-41"},{"bbox":[69,664,167,676],"type":"text","angle":0,"index":42,"merge_prev":false,"text":"函数 $u, v$ 的积分满足： ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":50,"id":"7002ffde-f09f-444f-b5e7-bec86031b872","page_size":[595,841],"block_position":"50-42"},{"bbox":[248,672,345,698],"type":"interline_equation","angle":0,"index":43,"text":"$$\n\\int u \\mathrm{d} v = u v - \\int v \\mathrm{d} u\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":50,"id":"0750b295-f5a1-4138-a0f1-cc598dfce5a1","page_size":[595,841],"block_position":"50-43"}],[{"bbox":[439,47,539,58],"type":"header","angle":0,"index":-99999,"text":"5.2 不定积分基本方法 ","color":{"line":"rgba(164, 164, 164, 1)","fill":"rgba(164, 164, 164, 1)"},"page_idx":51,"is_discarded":true,"id":"b800672c-0c06-47e0-ac42-559453a812ea","page_size":[595,841],"block_position":"51-0"},{"bbox":[73,73,151,86],"type":"title","angle":0,"index":1,"level":2,"text":"## 定理 5.2 (表格法)","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":51,"id":"28179de8-c9af-4d5e-927d-a8ab579f118f","page_size":[595,841],"block_position":"51-1"},{"bbox":[69,91,525,126],"type":"text","angle":0,"index":2,"merge_prev":false,"text":"函数 u, v 的积分 $\\int uv^{(n+1)}\\mathrm{d}x = uv^{(n)} - u'v^{(n-1)} + u''v^{(n-2)} - \\cdots + (-1)^{n}u^{(n)}v + (-1)^{n+1} \\int u^{(n+1)}v\\mathrm{d}x$ ，具体操作时可列以下表格： ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":51,"id":"84ce1623-76d9-4024-b114-df69b1746467","page_size":[595,841],"block_position":"51-2"},{"bbox":[138,132,455,165],"type":"table_body","angle":0,"index":3,"img_path":"/","id":"36af19bb-152c-4527-8da1-3606f8c24750","color":{"line":"rgba(103, 194, 63, 1)","fill":"rgba(103, 194, 63, 1)"},"table_body":"
u的各阶导数uu'u''u'''...u^{(n+1)}
v^{(n+1)}的各阶原函数v^{(n+1)}v^{(n)}v^{(n-1)}v^{(n-2)}...v
","page_idx":51,"page_size":[595,841],"block_position":"51-3"},{"bbox":[68,172,525,207],"type":"text","angle":0,"index":4,"merge_prev":false,"text":"计算方法：以 $u$ 为起点，左上右下相乘，符号先正后负交替，最后一项为最后一列两个的积分 $(-1)^{n + 1}\\int u^{(n + 1)}v\\mathrm{d}x$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":51,"id":"293a5205-a843-4447-8686-3019ff74bd5a","page_size":[595,841],"block_position":"51-4"},{"bbox":[33,217,422,243],"type":"text","angle":0,"index":5,"merge_prev":false,"text":"笔记 $\\int P_{n}(x)e^{kx}\\mathrm{d}x,\\int P_{n}(x)\\sin ax\\mathrm{d}x,\\int P_{n}(x)\\cos ax\\mathrm{d}x$ 都可以使用表格法快速算出 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":51,"id":"e8602fd9-dc52-47e3-94ab-ece3171245fa","page_size":[595,841],"block_position":"51-5"},{"bbox":[35,250,264,274],"type":"text","angle":0,"index":6,"merge_prev":false,"text":"练习 $5.1P_{n}(x)e^{kx}$ 型 计算 $\\int(x^{3}+2x+6)e^{2x}\\mathrm{d}x$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":51,"id":"7ebc4040-168c-4690-b7c6-6a2f05e1947e","page_size":[595,841],"block_position":"51-6"},{"bbox":[54,273,114,285],"type":"text","angle":0,"index":7,"merge_prev":false,"text":"解 先列表格: ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":51,"id":"64bbf304-6bdc-4d23-ba4d-a5ca174ffec8","page_size":[595,841],"block_position":"51-7"},{"bbox":[189,292,404,330],"type":"table_body","angle":0,"index":8,"img_path":"/","id":"d8ac2c34-455e-48ef-b91e-8cbbc885ebe6","color":{"line":"rgba(103, 194, 63, 1)","fill":"rgba(103, 194, 63, 1)"},"table_body":"
x^{3} + 2x + 63x^{2} + 26x60
e^{2x}\\frac{1}{2}e^{2x}\\frac{1}{4}e^{2x}\\frac{1}{8}e^{2x}\\frac{1}{16}e^{2x}
","page_idx":51,"page_size":[595,841],"block_position":"51-8"},{"bbox":[54,336,446,361],"type":"text","angle":0,"index":9,"merge_prev":false,"text":"得到结果为 $(x^{3} + 2x + 6)\\frac{1}{2} e^{2x} - (3x^{2} + 2)\\frac{1}{4} e^{2x} + 6x\\left(\\frac{1}{8} e^{2x}\\right) - 6\\left(\\frac{1}{16} e^{2x}\\right) + \\int 0\\cdot \\left(\\frac{1}{16} e^{2x}\\right)\\mathrm{d}x$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":51,"id":"809b48fc-6816-4202-ba59-81b499185f78","page_size":[595,841],"block_position":"51-9"},{"bbox":[74,370,180,383],"type":"title","angle":0,"index":10,"level":2,"text":"## 定理 5.3 (积分有理分解)","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":51,"id":"c1c04784-6c6e-445d-be95-2f71cbdd76f4","page_size":[595,841],"block_position":"51-10"},{"bbox":[68,388,524,426],"type":"text","angle":0,"index":11,"merge_prev":false,"text":"若被积函数为 $R(x)=\\frac{P(x)}{Q(x)}$ ，且 $Q(x)$ 的次数大于等于 3，若 $Q(x)=(x-a_{1})^{\\lambda_{1}}\\cdots(x-a_{k})^{\\lambda_{k}}(x^{2}+p_{1}x+q_{1})^{\\mu_{1}}\\cdots(x^{2}+p_{l}x+q_{l})^{\\mu_{l}}$ ，对于 $(x-a_{1})^{\\lambda_{1}}$ 和 $(x^{2}+p_{1}x+q_{1})^{\\mu_{1}}$ 项，对应展开式为： ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":51,"id":"1ad3552e-4642-4a5e-bd36-66053a9f7968","page_size":[595,841],"block_position":"51-11"},{"bbox":[198,428,383,476],"type":"interline_equation","angle":0,"index":12,"text":"$$\n\\left\\{ \\begin{array}{l} \\frac {A _ {1}}{x - a _ {1}} + \\frac {A _ {2}}{(x - a _ {1}) ^ {2}} + \\dots + \\frac {A _ {\\lambda_ {1}}}{(x - a) ^ {\\lambda_ {1}}} \\\\ \\frac {B _ {1} x + C _ {1}}{x ^ {2} + p _ {1} x + q _ {1}} + \\dots + \\frac {B _ {\\mu_ {1}} x + C _ {\\mu_ {1}}}{(x ^ {2} + p _ {1} x + q _ {1}) ^ {\\mu_ {1}}} \\end{array} \\right.\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":51,"id":"57cdf972-a56d-4c15-9f85-b14c9a729354","page_size":[595,841],"block_position":"51-12"},{"bbox":[68,477,525,539],"type":"text","angle":0,"index":13,"merge_prev":false,"text":"其中 $\\frac{A_{\\lambda_l}}{(x - a)^{\\lambda_l}}$ 的积分好算，而 $\\frac{B_{\\mu_l}x + C_{\\mu_l}}{(x^2 + p_1x + q_1)^{\\mu_l}}$ 可配方化为 $\\int \\frac{Lu + N}{(u^2 + r^2)^j}\\mathrm{d}u$ (因为判别式一定小于0)，第一部分 $L\\int \\frac{u}{(u^2 + r^2)^j}\\mathrm{d}u$ 好算，第二部分 $N\\int \\frac{1}{(u^2 + r^2)^j}\\mathrm{d}u$ 用 $u = r\\tan v$ 替换，变为形如 $\\int \\cos^k vv\\mathrm{d}v$ 的不定积分，有递推公式辅助求解。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":51,"id":"3caa1be8-3ffc-4882-a462-3fa9f8ae0415","page_size":[595,841],"block_position":"51-13"},{"bbox":[73,575,230,588],"type":"title","angle":0,"index":14,"level":2,"text":"## 定理 5.4 (三角有理式转化为有理式)","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":51,"id":"c2a5b02c-871b-490b-85dc-14be819a4490","page_size":[595,841],"block_position":"51-14"},{"bbox":[68,592,524,624],"type":"text","angle":0,"index":15,"merge_prev":false,"text":"若 $R(\\sin x, \\cos x)$ 为关于 $\\sin x, \\cos x$ 的有理式，记 $t = \\tan \\frac{x}{2}$ ，则 $x = 2\\arctan t, \\mathrm{d}x = \\frac{2}{1 + t^2}\\mathrm{d}t$ ，以及万能公式 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":51,"id":"01ac2f87-9274-43c5-b64d-631ad4e8e430","page_size":[595,841],"block_position":"51-15"},{"bbox":[196,622,394,649],"type":"interline_equation","angle":0,"index":17,"text":"$$\n\\sin x = \\frac {2 t}{1 + t ^ {2}}, \\cos x = \\frac {1 - t ^ {2}}{1 + t ^ {2}}, \\tan x = \\frac {2 t}{1 - t ^ {2}}\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":51,"id":"e7b6fef3-3e69-4b8a-9f1e-440ba46d2710","page_size":[595,841],"block_position":"51-16"},{"bbox":[69,653,92,665],"type":"text","angle":0,"index":18,"merge_prev":false,"text":"因此 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":51,"id":"b426374b-e391-4376-8cc7-27a4ed0a8559","page_size":[595,841],"block_position":"51-17"},{"bbox":[179,662,414,689],"type":"interline_equation","angle":0,"index":19,"text":"$$\n\\int R (\\sin x, \\cos x) \\mathrm{d} x = \\int R \\left(\\frac {2 t}{1 + t ^ {2}}, \\frac {1 - t ^ {2}}{1 + t ^ {2}}\\right) \\frac {2}{1 + t ^ {2}} \\mathrm{d} t\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":51,"id":"063fb113-4576-495f-98d8-f7f7f8959c5d","page_size":[595,841],"block_position":"51-18"},{"bbox":[291,803,303,811],"type":"page_number","angle":0,"index":1000015,"text":"44 ","color":{"line":"rgba(164, 164, 164, 1)","fill":"rgba(164, 164, 164, 1)"},"page_idx":51,"is_discarded":true,"id":"852bcecb-4e9c-4380-9abc-0914e421bd52","page_size":[595,841],"block_position":"51-19"}],[{"bbox":[449,47,539,58],"type":"header","angle":0,"index":-99999,"text":"5.3 不定积分表选例 ","color":{"line":"rgba(164, 164, 164, 1)","fill":"rgba(164, 164, 164, 1)"},"page_idx":52,"is_discarded":true,"id":"2f9f72a6-253c-4622-9fb9-688ce537d293","page_size":[595,841],"block_position":"52-0"},{"bbox":[73,73,231,86],"type":"title","angle":0,"index":1,"level":2,"text":"## 定理 5.5 (两类无理根式的不定积分)","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":52,"id":"f892707a-1bd8-4e47-9a37-2accf7e5089a","page_size":[595,841],"block_position":"52-1"},{"bbox":[67,90,527,134],"type":"text","angle":0,"index":2,"merge_prev":false,"text":"对于 $\\int R\\left(x,\\sqrt[n]{\\frac{ax + b}{cx + d}}\\right)\\mathrm{d}x$ ，只需要令 $t = \\sqrt[n]{\\frac{ax + b}{cx + d}}$ 即可化为有理函数。对于 $\\int R(x,\\sqrt{ax^2 + bx + c})$ 要么进行配方，要么用Euler变换。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":52,"id":"0a02855e-2cd5-44b2-a51d-1ae3aa40270d","page_size":[595,841],"block_position":"52-2"},{"bbox":[55,169,186,185],"type":"title","angle":0,"index":3,"level":2,"text":"## 5.3 不定积分表选例","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":52,"id":"5b775bfc-bbe3-45ed-a5ab-528dfcba6272","page_size":[595,841],"block_position":"52-3"},{"bbox":[74,198,267,211],"type":"text","angle":0,"index":4,"merge_prev":false,"text":"下面假设必背部分都会背，且不做计算证明 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":52,"id":"a98ce0cd-2679-4b7a-a77c-106ec65ebc0f","page_size":[595,841],"block_position":"52-4"},{"bbox":[55,230,174,244],"type":"title","angle":0,"index":5,"level":2,"text":"## 5.3.1 三角函数：重点","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":52,"id":"287341fc-e088-499a-9d06-99d3dc424ef2","page_size":[595,841],"block_position":"52-5"},{"bbox":[35,253,135,265],"type":"title","angle":0,"index":6,"level":2,"text":"## 练习 5.2 核心会算","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":52,"id":"e8ced741-b9c8-46cc-9e1a-bcd63b7c070d","page_size":[595,841],"block_position":"52-6"},{"bbox":[140,268,453,313],"type":"interline_equation","angle":0,"index":7,"text":"$$\n\\begin{array}{r l} \\int \\tan x \\mathrm{d} x = - \\ln | \\cos x | + c & \\int \\cot x \\mathrm{d} x = \\ln | \\sin x | + c \\\\ \\int \\sec x \\mathrm{d} x = \\ln | \\sec x + \\tan x | + c & \\int \\csc x \\mathrm{d} x = - \\ln | \\csc x + \\cot x | + c \\end{array}\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":52,"id":"74c9299a-ba05-46b3-8d2a-c1c7713d3411","page_size":[595,841],"block_position":"52-7"},{"bbox":[52,317,335,342],"type":"text","angle":0,"index":8,"merge_prev":false,"text":"解 $\\int \\tan x\\mathrm{d}x = -\\int \\frac{\\mathrm{d}\\cos x}{\\cos x},\\int \\cot x\\mathrm{d}x$ 同理。另外两个要记住 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":52,"id":"5115a67e-4cf3-43c2-b9ff-b809872f23e5","page_size":[595,841],"block_position":"52-8"},{"bbox":[114,343,478,370],"type":"interline_equation","angle":0,"index":9,"text":"$$\n\\int \\sec x \\mathrm{d} x = \\int \\frac {\\sec x (\\sec x + \\tan x)}{\\sec x + \\tan x} \\mathrm{d} x = \\int \\frac {\\mathrm{d} (\\sec x + \\tan x)}{\\sec x + \\tan x} = \\ln | \\sec x + \\tan x | + c\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":52,"id":"279d91cb-f939-463e-9e6b-bfe294af94a5","page_size":[595,841],"block_position":"52-9"},{"bbox":[35,378,395,414],"type":"text","angle":0,"index":10,"merge_prev":false,"text":"练习5.3简单平方(1) $\\int \\sin^2 x\\mathrm{d}x(2)$ 重点： $\\int \\tan^2 x\\mathrm{d}x$ (3)超重点： $\\int \\sec^3 x\\mathrm{d}x$ 解(1)直接用半角公式 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":52,"id":"82cb0576-d4be-4139-802d-693b30295e33","page_size":[595,841],"block_position":"52-10"},{"bbox":[72,412,349,437],"type":"text","angle":0,"index":11,"merge_prev":false,"text":"(2) 这个要记住， $\\int \\tan^2 x \\mathrm{d}x = \\int \\sec^2 x - 1 \\mathrm{d}x = \\tan x - x + c$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":52,"id":"4d31bf0f-59f8-4c1d-ab93-abfcb0aeabfd","page_size":[595,841],"block_position":"52-11"},{"bbox":[73,436,167,449],"type":"text","angle":0,"index":12,"merge_prev":false,"text":"(3) 这个经常会用到! ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":52,"id":"48fb0a74-4836-4000-875a-9d946410de66","page_size":[595,841],"block_position":"52-12"},{"bbox":[163,449,429,501],"type":"interline_equation","angle":0,"index":13,"text":"$$\n\\begin{array}{c} \\int \\sec^ {3} x \\mathrm{d} x = \\int \\sec x \\mathrm{d} \\tan x = \\sec x \\tan x - \\int \\tan^ {2} x \\sec x \\mathrm{d} x \\\\ = \\sec x \\tan x - \\int \\sec^ {3} x \\mathrm{d} x + \\int \\sec x \\mathrm{d} x \\end{array}\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":52,"id":"eed6db41-ab13-4746-9715-f1ec7d0d7dd4","page_size":[595,841],"block_position":"52-13"},{"bbox":[54,503,252,527],"type":"text","angle":0,"index":14,"merge_prev":false,"text":"因此 $I = \\frac{1}{2}\\sec x\\tan x + \\ln |\\sec x + \\tan x| + c$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":52,"id":"5ea91086-68ee-4b7b-8c10-c0a5d31fa5be","page_size":[595,841],"block_position":"52-14"},{"bbox":[35,533,404,559],"type":"text","angle":0,"index":15,"merge_prev":false,"text":"练习 $5.4\\sin x, \\cos x$ 加常数分式积分 (1) 重点: $\\int \\frac{\\mathrm{d}x}{1 \\pm \\cos x}$ (2) 重点: $\\int \\frac{\\mathrm{d}x}{1 \\pm \\sin x}$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":52,"id":"99968c2d-42c4-43cc-bdaf-b97f4ac3f1d1","page_size":[595,841],"block_position":"52-15"},{"bbox":[73,558,324,581],"type":"text","angle":0,"index":16,"merge_prev":false,"text":"(3) 极度重点，含参积分大量用到： $\\int \\frac{\\mathrm{d}x}{1 + a\\cos x}, |a| \\leq 1$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":52,"id":"a770bd7b-8a6a-42fd-94f7-3ac767d61763","page_size":[595,841],"block_position":"52-16"},{"bbox":[53,580,311,601],"type":"text","angle":0,"index":17,"merge_prev":false,"text":"解(1)根据 $1+\\cos x=2\\cos^{2}\\frac{x}{2},1-\\cos x=2\\sin^{2}\\frac{x}{2}$ ，因此 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":52,"id":"b21da90a-570d-4a5a-8ef1-b24a24ebc881","page_size":[595,841],"block_position":"52-17"},{"bbox":[193,602,391,650],"type":"interline_equation","angle":0,"index":18,"text":"$$\n\\left\\{ \\begin{array}{l} \\int \\frac {\\mathrm{d} x}{1 + \\cos x} = \\frac {1}{2} \\int \\sec^ {2} \\frac {x}{2} \\mathrm{d} x = \\tan \\frac {x}{2} + c \\\\ \\int \\frac {\\mathrm{d} x}{1 - \\cos x} = \\frac {1}{2} \\int \\csc^ {2} \\frac {x}{2} \\mathrm{d} x = - \\cot \\frac {x}{2} + c \\end{array} \\right.\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":52,"id":"4158b74a-5b16-4c9d-bf59-0a9b3e6ff422","page_size":[595,841],"block_position":"52-18"},{"bbox":[73,653,239,672],"type":"text","angle":0,"index":19,"merge_prev":false,"text":"(2) 根据诱导公式 $\\sin x = \\cos \\left(\\frac{\\pi}{2} - x\\right)$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":52,"id":"b1849ee5-3ead-4685-88d4-339a2f560001","page_size":[595,841],"block_position":"52-19"},{"bbox":[165,675,418,726],"type":"interline_equation","angle":0,"index":20,"text":"$$\n\\left\\{ \\begin{array}{l} \\int \\frac {\\mathrm{d} x}{1 + \\sin x} = \\int \\frac {\\mathrm{d} x}{1 + \\cos \\left(\\frac {\\pi}{2} - x\\right)} = - \\tan \\left(\\frac {\\pi}{4} - \\frac {x}{2}\\right) + c \\\\ \\int \\frac {\\mathrm{d} x}{1 - \\sin x} = \\int \\frac {\\mathrm{d} x}{1 - \\cos \\left(\\frac {\\pi}{2} - x\\right)} = \\cot \\left(\\frac {\\pi}{4} - \\frac {x}{2}\\right) + c \\end{array} \\right.\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":52,"id":"31aed820-cdbd-4fcb-adfa-c87d28af8b56","page_size":[595,841],"block_position":"52-20"},{"bbox":[291,803,302,811],"type":"page_number","angle":0,"index":1000020,"text":"45 ","color":{"line":"rgba(164, 164, 164, 1)","fill":"rgba(164, 164, 164, 1)"},"page_idx":52,"is_discarded":true,"id":"b0dc6dc9-9c82-4501-9ac8-fb65cc5d1972","page_size":[595,841],"block_position":"52-21"}],[{"bbox":[449,47,539,58],"type":"header","angle":0,"index":-99998,"text":"5.3 不定积分表选例 ","color":{"line":"rgba(164, 164, 164, 1)","fill":"rgba(164, 164, 164, 1)"},"page_idx":53,"is_discarded":true,"id":"35f2055b-f373-4bf8-9bff-aa0e009c8e69","page_size":[595,841],"block_position":"53-0"},{"bbox":[73,68,395,94],"type":"text","angle":0,"index":2,"merge_prev":false,"text":"(3) $a = \\pm 1$ 时即(1)。用万能公式，令 $t = \\tan \\frac{x}{2}$ ，则 $\\mathrm{dx} = \\frac{2}{1 + t^2}\\mathrm{dt}$ ，因此 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":53,"id":"6ea5c8b6-9c30-4a1f-b74a-0dda49da6a7f","page_size":[595,841],"block_position":"53-1"},{"bbox":[54,96,578,126],"type":"interline_equation","angle":0,"index":3,"text":"$$\nI = \\int \\frac {1}{1 + a \\frac {1 - t ^ {2}}{1 + t ^ {2}}} \\cdot \\frac {2}{1 + t ^ {2}} \\mathrm{d} t = \\int \\frac {2}{(1 + t ^ {2}) + a (1 - t ^ {2})} \\mathrm{d} t = \\int \\frac {2}{(1 - a) t ^ {2} + (1 + a)} \\mathrm{d} t = \\frac {2}{1 - a} \\sqrt {\\frac {1 - a}{1 + a}} \\arctan \\left(\\sqrt {\\frac {1 - a}{1 + a}} t\\right) + c\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":53,"id":"534d5817-ffcc-4ca7-83b6-25f492c418f5","page_size":[595,841],"block_position":"53-2"},{"bbox":[54,131,292,162],"type":"text","angle":0,"index":4,"merge_prev":false,"text":"得到结果为 $I = \\frac{2}{\\sqrt{1 - a^2}}\\arctan \\left(\\sqrt{\\frac{1 - a}{1 + a}}\\tan \\frac{x}{2}\\right) + c$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":53,"id":"8798450f-ef7b-40eb-90b5-915b0bd8bd52","page_size":[595,841],"block_position":"53-3"},{"bbox":[72,190,221,215],"type":"title","angle":0,"index":5,"level":2,"text":"## 定理 5.6 ( $\\frac{1}{a\\sin^{2}x + b\\cos^{2}x}$ 积分)","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":53,"id":"f4d6f496-8393-496f-8768-bb29adad3411","page_size":[595,841],"block_position":"53-4"},{"bbox":[68,218,362,243],"type":"text","angle":0,"index":6,"merge_prev":false,"text":"形如 $\\int \\frac{\\mathrm{d}x}{a\\sin^2x + b\\cos^2x}$ 的积分都采用上下同除以 $\\cos^2 x$ ，转换为 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":53,"id":"c470ac50-56a0-4358-993a-9b1826c69e89","page_size":[595,841],"block_position":"53-5"},{"bbox":[262,245,330,272],"type":"interline_equation","angle":0,"index":7,"text":"$$\n\\int {\\frac {\\sec^ {2} x \\mathrm{d} x}{a \\tan^ {2} x + b}}\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":53,"id":"3a9f4401-fb05-4737-82be-13127b27e431","page_size":[595,841],"block_position":"53-6"},{"bbox":[70,273,215,287],"type":"text","angle":0,"index":8,"merge_prev":false,"text":"直接用 $\\mathrm{d}(\\tan x) = \\sec^2 x\\mathrm{d}x$ 即可。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":53,"id":"59e1d8c7-bdcf-4b30-9cf7-0f0008b71278","page_size":[595,841],"block_position":"53-7"},{"bbox":[72,309,298,334],"type":"title","angle":0,"index":9,"level":2,"text":"## 定理5.7 $\\left(\\frac{1}{a\\sin^2x + b\\sin x\\cos x + c\\cos^2x + d}\\right.$ 积分）","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":53,"id":"bf1a5b3d-0fb5-4b06-aa60-575d49e46787","page_size":[595,841],"block_position":"53-8"},{"bbox":[68,337,420,362],"type":"text","angle":0,"index":10,"merge_prev":false,"text":"形如 $\\int \\frac{\\mathrm{d}x}{a\\sin^2x + b\\sin x\\cos x + c\\cos^2x + d}$ 的积分首先上下同除以 $\\cos^2 x$ ，得到 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":53,"id":"93580157-b244-4ead-b2a9-87369b4046ca","page_size":[595,841],"block_position":"53-9"},{"bbox":[220,364,371,391],"type":"interline_equation","angle":0,"index":11,"text":"$$\n\\int {\\frac {\\sec^ {2} x \\mathrm{d} x}{a \\tan^ {2} x + b \\tan x + c + d \\sec^ {2} x}}\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":53,"id":"5bc7a425-7138-48fd-a164-0ec73cefe7bd","page_size":[595,841],"block_position":"53-10"},{"bbox":[68,391,349,406],"type":"text","angle":0,"index":12,"merge_prev":false,"text":"再利用 $\\sec^2 x = \\tan^2 x + 1$ ，全部转换为 $\\tan x$ ，再分母配方即可 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":53,"id":"f19bb94e-4a8d-4662-85ba-c947ee2089d4","page_size":[595,841],"block_position":"53-11"},{"bbox":[35,431,520,480],"type":"text","angle":0,"index":13,"merge_prev":false,"text":"练习 $5.5\\sin x, \\cos x$ 无常数、无分子分式 (1) 重点: $\\int \\frac{dx}{\\sin^{2}x + 2\\cos^{2}x}$ (2) $\\int \\frac{1}{\\sin^{2}x\\cos x} dx$ (3) $\\int \\frac{dx}{\\sin x\\cos^{2}x}$ 解(1)同除以 $\\cos^{2}x$ ，得到 $\\int\\frac{\\sec^{2}x\\mathrm{d}x}{2+\\tan^{2}x}$ ，易算 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":53,"id":"60d9fd6d-1690-4abc-95b0-5ed01bc21837","page_size":[595,841],"block_position":"53-12"},{"bbox":[52,478,539,542],"type":"text","angle":0,"index":14,"merge_prev":false,"text":"(2) 同乘 $\\cos x$ ，得到 $I = \\int \\frac{\\mathrm{d}(\\sin x)}{\\sin^2 x (1 - \\sin^2 x)}$ ，即 $\\int \\frac{1}{\\sin^2 x} \\mathrm{d}(\\sin x) + \\int \\frac{1}{1 - \\sin^2 x} \\mathrm{d}x$ ，前者直接算，后者用 $\\int \\frac{1}{x^2 - a} \\mathrm{d}x = \\frac{1}{2a} \\ln \\left|\\frac{x - a}{x + a}\\right|$ 。\n(3) 与 (2) 同理 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":53,"id":"df620c16-06c4-4a34-a38c-56656733ed16","page_size":[595,841],"block_position":"53-13"},{"bbox":[35,553,530,602],"type":"text","angle":0,"index":15,"merge_prev":false,"text":"练习 5.6sin x, cos x 的高次 (1) $\\int(\\sin^{4}x + \\cos^{4}x)\\mathrm{d}x$ (2) 重点: $\\int\\frac{\\sin x\\cos x}{\\sin^{4}x + \\cos^{4}x}\\mathrm{d}x$ (3) 重点: $\\int\\frac{\\mathrm{d}x}{\\sin^{4}x + \\cos^{4}x}$ (4) $\\int\\sin^{6}x + \\cos^{6}x\\mathrm{d}x$ (5) $\\int\\frac{\\mathrm{d}x}{\\sin^{6}x + \\cos^{6}x}$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":53,"id":"c28b639e-3541-4deb-839f-ec0c5e11936b","page_size":[595,841],"block_position":"53-14"},{"bbox":[53,601,508,624],"type":"text","angle":0,"index":16,"merge_prev":false,"text":"解(1)降到一次： $\\sin^4 x + \\cos^4 x = (\\sin^2 x + \\cos^2 x)^2 - 2\\sin^2 x\\cos^2 x = 1 - \\frac{1}{2}\\sin^2 2x = 1 - \\frac{1}{4}(1 - \\cos 4x)$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":53,"id":"9c76ecb6-31e5-4444-9f22-82db594c1f09","page_size":[595,841],"block_position":"53-15"},{"bbox":[73,622,391,650],"type":"text","angle":0,"index":17,"merge_prev":false,"text":"(2) 降到二次，并切换 $\\sin x, \\cos x: \\int \\frac{\\frac{1}{2} \\sin 2x}{1 - \\frac{1}{2} \\sin^2 2x}$ ，要善于切换正余弦： ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":53,"id":"eef7a2bf-46a6-4adb-b679-066d274faa64","page_size":[595,841],"block_position":"53-16"},{"bbox":[158,654,435,679],"type":"interline_equation","angle":0,"index":18,"text":"$$\n\\int {\\frac {\\sin 2 x}{2 - \\sin^ {2} 2 x}} \\mathrm{d} x = - \\frac {1}{2} \\int {\\frac {\\mathrm{d} (\\cos 2 x)}{1 + \\cos^ {2} 2 x}} = - \\frac {1}{2} \\arctan (\\cos 2 x) + c\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":53,"id":"b85b1e92-f576-4c6a-9f40-a10d709109e6","page_size":[595,841],"block_position":"53-17"},{"bbox":[73,682,365,694],"type":"text","angle":0,"index":19,"merge_prev":false,"text":"(3) 直接降到二次，并转换为 $\\tan x, \\sec x$ ：使用 $\\sec^2 x = 1 + \\tan^2 x$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":53,"id":"8777ed4e-0d20-497c-b728-1d7371ece3af","page_size":[595,841],"block_position":"53-18"},{"bbox":[97,697,496,724],"type":"interline_equation","angle":0,"index":20,"text":"$$\nI = \\int {\\frac {\\mathrm{d} x}{1 - \\frac {1}{2} \\sin^ {2} 2 x}} = \\int {\\frac {\\sec^ {2} 2 x}{\\sec^ {2} 2 x - \\frac {1}{2} \\tan^ {2} 2 x}} = \\frac {1}{2} \\int {\\frac {\\mathrm{d} (\\tan 2 x)}{1 + \\frac {1}{2} \\tan^ {2} 2 x}} = \\frac {1}{\\sqrt {2}} \\arctan {\\frac {\\tan 2 x}{\\sqrt {2}}} + c\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":53,"id":"f32afb47-52a4-4492-b9f5-923c365550e1","page_size":[595,841],"block_position":"53-19"},{"bbox":[73,726,447,741],"type":"text","angle":0,"index":21,"merge_prev":false,"text":"(4) $\\sin^6 x + \\cos^6 x = (\\sin^2 x + \\cos^2 x)(\\sin^4 - \\sin^2 \\cos^2 x + \\cos^4 x)$ 转化为四次，再配方： ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":53,"id":"51372c15-45d2-4239-a717-4cffef8effe7","page_size":[595,841],"block_position":"53-20"},{"bbox":[143,743,449,767],"type":"interline_equation","angle":0,"index":22,"text":"$$\n(\\sin^ {2} x + \\cos^ {2} x) ^ {2} - 3 \\sin^ {2} x \\cos^ {2} x = 1 - \\frac {3}{4} \\sin^ {2} 2 x = 1 - \\frac {3}{8} (1 - \\cos 4 x)\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":53,"id":"7dc02b2f-ec92-438a-bfbf-cd43bad47d61","page_size":[595,841],"block_position":"53-21"},{"bbox":[291,803,303,812],"type":"page_number","angle":0,"index":999999,"text":"46 ","color":{"line":"rgba(164, 164, 164, 1)","fill":"rgba(164, 164, 164, 1)"},"page_idx":53,"is_discarded":true,"id":"49009ef7-cf46-4d99-b8a6-a9579afc2290","page_size":[595,841],"block_position":"53-22"}],[{"bbox":[449,47,539,58],"type":"header","angle":0,"index":-99999,"text":"5.3 不定积分表选例 ","color":{"line":"rgba(164, 164, 164, 1)","fill":"rgba(164, 164, 164, 1)"},"page_idx":54,"is_discarded":true,"id":"da6c542d-8823-4e86-91ae-2d799e6e0474","page_size":[595,841],"block_position":"54-0"},{"bbox":[54,68,172,92],"type":"text","angle":0,"index":1,"merge_prev":false,"text":"结果为 $\\frac{5}{8} x + \\frac{3}{32}\\sin 4x + c$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":54,"id":"2fdc96a4-5387-4134-841a-fb102db76680","page_size":[595,841],"block_position":"54-1"},{"bbox":[73,90,389,119],"type":"interline_equation","angle":0,"index":2,"text":"$$\n(5) I = \\int {\\frac {1}{1 - \\frac {3}{4} \\sin^ {2} 2 x}} \\mathrm{d} x = \\int {\\frac {\\sec^ {2} x}{\\sec^ {2} 2 x - \\frac {3}{4} \\tan^ {2} 2 x}} \\mathrm{d} x = 2 \\int {\\frac {\\mathrm{d} (\\tan 2 x)}{\\tan^ {2} 2 x + 4}}\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":54,"id":"e55cd707-4397-47bf-804e-3d74a3a4f1cf","page_size":[595,841],"block_position":"54-2"},{"bbox":[52,121,539,151],"type":"text","angle":0,"index":3,"merge_prev":false,"text":"笔记 高次问题，如果分母为 1 则降到一次直接积分，分子为 1 则降到二次转换为 $\\sec, \\tan$ ，均不为 1 则降到齐次转化为有理式。降次要用立方和公式，要么用配方。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":54,"id":"1a8a5783-c0ce-469a-a9f8-bb55a1f08749","page_size":[595,841],"block_position":"54-3"},{"bbox":[73,173,241,197],"type":"title","angle":0,"index":4,"level":2,"text":"## 定理 5.8 ( $\\frac{a\\sin x + b\\cos x}{c\\sin x + d\\cos x}$ 型计算技巧)","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":54,"id":"450c9621-34f5-47fc-966e-1b1565be38fe","page_size":[595,841],"block_position":"54-4"},{"bbox":[69,200,496,228],"type":"text","angle":0,"index":6,"merge_prev":false,"text":"$\\frac{a\\sin x + b\\cos x}{c\\sin x + d\\cos x}$ 型的积分把分子写为 $A$ 倍分母加上 $B$ 倍分母导数的形式，用待定系数法进行计算 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":54,"id":"285ae3cd-cd3e-4108-8289-c5399f5a3e60","page_size":[595,841],"block_position":"54-5"},{"bbox":[72,244,201,258],"type":"title","angle":0,"index":7,"level":2,"text":"## 推论 5.1 ( $\\tan x, \\cot x$ 分式型)","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":54,"id":"aa51026b-30e8-44f5-b4a0-880c11e172c8","page_size":[595,841],"block_position":"54-6"},{"bbox":[69,260,429,288],"type":"text","angle":0,"index":8,"merge_prev":false,"text":"类似于 $\\int \\frac{a\\mathrm{d}x}{b + c\\tan x},\\int \\frac{a\\mathrm{d}x}{b + c\\cot x}$ 都可以转换回 $\\frac{a\\sin x + b\\cos x}{c\\sin x + d\\cos x}$ 型积分进行计算 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":54,"id":"39962f8d-94c0-4f5f-a859-be0db42999e7","page_size":[595,841],"block_position":"54-7"},{"bbox":[35,312,424,338],"type":"text","angle":0,"index":9,"merge_prev":false,"text":"练习 5.7 $\\frac{a\\sin x+b\\cos x}{c\\sin x+d\\cos x}$ 型计算 (1) 重点: $\\int\\frac{\\sin x}{\\sin x+\\cos x}\\mathrm{d}x$ (2) 重点: $\\int\\frac{\\mathrm{d}x}{3+5\\tan x}$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":54,"id":"21f80920-55c3-44a9-9c0f-6942fe15cd37","page_size":[595,841],"block_position":"54-8"},{"bbox":[52,337,409,358],"type":"text","angle":0,"index":10,"merge_prev":false,"text":"解(1)设 $\\sin x = A(\\sin x + \\cos x) + B(\\cos x - \\sin x)$ ，得到 $A = \\frac{1}{2}, B = -\\frac{1}{2}$ ，因此 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":54,"id":"9c5bac92-12a2-47f9-a1e2-cb3db16e2c2a","page_size":[595,841],"block_position":"54-9"},{"bbox":[131,360,462,387],"type":"interline_equation","angle":0,"index":11,"text":"$$\nI = \\int {\\frac {{\\frac {1}{2}} (\\sin x + \\cos x) - {\\frac {1}{2}} \\mathrm{d} (\\sin x + \\cos x)}{\\sin x + \\cos x}} \\mathrm{d} x = {\\frac {1}{2}} x - {\\frac {1}{2}} \\ln | \\sin x + \\cos x | + c\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":54,"id":"72472997-e464-4321-accb-dd15f63b7d1f","page_size":[595,841],"block_position":"54-10"},{"bbox":[52,390,539,434],"type":"text","angle":0,"index":12,"merge_prev":false,"text":"(2) 由于 $\\frac{1}{3 + 5\\tan x} = \\frac{\\cos x}{5\\sin x + 3\\cos x}$ ，设 $\\cos x = A(5\\sin x + 3\\cos x) + B(5\\cos x - 3\\sin x)$ ，则 $5A = 3B, 3A + 5B = 1$ ，得到 $A = \\frac{3}{34}, B = \\frac{5}{34}$ ，得到 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":54,"id":"77d5a697-dbbb-4443-bad8-ebc05199f4bd","page_size":[595,841],"block_position":"54-11"},{"bbox":[214,435,380,460],"type":"interline_equation","angle":0,"index":13,"text":"$$\nI = \\frac {3}{3 4} x + \\frac {5}{3 4} \\ln | 5 \\sin x + 3 \\cos x | + c\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":54,"id":"b61a906b-e2bd-44cd-bf16-a789642f6bda","page_size":[595,841],"block_position":"54-12"},{"bbox":[73,481,227,505],"type":"title","angle":0,"index":14,"level":2,"text":"## 定理 5.9 ( $\\frac{1}{a\\sin x + b\\cos x + c}$ 积分)","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":54,"id":"c053304e-038e-4b01-a12f-9f3999685611","page_size":[595,841],"block_position":"54-13"},{"bbox":[68,508,382,533],"type":"text","angle":0,"index":15,"merge_prev":false,"text":"类似于 $\\int \\frac{\\mathrm{d}x}{a\\sin x + b\\cos x + c}$ 的积分采用二倍角公式转换为二次，得到 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":54,"id":"9024dfa6-dc1a-4cf3-a95b-36d115e7ce15","page_size":[595,841],"block_position":"54-14"},{"bbox":[208,535,385,562],"type":"interline_equation","angle":0,"index":16,"text":"$$\n\\int {\\frac {1}{2 a \\sin {\\frac {x}{2}} \\cos {\\frac {x}{2}} + b (2 \\cos^ {2} {\\frac {x}{2}} - 1) + c}} \\mathrm{d} x\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":54,"id":"ea2e0c85-0e54-4ce1-8317-d7784a72eff0","page_size":[595,841],"block_position":"54-15"},{"bbox":[67,564,452,587],"type":"text","angle":0,"index":17,"merge_prev":false,"text":"使用前面的积分方法，上下同时除以 $\\cos^2\\frac{x}{2}$ ，再根据 $\\sec^2 x = \\tan^2 x + 1$ 进行配方即可。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":54,"id":"077d10c2-f397-4c9f-9c3c-016b7bff99e2","page_size":[595,841],"block_position":"54-16"},{"bbox":[35,610,240,634],"type":"text","angle":0,"index":18,"merge_prev":false,"text":"练习 5.8 相关训练 (1) $\\int\\frac{dx}{\\sin x+2\\cos x+3}$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":54,"id":"1ab960ea-2af3-441d-a561-1afd5e2b9dca","page_size":[595,841],"block_position":"54-17"},{"bbox":[53,631,318,655],"type":"text","angle":0,"index":19,"merge_prev":false,"text":"解 (1) $\\sin x + 2\\cos x + 3 = 1 + 2\\sin \\frac{x}{2}\\cos \\frac{x}{2} +4\\cos^2{\\frac{x}{2}}$ ，因此 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":54,"id":"3230fb8e-510e-43a8-b499-82f037c7528e","page_size":[595,841],"block_position":"54-18"},{"bbox":[134,655,460,684],"type":"interline_equation","angle":0,"index":20,"text":"$$\nI = \\int {\\frac {\\sec^ {2} {\\frac {x}{2}}}{4 + 2 \\tan {\\frac {x}{2}} + \\sec^ {2} x}} = 2 \\int {\\frac {d (\\tan {\\frac {x}{2}} + 1)}{(\\tan {\\frac {x}{2}} + 1) ^ {2} + 4}} = \\arctan {\\frac {\\tan {\\frac {x}{2}} + 1}{2}} + c\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":54,"id":"a0fb9c90-b7aa-4654-b89c-dc310dc9abce","page_size":[595,841],"block_position":"54-19"},{"bbox":[73,703,217,727],"type":"title","angle":0,"index":21,"level":2,"text":"## 推论 5.2 ( $\\frac{p\\sin x + q\\cos x + s}{a\\sin x + b\\cos x + c}dx$ )","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":54,"id":"ce2f1eaa-bf3e-4d59-ab96-fea251119f79","page_size":[595,841],"block_position":"54-20"},{"bbox":[68,729,503,744],"type":"text","angle":0,"index":22,"merge_prev":false,"text":"这类积分可以将分子写为 $A$ 倍分母加 $B$ 倍分母导数加 $C$ 常数，其中常数项用前一个定理进行计算。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":54,"id":"b8f7c8a1-983d-49ab-a619-2676e6479410","page_size":[595,841],"block_position":"54-21"},{"bbox":[291,802,302,811],"type":"page_number","angle":0,"index":1000004,"text":"47 ","color":{"line":"rgba(164, 164, 164, 1)","fill":"rgba(164, 164, 164, 1)"},"page_idx":54,"is_discarded":true,"id":"6605e6d1-7760-493b-b7f6-3f245eced282","page_size":[595,841],"block_position":"54-22"}],[{"bbox":[449,47,539,58],"type":"header","angle":0,"index":-99999,"text":"5.3 不定积分表选例 ","color":{"line":"rgba(164, 164, 164, 1)","fill":"rgba(164, 164, 164, 1)"},"page_idx":55,"is_discarded":true,"id":"e9abdc6c-7ff2-4f9a-ae3a-16d768ca85de","page_size":[595,841],"block_position":"55-0"},{"bbox":[35,69,309,93],"type":"text","angle":0,"index":1,"merge_prev":false,"text":"练习5.9三角根号 (1) $\\int \\sin x\\cos x\\sqrt{a^2\\sin^2x + b^2\\cos^2x}\\mathrm{d}x$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":55,"id":"55714381-8095-4a45-851e-c9cdc0b38c87","page_size":[595,841],"block_position":"55-1"},{"bbox":[52,92,283,117],"type":"text","angle":0,"index":2,"merge_prev":false,"text":"解(1)写为 $\\frac{1}{2}\\int \\sqrt{(a^2 - b^2)\\sin^2x + b^2}\\mathrm{d}(\\sin^2 x)$ 即可 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":55,"id":"6f976cfd-e6ee-4de0-a9d7-04eef3abb799","page_size":[595,841],"block_position":"55-2"},{"bbox":[55,133,151,148],"type":"title","angle":0,"index":3,"level":2,"text":"## 5.3.2 对数与指数","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":55,"id":"ce2d61f2-4186-43c7-ace6-68a5faa89f83","page_size":[595,841],"block_position":"55-3"},{"bbox":[35,157,414,185],"type":"text","angle":0,"index":4,"merge_prev":false,"text":"练习 5.10 几道经典指数积分 (1) $\\int\\frac{dx}{e^{x}+e^{-x}}$ (2) $\\int\\frac{dx}{\\sqrt{1+e^{2x}}}$ (3) $\\int e^{x}\\left(\\frac{1-x}{1+x^{2}}\\right)^{2}dx$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":55,"id":"d49b047c-ae81-4521-a1a4-e64e5b481106","page_size":[595,841],"block_position":"55-4"},{"bbox":[52,185,341,207],"type":"text","angle":0,"index":5,"merge_prev":false,"text":"解(1)本质是 $e^x$ 的有理函数，故 $I = \\int \\frac{\\mathrm{d}(e^x)}{e^{2x} + 1} = \\arctan (e^x) + c$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":55,"id":"1006f50b-1c9e-4249-8c23-1f4661e73783","page_size":[595,841],"block_position":"55-5"},{"bbox":[72,207,370,233],"type":"text","angle":0,"index":6,"merge_prev":false,"text":"(2) 同乘 $e^{-x}$ ，故 $I = -\\int \\frac{\\mathrm{d}(e^{-x})}{\\sqrt{e^{-2x} + 1}} = -\\ln |e^{-x} + \\sqrt{e^{-2x} + 1}| + c$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":55,"id":"1fbcfcd8-2aef-4eaf-8789-451a8755b0bd","page_size":[595,841],"block_position":"55-6"},{"bbox":[54,232,556,267],"type":"text","angle":0,"index":7,"merge_prev":false,"text":"(3) $I=\\int\\frac{e^{x}}{1+x^{2}}dx-\\int\\frac{e^{x}}{(1+x^{2})^{2}}d(x^{2}+1)$ ，第一项实在是没法处理，只能期待第二项，第二项等于 $\\int e^{x}d(\\frac{1}{1+x^{2}})$ ，因此用分部积分得到 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":55,"id":"9b79097f-4604-421d-8a30-d560670075d4","page_size":[595,841],"block_position":"55-7"},{"bbox":[199,264,394,290],"type":"interline_equation","angle":0,"index":8,"text":"$$\n\\int {\\frac {e ^ {x}}{1 + x ^ {2}}} \\mathrm{d} x + \\int e ^ {x} \\mathrm{d} (\\frac {1}{1 + x ^ {2}}) = \\frac {e ^ {x}}{1 + x ^ {2}} + c\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":55,"id":"57b79c5a-0512-43ee-a338-0eb88ff316fe","page_size":[595,841],"block_position":"55-8"},{"bbox":[35,302,531,357],"type":"text","angle":0,"index":9,"merge_prev":false,"text":"练习5.11几道经典对数积分 (1) $\\int x^n\\ln x\\mathrm{d}x$ (2) $\\int \\left(\\frac{\\ln x}{x}\\right)^2\\mathrm{d}x$ (3) $\\int \\sqrt{x}\\ln^2 x\\mathrm{d}x$ (4) 特殊： $\\int \\frac{1}{1 - x^2}\\ln \\frac{1 + x}{1 - x}\\mathrm{d}x$ 解(1)列表法得到： $I = \\frac{x^{n + 1}\\ln x}{n + 1} -\\int \\frac{x^n}{n + 1}\\mathrm{d}x = \\frac{x^{n + 1}}{1 + n} (\\ln x - \\frac{1}{n + 1}) + c$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":55,"id":"00df200e-c0a8-43a0-b775-56a72763e0c2","page_size":[595,841],"block_position":"55-9"},{"bbox":[252,363,342,407],"type":"table_body","angle":0,"index":10,"img_path":"/","id":"1d2718ed-0594-4840-96b7-ebba84609dde","color":{"line":"rgba(103, 194, 63, 1)","fill":"rgba(103, 194, 63, 1)"},"table_body":"
$ \\ln x $$ \\frac{1}{x} $
$ x^{n} $$ \\frac{1}{n+1}x^{n+1} $
","page_idx":55,"page_size":[595,841],"block_position":"55-10"},{"bbox":[72,414,384,426],"type":"text","angle":0,"index":11,"merge_prev":false,"text":"(2) 无法直接列表，因为中间项会交叉，因此可以用表格法一步一步列， ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":55,"id":"c8a03613-53dc-4376-9df2-11c98d246bb3","page_size":[595,841],"block_position":"55-11"},{"bbox":[112,429,480,454],"type":"interline_equation","angle":0,"index":12,"text":"$$\nI = - \\frac {\\ln^ {2} x}{x} + \\int \\frac {2 \\ln x}{x ^ {2}} \\mathrm{d} x = - \\frac {\\ln^ {2} x}{x} - \\frac {2 \\ln x}{x} + \\int \\frac {2}{x ^ {2}} \\mathrm{d} x = - \\frac {1}{x} (\\ln^ {2} x + 2 \\ln x + 2) + c\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":55,"id":"e64af7b9-6249-47a5-965e-4a67afb6a294","page_size":[595,841],"block_position":"55-12"},{"bbox":[73,458,205,470],"type":"text","angle":0,"index":13,"merge_prev":false,"text":"(3) 同理用表格法一步一步列， ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":55,"id":"e26717f9-f602-43c9-9692-c0b590da3e4f","page_size":[595,841],"block_position":"55-13"},{"bbox":[136,471,455,498],"type":"interline_equation","angle":0,"index":14,"text":"$$\nI = \\frac {2}{3} x ^ {\\frac {3}{2}} \\ln^ {2} x - \\frac {4}{3} \\int x ^ {\\frac {1}{2}} \\ln x \\mathrm{d} x = \\frac {2}{3} x ^ {\\frac {3}{2}} \\ln^ {2} x - \\frac {4}{3} \\left(\\frac {2}{3} \\ln x x ^ {\\frac {3}{2}} - \\frac {2}{3} \\int x ^ {\\frac {1}{2}} \\mathrm{d} x\\right)\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":55,"id":"05c423cd-f90d-4493-8b0d-dbcf2acf85e2","page_size":[595,841],"block_position":"55-14"},{"bbox":[72,501,351,527],"type":"text","angle":0,"index":16,"merge_prev":false,"text":"(4)一看就是特殊配凑的，要注意到 $\\frac{1}{1 - x^2}\\mathrm{d}x = \\frac{1}{2}\\mathrm{d}\\left(\\ln \\frac{1 + x}{1 - x}\\right)$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":55,"id":"2514126e-ba5d-4f23-bf60-3f71409a9154","page_size":[595,841],"block_position":"55-15"},{"bbox":[55,543,139,557],"type":"title","angle":0,"index":17,"level":2,"text":"## 5.3.3 分式问题","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":55,"id":"6d8eaeef-4b68-4a5c-97e0-9ce2880c2da1","page_size":[595,841],"block_position":"55-16"},{"bbox":[35,565,420,592],"type":"text","angle":0,"index":18,"merge_prev":false,"text":"练习5.12几个基础型(1) $\\int \\frac{\\mathrm{d}x}{ax^2 + b} (a > 0)$ (2)重点： $\\int \\frac{\\mathrm{d}x}{(ax^2 + b)^2}$ (3) $\\int \\frac{\\mathrm{d}x}{(ax^2 + b)^n}$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":55,"id":"bad9b4ce-480b-419c-a491-5610fe1ae316","page_size":[595,841],"block_position":"55-17"},{"bbox":[73,590,317,615],"type":"text","angle":0,"index":19,"merge_prev":false,"text":"(4) 重点: $\\int \\frac{\\mathrm{d}x}{ax^2 + bx + c}$ (5) 重点: $\\int \\frac{x}{ax^2 + bx + c}\\mathrm{d}x$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":55,"id":"b961402a-4a76-4f93-872f-64c81e75ceba","page_size":[595,841],"block_position":"55-18"},{"bbox":[52,613,539,677],"type":"text","angle":0,"index":20,"merge_prev":false,"text":"解(1)原式 $= \\frac{1}{a}\\int \\frac{\\mathrm{d}x}{x^2 + \\frac{b}{a}}$ ，需要分类讨论： $b > 0$ 时变为 $\\frac{1}{a}\\int \\frac{\\mathrm{d}x}{x^2 + (\\sqrt{\\frac{b}{a}})^2} = \\frac{1}{a}\\cdot \\sqrt{\\frac{a}{b}}\\arctan (\\sqrt{\\frac{a}{b}} x)$ 。 $b < 0$ 时变为 $\\frac{1}{a}\\int \\frac{\\mathrm{d}x}{x^2 + (\\sqrt{-\\frac{b}{a}})^2}$ ，后面同理 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":55,"id":"238b428a-2b48-4cf3-ad5f-dcd3b4e00896","page_size":[595,841],"block_position":"55-19"},{"bbox":[54,677,539,729],"type":"text","angle":0,"index":21,"merge_prev":false,"text":"(2) 原式 $= -\\int \\frac{1}{2ax} \\mathrm{d}\\frac{1}{ax^2 + b} = -\\frac{1}{2ax} \\cdot \\frac{1}{ax^2 + b} + \\int \\frac{1}{ax^2 + b} \\mathrm{d}\\frac{1}{2ax} = -\\frac{1}{2ax} \\cdot \\frac{1}{ax^2 + b} - \\int \\frac{1}{ax^2 + b} \\cdot \\frac{1}{2ax^2} \\mathrm{d}x,$ 再对 $\\frac{1}{2ax^2(ax^2 + b)} = \\frac{\\frac{1}{b}}{2ax^2} + \\frac{-\\frac{1}{2b}}{ax^2 + b}$ 有理分解，代入后全变为基础型1。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":55,"id":"4f2c550b-5f5c-4a7e-9a13-fe17bffd893c","page_size":[595,841],"block_position":"55-20"},{"bbox":[72,729,370,754],"type":"text","angle":0,"index":22,"merge_prev":false,"text":"(3) 同理转化为 $\\int \\frac{1}{2ax(1 - n)}\\mathrm{d}\\left(\\frac{1}{(ax^2 + b)^{n - 1}}\\right)$ ，再用分部积分去做。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":55,"id":"7aeb5d0f-dd56-4eb1-9900-464fe7919f6c","page_size":[595,841],"block_position":"55-21"},{"bbox":[291,802,302,812],"type":"page_number","angle":0,"index":1000014,"text":"48 ","color":{"line":"rgba(164, 164, 164, 1)","fill":"rgba(164, 164, 164, 1)"},"page_idx":55,"is_discarded":true,"id":"7f02763d-291a-4ca2-8dbe-1e0ded754b39","page_size":[595,841],"block_position":"55-22"}],[{"bbox":[449,47,539,58],"type":"header","angle":0,"index":-99998,"text":"5.3 不定积分表选例 ","color":{"line":"rgba(164, 164, 164, 1)","fill":"rgba(164, 164, 164, 1)"},"page_idx":56,"is_discarded":true,"id":"fb8d6c23-9f4e-4ddb-8738-24a6133f32e7","page_size":[595,841],"block_position":"56-0"},{"bbox":[72,68,471,98],"type":"text","angle":0,"index":2,"merge_prev":false,"text":"(4) 原式 $= \\frac{1}{a} \\int \\frac{\\mathrm{d}x}{x^2 + \\frac{b}{a} x + \\frac{c}{a}} = \\frac{1}{a} \\int \\frac{\\mathrm{d}x}{(x + \\frac{b}{2a})^2 + \\frac{c}{a} - \\frac{b^2}{4a^2}}$ ，将 $\\frac{c}{a} - \\frac{b^2}{4a^2}$ 看作整体，变为 (1) ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":56,"id":"aeb6a299-9e1c-4383-9585-79f983525fc9","page_size":[595,841],"block_position":"56-1"},{"bbox":[54,96,581,134],"type":"text","angle":0,"index":3,"merge_prev":false,"text":"(5) 原式 = $\\frac{1}{2a}\\int\\frac{2ax}{ax^{2}+bx+c}=\\frac{1}{2a}\\int\\frac{2ax+b}{ax^{2}+bx+c}-\\frac{b}{ax^{2}+bx+c}dx=\\frac{1}{2a}\\left(\\int\\frac{\\mathrm{d}(ax^{2}+bx+c)}{ax^{2}+bx+c}-\\int\\frac{b}{ax^{2}+bx+c}\\mathrm{d}x\\right)$ ，变为 (4). ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":56,"id":"ba884457-2557-4630-a211-d1e1f9778a52","page_size":[595,841],"block_position":"56-2"},{"bbox":[35,152,258,163],"type":"text","angle":0,"index":4,"merge_prev":false,"text":"练习 5.13 次方和差 尽量记住下面的分解方法: ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":56,"id":"9b199cfe-cd63-4209-bd72-30cf3733b086","page_size":[595,841],"block_position":"56-3"},{"bbox":[73,163,185,186],"type":"text","angle":0,"index":5,"merge_prev":false,"text":"(1) $\\int \\frac{\\mathrm{d}x}{x^3 + 1}$ (2) $\\int \\frac{\\mathrm{d}x}{x^3 - 1}$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":56,"id":"f0054d96-f530-4132-8127-9906c33aff96","page_size":[595,841],"block_position":"56-4"},{"bbox":[73,185,286,210],"type":"text","angle":0,"index":6,"merge_prev":false,"text":"(3) $\\int\\frac{dx}{x^{4}-1}$ (4)重点： $\\int\\frac{dx}{x^{4}+1}$ (一定要记住！) ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":56,"id":"74695e56-014c-419d-9ec4-f329710a3d13","page_size":[595,841],"block_position":"56-5"},{"bbox":[74,209,186,233],"type":"text","angle":0,"index":7,"merge_prev":false,"text":"(5) $\\int\\frac{dx}{x^{6}-1}$ (6) $\\int\\frac{dx}{x^{6}+1}$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":56,"id":"abf0e024-ef78-488b-85d3-afb72f112a22","page_size":[595,841],"block_position":"56-6"},{"bbox":[53,230,359,257],"type":"text","angle":0,"index":8,"merge_prev":false,"text":"解(1)写为 $\\frac{1}{x^3 + 1} = \\frac{1}{3(x + 1)} + \\frac{-x + 2}{3(x^2 - x + 1)}$ ，后者给分母配方即可 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":56,"id":"28d49afb-c7a7-4d22-a1a9-ed3d9358bb32","page_size":[595,841],"block_position":"56-7"},{"bbox":[73,255,150,267],"type":"text","angle":0,"index":9,"merge_prev":false,"text":"(2) 用立方差即可 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":56,"id":"d5bfe82a-55c3-4381-bd79-a05d644564d4","page_size":[595,841],"block_position":"56-8"},{"bbox":[73,266,242,290],"type":"text","angle":0,"index":10,"merge_prev":false,"text":"(3) 用 $\\frac{1}{x^4 - 1} = \\frac{1}{2}\\left(\\frac{1}{x^2 - 1} - \\frac{1}{x^2 + 1}\\right)$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":56,"id":"e7864c21-8827-47a0-9e0a-20552808eedf","page_size":[595,841],"block_position":"56-9"},{"bbox":[73,291,289,317],"type":"text","angle":0,"index":11,"merge_prev":false,"text":"(4) 构造 $M = \\int \\frac{x^2 - 1}{x^4 + 1} \\, \\mathrm{d}x, N = \\int \\frac{x^2 + 1}{x^4 + 1}$ ，得到 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":56,"id":"3dc9e0ea-5c41-48fb-a84e-1da584120e18","page_size":[595,841],"block_position":"56-10"},{"bbox":[171,319,411,375],"type":"interline_equation","angle":0,"index":12,"text":"$$\n\\left\\{ \\begin{array}{l} M = \\int \\frac {1 - \\frac {1}{x ^ {2}}}{x ^ {2} + \\frac {1}{x ^ {2}}} = \\int \\frac {\\mathrm{d} (x + \\frac {1}{x})}{(x + \\frac {1}{x}) ^ {2} - 2} \\\\ N = \\int \\frac {x ^ {2} + 1}{x ^ {4} + 1} \\mathrm{d} x = \\int \\frac {1 + \\frac {1}{x ^ {2}}}{x ^ {2} + \\frac {1}{x ^ {2}}} \\mathrm{d} x = \\int \\frac {\\mathrm{d} (x - \\frac {1}{x})}{(x - \\frac {1}{x}) ^ {2} + 2} \\end{array} \\right.\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":56,"id":"6e11cc86-fa52-4d7e-a125-a7fef69debf2","page_size":[595,841],"block_position":"56-11"},{"bbox":[54,377,157,401],"type":"text","angle":0,"index":13,"merge_prev":false,"text":"因此所求即 $\\frac{1}{2} (N - M)$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":56,"id":"4024d0eb-ebee-41b3-ab6f-f6e8842a02d8","page_size":[595,841],"block_position":"56-12"},{"bbox":[73,396,406,422],"type":"text","angle":0,"index":14,"merge_prev":false,"text":"(5) 用平方差 (不要用立方差): $\\frac{1}{x^6 - 1} = \\frac{1}{2}\\left(\\frac{1}{x^3 - 1} + \\frac{1}{x^3 + 1}\\right)$ , 转换为 (1)(2) ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":56,"id":"6b33ad6f-65dc-4725-b63c-b0bf81d4afd2","page_size":[595,841],"block_position":"56-13"},{"bbox":[73,421,212,432],"type":"text","angle":0,"index":15,"merge_prev":false,"text":"(6) 类似四次方的讨论 + 立方和 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":56,"id":"03faf74f-efa0-4726-8f7c-eb7900bdb6d6","page_size":[595,841],"block_position":"56-14"},{"bbox":[159,435,434,546],"type":"interline_equation","angle":0,"index":16,"text":"$$\n\\begin{array}{l} I = \\frac {1}{2} \\int \\frac {x ^ {4} + 1}{x ^ {6} + 1} \\mathrm{d} x - \\frac {1}{2} \\int \\frac {x ^ {4} - 1}{x ^ {6} + 1} \\mathrm{d} x \\\\ = \\frac {1}{2} \\int \\frac {x ^ {2} + (x ^ {4} - x ^ {2} + 1)}{x ^ {6} + 1} \\mathrm{d} x - \\frac {1}{2} \\int \\frac {(x ^ {2} - 1) (x ^ {2} + 1)}{(x ^ {2} + 1) (x ^ {4} - x ^ {2} + 1)} \\mathrm{d} x \\\\ = \\frac {1}{2} \\int \\frac {x ^ {2}}{x ^ {6} + 1} \\mathrm{d} x + \\frac {1}{2} \\int \\frac {\\mathrm{d} x}{x ^ {2} + 1} - \\frac {1}{2} \\int \\frac {x ^ {2} - 1}{x ^ {4} - x ^ {2} + 1} \\mathrm{d} x \\\\ = \\frac {1}{6} \\int \\frac {\\mathrm{d} (x ^ {3})}{(x ^ {3}) ^ {2} + 1} + \\frac {1}{2} \\arctan x - \\frac {1}{2} \\int \\frac {\\mathrm{d} (x + \\frac {1}{x})}{(x + \\frac {1}{x}) ^ {2} + 3} \\mathrm{d} x \\end{array}\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":56,"id":"de2067cd-45e0-4098-a4d5-a8e00a5f2e11","page_size":[595,841],"block_position":"56-15"},{"bbox":[35,553,333,581],"type":"text","angle":0,"index":17,"merge_prev":false,"text":"练习5.14几个经典题 (1) $\\int \\frac{x^2}{(1 - x)^{100}}\\mathrm{d}x$ (2) $\\int \\frac{\\mathrm{d}x}{(x + a)^2(x + b)^2}$ 解(1)显然待定系数法有理分解不可行,因此用技巧 $x^{2}=[(1-x)-1]^{2}$ 得到 $\\frac{x^{2}}{(1-x)^{100}}=\\frac{(1-x)^{2}-2(1-x)+1}{(1-x)^{100}}$ ,做换元即可 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":56,"id":"c8488733-2c81-42aa-953c-562a2f28ba9f","page_size":[595,841],"block_position":"56-16"},{"bbox":[52,580,539,615],"type":"text","angle":0,"index":18,"merge_prev":false,"lines_deleted":true,"text":"","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":56,"id":"c8488733-2c81-42aa-953c-562a2f28ba9f","page_size":[595,841],"block_position":"56-17"},{"bbox":[73,619,255,631],"type":"text","angle":0,"index":19,"merge_prev":false,"text":"(2) 这种有理分解比较常见，最好记下来， ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":56,"id":"bc21f27b-7faf-4743-bdc4-37322886d534","page_size":[595,841],"block_position":"56-18"},{"bbox":[80,634,511,664],"type":"interline_equation","angle":0,"index":20,"text":"$$\n{\\frac {1}{(x + a) ^ {2} (x + b) ^ {2}}} = {\\frac {1}{(b - a) ^ {2}}} \\left({\\frac {1}{x + a}} - {\\frac {1}{x + b}}\\right) ^ {2} = {\\frac {1}{(b - a) ^ {2}}} \\left[ {\\frac {1}{(x + a) ^ {2}}} + {\\frac {1}{(x + b) ^ {2}}} - {\\frac {2}{(x + a) (x + b)}} \\right]\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":56,"id":"905c9353-303f-4720-b360-5c17ce1037c9","page_size":[595,841],"block_position":"56-19"},{"bbox":[55,678,139,692],"type":"title","angle":0,"index":21,"level":2,"text":"## 5.3.4 根式问题","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":56,"id":"bd14dc15-d0ca-44c4-817f-b98046dfaa47","page_size":[595,841],"block_position":"56-20"},{"bbox":[35,699,539,751],"type":"text","angle":0,"index":22,"merge_prev":false,"text":"练习5.15三个基础公式的证明(1) $\\int \\sqrt{a^2 - x^2}\\mathrm{d}x$ (2)重点： $\\int \\sqrt{x^2 + a^2}\\mathrm{d}x$ (3)重点： $\\int \\sqrt{x^2 - a^2}\\mathrm{d}x$ 解(1)设 $x = a\\sin t$ ，则 $I = \\int a\\cos t\\cdot a\\cos t\\mathrm{d}t = \\frac{a^2}{2}\\int (1 + \\cos 2t)\\mathrm{d}t = \\frac{a^2}{2} (t + \\sin t\\cos t) + C$ 。画一个三角形做 辅助，角 $t$ 对着 $x$ ，斜边为 $a$ ，因此 $\\sin t = \\frac{x}{a},\\cos t = \\frac{\\sqrt{a^2 - x^2}}{a}$ ，带回得到： ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":56,"id":"89ba9e4f-4c7e-4395-95c3-88e85c3a8bef","page_size":[595,841],"block_position":"56-21"},{"bbox":[291,802,302,811],"type":"page_number","angle":0,"index":999999,"text":"49 ","color":{"line":"rgba(164, 164, 164, 1)","fill":"rgba(164, 164, 164, 1)"},"page_idx":56,"is_discarded":true,"id":"7fcc3712-fd62-4927-91d8-da38f513107a","page_size":[595,841],"block_position":"56-22"}],[{"bbox":[449,47,539,58],"type":"header","angle":0,"index":-99999,"text":"5.3 不定积分表选例 ","color":{"line":"rgba(164, 164, 164, 1)","fill":"rgba(164, 164, 164, 1)"},"page_idx":57,"is_discarded":true,"id":"da5f947d-9ba6-402c-adcb-724708096e53","page_size":[595,841],"block_position":"57-0"},{"bbox":[54,70,386,96],"type":"text","angle":0,"index":1,"merge_prev":false,"lines_deleted":true,"text":"","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":57,"id":"89ba9e4f-4c7e-4395-95c3-88e85c3a8bef","page_size":[595,841],"block_position":"57-1"},{"bbox":[214,98,379,120],"type":"interline_equation","angle":0,"index":2,"text":"$$\nI = \\frac {1}{2} (x \\sqrt {a ^ {2} - x ^ {2}} + a ^ {2} \\arcsin {\\frac {x}{a}}) + C\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":57,"id":"8344051d-f870-44e9-8e65-0da8669d6631","page_size":[595,841],"block_position":"57-2"},{"bbox":[53,121,540,159],"type":"text","angle":0,"index":3,"merge_prev":false,"text":"(2) $x = a\\tan t$ ， $I = \\int a^2\\sec^3 x\\mathrm{d}t = \\frac{a^2}{2} (\\sec x\\tan x) + \\frac{a^2}{2}\\ln |\\tan x + \\sec x|$ （一定要记住 $\\sec x,\\sec^3 x$ 的积分，如果一次方都没记住，就死翘翘了。。），因此得到结果为 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":57,"id":"a699ec7a-b87c-42b0-809f-762563bf5153","page_size":[595,841],"block_position":"57-3"},{"bbox":[203,161,390,184],"type":"interline_equation","angle":0,"index":4,"text":"$$\n\\frac {1}{2} (x \\sqrt {x ^ {2} + a ^ {2}} + a ^ {2} \\ln | x + \\sqrt {x ^ {2} + a ^ {2}} |) + C\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":57,"id":"57d10bed-bc79-4300-96cd-0afad5294cfc","page_size":[595,841],"block_position":"57-4"},{"bbox":[73,185,497,211],"type":"text","angle":0,"index":5,"merge_prev":false,"text":"(3) 设 $x = a \\sec t$ ，得到 $I = a^2 \\int \\tan^2 t \\sec t \\, \\mathrm{d}t = a^2 \\int \\sec^3 t - \\sec t \\, \\mathrm{d}t$ ，背公式以及画三角形得到： ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":57,"id":"0d3a8202-99a3-4f43-b5f8-a16cfe4d5ec0","page_size":[595,841],"block_position":"57-5"},{"bbox":[107,212,486,238],"type":"interline_equation","angle":0,"index":6,"text":"$$\nI = \\frac {a ^ {2}}{2} (\\sec t \\tan t - \\ln | \\tan t + \\sec t |) + C = \\frac {1}{2} (x \\sqrt {x ^ {2} - a ^ {2}} - a ^ {2} \\ln | x + \\sqrt {x ^ {2} - a ^ {2}} |) + C\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":57,"id":"46134a42-2ab3-449c-a943-beeb0edb80c6","page_size":[595,841],"block_position":"57-6"},{"bbox":[35,248,454,275],"type":"text","angle":0,"index":7,"merge_prev":false,"text":"练习5.16 $(x - a)(b - x)$ 型(1) $\\int \\frac{\\mathrm{d}x}{\\sqrt{x(1 - x)}}$ (2) $\\int \\frac{\\mathrm{d}x}{\\sqrt{(x - a)(b - x)}}$ (3) $\\int \\sqrt{(x - a)(b - x)}\\mathrm{d}x$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":57,"id":"198c1791-ba9b-4534-b1d2-bdd26138b5ae","page_size":[595,841],"block_position":"57-7"},{"bbox":[52,275,447,299],"type":"text","angle":0,"index":9,"merge_prev":false,"text":"解(1)设 $x = \\sin^2 t$ ，则 $I = \\int \\frac{2\\sin t\\cos t}{\\sin t\\cos t}\\mathrm{d}t = 2t + C$ ，直接带回得到 $I = 2\\arcsin \\sqrt{x} +C$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":57,"id":"1edb5fec-a0a7-43df-9302-f6cc9ae78802","page_size":[595,841],"block_position":"57-8"},{"bbox":[72,299,395,324],"type":"text","angle":0,"index":10,"merge_prev":false,"text":"(2) 令 $t = x - a$ ，则变为 $\\int \\frac{\\mathrm{d}t}{\\sqrt{t(b - a - t)}}$ ，令 $t = (b - a)\\sin^2 s$ 即可，故 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":57,"id":"a6f15b4f-79f4-4606-934c-d29208e96141","page_size":[595,841],"block_position":"57-9"},{"bbox":[127,329,465,358],"type":"interline_equation","angle":0,"index":11,"text":"$$\nI = \\int {\\frac {2 (b - a) \\sin s \\cos s \\mathrm{d} s}{(b - a) \\sin s \\cos s}} = 2 s = 2 \\arcsin {\\sqrt {\\frac {t}{(b - a)}}} = 2 \\arcsin {\\sqrt {\\frac {x - a}{b - a}}} + C\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":57,"id":"12813d3d-6972-4c37-8dda-dd98f38f7977","page_size":[595,841],"block_position":"57-10"},{"bbox":[73,362,423,389],"type":"text","angle":0,"index":12,"merge_prev":false,"text":"(3) 同理得到 $I = \\int 2(b - a)^2 \\sin^2 s \\cos^2 s \\, \\mathrm{d}s = \\frac{(b - a)^2}{4} \\int (1 - \\cos 4t) \\, \\mathrm{d}t$ ，后续略 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":57,"id":"9ec47fb3-e7f5-4542-99d5-69d53fba50fb","page_size":[595,841],"block_position":"57-11"},{"bbox":[35,394,370,425],"type":"text","angle":0,"index":13,"merge_prev":false,"text":"练习5.17其他特别经典的根式积分问题(1) $\\int \\left(\\sqrt{\\frac{1 + x}{1 - x}} +\\sqrt{\\frac{1 - x}{1 + x}}\\right)\\mathrm{d}x$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":57,"id":"37b82843-6896-47b7-a45d-eee1976ca8ca","page_size":[595,841],"block_position":"57-12"},{"bbox":[88,424,171,450],"type":"interline_equation","angle":0,"index":14,"text":"$$\n\\int \\frac {\\mathrm{d} x}{(\\sqrt {x} + 1) (x + 3)}\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":57,"id":"b84500e5-ae90-4ccf-8512-47b1482160a7","page_size":[595,841],"block_position":"57-13"},{"bbox":[74,449,170,479],"type":"text","angle":0,"index":15,"merge_prev":false,"text":"(3) $\\int \\frac{\\mathrm{d}x}{x\\sqrt{x^2 - 2x - 3}}$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":57,"id":"7baf4ed0-0255-4b00-87c5-0490e8f07efa","page_size":[595,841],"block_position":"57-14"},{"bbox":[88,478,180,497],"type":"interline_equation","angle":0,"index":16,"text":"$$\nI = \\frac {\\mathrm{d} x}{x + \\sqrt {x ^ {2} - x + 1}}\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":57,"id":"a1549161-ecae-4949-801b-b03978b11e9f","page_size":[595,841],"block_position":"57-15"},{"bbox":[52,494,421,522],"type":"text","angle":0,"index":17,"merge_prev":false,"text":"解(1)让分母一样： $\\frac{1 + x}{\\sqrt{1 - x^2}} +\\frac{1 - x}{\\sqrt{1 - x^2}} = \\frac{2}{\\sqrt{1 - x^2}}$ ，直接转换为了 $\\arcsin x$ 的积分 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":57,"id":"b09072d6-eaa2-414e-8031-6cd4d6542d3d","page_size":[595,841],"block_position":"57-16"},{"bbox":[54,518,537,581],"type":"text","angle":0,"index":18,"merge_prev":false,"text":"(2) 令 $t = \\sqrt{x}$ ，则 $I = \\int \\frac{2t}{(t + 1)(t^2 + 3)}\\mathrm{d}t = \\int \\frac{A}{t + 1}\\mathrm{d}t + \\int \\frac{Bt + C}{t^2 + 3}\\mathrm{d}t$ ，解出 $A = -\\frac{1}{2}, B = \\frac{1}{2}, C = \\frac{3}{2}$ ，从而 $I = -\\frac{1}{2}\\ln |t + 1| + \\int \\frac{1}{4}\\ln |t^2 + 3| + \\frac{\\sqrt{3}}{2}\\arctan \\frac{t}{\\sqrt{3}} + C$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":57,"id":"83596462-8a01-4670-8a86-51ae4c34f674","page_size":[595,841],"block_position":"57-17"},{"bbox":[53,585,539,622],"type":"text","angle":0,"index":19,"merge_prev":false,"text":"(3.1) 方法 1: 使用 Euler 变换。设 $\\sqrt{x^2 - 2x + 3} = x - t$ ，则得到 $x = \\frac{t^2 + 3}{2(t - 1)}$ ，进而 $\\mathrm{d}x = \\frac{t^2 - 2t + 3}{2(t - 1)^2}\\mathrm{d}t$ ，此时积分变为 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":57,"id":"4925c565-2dde-4d0b-bbf0-6d4a3b15db7d","page_size":[595,841],"block_position":"57-18"},{"bbox":[204,619,389,646],"type":"interline_equation","angle":0,"index":20,"text":"$$\nI = - \\int {\\frac {2}{t ^ {2} + 3}} \\mathrm{d} t = - \\frac {2}{\\sqrt {3}} \\arctan {\\frac {t}{\\sqrt {3}}} + C\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":57,"id":"54521efd-d5b6-4468-8e04-eeb1a7049b80","page_size":[595,841],"block_position":"57-19"},{"bbox":[54,653,539,690],"type":"text","angle":0,"index":21,"merge_prev":false,"text":"(3.2) 方法 2: 配方 + 三角函数 (一般还是推荐三角函数)。 $I = \\int \\frac{\\mathrm{d}x}{x[(x-1)^{2}-4]^{\\frac{1}{2}}}$ ，做换元 $x - 1 = 2 \\sec t$ ，因此 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":57,"id":"88c7b9e0-c4bd-4011-8338-26c4c40e7ef1","page_size":[595,841],"block_position":"57-20"},{"bbox":[205,687,388,714],"type":"interline_equation","angle":0,"index":22,"text":"$$\nI = \\int {\\frac {2 \\sec {t} \\tan {t}}{(2 \\sec {t} + 1) 2 \\tan {t}}} \\mathrm{d} t = \\int {\\frac {\\mathrm{d} t}{2 + \\cos {t}}}\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":57,"id":"051a0279-9130-400e-b08c-9cc1b4cef742","page_size":[595,841],"block_position":"57-21"},{"bbox":[53,721,540,764],"type":"text","angle":0,"index":23,"merge_prev":false,"text":"背公式即 $\\frac{2}{\\sqrt{1 - a^2}}\\arctan \\left(\\sqrt{\\frac{1 - a}{1 + a}}\\tan \\frac{x}{2}\\right) + C$ ，代入 $a = \\frac{1}{2}$ 。注意带回时需要用半角公式， $\\tan \\frac{t}{2} = \\frac{\\sin t}{1 + \\cos t}$ 再画个三角形进行辅助。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":57,"id":"42aedff7-74fe-4601-b9ae-6a112ce1c492","page_size":[595,841],"block_position":"57-22"},{"bbox":[291,802,303,812],"type":"page_number","angle":0,"index":1000007,"text":"50 ","color":{"line":"rgba(164, 164, 164, 1)","fill":"rgba(164, 164, 164, 1)"},"page_idx":57,"is_discarded":true,"id":"918af4e1-a392-4bcd-9ab1-9cd3cbca2d97","page_size":[595,841],"block_position":"57-23"}],[{"bbox":[449,47,539,58],"type":"header","angle":0,"index":-99999,"text":"5.3 不定积分表选例 ","color":{"line":"rgba(164, 164, 164, 1)","fill":"rgba(164, 164, 164, 1)"},"page_idx":58,"is_discarded":true,"id":"60de0683-1ec2-4b38-863e-86779ee0dfc4","page_size":[595,841],"block_position":"58-0"},{"bbox":[73,72,252,84],"type":"text","angle":0,"index":1,"merge_prev":false,"text":"(4) 可以用 Euler 变换或者配方三角替换。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":58,"id":"a45b5ea1-d827-446e-bd72-ac1cd35562dd","page_size":[595,841],"block_position":"58-1"},{"bbox":[55,102,151,116],"type":"title","angle":0,"index":2,"level":2,"text":"## 5.3.5 配对积分法","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":58,"id":"52cae724-dab2-48b3-9f53-bf44273f973b","page_size":[595,841],"block_position":"58-2"},{"bbox":[35,124,539,170],"type":"text","angle":0,"index":3,"merge_prev":false,"text":"练习5.18经典配对积分法 (1) 计算 $I_{1} = \\int e^{ax}\\cos bx\\mathrm{d}x,I_{2} = \\int e^{ax}\\sin bx\\mathrm{d}x$ (2) 计算 $I_{1} = \\sin (\\ln x)\\mathrm{d}x,I_{2} = \\int \\cos (\\ln x)\\mathrm{d}x$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":58,"id":"38e356a2-51e5-4de1-b964-b686e34a0c53","page_size":[595,841],"block_position":"58-3"},{"bbox":[52,171,474,195],"type":"text","angle":0,"index":4,"merge_prev":false,"text":"解(1)用分部积分得到 $I_{1} = \\frac{1}{a}\\left(e^{ax}\\cos bx + bI_{2}\\right)$ ，同理 $I_{2} = \\frac{1}{a}\\left(e^{ax}\\sin x - bI_{1}\\right)$ ，列出线性方程组： ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":58,"id":"5af1c015-39b6-4d9f-bf2b-39d8933d1661","page_size":[595,841],"block_position":"58-4"},{"bbox":[142,197,442,240],"type":"interline_equation","angle":0,"index":5,"text":"$$\n\\left\\{ \\begin{array}{l} a I _ {1} - b I _ {2} = e ^ {a x} \\cos b x \\\\ b I _ {1} + a I _ {2} = e ^ {a x} \\sin b x \\end{array} \\right. \\Rightarrow \\left\\{ \\begin{array}{l} I _ {1} = \\frac {e ^ {a x}}{a ^ {2} + b ^ {2}} (a \\cos b x + b \\sin b x) + C \\\\ I _ {2} = \\frac {e ^ {a x}}{a ^ {2} + b ^ {2}} (a \\sin b x - b \\cos b x) + C \\end{array} \\right.\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":58,"id":"c8dd4a83-5c8c-4328-8537-ee79f98037a9","page_size":[595,841],"block_position":"58-5"},{"bbox":[52,243,539,291],"type":"text","angle":0,"index":6,"merge_prev":false,"text":"(2) 直接用分部积分： $\\int \\sin (\\ln x)\\mathrm{d}x = x\\sin (\\ln x) - \\int \\cos (\\ln x)\\mathrm{d}x$ ，同理 $\\int \\cos (\\ln x)\\mathrm{d}x = x\\cos (\\ln x) + \\int \\sin (\\ln x)\\mathrm{d}x$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":58,"id":"5ebfc3e4-4d43-42cf-959f-299422b4f357","page_size":[595,841],"block_position":"58-6"},{"bbox":[55,306,139,321],"type":"title","angle":0,"index":7,"level":2,"text":"## 5.3.6 递推问题","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":58,"id":"a9f2e0fd-a80f-4d1a-b939-4aec46595d77","page_size":[595,841],"block_position":"58-7"},{"bbox":[35,329,446,356],"type":"text","angle":0,"index":8,"merge_prev":false,"text":"练习 5.19 几个最常见递推 (1) $I_{n}=\\int\\frac{\\sin nx}{\\sin x}dx$ (2) $I_{n}=\\int\\sec^{n}x\\mathrm{d}x$ (3) $I_{n}=\\int e^{ax}\\sin^{n}x\\mathrm{d}x$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":58,"id":"90879e42-9742-4e49-8a64-62a3681158b9","page_size":[595,841],"block_position":"58-8"},{"bbox":[52,353,218,365],"type":"text","angle":0,"index":9,"merge_prev":false,"text":"解(1)首先根据和差公式和积化和差: ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":58,"id":"f73ca542-f454-4eee-a2b0-2d721b840971","page_size":[595,841],"block_position":"58-9"},{"bbox":[55,367,536,392],"type":"interline_equation","angle":0,"index":10,"text":"$$\n\\sin n x = \\sin [ (n - 1) x + x ] = \\sin (n - 1) x \\cos x + \\cos (n - 1) x \\sin x = \\frac {1}{2} [ \\sin n x + \\sin (n - 2) x ] + \\cos (n - 1) x \\sin x\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":58,"id":"e97363e1-db51-46db-a7cf-3715986f0986","page_size":[595,841],"block_position":"58-10"},{"bbox":[55,392,77,404],"type":"text","angle":0,"index":11,"merge_prev":false,"text":"因此 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":58,"id":"4f7a58d8-483e-43ef-be35-4bc9250a5cad","page_size":[595,841],"block_position":"58-11"},{"bbox":[136,401,456,426],"type":"interline_equation","angle":0,"index":12,"text":"$$\nI _ {n} = \\frac {1}{2} (I _ {n} + I _ {n - 2}) + \\frac {1}{n - 1} \\sin (n - 1) x \\Rightarrow I _ {n} = \\frac {2}{n - 1} \\sin (n - 1) x + I _ {n - 2}\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":58,"id":"912d5c04-75cc-4050-a5b0-af28c578e3af","page_size":[595,841],"block_position":"58-12"},{"bbox":[73,432,229,444],"type":"text","angle":0,"index":13,"merge_prev":false,"text":"(2) 先凑出 $\\mathrm{d}\\tan x$ 并使用分部积分: ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":58,"id":"26400235-792b-4251-ac91-256f147ba580","page_size":[595,841],"block_position":"58-13"},{"bbox":[120,446,473,471],"type":"interline_equation","angle":0,"index":14,"text":"$$\n\\int \\sec^ {n} x \\mathrm{d} x = \\int \\sec^ {n - 2} x \\mathrm{d} (\\tan x) = \\tan x \\sec^ {n - 2} x - (n - 2) \\int \\tan^ {2} x \\sec^ {n - 2} x \\mathrm{d} x\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":58,"id":"59073eef-2da8-4c12-aac7-83d1d004b9be","page_size":[595,841],"block_position":"58-14"},{"bbox":[55,473,175,486],"type":"text","angle":0,"index":15,"merge_prev":false,"text":"再代入 $\\tan^2 x = \\sec^2 x - 1$ : ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":58,"id":"dab1a969-80bf-47f2-a5d7-928e62e5b9b2","page_size":[595,841],"block_position":"58-15"},{"bbox":[127,488,465,513],"type":"interline_equation","angle":0,"index":16,"text":"$$\nI _ {n} = \\tan x \\sec^ {n - 2} x - (n - 2) (I _ {n} - I _ {n - 2}) \\Rightarrow I _ {n} = \\frac {\\tan x \\sec^ {n - 2} x}{n - 1} + \\frac {n - 2}{n - 1} I _ {n - 2}\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":58,"id":"8be9eb6b-eb09-482e-9b00-83bb4f94050b","page_size":[595,841],"block_position":"58-16"},{"bbox":[33,516,539,560],"type":"text","angle":0,"index":17,"merge_prev":false,"text":"笔记 $\\int^{\\text{(3)}}\\sec x\\mathrm{d}x = \\ln |\\tan x + \\sec x|$ 和 $\\int \\sec^3 x\\mathrm{d}x = \\frac{1}{2}\\tan x\\sec x + \\frac{1}{2}\\ln |\\tan x + \\sec x|$ 在根式中经常用到，要记住，具体证明过程见前面部分。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":58,"id":"c610ef47-86a6-4ef6-919e-436e72803e03","page_size":[595,841],"block_position":"58-17"},{"bbox":[291,803,302,812],"type":"page_number","angle":0,"index":1000017,"text":"51 ","color":{"line":"rgba(164, 164, 164, 1)","fill":"rgba(164, 164, 164, 1)"},"page_idx":58,"is_discarded":true,"id":"9456859d-81c8-4702-ac24-cd210c6884dd","page_size":[595,841],"block_position":"58-18"}],[{"bbox":[244,74,359,94],"type":"title","angle":0,"index":0,"level":2,"text":"## 第6章 定积分","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":59,"id":"606cff43-6158-444d-995f-418e2255a181","page_size":[595,841],"block_position":"59-0"},{"bbox":[55,122,158,139],"type":"title","angle":0,"index":1,"level":2,"text":"## 6.1 定积分理论","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":59,"id":"0ddf66b0-c89b-4c35-887b-8221bae1fcf8","page_size":[595,841],"block_position":"59-1"},{"bbox":[55,155,127,169],"type":"title","angle":0,"index":2,"level":2,"text":"## 6.1.1 可积性","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":59,"id":"9f82512b-612a-4066-959a-164c7226d3ec","page_size":[595,841],"block_position":"59-2"},{"bbox":[73,183,183,195],"type":"title","angle":0,"index":3,"level":2,"text":"## 定义 6.1 (Riemann 积分)","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":59,"id":"8e459f0e-2b11-4a77-9ace-4aa6c2cdbc47","page_size":[595,841],"block_position":"59-3"},{"bbox":[67,199,525,258],"type":"text","angle":0,"index":4,"merge_prev":false,"text":"$f(x)$ 是 $[a,b]$ 上的函数，将 $[a,b]$ 划分为 $n$ 等份 $T$ ，若 $\\lim_{||T||\\to 0}\\sum_{i = 1}^{n}f(\\xi_i)\\Delta x_i = J$ ，这里 $J$ 为常数，则称 $f(x)$ 在 $[a,b]$ 可积，且记 $\\int_{a}^{b}f(x)\\mathrm{d}x = J$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":59,"id":"d451b34e-a3bc-4960-9232-1c472f4cb565","page_size":[595,841],"block_position":"59-4"},{"bbox":[73,278,192,290],"type":"title","angle":0,"index":5,"level":2,"text":"## 定义 6.2 (Darboux 上下和)","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":59,"id":"9d68946d-560d-487a-b7bc-e00694c56f06","page_size":[595,841],"block_position":"59-5"},{"bbox":[68,294,493,315],"type":"text","angle":0,"index":6,"merge_prev":false,"text":"在每个区间上取 $M_{i} = \\sup_{x\\in \\Delta_{i}}f(x),m_{i} = \\inf_{x\\in \\Delta_{i}}f(x)$ ，记振幅 $\\omega_{i} = M_{i} - m_{i}$ ，定义Darboux上下和为 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":59,"id":"a14fec06-ec62-4767-8257-202e723a61ee","page_size":[595,841],"block_position":"59-6"},{"bbox":[212,318,381,350],"type":"interline_equation","angle":0,"index":7,"text":"$$\nS (T) = \\sum_ {i = 1} ^ {n} M _ {i} \\Delta x _ {i}, s (T) = \\sum_ {i = 1} ^ {n} m _ {i} \\Delta x _ {i}\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":59,"id":"7bd0135d-aed1-42a7-9dc7-e3f6d143edde","page_size":[595,841],"block_position":"59-7"},{"bbox":[69,352,215,365],"type":"text","angle":0,"index":8,"merge_prev":false,"text":"定义 $f(x)$ 在 $[a,b]$ 的上下积分为： ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":59,"id":"5c22fcb1-f2ff-4f0e-8c0d-74755137571e","page_size":[595,841],"block_position":"59-8"},{"bbox":[237,367,355,386],"type":"interline_equation","angle":0,"index":9,"text":"$$\nS = \\inf _ {T} S (T), s = \\sup _ {T} s (T)\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":59,"id":"fc2c993a-0286-4544-acbc-790171778692","page_size":[595,841],"block_position":"59-9"},{"bbox":[73,409,190,422],"type":"title","angle":0,"index":10,"level":2,"text":"## 定理 6.1 (可积函数的性质)","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":59,"id":"781a1fff-497a-4509-b02d-369f26f7c769","page_size":[595,841],"block_position":"59-10"},{"bbox":[69,426,195,439],"type":"text","angle":0,"index":11,"merge_prev":false,"text":"若 $f(x)$ 在 $[a,b]$ 上可积，则： ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":59,"id":"3eb59210-fa79-4cf5-86b1-5931972c8870","page_size":[595,841],"block_position":"59-11"},{"bbox":[82,441,452,454],"type":"text","angle":0,"index":12,"merge_prev":false,"text":"- 有界性： $f(x)$ 在 $[a, b]$ 上有界 (反之不正确，有界不一定可积，例如 Dirichlet 函数) ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":59,"id":"0cde0f72-17c8-4535-9d93-a482935077e2","page_size":[595,841],"block_position":"59-12"},{"bbox":[53,470,508,484],"type":"text","angle":0,"index":13,"merge_prev":false,"text":"证明 反设 $f$ 无界，则存在某个小区间 $\\Delta_k$ ，使得 $\\Delta_k$ 上 $f$ 无界，假设其他区间都有界，此时 Riemann 和 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":59,"id":"d08eb075-ed37-42b2-9ebb-da19f9c5738d","page_size":[595,841],"block_position":"59-13"},{"bbox":[195,486,398,518],"type":"interline_equation","angle":0,"index":14,"text":"$$\n\\left| \\sum_ {i = 1} ^ {n} f \\left(\\xi_ {i}\\right) \\Delta x _ {i} \\right| \\geq \\left| f \\left(\\xi_ {k}\\right) \\Delta x _ {k} \\right| - \\left| \\sum_ {i \\neq k} f \\left(\\xi_ {i}\\right) \\Delta x _ {i} \\right|\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":59,"id":"292f30e1-aec7-4d0a-af7c-bda6bb7366bd","page_size":[595,841],"block_position":"59-14"},{"bbox":[55,522,370,534],"type":"text","angle":0,"index":15,"merge_prev":false,"text":"对 $\\forall M > 0$ ，无论如何分割，Riemann和都能大于 $M$ ，因此无界不可积。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":59,"id":"2e176a80-a70e-4c8c-9465-77edb6ef8ea9","page_size":[595,841],"block_position":"59-15"},{"bbox":[73,543,190,555],"type":"title","angle":0,"index":16,"level":2,"text":"## 定理 6.2 (可积性充要条件)","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":59,"id":"01672ffc-ef3d-4084-93be-a313ff6ebf50","page_size":[595,841],"block_position":"59-16"},{"bbox":[68,560,188,572],"type":"text","angle":0,"index":17,"merge_prev":false,"text":"$f(x)$ 在 $[a,b]$ 可积当且仅当 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":59,"id":"db9848b8-e275-464c-85b3-26b2e6e4bdfa","page_size":[595,841],"block_position":"59-17"},{"bbox":[83,575,332,587],"type":"text","angle":0,"index":18,"merge_prev":false,"text":"- 上下积分角度： $f(x)$ 在 $[a, b]$ 的上积分 $s$ 等于下积分 $S$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":59,"id":"284d7eba-950d-4f5e-8f12-4e71035cc764","page_size":[595,841],"block_position":"59-18"},{"bbox":[83,588,427,613],"type":"text","angle":0,"index":19,"merge_prev":false,"text":"- 上下和角度 (最常用！): $\\forall \\epsilon$ ，存在分割 $T$ 使得 $S(T) - s(T) = \\sum_{T} \\omega_i \\Delta x_i < \\epsilon$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":59,"id":"5f455afc-fc49-4706-843b-36b7ff87568d","page_size":[595,841],"block_position":"59-19"},{"bbox":[83,614,465,629],"type":"text","angle":0,"index":20,"merge_prev":false,"text":"- 振幅角度：对 $\\forall \\epsilon, \\eta$ ，总存在分割 $T$ ，使得振幅 $\\omega_{k'} \\geq \\epsilon$ 的那些小区间总长 $\\sum \\omega_{k'} < \\eta$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":59,"id":"43a178e2-c130-4ea9-91d1-3a74ad9a5ad2","page_size":[595,841],"block_position":"59-20"},{"bbox":[83,629,263,644],"type":"text","angle":0,"index":21,"merge_prev":false,"text":"- 实变角度： $f(x)$ 在 $[a, b]$ 几乎处处连续 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":59,"id":"0f2cbf8e-f1ba-4816-bdc7-a46b107bf9a1","page_size":[595,841],"block_position":"59-21"},{"bbox":[35,676,352,688],"type":"text","angle":0,"index":22,"merge_prev":false,"text":"练习6.1Dirichlet函数的不可积性证明Dirichlet函数在 $[a,b]$ 不可积 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":59,"id":"7935265d-885d-4e4a-a570-8b834ac5e802","page_size":[595,841],"block_position":"59-22"},{"bbox":[52,692,441,719],"type":"text","angle":0,"index":23,"merge_prev":false,"text":"证明 由于任意分割 T，振幅 $\\omega = 1$ ，因此 $S(T) - s(T) = \\sum_{i=1}^{n} \\Delta x_i = b - a$ ，因此不可积。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":59,"id":"b3fb2855-fa29-4d99-85ba-285d316c5926","page_size":[595,841],"block_position":"59-23"}],[{"bbox":[469,47,539,58],"type":"header","angle":0,"index":-99999,"text":"6.1 定积分理论 ","color":{"line":"rgba(164, 164, 164, 1)","fill":"rgba(164, 164, 164, 1)"},"page_idx":60,"is_discarded":true,"id":"c3a180ab-7dea-4587-af32-26434ba4798c","page_size":[595,841],"block_position":"60-0"},{"bbox":[73,74,213,86],"type":"title","angle":0,"index":1,"level":2,"text":"## 定理 6.3 (Riemann 可积函数类)","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":60,"id":"24912369-d186-44a4-909b-a67f943c1894","page_size":[595,841],"block_position":"60-1"},{"bbox":[68,90,311,103],"type":"text","angle":0,"index":2,"merge_prev":false,"text":"$f(x)$ 定义于 $[a,b]$ ，常见的有三个Riemann可积函数类： ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":60,"id":"500d0f5e-ad7b-48ba-9e02-a9fac1f5a398","page_size":[595,841],"block_position":"60-2"},{"bbox":[83,106,159,119],"type":"text","angle":0,"index":3,"merge_prev":false,"text":"• $[a,b]$ 连续函数 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":60,"id":"a2266b3c-6f86-47c2-9d52-fd65205dfe73","page_size":[595,841],"block_position":"60-3"},{"bbox":[83,121,261,134],"type":"text","angle":0,"index":4,"merge_prev":false,"text":"- 在 $[a, b]$ 只有有限个间断点的有界函数 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":60,"id":"84f7928f-f600-4d47-8020-2dbf79d7b683","page_size":[595,841],"block_position":"60-4"},{"bbox":[83,137,169,149],"type":"text","angle":0,"index":5,"merge_prev":false,"text":"- $[a, b]$ 上单调函数 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":60,"id":"aae3b0f1-9f87-445f-aaf1-2df23cf86517","page_size":[595,841],"block_position":"60-5"},{"bbox":[54,165,379,179],"type":"text","angle":0,"index":6,"merge_prev":false,"text":"证明 (1) 根据一致连续性，每个 $\\omega_{i}$ 都可以足够小，因此 $\\sum \\omega_{i} \\Delta x_{i}$ 足够小 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":60,"id":"c0071b4f-91d2-456a-9d58-df854b4d3940","page_size":[595,841],"block_position":"60-6"},{"bbox":[73,183,323,195],"type":"text","angle":0,"index":7,"merge_prev":false,"text":"(2) 将间断点处的 $\\Delta x_{i}$ 设置得足够小，控制住间断点差距 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":60,"id":"fb828b52-d9f8-4302-aa8f-99015581e596","page_size":[595,841],"block_position":"60-7"},{"bbox":[73,198,284,211],"type":"text","angle":0,"index":8,"merge_prev":false,"text":"(3) $\\omega_{i}=f(x_{i})-f(x_{i-1})$ ，只需要 $\\omega_{i}$ 取得足够小 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":60,"id":"868cf663-2a26-4a82-aa0b-c94ab1fe3238","page_size":[595,841],"block_position":"60-8"},{"bbox":[35,222,415,249],"type":"text","angle":0,"index":9,"merge_prev":false,"text":"练习6.2可积性判断(1)证明Riemann函数 $R(x)$ 在 $[0,1]$ 可积，且 $\\int_0^1 R(x)\\mathrm{d}x = 0$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":60,"id":"aedb2a8e-5f0a-422c-ab02-cdb8aedbaa6f","page_size":[595,841],"block_position":"60-9"},{"bbox":[73,248,235,261],"type":"text","angle":0,"index":10,"merge_prev":false,"text":"(2) 证明下面的 $f(x)$ 在 $[0,1]$ 上可积: ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":60,"id":"d8985c42-98eb-4797-afa0-2581a8c924b8","page_size":[595,841],"block_position":"60-10"},{"bbox":[226,264,365,308],"type":"interline_equation","angle":0,"index":11,"text":"$$\nf (x) = \\left\\{ \\begin{array}{l l} 0, & x = 0 \\\\ \\frac {1}{x} - \\left[ \\frac {1}{x} \\right], & x \\in (0, 1 ] \\end{array} \\right.\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":60,"id":"a55507de-062b-48f5-a6d0-4d9e98997cf1","page_size":[595,841],"block_position":"60-11"},{"bbox":[53,311,539,372],"type":"text","angle":0,"index":12,"merge_prev":false,"text":"证明 (1) 只需证 $\\sum \\omega_{i} \\Delta x_{i} < \\epsilon$ ，给定 $\\forall \\epsilon$ ，满足 $\\frac{1}{q} > \\frac{\\epsilon}{2}$ 的只有有限个，且它们的函数值均小于等于 $\\frac{1}{2}$ 。把这有限个全部列出 $x_{1}, \\cdots, x_{k}$ ，这些点每个最多落入两个小区间中，取 $\\|T\\| < \\frac{\\epsilon}{2k}$ 即可，则该部分 $\\sum \\omega_{i} \\Delta x_{i} < \\frac{\\epsilon}{2}$ 。另一侧直接 $\\sum \\omega_{i} \\Delta x_{i} < \\frac{\\epsilon}{2} \\sum \\Delta x_{i} \\leq \\frac{\\epsilon}{2}$ 就行。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":60,"id":"eb515963-643c-4d01-b1dc-d90e99c4ed85","page_size":[595,841],"block_position":"60-12"},{"bbox":[52,373,539,412],"type":"text","angle":0,"index":13,"merge_prev":false,"text":"(2) $f(x)$ 振幅小于等于1，不连续点为 $0, \\frac{1}{2}, \\frac{1}{3}, \\dots, \\frac{1}{n}, \\dots$ 。对 $\\forall \\epsilon, \\exists \\delta_1$ ，当 $|\\Delta x_i| < \\delta_1$ 时， $(\\epsilon, 1)$ 分割满足 $\\sum \\omega_i \\Delta x_i < \\epsilon$ 。而 $(0, \\epsilon)$ 的任意分割也满足 $\\sum \\omega_i \\Delta x_i < \\epsilon$ ，因此取 $\\delta = \\min\\{\\delta_1, \\epsilon\\}$ 时，整体的分割满足： ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":60,"id":"e0f3e24e-934c-458c-8e5d-a4402209a5b1","page_size":[595,841],"block_position":"60-13"},{"bbox":[264,414,329,431],"type":"interline_equation","angle":0,"index":14,"text":"$$\n\\sum \\omega_ {i} \\Delta x _ {i} < \\epsilon\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":60,"id":"190b8bc1-b141-42fe-875c-a5b71121363c","page_size":[595,841],"block_position":"60-14"},{"bbox":[35,434,539,463],"type":"text","angle":0,"index":15,"merge_prev":false,"text":"练习 6.3 可积性相关证明 (1) $f(x)$ 在 $[a, b]$ 有定义， $\\forall \\epsilon$ ，存在 $[a, b]$ 上可积函数 $g(x)$ ，使得 $\\forall x \\in [a, b], |f(x) - g(x)| < \\epsilon$ ，证明 $f(x)$ 在 $[a, b]$ 可积 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":60,"id":"e77ec4a8-efd9-4e2c-9818-08f797ce0862","page_size":[595,841],"block_position":"60-15"},{"bbox":[52,464,539,502],"type":"text","angle":0,"index":17,"merge_prev":false,"text":"证明 (1) 由于 $g(x)$ 可以，存在分割 $T$ 使得 $\\sum_{T} \\omega_{i}^{g} \\Delta x_{i} < \\epsilon$ ，而 $\\omega_{i}^{f} \\leq \\omega_{i}^{f - g} + \\omega_{i}^{g}$ ，以及 $\\omega_{i}^{f - g} \\leq 2\\epsilon$ （因为 $g$ 本身可能 $\\epsilon$ 震荡， $|f(x) - g(x)|$ 又可能 $\\epsilon$ 震荡），得到 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":60,"id":"ecaf4efb-372a-4515-9041-6237ddebcdc2","page_size":[595,841],"block_position":"60-16"},{"bbox":[174,505,418,530],"type":"interline_equation","angle":0,"index":18,"text":"$$\n\\sum_ {T} \\omega_ {i} ^ {f} \\Delta x _ {i} \\leq \\sum_ {T} \\omega_ {i} ^ {f - g} \\Delta x _ {i} + \\sum_ {T} \\omega_ {i} ^ {g} \\Delta x _ {i} < 2 \\epsilon (b - a) + \\epsilon\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":60,"id":"fa76cdec-1461-40a3-9653-0390b00423f5","page_size":[595,841],"block_position":"60-17"},{"bbox":[56,551,199,565],"type":"title","angle":0,"index":19,"level":2,"text":"## 6.1.2 Newton-Leibniz 公式","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":60,"id":"da4a0694-8144-428f-91cd-4ec72f6af5bf","page_size":[595,841],"block_position":"60-18"},{"bbox":[72,578,212,591],"type":"title","angle":0,"index":20,"level":2,"text":"## 定理 6.4 (Newton-Leibniz 公式)","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":60,"id":"81c6aa1d-21c7-45f5-96ac-1a23efbc53bd","page_size":[595,841],"block_position":"60-19"},{"bbox":[68,595,412,608],"type":"text","angle":0,"index":21,"merge_prev":false,"text":"$f(x)$ 在 $[a,b]$ 可积， $F(x)$ 在 $[a,b]$ 上连续，且除有限个点外， $F^{\\prime}(x) = f(x)$ ，则 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":60,"id":"93976f1d-7a3b-4bb7-8654-8f118f5904de","page_size":[595,841],"block_position":"60-20"},{"bbox":[238,611,355,640],"type":"interline_equation","angle":0,"index":22,"text":"$$\n\\int_ {a} ^ {b} f (x) \\mathrm{d} x = F (b) - F (a)\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":60,"id":"d720e532-e9bc-46ab-a36b-ea28cfcee346","page_size":[595,841],"block_position":"60-21"},{"bbox":[52,655,539,683],"type":"text","angle":0,"index":23,"merge_prev":false,"text":"笔记 Newton-Leibniz 公式的意义在于将定积分的计算转换为了原函数的计算，不过可积与存在原函数也没有直接关系，具体见下面题目。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":60,"id":"d752936e-f3a4-47de-b601-17e649bc8228","page_size":[595,841],"block_position":"60-22"},{"bbox":[35,702,368,714],"type":"text","angle":0,"index":24,"merge_prev":false,"text":"练习 6.4 可积与原函数 举例说明可积未必有原函数，有原函数未必可积 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":60,"id":"c1cbf45f-206a-4ca7-bb2b-fe50a9f16257","page_size":[595,841],"block_position":"60-23"},{"bbox":[53,717,276,729],"type":"text","angle":0,"index":25,"merge_prev":false,"text":"解(1)例如 $[-1, 1]$ 上的 $\\operatorname{sgn}(x)$ ，其可积但无原函数 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":60,"id":"84a812ff-0758-43d8-a59a-a15f225d8324","page_size":[595,841],"block_position":"60-24"},{"bbox":[291,803,302,812],"type":"page_number","angle":0,"index":1000015,"text":"53 ","color":{"line":"rgba(164, 164, 164, 1)","fill":"rgba(164, 164, 164, 1)"},"page_idx":60,"is_discarded":true,"id":"d59bcc3b-f3d4-483e-9594-ecfab35c0d45","page_size":[595,841],"block_position":"60-25"}],[{"bbox":[469,47,539,58],"type":"header","angle":0,"index":-99998,"text":"6.1 定积分理论 ","color":{"line":"rgba(164, 164, 164, 1)","fill":"rgba(164, 164, 164, 1)"},"page_idx":61,"is_discarded":true,"id":"3523f6b7-6b1f-4c47-a8e9-637306b73a23","page_size":[595,841],"block_position":"61-0"},{"bbox":[73,72,315,84],"type":"text","angle":0,"index":2,"merge_prev":false,"text":"(2)例如原函数如下，其在 $[0,1]$ 可导，但是 $F^{\\prime}(x)$ 无界 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":61,"id":"228d1bcb-56df-4910-9694-568b789f9f6e","page_size":[595,841],"block_position":"61-1"},{"bbox":[227,89,364,128],"type":"interline_equation","angle":0,"index":3,"text":"$$\nF (x) = \\left\\{ \\begin{array}{l l} 0, & x = 0 \\\\ x ^ {\\frac {3}{2}} \\sin \\frac {1}{x}, & x \\in (0, 1 ] \\end{array} \\right.\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":61,"id":"dd5f1f11-1c7e-4aa7-8a92-ffdee9d8bce7","page_size":[595,841],"block_position":"61-2"},{"bbox":[35,134,539,205],"type":"text","angle":0,"index":4,"merge_prev":false,"text":"练习6.5使用Newton-Leibniz公式(1)已知 $f(x) = \\left\\{ \\begin{array}{ll}2022, & x = 0\\\\ 2x\\sin \\frac{1}{x} -\\cos \\frac{1}{x}, & x\\in (0,1] \\end{array} \\right.$ ，证明 $f(x)$ 在[0,1]可积，求 $\\int_0^1 f(x)\\mathrm{d}x$ 解(1) $f(x)$ 在0处为跳跃间断点，只有一处，而[0,1]上 $f(x)$ 有界，故可积。构造 $F(x)$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":61,"id":"511a6ef8-c3fb-4706-835c-0270eb938403","page_size":[595,841],"block_position":"61-3"},{"bbox":[52,196,439,210],"type":"text","angle":0,"index":5,"merge_prev":false,"text":"解(1) $f(x)$ 在0处为跳跃间断点，只有一处，而 $[0,1]$ 上 $f(x)$ 有界，故可积。构造 $F(x)$ ： ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":61,"id":"c3eecf26-ff37-4831-ad16-e6333950c5aa","page_size":[595,841],"block_position":"61-4"},{"bbox":[229,215,362,253],"type":"interline_equation","angle":0,"index":6,"text":"$$\nF (x) = \\left\\{ \\begin{array}{l l} 0, & x = 0 \\\\ x ^ {2} \\sin \\frac {1}{x}, & x \\in (0, 1 ] \\end{array} \\right.\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":61,"id":"e183753f-cc1d-4929-ac5b-d566d353aab0","page_size":[595,841],"block_position":"61-5"},{"bbox":[54,254,387,285],"type":"text","angle":0,"index":7,"merge_prev":false,"text":"$F(x)$ 除了0处外满足 $F^{\\prime}(x) = f(x)$ ，因此 $\\int_0^1 f(x)\\mathrm{d}x = F(1) - F(0) = \\sin 1$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":61,"id":"9131e188-5a66-4db4-afbb-c46ce55f153f","page_size":[595,841],"block_position":"61-6"},{"bbox":[73,298,191,311],"type":"title","angle":0,"index":8,"level":2,"text":"## 定理 6.5 (微积分基本定理)","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":61,"id":"b400c646-e50f-4d33-a469-dce0ade09027","page_size":[595,841],"block_position":"61-7"},{"bbox":[67,315,340,342],"type":"text","angle":0,"index":9,"merge_prev":false,"text":"$f(x)$ 为 $[a,b]$ 可积函数，定义 $F(x) = \\int_{a}^{x}f(t)\\mathrm{d}t,x\\in [a,b]$ ，则 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":61,"id":"d5e1e588-1c38-4465-82cd-706540c59d44","page_size":[595,841],"block_position":"61-8"},{"bbox":[83,339,215,352],"type":"text","angle":0,"index":10,"merge_prev":false,"text":"- $F(x)$ 是 $[a, b]$ 上的连续函数 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":61,"id":"12a159ad-5c8f-4822-b9a4-a982e9d04a21","page_size":[595,841],"block_position":"61-9"},{"bbox":[83,354,396,368],"type":"text","angle":0,"index":11,"merge_prev":false,"text":"若 $x_0 \\in [a, b]$ 为 $f(x)$ 连续点，则 $F(x)$ 在 $x_0$ 可导，且 $F'(x_0) = f(x_0)$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":61,"id":"d2099781-d955-4772-81e9-4ac10a293e46","page_size":[595,841],"block_position":"61-10"},{"bbox":[83,370,445,384],"type":"text","angle":0,"index":12,"merge_prev":false,"text":"若 $f(x)$ 为 $[a,b]$ 上连续函数，则 $F(x)$ 为 $[a,b]$ 上连续可微函数，且 $F'(x) = f(x)$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":61,"id":"3a0d34c8-d6a7-42ef-b180-ea4b0b843977","page_size":[595,841],"block_position":"61-11"},{"bbox":[35,416,515,428],"type":"text","angle":0,"index":13,"merge_prev":false,"text":"练习 6.6 使用微积分基本定理 (1) 重点: $f(x)$ 在 $[A, B]$ 上可积, $a, b \\in (A, B)$ 是 $f(x)$ 的两个连续点, 证明 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":61,"id":"7c2bf7e4-5c79-41ec-b75f-553675f59f31","page_size":[595,841],"block_position":"61-12"},{"bbox":[204,432,387,459],"type":"interline_equation","angle":0,"index":14,"text":"$$\n\\lim _ {h \\rightarrow 0} \\int_ {a} ^ {b} \\frac {f (x + h) - f (x)}{h} \\mathrm{d} x = f (b) - f (a)\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":61,"id":"2498b3f4-826c-43e1-b69f-8a63b9ff67c7","page_size":[595,841],"block_position":"61-13"},{"bbox":[53,461,181,473],"type":"text","angle":0,"index":15,"merge_prev":false,"text":"证明 (1) 对左侧做如下化简: ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":61,"id":"25357826-478a-4b16-9948-ad21e3055f03","page_size":[595,841],"block_position":"61-14"},{"bbox":[162,476,429,574],"type":"interline_equation","angle":0,"index":16,"text":"$$\n\\begin{array}{l} \\int_ {a} ^ {b} \\frac {f (x + h) - f (x)}{h} \\mathrm{d} x = \\frac {1}{h} \\left(\\int_ {a} ^ {b} f (x + h) \\mathrm{d} x - \\int_ {a} ^ {b} f (x) \\mathrm{d} x\\right) \\\\ \\qquad = \\frac {1}{h} \\left(\\int_ {a + h} ^ {b + h} f (x) \\mathrm{d} x - \\int_ {a} ^ {b} f (x) \\mathrm{d} x\\right) \\\\ \\qquad = \\frac {1}{h} \\left(\\int_ {b} ^ {b + h} f (x) \\mathrm{d} x - \\int_ {a} ^ {a + h} f (x) \\mathrm{d} x\\right) \\end{array}\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":61,"id":"53623c0b-023d-4ad2-991f-9608035130b5","page_size":[595,841],"block_position":"61-15"},{"bbox":[54,577,446,605],"type":"text","angle":0,"index":17,"merge_prev":false,"text":"令 $F(x) = \\int_{A}^{x}f(t)\\mathrm{d}t$ ，则显然 $F^{\\prime}(b) = \\lim_{h\\to 0^{+}}\\frac{F(b + h) - F(b)}{h} = f(b),F^{\\prime}(a) = f(a)$ 同理。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":61,"id":"a971e33e-a04a-4990-a11b-1826c1bb88cf","page_size":[595,841],"block_position":"61-16"},{"bbox":[55,619,174,634],"type":"title","angle":0,"index":18,"level":2,"text":"## 6.1.3 四则运算与复合","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":61,"id":"b22332d0-b142-4417-9828-68c51b91bf2b","page_size":[595,841],"block_position":"61-17"},{"bbox":[73,648,201,661],"type":"title","angle":0,"index":19,"level":2,"text":"## 定理 6.6 (四则运算的可积性)","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":61,"id":"b1216a76-e74e-4d02-b413-cb50ebeee89f","page_size":[595,841],"block_position":"61-18"},{"bbox":[68,663,533,704],"type":"text","angle":0,"index":20,"merge_prev":false,"text":"若 $f(x), g(x)$ 在 $[a, b]$ 可积，则 $kf(x), f(x) \\pm g(x), f(x)g(x)$ 在 $[a, b]$ 可积，但 $g(x) \\neq 0$ 时 $\\frac{f(x)}{g(x)}$ 也不一定可积 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":61,"id":"853fc387-543a-4303-be0a-ba3d2d997f4f","page_size":[595,841],"block_position":"61-19"},{"bbox":[54,716,167,729],"type":"text","angle":0,"index":21,"merge_prev":false,"text":"证明 (1) 数乘、加减显然 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":61,"id":"e6e288d4-df67-469e-b286-24afc4f303e6","page_size":[595,841],"block_position":"61-20"},{"bbox":[73,729,539,756],"type":"text","angle":0,"index":22,"merge_prev":false,"text":"(2) 乘法：根据可积可知有界，设 $|f(x)| \\leq M, |g(x)| \\leq N$ ， $\\sum_{T} \\omega_{i}^{fg} \\Delta x_{i} < M \\sum_{T} \\omega_{i}^{g} \\Delta x_{i} + N \\sum_{T} \\omega_{i}^{f} \\Delta x_{i}$ 即可 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":61,"id":"bfa41003-7b48-4f80-aa2b-aeaa9d5774ad","page_size":[595,841],"block_position":"61-21"},{"bbox":[291,803,303,811],"type":"page_number","angle":0,"index":999999,"text":"54 ","color":{"line":"rgba(164, 164, 164, 1)","fill":"rgba(164, 164, 164, 1)"},"page_idx":61,"is_discarded":true,"id":"ef88150b-57e4-4de7-ae30-dc95df4c13ac","page_size":[595,841],"block_position":"61-22"}],[{"bbox":[469,47,539,58],"type":"header","angle":0,"index":-99998,"text":"6.1 定积分理论 ","color":{"line":"rgba(164, 164, 164, 1)","fill":"rgba(164, 164, 164, 1)"},"page_idx":62,"is_discarded":true,"id":"7b85f3fe-9bf4-4dc0-93ee-45fcc427cd4d","page_size":[595,841],"block_position":"62-0"},{"bbox":[72,69,398,108],"type":"text","angle":0,"index":2,"merge_prev":false,"text":"(3)除法：例如 $f(x) = 1,g(x) = \\left\\{ \\begin{array}{ll}x, & x\\in (0,1]\\\\ 1, & x = 0 \\end{array} \\right.$ ， $\\frac{1}{g(x)}$ 因为无界而不可积 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":62,"id":"f536e5c7-49d0-4ab9-a071-70f4e20bb41c","page_size":[595,841],"block_position":"62-1"},{"bbox":[35,115,308,128],"type":"text","angle":0,"index":3,"merge_prev":false,"text":"练习6.7除法增加条件可积设 $f(x)$ 在 $[a,b]$ 上函数，证明 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":62,"id":"1ebeff98-0147-442c-9e43-f6544d3a5c71","page_size":[595,841],"block_position":"62-2"},{"bbox":[72,130,335,152],"type":"text","angle":0,"index":4,"merge_prev":false,"text":"(1) 若 $f(x)$ 可积，且 $|f(x)| \\geq m > 0$ ，则 $\\frac{1}{f(x)}$ 在 $[a, b]$ 可积 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":62,"id":"a2eeafa2-c932-4b08-97d3-b7e3cdce39bb","page_size":[595,841],"block_position":"62-3"},{"bbox":[72,153,332,175],"type":"text","angle":0,"index":5,"merge_prev":false,"text":"(2) 若 $f(x)$ 在 $[a, b]$ 连续且恒不为零，则 $\\frac{1}{f(x)}$ 在 $[a, b]$ 可积 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":62,"id":"abac1600-b494-4237-b232-83ef393a2627","page_size":[595,841],"block_position":"62-4"},{"bbox":[54,174,161,186],"type":"text","angle":0,"index":6,"merge_prev":false,"text":"证明 (1) 考虑振幅放缩: ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":62,"id":"61266cf4-4fdc-46b0-87b1-3750030df1c8","page_size":[595,841],"block_position":"62-5"},{"bbox":[198,184,394,211],"type":"interline_equation","angle":0,"index":7,"text":"$$\n\\left| \\frac {1}{f (x ^ {\\prime})} - \\frac {1}{f (x ^ {\\prime \\prime})} \\right| = \\left| \\frac {f (x ^ {\\prime \\prime}) - f (x ^ {\\prime})}{f (x ^ {\\prime}) f (x ^ {\\prime \\prime})} \\right| \\leq \\frac {1}{m ^ {2}} \\omega_ {i} ^ {f}\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":62,"id":"9271b73d-4f72-4259-b764-d1e11cf67938","page_size":[595,841],"block_position":"62-6"},{"bbox":[73,217,265,230],"type":"text","angle":0,"index":8,"merge_prev":false,"text":"(2) 由于连续，能找到最小值，可转换为 (1) ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":62,"id":"e63c955c-2984-493e-9329-a895e9ea4394","page_size":[595,841],"block_position":"62-7"},{"bbox":[73,253,210,265],"type":"title","angle":0,"index":9,"level":2,"text":"## 定理 6.7 (复合 (较难，重点！))","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":62,"id":"64c4b593-5506-4e8e-a659-302d66cbb6d5","page_size":[595,841],"block_position":"62-8"},{"bbox":[66,266,537,287],"type":"text","angle":0,"index":10,"merge_prev":false,"text":"$f(x)$ 在 $[a,b]$ 连续， $\\varphi(t)$ 在 $[\\alpha,\\beta]$ 可积，且 $a \\leq \\varphi(t) \\leq b, t \\in [\\alpha,\\beta]$ ，则 $f(\\varphi(t))$ 在 $[\\alpha,\\beta]$ 可积 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":62,"id":"44ede523-6aa2-4728-ac7b-adc0e3a14164","page_size":[595,841],"block_position":"62-9"},{"bbox":[53,297,539,345],"type":"text","angle":0,"index":11,"merge_prev":false,"text":"证明 $f(x)$ 在 $[a,b]$ 连续，因此有界且一致连续。设 $M = \\max_{x\\in [a,b]}f(x)$ ， $\\forall \\epsilon, \\exists \\delta, \\forall x', x'' \\in [a,b]$ ，若 $|x' - x''| < \\delta$ 就有 $|f(x') - f(x'')| < \\epsilon$ 。由于 $\\varphi$ 可积，存在分割 $T$ ，使得 $\\sum_{\\omega_i^\\varphi \\geq \\delta} \\Delta t_i < \\epsilon$ ，因此 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":62,"id":"d5c2c686-2bf5-4e00-81f8-186ba0ad9576","page_size":[595,841],"block_position":"62-10"},{"bbox":[196,349,396,427],"type":"interline_equation","angle":0,"index":12,"text":"$$\n\\begin{array}{r l} & {\\sum_ {T} \\omega_ {i} ^ {f \\circ \\varphi} \\Delta t _ {i} = \\sum_ {\\omega_ {i} ^ {\\varphi} \\geq \\delta} \\omega_ {i} ^ {f \\circ \\varphi} \\Delta t _ {i} + \\sum_ {\\omega_ {i} ^ {\\varphi} < \\delta} \\omega_ {i} ^ {f \\circ \\varphi} \\Delta t _ {i}} \\\\ & {\\qquad \\leq 2 M \\sum_ {\\omega_ {i} ^ {\\varphi} \\geq \\delta} \\Delta t _ {i} + \\epsilon \\sum_ {\\omega_ {i} ^ {\\varphi} < \\delta} \\Delta t _ {i}} \\\\ & {\\qquad < 2 M \\epsilon + (\\beta - \\alpha) \\epsilon = \\epsilon^ {\\prime}} \\end{array}\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":62,"id":"6b3f934b-4a4b-44b6-975e-22df82e197db","page_size":[595,841],"block_position":"62-11"},{"bbox":[55,449,175,463],"type":"title","angle":0,"index":13,"level":2,"text":"## 6.1.4 定积分的保号性","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":62,"id":"6681fbb5-c877-4ed5-a0d2-ceef4dd831f9","page_size":[595,841],"block_position":"62-12"},{"bbox":[73,476,182,488],"type":"title","angle":0,"index":14,"level":2,"text":"## 定理 6.8 (定积分保号性)","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":62,"id":"eff1d457-5da4-43d7-94a0-431ee180fb8d","page_size":[595,841],"block_position":"62-13"},{"bbox":[66,491,453,522],"type":"text","angle":0,"index":15,"merge_prev":false,"text":"$f(x)$ 在 $[a,b]$ 连续， $f(x)\\geq 0$ ，则 $\\int_{a}^{b}f(x)\\mathrm{d}x\\geq 0$ ，且 $\\int_{a}^{b}f(x)\\mathrm{d}x = 0$ 当且仅当 $f(x)\\equiv 0$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":62,"id":"22ac191a-bf3f-48ec-8f88-234221a76e87","page_size":[595,841],"block_position":"62-14"},{"bbox":[36,546,552,585],"type":"text","angle":0,"index":16,"merge_prev":false,"text":"练习6.8等于0的推广(1)重点: $f(x)$ 在 $[a,b]$ 连续,且对任意满足 $\\int_{a}^{b}g(x)\\mathrm{d}x=0$ 的连续函数 $g(x)$ 有 $\\int_{a}^{b}f(x)g(x)\\mathrm{d}x=0$ ,证明: $f(x)$ 为常值函数 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":62,"id":"6a882883-2125-49c2-b5cd-2551905de69d","page_size":[595,841],"block_position":"62-15"},{"bbox":[52,587,515,628],"type":"text","angle":0,"index":17,"merge_prev":false,"text":"(2) 重点 (ZJU2020): $f(x)$ 在 $[a, b]$ 可导, $f(a) = 0$ , 对 $\\forall x \\in [a, b]$ 有 $|f(x)| \\geq |f'(x)|$ , 证明: $f(x) \\equiv 0$ 证明 (1) 取 $g(x) = f(x) - \\frac{1}{b - a} \\int_{a}^{b} f(t) \\mathrm{d}t$ , 显然 $\\int_{a}^{b} g(x) \\mathrm{d}x = 0$ 满足条件, 此时 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":62,"id":"889c7ecf-e8c5-4ce7-9d83-1808c763dd6d","page_size":[595,841],"block_position":"62-16"},{"bbox":[176,629,418,660],"type":"interline_equation","angle":0,"index":18,"text":"$$\n\\int_ {a} ^ {b} g ^ {2} (x) \\mathrm{d} x = \\int_ {a} ^ {b} g (x) \\left(f (x) - \\frac {1}{b - a} \\int_ {a} ^ {b} f (x) \\mathrm{d} x\\right) = 0\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":62,"id":"ba517473-e5df-4ff3-bf46-3cc13cf27bac","page_size":[595,841],"block_position":"62-17"},{"bbox":[73,663,376,690],"type":"text","angle":0,"index":19,"merge_prev":false,"text":"(2) 可证明 $\\int_{a}^{x} f(t) \\mathrm{d}t = 0, \\forall x \\in [a, b]$ ，则考虑 $g(x) = \\int_{a}^{x} |f(t)| \\mathrm{d}t$ ，有 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":62,"id":"7d2b3a71-d140-41ce-9116-5987580fc064","page_size":[595,841],"block_position":"62-18"},{"bbox":[182,692,409,719],"type":"interline_equation","angle":0,"index":20,"text":"$$\ng (x) \\geq \\int_ {a} ^ {x} | f ^ {\\prime} (t) | \\mathrm{d} t \\geq \\left| \\int_ {a} ^ {x} f ^ {\\prime} (t) \\mathrm{d} t \\right| = | f (x) | = g ^ {\\prime} (x)\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":62,"id":"d926266d-e81d-4e50-9dfb-d3fd9c6064d6","page_size":[595,841],"block_position":"62-19"},{"bbox":[52,720,539,751],"type":"text","angle":0,"index":21,"merge_prev":false,"text":"因此得到 $(e^{-x}g(x))' = e^{-x}(g'(x) - g(x))\\leq 0$ ，即 $e^{-x}g(x)$ 单调减，而 $e^{-a}g(a) = 0$ ，且 $e^{-x}g(x)\\geq 0$ ，综上得到 $g(x)\\equiv 0$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":62,"id":"1d84cd06-cc78-4520-84b4-fe7939fd1948","page_size":[595,841],"block_position":"62-20"},{"bbox":[291,803,302,812],"type":"page_number","angle":0,"index":999999,"text":"55 ","color":{"line":"rgba(164, 164, 164, 1)","fill":"rgba(164, 164, 164, 1)"},"page_idx":62,"is_discarded":true,"id":"395722d5-cf59-4242-b918-017e49336763","page_size":[595,841],"block_position":"62-21"}],[{"bbox":[469,47,539,58],"type":"header","angle":0,"index":-99999,"text":"6.1 定积分理论 ","color":{"line":"rgba(164, 164, 164, 1)","fill":"rgba(164, 164, 164, 1)"},"page_idx":63,"is_discarded":true,"id":"7cc22a2c-4453-4d7d-b873-27ad3241d509","page_size":[595,841],"block_position":"63-0"},{"bbox":[35,69,539,202],"type":"text","angle":0,"index":1,"merge_prev":false,"text":"练习 6.9 零点问题 (1) $f(x)$ 为 $[0, \\pi]$ 上连续函数，且 $\\int_{0}^{\\pi} f(x) \\sin x \\, \\mathrm{d}x = \\int_{0}^{\\pi} f(x) \\cos x \\, \\mathrm{d}x = 0$ ，证明 $f(x)$ 在 $(0, \\pi)$ 上至少有两个零点。\n\n(2) $f(x)$ 为 $[a, b]$ 上连续函数， $\\int_{a}^{b} f(x) \\, \\mathrm{d}x = \\int_{a}^{b} x f(x) \\, \\mathrm{d}x = \\cdots = \\int_{a}^{b} x^n f(x) \\, \\mathrm{d}x = 0$ ，证明 $f(x)$ 在 $(a, b)$ 上至少有 $n + 1$ 个零点\n\n证明 (1) 反设无零点，则 $f(x)$ 在 $(0, \\pi)$ 不变号，而 $\\sin x$ 也不变号，因此与 $\\int_{0}^{\\pi} f(x) \\sin x \\, \\mathrm{d}x = 0$ 矛盾，因此至少有一个零点。若 $f(x)$ 只有一个零点 $\\xi$ ，则 $f(x)$ 在 $(0, \\xi), (\\xi, \\pi)$ 内不变号，由于 $\\int_{0}^{\\pi} f(x) \\sin x \\, \\mathrm{d}x = 0$ ，因此在上述两个区间符号相反，而 $\\sin (x - \\xi)$ 也仅在 $\\xi$ 为 0， $(0, \\xi), (\\xi, \\pi)$ 符号相反，因此 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":63,"id":"a428f6d8-a121-4b2c-9b61-e0add927cc40","page_size":[595,841],"block_position":"63-1"},{"bbox":[136,204,457,230],"type":"interline_equation","angle":0,"index":2,"text":"$$\n\\int_ {0} ^ {\\pi} f (x) \\sin (x - \\xi) \\mathrm{d} x = \\cos \\xi \\int_ {0} ^ {\\pi} f (x) \\sin x \\mathrm{d} x - \\sin \\xi \\int_ {0} ^ {\\pi} f (x) \\cos x \\mathrm{d} x = 0\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":63,"id":"16e8a5f8-9304-4367-9ff3-0c4aca520ddf","page_size":[595,841],"block_position":"63-2"},{"bbox":[53,232,310,244],"type":"text","angle":0,"index":3,"merge_prev":false,"text":"这与 $f(x)\\sin (x - \\xi)$ 非零矛盾，因此 $f(x)$ 至少有两个零点 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":63,"id":"fbf8a943-4049-4a98-a71f-1c8cf21023e0","page_size":[595,841],"block_position":"63-3"},{"bbox":[53,245,539,325],"type":"text","angle":0,"index":4,"merge_prev":false,"text":"(2) 反设 $f(x)$ 在 $(a, b)$ 上至多 n 个零点，则其最多改变 n 次符号。由于 $\\int_{a}^{b} f(x) \\, dx = 0$ ，得到 $f(x)$ 在 $(a, b)$ 至少改变一次符号，一定有能从 n 个零点中找出 k 个零点 $x_{1}, x_{2}, \\cdots, x_{k}$ 满足 $f(x)$ 在 $(a, x_{1}), (x_{1}, x_{2}), \\cdots, (x_{k}, b)$ 上不变号，且在相邻区间变号，而 $g(x) = (x - x_{1})(x - x_{2}) \\cdots (x - x_{k})$ 有同样的性质，于是 $f(x)g(x)$ 不变号，但 $\\int_{a}^{b} f(x)g(x) \\, dx = 0$ ，这显然矛盾。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":63,"id":"d8026f7f-450f-44a0-aca9-2b5dba298576","page_size":[595,841],"block_position":"63-4"},{"bbox":[55,339,163,353],"type":"title","angle":0,"index":5,"level":2,"text":"## 6.1.5 积分中值定理","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":63,"id":"7e1e9918-c16b-466e-a598-86a008de5bb4","page_size":[595,841],"block_position":"63-5"},{"bbox":[72,366,201,380],"type":"title","angle":0,"index":6,"level":2,"text":"## 定理 6.9 (积分第一中值定理)","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":63,"id":"85c3e4b6-90ec-4546-9f65-befaf166f7b0","page_size":[595,841],"block_position":"63-6"},{"bbox":[66,381,503,413],"type":"text","angle":0,"index":7,"merge_prev":false,"text":"$f(x), g(x)$ 在 $[a, b]$ 连续， $g(x)$ 在 $[a, b]$ 不变号，则 $\\exists \\xi \\in (a, b)$ 使得 $\\int_{a}^{b} f(x) g(x) \\mathrm{d}x = f(\\xi) \\int_{a}^{b} g(x) \\mathrm{d}x$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":63,"id":"3d3c4c93-10fe-47ae-9f51-0c5cf3148987","page_size":[595,841],"block_position":"63-7"},{"bbox":[53,426,537,469],"type":"text","angle":0,"index":8,"merge_prev":false,"text":"证明 不妨设 $g(x) \\geq 0$ ，由于 $f(x)$ 在 $[a, b]$ 连续，设 $M, m$ 分别是最大值和最小值，则 $mg(x) \\leq f(x)g(x) \\leq Mg(x)$ ，因此得到 $m\\int_{a}^{b} g(x) \\mathrm{d}x \\leq \\int_{a}^{b} f(x)g(x) \\mathrm{d}x \\leq M\\int_{a}^{b} g(x) \\mathrm{d}x$ ，根据介值性可知（具体讨论略去）。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":63,"id":"15e5b18d-b1b2-4873-9e8c-ad62039d2dd3","page_size":[595,841],"block_position":"63-8"},{"bbox":[73,476,211,488],"type":"title","angle":0,"index":9,"level":2,"text":"## 推论 6.1 (第一中值定理的推广)","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":63,"id":"d823c77c-6e77-4581-a97c-1228ce8fa714","page_size":[595,841],"block_position":"63-9"},{"bbox":[67,491,533,535],"type":"text","angle":0,"index":10,"merge_prev":false,"text":"若 $f(x), g(x)$ 在 $[a, b]$ 可积 (有上下确界但不一定连续)， $g(x)$ 在 $[a, b]$ 不变号， $M, m$ 为 $f(x)$ 在 $[a, b]$ 的上下确界，则 $\\exists \\eta \\in [m, M]$ (不一定能为函数值 $f(\\xi)$ ) 使得 $\\int_{a}^{b} f(x) g(x) \\mathrm{d}x = \\eta \\int_{a}^{b} g(x) \\mathrm{d}x$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":63,"id":"ab12f831-1f35-4fee-bce9-0cc9dcc8b370","page_size":[595,841],"block_position":"63-10"},{"bbox":[35,555,539,664],"type":"text","angle":0,"index":11,"merge_prev":false,"text":"练习6.10第一中值定理(技巧较强) $f(x)$ 是[0,1]连续函数， $\\int_0^1 f(x)\\mathrm{d}x = \\int_0^1 xf(x)\\mathrm{d}x$ ，证明： $\\exists \\xi \\in (0,1)$ 使得 $\\int_0^\\xi f(x)\\mathrm{d}x = 0$ 证明考虑 $F(x) = \\int_0^x f(t)\\mathrm{d}t$ ，想使用积分第一中值定理可证 $\\int_0^1 F(x)\\mathrm{d}x = 0$ ，则可推出 $F(\\xi) = 0$ 。考虑 $\\int_0^1 F(x)\\mathrm{d}x = \\int_0^1\\mathrm{d}x\\int_0^x f(t)\\mathrm{d}t$ ，交换积分顺序得到 $\\int_0^1\\mathrm{d}t\\int_t^1 f(t)\\mathrm{d}x = \\int_0^1 (1 - t)f(t)\\mathrm{d}t = 0$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":63,"id":"5c3b6d87-d5aa-4919-b8c3-18792efce8bd","page_size":[595,841],"block_position":"63-11"},{"bbox":[73,677,206,689],"type":"title","angle":0,"index":12,"level":2,"text":"## 定理 6.10 (积分第二中值定理)","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":63,"id":"8655e7df-43a1-4496-850a-38bb52dba46f","page_size":[595,841],"block_position":"63-12"},{"bbox":[69,692,404,762],"type":"text","angle":0,"index":14,"merge_prev":false,"text":"设 $f(x)$ 在 $[a,b]$ 可积， $g(x)$ 在 $[a,b]$ 单调，则：(记忆方式：非负情况取最大)\n若 $g(x)$ 单增非负： $\\exists \\xi \\in [a,b]$ 使得 $\\int_{a}^{b} f(x) g(x) \\mathrm{d}x = g(b) \\int_{\\xi}^{b} f(x) \\mathrm{d}x$ 若 $g(x)$ 单减非负： $\\exists \\xi \\in [a,b]$ 使得 $\\int_{a}^{b} f(x) g(x) \\mathrm{d}x = g(a) \\int_{a}^{\\xi} f(x) \\mathrm{d}x$ 若 $g(x)$ 单调： $\\exists \\xi \\in [a,b]$ 使得 $\\int_{a}^{b}f(x)g(x)\\mathrm{d}x = g(a)\\int_{a}^{\\xi}f(x)\\mathrm{d}x + g(b)\\int_{\\xi}^{b}f(x)\\mathrm{d}x$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":63,"id":"c3452e7b-96d5-45d7-b887-53b95111c19c","page_size":[595,841],"block_position":"63-13"},{"bbox":[291,803,303,812],"type":"page_number","angle":0,"index":1000012,"text":"56 ","color":{"line":"rgba(164, 164, 164, 1)","fill":"rgba(164, 164, 164, 1)"},"page_idx":63,"is_discarded":true,"id":"45d94467-47be-4948-85cd-f7645e731448","page_size":[595,841],"block_position":"63-14"}],[{"bbox":[429,47,539,58],"type":"header","angle":0,"index":-99999,"text":"6.2 三角函数定积分计算 ","color":{"line":"rgba(164, 164, 164, 1)","fill":"rgba(164, 164, 164, 1)"},"page_idx":64,"is_discarded":true,"id":"456bff7d-c8e7-47fc-a15a-9ca7e82930b5","page_size":[595,841],"block_position":"64-0"},{"bbox":[83,79,449,111],"type":"text","angle":0,"index":1,"merge_prev":false,"lines_deleted":true,"text":"","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":64,"id":"c3452e7b-96d5-45d7-b887-53b95111c19c","page_size":[595,841],"block_position":"64-1"},{"bbox":[54,126,345,139],"type":"text","angle":0,"index":2,"merge_prev":false,"text":"证明 只证明 (3) 的特殊情况：设 $f(x)$ 连续， $g(x)$ 连续可微且单调 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":64,"id":"56d7cf90-2477-4ae5-8bbf-0253c2db8aa3","page_size":[595,841],"block_position":"64-2"},{"bbox":[149,141,443,248],"type":"interline_equation","angle":0,"index":3,"text":"$$\n\\begin{array}{l} \\int_ {a} ^ {b} f (x) g (x) \\mathrm{d} x = \\int_ {a} ^ {b} g (x) \\mathrm{d} (F (x)) = g (x) F (x) | _ {a} ^ {b} - \\int_ {a} ^ {b} F (x) g ^ {\\prime} (x) \\mathrm{d} x \\\\ \\qquad = g (b) F (b) - g (a) F (a) - F (\\xi) \\int_ {a} ^ {b} g ^ {\\prime} (x) \\mathrm{d} x \\\\ \\qquad = g (a) [ F (\\xi) - F (a) ] + g (b) [ F (b) - F (\\xi) ] \\\\ \\qquad = g (a) \\int_ {a} ^ {\\xi} f (x) \\mathrm{d} x + g (b) \\int_ {\\xi} ^ {b} f (x) \\mathrm{d} x \\end{array}\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":64,"id":"861ca34c-6cc1-4267-965a-638af65aa83d","page_size":[595,841],"block_position":"64-3"},{"bbox":[52,255,539,284],"type":"text","angle":0,"index":4,"merge_prev":false,"text":"笔记 第二中值定理证明过于繁琐，这里不给出全部，只给出最后一种特殊情况的证明。其最大的应用是证明反常积分的 Abel 和 Dirichlet 判别法。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":64,"id":"aeb8ff94-d10c-40f4-bf6f-288ba90132d1","page_size":[595,841],"block_position":"64-4"},{"bbox":[55,306,215,322],"type":"title","angle":0,"index":5,"level":2,"text":"## 6.2 三角函数定积分计算","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":64,"id":"91d4b05a-d804-4ab7-a7ce-63966e8e26cb","page_size":[595,841],"block_position":"64-5"},{"bbox":[55,334,229,361],"type":"text","angle":0,"index":6,"merge_prev":false,"text":"6.2.1 $[0, \\frac{\\pi}{2}]$ 上三角函数积分公式 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":64,"id":"73543db4-001e-4ef6-abfe-1b83ced0c9ed","page_size":[595,841],"block_position":"64-6"},{"bbox":[73,370,212,390],"type":"title","angle":0,"index":7,"level":2,"text":"## 定理6.11 $\\left[\\frac{0}{2},\\frac{\\pi}{2}\\right]$ 三角函数积分）","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":64,"id":"212bfbe1-ef8b-4d9e-91b0-5ef557aaf0c5","page_size":[595,841],"block_position":"64-7"},{"bbox":[68,395,370,424],"type":"text","angle":0,"index":8,"merge_prev":false,"text":"设 $f(x)$ 是连续函数，则 $\\int_0^{\\frac{\\pi}{2}}f(\\sin x,\\cos x)\\mathrm{d}x = \\int_0^{\\frac{\\pi}{2}}f(\\cos x,\\sin x)\\mathrm{d}x$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":64,"id":"1340230e-c7da-4885-bbab-72a4ae5cca0e","page_size":[595,841],"block_position":"64-8"},{"bbox":[53,432,381,462],"type":"text","angle":0,"index":9,"merge_prev":false,"text":"证明 令 $x = \\frac{\\pi}{2} - t$ ，则左侧为 $\\int_{\\frac{\\pi}{2}}^{0} f(\\cos t, \\sin t)(-\\mathrm{d}t) = \\int_{0}^{\\frac{\\pi}{2}} f(\\cos t, \\sin t) \\mathrm{d}t$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":64,"id":"caa1d876-c980-4211-a50a-4c7c39e32652","page_size":[595,841],"block_position":"64-9"},{"bbox":[36,465,531,494],"type":"text","angle":0,"index":10,"merge_prev":false,"text":"练习6.11重要练习结论(1)证明： $\\forall k$ ， $\\int_0^{\\frac{\\pi}{2}}\\frac{\\sin^k x}{\\sin^k x + \\cos^k x}\\mathrm{d}x = \\int_0^{\\frac{\\pi}{2}}\\frac{\\cos^k x}{\\sin^k x + \\cos^k x}\\mathrm{d}x = \\int_0^{\\frac{\\pi}{2}}\\frac{x}{1 + \\tan^k x} = \\frac{\\pi}{4}$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":64,"id":"af22ad4e-8adb-4ee6-a54c-acc0766ad79b","page_size":[595,841],"block_position":"64-10"},{"bbox":[73,491,177,518],"type":"text","angle":0,"index":11,"merge_prev":false,"text":"(2) 计算 $\\int_0^{\\frac{\\pi}{2}}\\ln \\tan x\\mathrm{d}x$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":64,"id":"34027d58-31b1-4787-be4d-e3d2acd2e5f6","page_size":[595,841],"block_position":"64-11"},{"bbox":[73,518,224,545],"type":"text","angle":0,"index":12,"merge_prev":false,"text":"(3) 计算 $\\int_{0}^{+\\infty}\\frac{\\mathrm{d}x}{(1+x^{2})(1+x^{2022})}$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":64,"id":"11d67437-9a6e-4878-ad03-4fa1485f60c1","page_size":[595,841],"block_position":"64-12"},{"bbox":[74,544,184,570],"type":"text","angle":0,"index":13,"merge_prev":false,"text":"(4) 计算 $\\int_0^{+\\infty} \\frac{\\ln x}{4 + x^2} \\, \\mathrm{d}x$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":64,"id":"c2753289-7643-4ae0-b2f9-e4391fb7e6e5","page_size":[595,841],"block_position":"64-13"},{"bbox":[53,569,306,595],"type":"text","angle":0,"index":14,"merge_prev":false,"text":"证明 (1) 根据对称性可知 $2S = \\int_{0}^{\\frac{\\pi}{2}} \\mathrm{d}x = \\frac{\\pi}{2}$ ，因此 $S = \\frac{\\pi}{4}$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":64,"id":"40a29ead-9731-4178-847c-72bc4955fd9b","page_size":[595,841],"block_position":"64-14"},{"bbox":[73,594,241,622],"type":"text","angle":0,"index":15,"merge_prev":false,"text":"(2) $\\int_{0}^{\\frac{\\pi}{2}}\\ln \\sin x\\mathrm{d}x - \\int_{0}^{\\frac{\\pi}{2}}\\ln \\cos x\\mathrm{d}x = 0$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":64,"id":"5e8c2f9e-657a-4aed-b1ea-a3c8196f1447","page_size":[595,841],"block_position":"64-15"},{"bbox":[73,622,358,647],"type":"text","angle":0,"index":16,"merge_prev":false,"text":"(3) 设 $x = \\tan t$ ，得到 $\\int_0^{\\frac{\\pi}{2}} \\frac{1}{1 + \\tan^{2022} x} \\mathrm{d}t$ ，用 (1) 结论可知为 $\\frac{\\pi}{4}$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":64,"id":"0efa1619-36a3-4e0a-8801-23db903d6c15","page_size":[595,841],"block_position":"64-16"},{"bbox":[73,647,464,674],"type":"text","angle":0,"index":17,"merge_prev":false,"text":"(4) 令 $x = 2 \\tan t$ ，得到 $I = \\int_{0}^{\\frac{\\pi}{2}} \\frac{\\ln(2 \\tan t)}{4 \\sec^{2} t} 2 \\sec^{2} t \\, \\mathrm{d}t = \\frac{\\pi}{4} \\ln 2 + \\frac{1}{2} \\int_{0}^{\\frac{\\pi}{2}} \\ln(\\tan t) \\, \\mathrm{d}t = \\frac{\\pi}{4} \\ln 2$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":64,"id":"26db5871-be87-43bc-9522-5a9380ee9e52","page_size":[595,841],"block_position":"64-17"},{"bbox":[291,802,302,811],"type":"page_number","angle":0,"index":1000017,"text":"57 ","color":{"line":"rgba(164, 164, 164, 1)","fill":"rgba(164, 164, 164, 1)"},"page_idx":64,"is_discarded":true,"id":"3edcb6ad-fb06-4876-858b-6a2000ab062d","page_size":[595,841],"block_position":"64-18"}],[{"bbox":[429,46,539,58],"type":"header","angle":0,"index":-99999,"text":"6.2 三角函数定积分计算 ","color":{"line":"rgba(164, 164, 164, 1)","fill":"rgba(164, 164, 164, 1)"},"page_idx":65,"is_discarded":true,"id":"675187a5-8d02-4d0a-a5ce-92dafedbd06f","page_size":[595,841],"block_position":"65-0"},{"bbox":[73,73,176,86],"type":"title","angle":0,"index":1,"level":2,"text":"## 定理 6.12 (Wallis 推广)","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":65,"id":"8da1c372-1861-483b-82bd-1c3f82271337","page_size":[595,841],"block_position":"65-1"},{"bbox":[67,89,500,173],"type":"interline_equation","angle":0,"index":2,"text":"$$\n\\begin{array}{r l} & {I (m, n) = \\int_ {0} ^ {\\frac {\\pi}{2}} \\cos^ {m} x \\sin^ {n} x \\mathrm{d} x, \\text {则} I (m, n) = I (n, m), \\text {且}} \\\\ & {\\qquad I (m, n) = \\frac {m - 1}{m + n} I (m - 2, n) = \\frac {n - 1}{m + n} I (m, n - 2) = \\left\\{ \\begin{array}{l l} {\\frac {(m - 1) ! ! (n - 1) ! !}{(m + n) ! !},} & {m, n \\text {不全为偶数}} \\\\ {\\frac {(m - 1) ! ! (n - 1) ! !}{(m + n) ! !} \\frac {\\pi}{2},} & {m, n \\text {均为偶数}} \\end{array} \\right.} \\end{array}\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":65,"id":"dd8e7e0e-7921-4112-b0c8-37ba8248d59d","page_size":[595,841],"block_position":"65-2"},{"bbox":[35,195,238,224],"type":"text","angle":0,"index":3,"merge_prev":false,"text":"练习6.12基础练习 $(1)\\int_{0}^{\\frac{\\pi}{2}}\\cos^{3}x\\sin^{2}x\\mathrm{d}x$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":65,"id":"35a65ffe-aa72-43d9-8e8d-cb925bdb698b","page_size":[595,841],"block_position":"65-3"},{"bbox":[74,250,165,275],"type":"interline_equation","angle":0,"index":4,"text":"$$\n(3) \\int_ {0} ^ {\\frac {\\pi}{2}} \\sin^ {3} \\cos^ {5} x \\mathrm{d} x\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":65,"id":"a39eca78-7aac-4f13-925e-626c785356c7","page_size":[595,841],"block_position":"65-4"},{"bbox":[74,275,170,301],"type":"interline_equation","angle":0,"index":5,"text":"$$\n(4) \\int_ {0} ^ {\\frac {\\pi}{2}} \\sin^ {2} x \\cos^ {4} x \\mathrm{d} x\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":65,"id":"c5555ed9-84d4-4bcf-bc5b-cbbe28347dca","page_size":[595,841],"block_position":"65-5"},{"bbox":[54,299,223,329],"type":"interline_equation","angle":0,"index":6,"text":"$$\n\\text {解} (1) I (3, 2) = \\frac {2 ! ! \\cdot 1 ! !}{5 ! ! _ {3 \\cdot 2}} = \\frac {2 \\cdot 1}{5 \\cdot 3 \\cdot 1} = \\frac {2}{1 5}\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":65,"id":"cacb0fe8-d564-41c1-9835-7c52c3dde7cc","page_size":[595,841],"block_position":"65-6"},{"bbox":[73,326,196,352],"type":"interline_equation","angle":0,"index":7,"text":"$$\n\\begin{array}{l} (2) I (4, 3) = \\frac {5 \\cdot 2}{7 \\cdot 5 \\cdot 3} = \\frac {2}{3 5} \\\\ (2) I (3, 5) = \\frac {2 \\cdot 4 \\cdot 2}{1} \\end{array}\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":65,"id":"8d27ad2c-1d84-463d-b96c-8e21036e8adf","page_size":[595,841],"block_position":"65-7"},{"bbox":[73,347,198,368],"type":"interline_equation","angle":0,"index":8,"text":"$$\n(3) I (3, 5) = \\frac {2 \\cdot 4 \\cdot 2}{8 \\cdot 6 \\cdot 4 \\cdot 2} = \\frac {1}{2 4}\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":65,"id":"75a8b582-361f-486b-a39a-e315ab577ce0","page_size":[595,841],"block_position":"65-8"},{"bbox":[73,357,198,387],"type":"interline_equation","angle":0,"index":9,"text":"$$\n(4) I (2, 4) = \\frac {8 \\cdot 6 \\cdot 4 \\cdot 2}{1 \\cdot 3 \\cdot 1} \\frac {\\pi}{2} = \\frac {2 4}{\\pi}\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":65,"id":"c41044cc-e27b-4184-aeaa-6bcfb84c3b4c","page_size":[595,841],"block_position":"65-9"},{"bbox":[55,401,238,417],"type":"title","angle":0,"index":10,"level":2,"text":"## 6.2.2 $[0, \\pi], [0, 2\\pi]$ 上三角函数积分","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":65,"id":"ed334bd8-8acb-4da4-80a6-f226bba30ba9","page_size":[595,841],"block_position":"65-10"},{"bbox":[73,429,186,441],"type":"title","angle":0,"index":11,"level":2,"text":"## 定理 6.13 (对称性的积分)","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":65,"id":"168383c8-4293-4cb7-ac4f-dc9708fd51d7","page_size":[595,841],"block_position":"65-11"},{"bbox":[67,444,526,503],"type":"text","angle":0,"index":12,"merge_prev":false,"text":"若 $f(x)$ 关于 $(x_0,0)$ 中心对称，则 $\\int_{x_0 - a}^{x_0 + a}f(x)\\mathrm{d}x = 0$ 。若 $f(x)$ 关于 $x = x_0$ 轴对称，则 $\\int_{x_0 - a}^{x_0 + a}f(x)\\mathrm{d}x = 2\\int_{x_0 - a}^{x_0}f(x)\\mathrm{d}x = 2\\int_{x_0}^{x_0 + a}f(x)\\mathrm{d}x$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":65,"id":"8594b9f0-2906-4769-aab1-5bfe4738ee6d","page_size":[595,841],"block_position":"65-12"},{"bbox":[53,517,539,546],"type":"text","angle":0,"index":13,"merge_prev":false,"text":"笔记 若 $f(x), g(x)$ 都关于 $x = x_0$ 轴对称，则 $f(x)g(x)$ 也关于 $x = x_0$ 轴对称。 $f(x)$ 轴对称， $g(x)$ 中心对称，则 $f(x)g(x)$ 中心对称。 $f(x), g(x)$ 都中心对称，则 $f(x)g(x)$ 轴对称。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":65,"id":"05e6d395-540a-4ca5-83cc-755707b71850","page_size":[595,841],"block_position":"65-13"},{"bbox":[73,555,231,568],"type":"title","angle":0,"index":14,"level":2,"text":"## 定理 6.14 ([0, π] 上 Wallis 公式推广)","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":65,"id":"acfc8e74-9e92-4a63-a92e-80de156ff64f","page_size":[595,841],"block_position":"65-14"},{"bbox":[69,572,183,585],"type":"text","angle":0,"index":15,"merge_prev":false,"text":"对任意非负整数 $m, n$ ，有 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":65,"id":"64435221-86ac-41b8-a5b9-7e0e279cdb82","page_size":[595,841],"block_position":"65-15"},{"bbox":[176,589,417,626],"type":"interline_equation","angle":0,"index":16,"text":"$$\nJ (m, n) = \\int_ {0} ^ {\\pi} \\sin^ {m} x \\cos^ {n} x \\mathrm{d} x = {\\left\\{ \\begin{array}{l l} {0,} & {n \\text {为奇数}} \\\\ {2 I (m, n),} & {n \\text {为偶数}} \\end{array} \\right.}\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":65,"id":"09a654e3-9c2a-4f81-a691-635afdb2ab65","page_size":[595,841],"block_position":"65-16"},{"bbox":[69,629,193,642],"type":"text","angle":0,"index":17,"merge_prev":false,"text":"注意这里 $J(m,n)\\neq J(n,m)$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":65,"id":"a6e9c8a4-37bb-40f7-b1d2-154c49975ec7","page_size":[595,841],"block_position":"65-17"},{"bbox":[54,655,359,677],"type":"text","angle":0,"index":18,"merge_prev":false,"text":"证明 根据 $\\sin^m x\\cos^n x$ 在 $n$ 为偶关于 $\\frac{\\pi}{2}$ 轴对称， $n$ 为奇时中心对称。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":65,"id":"3fd63b04-d201-45b2-9226-bceb7c61b63e","page_size":[595,841],"block_position":"65-18"},{"bbox":[73,686,236,699],"type":"title","angle":0,"index":20,"level":2,"text":"## 定理 6.15 ( $[0, 2\\pi]$ 上 Wallis 公式推广)","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":65,"id":"7335a2c7-08e7-4405-8252-608cf70bec37","page_size":[595,841],"block_position":"65-19"},{"bbox":[69,704,163,716],"type":"text","angle":0,"index":21,"merge_prev":false,"text":"对任意非负 $m, n$ ，有 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":65,"id":"15e960c6-333a-4b86-8b43-d5e90395d120","page_size":[595,841],"block_position":"65-20"},{"bbox":[155,719,437,758],"type":"interline_equation","angle":0,"index":22,"text":"$$\nK (m, n) = \\int_ {0} ^ {2 \\pi} \\sin^ {m} x \\cos^ {n} x \\mathrm{d} x = {\\left\\{ \\begin{array}{l l} {0,} & {m, n \\text {不全为偶数}} \\\\ {4 I (m, n),} & {m, n \\text {均为偶数}} \\end{array} \\right.}\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":65,"id":"60c29ea4-36b3-4a19-816e-5af894361709","page_size":[595,841],"block_position":"65-21"},{"bbox":[291,803,302,812],"type":"page_number","angle":0,"index":1000018,"text":"58 ","color":{"line":"rgba(164, 164, 164, 1)","fill":"rgba(164, 164, 164, 1)"},"page_idx":65,"is_discarded":true,"id":"bf9af960-8128-475e-8ca4-eca08f831191","page_size":[595,841],"block_position":"65-22"}],[{"bbox":[429,47,539,58],"type":"header","angle":0,"index":-99999,"text":"6.2 三角函数定积分计算 ","color":{"line":"rgba(164, 164, 164, 1)","fill":"rgba(164, 164, 164, 1)"},"page_idx":66,"is_discarded":true,"id":"b92feb37-ca37-4bc7-815c-38ca35e59ae6","page_size":[595,841],"block_position":"66-0"},{"bbox":[69,79,180,94],"type":"text","angle":0,"index":1,"merge_prev":false,"text":"这里 $K(m,n) = K(n,m)$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":66,"id":"0e13241b-ddec-4ede-98e0-4557e7e4bec0","page_size":[595,841],"block_position":"66-1"},{"bbox":[35,118,540,148],"type":"text","angle":0,"index":2,"merge_prev":false,"text":"练习6.13(0～π)上的计算(1) $\\int_{0}^{\\pi}\\cos^{3}x\\sin^{2}xdx(2)\\int_{0}^{\\pi}\\cos^{4}x\\sin^{3}xdx(3)\\int_{0}^{\\pi}\\sin^{3}x\\cos^{5}xdx(4)\\int_{0}^{\\pi}\\sin^{2}x\\cos^{4}xdx$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":66,"id":"06f7022d-a13e-4f94-9da2-ae4ae687efb1","page_size":[595,841],"block_position":"66-2"},{"bbox":[53,157,193,169],"type":"text","angle":0,"index":3,"merge_prev":false,"text":"解(1) $\\cos x$ 是奇数幂，因此为0 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":66,"id":"90a6d38d-ddce-4e49-862a-19ef73695077","page_size":[595,841],"block_position":"66-3"},{"bbox":[72,167,262,191],"type":"text","angle":0,"index":4,"merge_prev":false,"text":"(2) 偶数幂，因此 $2I(4,3) = 2\\frac{3 \\cdot 2}{7 \\cdot 5 \\cdot 3} = \\frac{4}{35}$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":66,"id":"fd5e5d96-9608-4c84-9162-c6cbedc499e0","page_size":[595,841],"block_position":"66-4"},{"bbox":[73,188,140,200],"type":"text","angle":0,"index":5,"merge_prev":false,"text":"(3) 奇数，为 0 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":66,"id":"7250e20c-8bdd-46ba-977f-c2ed0862f1bf","page_size":[595,841],"block_position":"66-5"},{"bbox":[73,200,280,222],"type":"text","angle":0,"index":6,"merge_prev":false,"text":"(4) 偶数幂，但 $m, n$ 均为偶数，因此结果为 $\\frac{\\pi}{16}$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":66,"id":"bb767c18-fba5-4937-8528-fe858e1287cd","page_size":[595,841],"block_position":"66-6"},{"bbox":[35,226,471,275],"type":"text","angle":0,"index":7,"merge_prev":false,"text":"练习 6.14(0 ~ 2π) 上的计算 (1) $\\int_{0}^{2\\pi}\\cos^{3}x\\sin^{2}xdx$ (2) $\\int_{0}^{2\\pi}\\cos^{2}x\\sin^{3}xdx$ (3) $\\int_{0}^{2\\pi}\\sin^{2}x\\cos^{4}xdx$ 解 (1) 不全为偶，因此为 0 (2)0 (3) 均为偶数，结果为 $\\frac{\\pi}{8}$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":66,"id":"894dbf4c-2d8d-4a6b-a08a-d2fdc7531962","page_size":[595,841],"block_position":"66-7"},{"bbox":[33,285,393,300],"type":"text","angle":0,"index":8,"merge_prev":false,"text":"笔记 周期函数的积分区域可以转换至任意周期区间，因此其他区间也可计算。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":66,"id":"6fd3d324-6cb2-400c-96cd-793dbaf83716","page_size":[595,841],"block_position":"66-8"},{"bbox":[35,298,539,370],"type":"text","angle":0,"index":9,"merge_prev":false,"text":"练习6.15非标准区间 (1) $\\int_{-\\pi}^{\\pi}\\cos^4 x\\sin^4 x\\mathrm{d}x$ (2) $\\int_{-\\pi}^{3\\pi}\\sin^5 x\\cos^6 x\\mathrm{d}x$ (3) $\\int_{-\\pi}^{0}\\cos^3 x\\sin^2 x\\mathrm{d}x$ (4) $\\int_{-\\pi}^{0}\\cos^4 x\\sin^3 x\\mathrm{d}x$ (5) $\\int_{\\pi}^{2\\pi}\\sin^2 x\\cos^4 x\\mathrm{d}x$ (6) $\\int_0^\\pi |\\cos^5 x|\\mathrm{d}x$ 解(1) $2J(4,4) = 4I(4,4) = \\frac{3}{64}\\pi$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":66,"id":"6a2d2dc8-168d-4b58-9dc4-e03bef939ac2","page_size":[595,841],"block_position":"66-9"},{"bbox":[72,370,376,384],"type":"text","angle":0,"index":10,"merge_prev":false,"text":"(2) 先转换到 $[0, 4\\pi]$ ，再变成 2 倍的 $[0, 2\\pi]$ ，非全为偶，因此结果为 0 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":66,"id":"45f05d64-ce5e-4ab3-adbf-cf34c066c225","page_size":[595,841],"block_position":"66-10"},{"bbox":[73,383,262,407],"type":"text","angle":0,"index":11,"merge_prev":false,"text":"(3) 偶函数，转换为 $\\int_0^\\pi \\cos^3 x\\sin^2 x\\mathrm{d}x = 0$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":66,"id":"623f6eba-24ca-4708-b049-6acaa80fcef6","page_size":[595,841],"block_position":"66-11"},{"bbox":[73,407,339,432],"type":"text","angle":0,"index":12,"merge_prev":false,"text":"(4) 奇函数，转换为 $-\\int_{0}^{\\pi}\\cos^{4}x\\sin^{2}xdx=-2I(4,3)=-\\frac{4}{35}$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":66,"id":"97dd8ec6-b61b-499e-957f-3d4d95f97a7e","page_size":[595,841],"block_position":"66-12"},{"bbox":[73,432,445,456],"type":"text","angle":0,"index":13,"merge_prev":false,"text":"(5) 要看 $\\sin$ 在 $x = \\pi$ 的对称性，其是中心对称，因此等价于 $\\int_{0}^{\\pi} \\sin^{2} x \\cos^{4} x dx = \\frac{\\pi}{16}$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":66,"id":"a907a016-e183-4545-971b-c387d5ab379b","page_size":[595,841],"block_position":"66-13"},{"bbox":[73,454,262,478],"type":"text","angle":0,"index":14,"merge_prev":false,"text":"(6) 绝对值肯定轴对称，因此 $2I(5,0) = \\frac{16}{15}$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":66,"id":"1e51b672-8cdc-40f2-9934-b9282ca9ddef","page_size":[595,841],"block_position":"66-14"},{"bbox":[35,483,561,569],"type":"text","angle":0,"index":15,"merge_prev":false,"text":"练习6.16其他题目证明： $\\int_0^{2\\pi}f(a\\sin x + b\\cos x)\\mathrm{d}x = 2\\int_0^\\pi f(\\sqrt{a^2 + b^2}\\cos x)\\mathrm{d}x$ 证明根据 $a\\sin x + b\\cos x = \\sqrt{a^2 + b^2}\\cos (x + \\varphi)$ ，这里 $\\cos \\varphi = \\frac{b}{\\sqrt{a^2 + b^2}}$ 。因此 $\\int_0^{2\\pi}f(\\sqrt{a^2 + b^2}\\cos (x + y))\\mathrm{d}x = \\int_y^{2\\pi +y}f(\\sqrt{a^2 + b^2}\\cos t)\\mathrm{d}t = \\int_0^{2\\pi}f(\\sqrt{a^2 + b^2}\\cos x)\\mathrm{d}x = 2\\int_0^\\pi f(\\sqrt{a^2 + b^2}\\cos x)\\mathrm{d}x$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":66,"id":"3c4305d8-2d03-413e-b0ed-9d05e6c6795d","page_size":[595,841],"block_position":"66-15"},{"bbox":[56,596,163,610],"type":"title","angle":0,"index":16,"level":2,"text":"## 6.2.3 区间再现公式","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":66,"id":"85013f2c-3d17-4f57-aa8f-c9864036ed05","page_size":[595,841],"block_position":"66-16"},{"bbox":[73,624,166,636],"type":"title","angle":0,"index":17,"level":2,"text":"## 定理 6.16 (区间再现)","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":66,"id":"56c92324-a392-4bd8-99f6-a6c8284b40ec","page_size":[595,841],"block_position":"66-17"},{"bbox":[67,640,390,669],"type":"text","angle":0,"index":18,"merge_prev":false,"text":"$f(x)$ 为连续函数， $\\int_0^\\pi xf(\\sin x)\\mathrm{d}x = \\frac{\\pi}{2}\\int_0^\\pi f(\\sin x)\\mathrm{d}x = \\pi \\int_0^{\\frac{\\pi}{2}}f(\\sin x)\\mathrm{d}x$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":66,"id":"3db4f69e-14c2-4f52-9b04-0d28ab50ca87","page_size":[595,841],"block_position":"66-18"},{"bbox":[53,677,539,730],"type":"text","angle":0,"index":19,"merge_prev":false,"text":"证明 令 $x = \\pi - t$ ，得到 $I = \\int_{\\pi}^{0} (\\pi - t) f(\\sin t)(-\\mathrm{d}t) = \\pi \\int_{0}^{\\pi} f(\\sin t) \\mathrm{d}t - \\int_{0}^{\\pi} t f(\\sin t) \\mathrm{d}t$ ，因此得到 $I = \\frac{\\pi}{2} \\int_{0}^{\\pi} f(\\sin t) \\mathrm{d}t$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":66,"id":"870d903e-819f-4a67-88f1-c8e612a8c0bf","page_size":[595,841],"block_position":"66-19"},{"bbox":[35,734,445,761],"type":"text","angle":0,"index":20,"merge_prev":false,"text":"练习6.17区间再现公式的应用 (1) $\\int_0^\\pi x\\ln \\sin x\\mathrm{d}x$ (2) $\\int_0^\\pi \\frac{x\\sin x}{1 + \\cos^2x}\\mathrm{d}x$ (3) $\\int_0^\\pi \\frac{x}{1 + \\cos^2x}\\mathrm{d}x$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":66,"id":"e6b9277e-18a8-41da-a52f-e44be51a9365","page_size":[595,841],"block_position":"66-20"},{"bbox":[291,802,302,811],"type":"page_number","angle":0,"index":1000020,"text":"59 ","color":{"line":"rgba(164, 164, 164, 1)","fill":"rgba(164, 164, 164, 1)"},"page_idx":66,"is_discarded":true,"id":"2d09cce9-9c21-4567-a05d-dff8b8bb6e03","page_size":[595,841],"block_position":"66-21"}],[{"bbox":[429,47,539,58],"type":"header","angle":0,"index":-99999,"text":"6.2 三角函数定积分计算 ","color":{"line":"rgba(164, 164, 164, 1)","fill":"rgba(164, 164, 164, 1)"},"page_idx":67,"is_discarded":true,"id":"c6c67921-57c4-45a3-9f07-05af771a99e4","page_size":[595,841],"block_position":"67-0"},{"bbox":[53,69,244,95],"type":"text","angle":0,"index":1,"merge_prev":false,"text":"解(1)变为 $I = \\frac{\\pi}{2}\\int_0^\\pi \\ln \\sin x dx = -\\frac{\\pi^2}{2}\\ln 2$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":67,"id":"d49d099e-2176-40bf-ad2d-c7ba754f4d72","page_size":[595,841],"block_position":"67-1"},{"bbox":[73,95,306,120],"type":"text","angle":0,"index":2,"merge_prev":false,"text":"(2) $I=\\frac{\\pi}{2}\\int_{0}^{\\pi}\\frac{\\sin x}{1+\\cos^{2}x}dx=-\\frac{\\pi}{2}\\arctan\\cos x|_{0}^{\\pi}=\\frac{\\pi^{2}}{4}$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":67,"id":"118791d2-3b0e-45f3-8ade-212796f9d32a","page_size":[595,841],"block_position":"67-2"},{"bbox":[73,119,491,148],"type":"interline_equation","angle":0,"index":3,"text":"$$\nI = \\int_ {0} ^ {\\pi} \\frac {x}{1 + \\cos^ {2} x} d x = \\frac {\\pi}{2} \\int_ {0} ^ {\\pi} \\frac {1}{1 + \\cos^ {2} x} d x = \\pi \\int_ {0} ^ {\\frac {\\pi}{2}} \\frac {\\sec^ {2} x}{\\sec^ {2} x + 1} d x = \\pi \\int_ {0} ^ {\\frac {\\pi}{2}} \\frac {d (\\tan x)}{\\tan^ {2} x + 2} = \\frac {\\pi^ {2}}{2 \\sqrt {2}}\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":67,"id":"ffcdd2d1-ceb3-45bd-b973-877ffb53fe30","page_size":[595,841],"block_position":"67-3"},{"bbox":[54,148,539,183],"type":"text","angle":0,"index":4,"merge_prev":false,"text":"(4)本题区间比较麻烦，不能用区间再现。可尝试用三角变换： $I = \\int_{0}^{\\frac{1}{4}}\\frac{x}{2\\cos^{2}x} dx = \\frac{1}{2}\\int_{0}^{\\frac{1}{4}}xd(\\tan x)$ ，用分部积分得到 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":67,"id":"3e9f83f9-2630-47ea-95ab-e5656bcc214e","page_size":[595,841],"block_position":"67-4"},{"bbox":[205,181,387,211],"type":"interline_equation","angle":0,"index":5,"text":"$$\n\\frac {1}{2} x \\tan x \\bigg | _ {0} ^ {\\frac {\\pi}{4}} + \\frac {1}{2} \\ln \\cos x \\bigg | _ {0} ^ {\\frac {\\pi}{4}} = \\frac {\\pi}{8} + \\frac {1}{2} \\ln \\frac {\\sqrt {2}}{2}\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":67,"id":"e768f9e0-e7e1-49be-b823-b2d145266cc8","page_size":[595,841],"block_position":"67-5"},{"bbox":[35,220,377,235],"type":"text","angle":0,"index":6,"merge_prev":false,"text":"笔记上面几个虽然并非严格的 $xf(\\sin x)$ 形式，但是也可以使用区间再现。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":67,"id":"8a87b0bc-f8c1-462c-9085-2e08d4a092fa","page_size":[595,841],"block_position":"67-6"},{"bbox":[55,253,146,267],"type":"title","angle":0,"index":7,"level":2,"text":"## 6.2.4 Euler 积分","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":67,"id":"4519f7ad-ab7b-4312-8387-1e62cb39347b","page_size":[595,841],"block_position":"67-7"},{"bbox":[35,275,548,326],"type":"text","angle":0,"index":8,"merge_prev":false,"text":"练习6.18Euler积分(1)Euler积分: $I=\\int_{0}^{\\frac{\\pi}{2}}\\ln\\sin x\\mathrm{d}x=-\\frac{\\pi}{2}\\ln2$ (2)推论: $\\int_{0}^{\\frac{\\pi}{2}}\\ln\\cos x\\mathrm{d}x=-\\frac{\\pi}{2}\\ln2,\\int_{0}^{\\frac{\\pi}{2}}\\ln\\tan x\\mathrm{d}x=0,\\int_{0}^{\\pi}\\ln\\sin x\\mathrm{d}x=-\\pi\\ln2$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":67,"id":"d835c99e-f013-4753-8ed3-ef3b4cadefec","page_size":[595,841],"block_position":"67-8"},{"bbox":[52,324,540,364],"type":"text","angle":0,"index":9,"merge_prev":false,"text":"解(1)该积分是瑕积分, $I=\\int_{0}^{\\frac{\\pi}{2}}\\ln\\sin xdx=\\int_{0}^{\\frac{\\pi}{2}}\\ln\\cos xdx=\\frac{1}{2}\\int_{0}^{\\frac{\\pi}{2}}\\ln\\sin x+\\ln\\cos xdx=\\frac{1}{2}\\int_{0}^{\\frac{\\pi}{2}}\\ln\\sin x\\cos xdx$ ,\n根据倍角公式得到 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":67,"id":"21f5ad92-c194-4fe3-a1be-70a14b9e7f63","page_size":[595,841],"block_position":"67-9"},{"bbox":[93,365,500,394],"type":"interline_equation","angle":0,"index":10,"text":"$$\nI = \\frac {1}{2} \\int_ {0} ^ {\\frac {\\pi}{2}} \\ln \\frac {1}{2} \\sin 2 x \\mathrm{d} x = \\frac {1}{2} \\int_ {0} ^ {\\frac {\\pi}{2}} \\ln \\sin 2 x \\mathrm{d} x - \\frac {\\pi}{4} \\ln 2 = \\frac {1}{4} \\int_ {0} ^ {\\pi} \\ln \\sin t \\mathrm{d} t - \\frac {\\pi}{4} \\ln 2 = \\frac {1}{2} I - \\frac {\\pi}{4} \\ln 2\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":67,"id":"73abd409-d618-4fb0-a934-8264170248ef","page_size":[595,841],"block_position":"67-10"},{"bbox":[55,396,152,417],"type":"text","angle":0,"index":11,"merge_prev":false,"text":"因此得到 $I = -\\frac{\\pi}{2}\\ln 2$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":67,"id":"f28ac38b-dccf-4471-b92e-45f82d121750","page_size":[595,841],"block_position":"67-11"},{"bbox":[33,417,308,436],"type":"text","angle":0,"index":12,"merge_prev":false,"text":"笔记要证明Euler积分收敛，只需证明 $\\lim_{x\\to 0^{+}}x\\ln \\sin x = 0$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":67,"id":"f47c252e-cd22-4f37-a786-ba7c59389362","page_size":[595,841],"block_position":"67-12"},{"bbox":[35,444,487,471],"type":"text","angle":0,"index":13,"merge_prev":false,"text":"练习6.19Euler积分的应用 (1) $\\int_0^{\\frac{\\pi}{2}}x\\cot x\\mathrm{d}x$ (2) $\\int_0^1\\frac{\\ln x}{\\sqrt{1 - x^2}}\\mathrm{d}x$ (3) $\\int_0^1\\frac{\\arcsin x}{x}\\mathrm{d}x$ (4) $\\int_0^\\pi \\frac{x\\sin x}{1 - \\cos x}\\mathrm{d}x$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":67,"id":"359f82a0-b81b-412e-b568-ad8ac0eefaef","page_size":[595,841],"block_position":"67-13"},{"bbox":[66,470,337,497],"type":"interline_equation","angle":0,"index":14,"text":"$$\nI = \\int_ {0} ^ {\\frac {\\pi}{2}} x \\mathrm{d} (\\ln \\sin x) = x \\ln \\sin x \\big | _ {0} ^ {\\frac {\\pi}{2}} - \\int_ {0} ^ {\\frac {\\pi}{2}} \\ln \\sin x \\mathrm{d} x = \\frac {\\pi}{2} \\ln 2\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":67,"id":"4788e327-cece-4688-bc17-8e288294f902","page_size":[595,841],"block_position":"67-14"},{"bbox":[72,497,317,523],"type":"text","angle":0,"index":15,"merge_prev":false,"text":"(2) 令 $x = \\sin t$ ，得到 $I = \\int_{0}^{\\frac{\\pi}{2}} \\frac{\\ln \\sin t}{\\cos t} \\cos t \\, \\mathrm{d}t = -\\frac{\\pi}{2} \\ln 2$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":67,"id":"349f8db8-80ec-4ddf-ad57-d14c1c7b7cb7","page_size":[595,841],"block_position":"67-15"},{"bbox":[73,523,340,550],"type":"text","angle":0,"index":16,"merge_prev":false,"text":"(3) 令 $x = \\sin t$ ，得到 $\\int_{0}^{\\frac{\\pi}{2}} \\frac{t}{\\sin t} \\cos t \\, \\mathrm{d}t = \\int_{0}^{\\frac{\\pi}{2}} t \\cot t \\, \\mathrm{d}t$ ，即 (1) ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":67,"id":"8169350c-2f26-4375-80ca-c076cf902f58","page_size":[595,841],"block_position":"67-16"},{"bbox":[54,549,537,587],"type":"text","angle":0,"index":17,"merge_prev":false,"text":"(4) 注意 $\\cot \\frac{x}{2} = \\frac{\\sin x}{1 - \\cos x}$ (或者可以当场对 $1 - \\cos x$ 展开)，因此 $I = \\int_{0}^{\\pi} x \\cot \\frac{x}{2} \\mathrm{d}x = 4 \\int_{0}^{\\frac{\\pi}{2}} t \\cot(t) \\mathrm{d}t$ ，变为 (1) ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":67,"id":"eb81cbb1-7b90-401d-8f26-cafcb84ca7c4","page_size":[595,841],"block_position":"67-17"},{"bbox":[56,621,163,634],"type":"title","angle":0,"index":18,"level":2,"text":"## 6.2.5 Dirichlet 积分","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":67,"id":"75593e04-ece1-48ed-b7a2-8a0ac4a26583","page_size":[595,841],"block_position":"67-18"},{"bbox":[73,648,186,660],"type":"title","angle":0,"index":19,"level":2,"text":"## 定理 6.17 (Dirichlet 积分)","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":67,"id":"273c6d8b-eb6a-4299-802b-5959f77db2dc","page_size":[595,841],"block_position":"67-19"},{"bbox":[68,665,229,692],"type":"text","angle":0,"index":20,"merge_prev":false,"text":"Dirichlet 积分 $\\int_{0}^{\\pi}\\frac{\\sin\\frac{2n+1}{2}x}{2\\sin\\frac{x}{2}}=\\frac{\\pi}{2}dx$ 。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":67,"id":"b9e28707-85de-4660-9775-d7f674e9f20b","page_size":[595,841],"block_position":"67-20"},{"bbox":[69,692,506,720],"type":"interline_equation","angle":0,"index":21,"text":"$$\n\\mathrm{推论} \\colon \\int_ {0} ^ {\\pi} {\\frac {\\sin {\\frac {2 n + 1}{2}} x}{\\sin {\\frac {x}{2}}}} \\mathrm{d} x = \\pi , \\int_ {0} ^ {\\frac {\\pi}{2}} {\\frac {\\sin (2 n + 1) x}{\\sin x}} \\mathrm{d} x = {\\frac {\\pi}{2}}, \\int_ {0} ^ {\\frac {\\pi}{2}} {\\frac {\\sin (2 n) x}{\\sin x}} \\mathrm{d} x = 0, \\int_ {0} ^ {\\pi} {\\frac {\\sin (2 n + 1) x}{\\sin x}} \\mathrm{d} x = \\pi\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":67,"id":"edc71521-a4a3-41fc-a985-ef106f81296c","page_size":[595,841],"block_position":"67-21"},{"bbox":[291,803,303,812],"type":"page_number","angle":0,"index":1000021,"text":"60 ","color":{"line":"rgba(164, 164, 164, 1)","fill":"rgba(164, 164, 164, 1)"},"page_idx":67,"is_discarded":true,"id":"7802f143-f4eb-4998-927f-00ca24dd0e2d","page_size":[595,841],"block_position":"67-22"}],[{"bbox":[457,47,539,58],"type":"header","angle":0,"index":-99999,"text":"6.3 积分 (不) 等式 ","color":{"line":"rgba(164, 164, 164, 1)","fill":"rgba(164, 164, 164, 1)"},"page_idx":68,"is_discarded":true,"id":"0af5d4ed-9dfc-4e3d-b55b-55a3d6ced28b","page_size":[595,841],"block_position":"68-0"},{"bbox":[54,68,533,132],"type":"text","angle":0,"index":1,"merge_prev":false,"text":"证明 (1)Dirichlet: 关键在于 $\\cos x + \\cos 2x + \\cdots + \\cos nx = \\frac{\\sin(n + \\frac{1}{2})x - \\sin\\frac{x}{2}}{2\\sin\\frac{x}{2}} = \\frac{\\sin\\frac{2n+1}{2}x}{2\\sin\\frac{x}{2}} - \\frac{1}{2}$ ，从而得到： $\\int_{0}^{\\pi}\\frac{\\sin\\frac{2n+1}{2}x}{2\\sin\\frac{x}{2}}dx=\\sum_{k=1}^{n}\\int_{0}^{\\pi}\\cos kxdx+\\frac{\\pi}{2}=\\sum_{k=1}^{n}\\left.\\frac{1}{k}\\sin kx\\right|_{0}^{\\pi}+\\frac{\\pi}{2}=\\frac{\\pi}{2}$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":68,"id":"f2c1e3e8-5a7c-473c-aad7-b574af19173c","page_size":[595,841],"block_position":"68-1"},{"bbox":[72,135,461,148],"type":"text","angle":0,"index":2,"merge_prev":false,"text":"(2) 推论：第一个就是把分母 2 拿走，第二个 (分子奇数情况) 取 $t = 2x$ ，第四个用对称性 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":68,"id":"00c1b10a-e240-4c43-81d1-763e11ea0ba2","page_size":[595,841],"block_position":"68-2"},{"bbox":[73,147,222,171],"type":"text","angle":0,"index":3,"merge_prev":false,"text":"(3) 第三个设 $I_{n} = \\int_{0}^{\\pi}\\frac{\\sin 2nx}{\\sin x}\\mathrm{d}x,$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":68,"id":"1eb33a97-51f3-4439-9f25-a743fcc0dcdd","page_size":[595,841],"block_position":"68-3"},{"bbox":[97,172,495,228],"type":"interline_equation","angle":0,"index":4,"text":"$$\n\\begin{array}{c} I _ {n} - I _ {n - 1} = \\int_ {0} ^ {\\pi} \\frac {\\sin 2 n x - \\sin (2 n - 2) x}{\\sin x} d x = \\int_ {0} ^ {\\pi} \\frac {2}{\\sin x} (\\cos \\frac {4 n - 2}{2} x \\cdot \\sin \\frac {2 n - (2 n - 2)}{2} x) \\mathrm{d} x \\\\ = 2 \\int_ {0} ^ {\\pi} \\cos (2 n - 1) x \\mathrm{d} x = \\frac {2}{2 n - 1} \\sin (2 n - 1) x \\bigg | _ {0} ^ {\\pi} = 0 \\end{array}\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":68,"id":"a112f4ba-c81d-41ae-b056-65388a36e1c9","page_size":[595,841],"block_position":"68-4"},{"bbox":[54,231,158,243],"type":"text","angle":0,"index":6,"merge_prev":false,"text":"而 $I_{1} = 0$ ，因此 $I_{n} = 0$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":68,"id":"d8b3fc4c-a26f-422c-80ef-3cca9c93c298","page_size":[595,841],"block_position":"68-5"},{"bbox":[55,261,211,275],"type":"title","angle":0,"index":7,"level":2,"text":"## 6.2.6 其他几个常见三角积分","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":68,"id":"b55bc726-b9cd-46b8-8fa4-ee5d207c8020","page_size":[595,841],"block_position":"68-6"},{"bbox":[35,284,565,364],"type":"text","angle":0,"index":8,"merge_prev":false,"text":"练习6.20几个常见三角分数积分(1) $\\int_0^\\pi \\frac{dx}{1 + \\sin^2x} dx$ (2) $\\int_0^{2\\pi}\\frac{dx}{1 + a\\cos x},(-1 < a < 1)$ 解(1)上下同乘 $\\sec^2 x$ ，得到 $\\int \\frac{\\sec^2x}{\\sec^2x + \\tan^2x} dx = \\int \\frac{d(\\tan x)}{2\\tan^2x + 1} = \\frac{1}{\\sqrt{2}}\\arctan (\\sqrt{2}\\tan x) + c,$ 注意 $\\tan x$ 在 $\\frac{\\pi}{2}$ 是间断的！根据对称性转换为 $I = 2\\int_0^{\\frac{\\pi}{2}}\\frac{dx}{1 + \\sin^2x} = \\sqrt{2}\\arctan (\\sqrt{2}\\tan x)|_0^{\\frac{\\pi}{2}} = \\sqrt{2} (\\frac{\\pi}{2} -0) = 2\\frac{\\pi}{\\sqrt{2}}$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":68,"id":"e04e9b06-2110-40d5-85e8-231c7d58f4f6","page_size":[595,841],"block_position":"68-7"},{"bbox":[54,363,581,414],"type":"text","angle":0,"index":9,"merge_prev":false,"text":"(2)具体的不定积分计算见不定积分一节,是非常经典的积分。 $I=2\\int_{0}^{\\pi}\\frac{dx}{1+a\\cos x}=\\frac{4}{\\sqrt{1-a^{2}}}\\arctan\\sqrt{\\frac{1-a}{1+a}}\\tan\\frac{x}{2}|_{0}^{\\pi}$ ，显然 $x=\\pi$ 为间断点，使用对称性非常正确！因此结果为 $\\frac{2\\pi}{\\sqrt{1-a^{2}}}$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":68,"id":"e451fcdf-77e7-4ef8-ade6-396daf397dbb","page_size":[595,841],"block_position":"68-8"},{"bbox":[52,416,539,444],"type":"text","angle":0,"index":10,"merge_prev":false,"text":"笔记 一定要注意函数的连续性，用对称性转换。因此能用对称性的一定要优先用对称性。上面两个题目也是重要的结论。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":68,"id":"83d9d5b9-be31-4e3d-80fc-d339e865befd","page_size":[595,841],"block_position":"68-9"},{"bbox":[55,482,174,498],"type":"title","angle":0,"index":11,"level":2,"text":"## 6.3 积分 (不) 等式","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":68,"id":"efd8f625-d08f-4b56-b64c-714cbc3ea6b3","page_size":[595,841],"block_position":"68-10"},{"bbox":[55,515,139,529],"type":"title","angle":0,"index":12,"level":2,"text":"## 6.3.1 整体放缩","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":68,"id":"c252991e-2c95-489e-a7b5-ee378046e0e2","page_size":[595,841],"block_position":"68-11"},{"bbox":[35,537,537,590],"type":"text","angle":0,"index":13,"merge_prev":false,"text":"练习6.21几个简单积分放缩证明(1) $1<\\int_{0}^{\\frac{\\pi}{2}}\\frac{\\sin x}{x}\\mathrm{d}x<\\frac{\\pi}{2}$ (2) $\\frac{2}{9}\\pi^{2}<\\int_{\\frac{\\pi}{6}}^{\\frac{\\pi}{2}}\\frac{2x}{\\sin x}\\mathrm{d}x<\\frac{1}{3}\\pi^{2}$ (3) $3\\sqrt{e}<\\int_{e}^{4e}\\frac{\\ln x}{\\sqrt{x}}<6$ 证明(1)根据 $f'(x)=\\frac{x-\\tan x}{x^{2}\\sec x}$ 可知函数单减，因此 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":68,"id":"284a9f67-e50c-4266-8f2d-bae47948b69b","page_size":[595,841],"block_position":"68-12"},{"bbox":[226,591,367,615],"type":"interline_equation","angle":0,"index":14,"text":"$$\nf \\left(\\frac {\\pi}{2}\\right) = \\frac {2}{\\pi} \\leq \\frac {\\sin x}{x} < f (0) = 1\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":68,"id":"7ab8efa9-4552-4345-808b-5eda2f3d4b0e","page_size":[595,841],"block_position":"68-13"},{"bbox":[73,615,212,628],"type":"text","angle":0,"index":15,"merge_prev":false,"text":"(2) 根据导数可知单调增，因此 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":68,"id":"62d5bea7-ec42-4640-bb70-09a357a12362","page_size":[595,841],"block_position":"68-14"},{"bbox":[218,629,374,653],"type":"interline_equation","angle":0,"index":16,"text":"$$\nf \\left(\\frac {\\pi}{6}\\right) = \\frac {2}{3} \\pi \\leq \\frac {2 x}{\\sin x} \\leq f \\left(\\frac {\\pi}{2}\\right) = \\pi\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":68,"id":"95e1f1e2-faf8-461d-852a-fe4e2b61fb3e","page_size":[595,841],"block_position":"68-15"},{"bbox":[72,654,329,681],"type":"text","angle":0,"index":17,"merge_prev":false,"text":"(3) $f'(x)=\\frac{2-\\ln x}{2x\\sqrt{x}}$ ，因此 $\\frac{1}{\\sqrt{e}}=f(e)\\leq f(x)\\leq f(e^{2})=\\frac{2}{e}$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":68,"id":"ccacd19b-48df-4f47-b238-6e0e625ab92e","page_size":[595,841],"block_position":"68-16"},{"bbox":[55,697,188,710],"type":"title","angle":0,"index":18,"level":2,"text":"## 6.3.2 同区域单调性放缩","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":68,"id":"4bbd74b9-de63-4fd4-b1d1-26bbc2b9cded","page_size":[595,841],"block_position":"68-17"},{"bbox":[35,719,528,776],"type":"text","angle":0,"index":19,"merge_prev":false,"text":"练习6.22几道经典同区域相减(1) $f(x)$ 在 $[a,b]$ 上严格增的连续函数，证明： $\\int_{a}^{b}xf(x)\\mathrm{d}x > \\frac{a + b}{2}\\int_{a}^{b}f(x)\\mathrm{d}x$ (2)证明 $\\int_0^{\\frac{\\pi}{2}}\\frac{\\cos x}{1 + x^2}\\mathrm{d}x > \\int_0^{\\frac{\\pi}{2}}\\frac{\\sin x}{1 + x^2}\\mathrm{d}x$ (3)证明 $\\int_0^{\\sqrt{2\\pi}}\\sin x^2\\mathrm{d}x > 0$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":68,"id":"81e21e7b-12d5-45d4-b4a9-5074668e4d5e","page_size":[595,841],"block_position":"68-18"},{"bbox":[290,803,302,812],"type":"page_number","angle":0,"index":1000004,"text":"61 ","color":{"line":"rgba(164, 164, 164, 1)","fill":"rgba(164, 164, 164, 1)"},"page_idx":68,"is_discarded":true,"id":"6a62687a-44e9-421d-85ec-5ae62b96101d","page_size":[595,841],"block_position":"68-19"}],[{"bbox":[457,47,539,58],"type":"header","angle":0,"index":-99999,"text":"6.3 积分 (不) 等式 ","color":{"line":"rgba(164, 164, 164, 1)","fill":"rgba(164, 164, 164, 1)"},"page_idx":69,"is_discarded":true,"id":"14064c5e-a191-4021-8c3f-509ebe28a07b","page_size":[595,841],"block_position":"69-0"},{"bbox":[54,69,538,108],"type":"text","angle":0,"index":1,"merge_prev":false,"text":"证明 (1) 等价于证明 $\\int_{a}^{b}\\left(x - \\frac{a + b}{2}\\right)f(x)\\mathrm{d}x > 0$ ，进行有目的性的换元，分为两段，用 $t$ 表示 $x$ 与 $\\frac{a + b}{2}$ 之间的距离： ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":69,"id":"dd3aea60-b8b2-4796-9518-a3bbbd833c62","page_size":[595,841],"block_position":"69-1"},{"bbox":[116,111,477,206],"type":"interline_equation","angle":0,"index":2,"text":"$$\n\\begin{array}{l} \\int_ {a} ^ {b} \\left(x - \\frac {a + b}{2}\\right) f (x) \\mathrm{d} x = \\int_ {a} ^ {\\frac {a + b}{2}} \\left(x - \\frac {a + b}{2}\\right) f (x) \\mathrm{d} x + \\int_ {\\frac {a + b}{2}} ^ {b} \\left(x - \\frac {a + b}{2}\\right) f (x) \\mathrm{d} x \\\\ = \\int_ {\\frac {b - a}{2}} ^ {0} (- t) f (\\frac {a + b}{2} - t) (- \\mathrm{d} t) + \\int_ {0} ^ {\\frac {b - a}{2}} t f (t + \\frac {a + b}{2}) \\mathrm{d} t \\\\ = \\int_ {0} ^ {\\frac {b - a}{2}} t \\left[ f (t + \\frac {a + b}{2}) - f (\\frac {a + b}{2} - t) \\right] \\mathrm{d} t > 0 \\end{array}\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":69,"id":"e116c4f5-09a4-4fa5-9c10-dfe003159cc6","page_size":[595,841],"block_position":"69-2"},{"bbox":[53,209,538,253],"type":"text","angle":0,"index":3,"merge_prev":false,"text":"(2) 等价于 $\\int_{0}^{\\frac{\\pi}{2}} \\frac{\\sin x - \\cos x}{1 + x^2} \\mathrm{d}x < 0$ ，而 $\\sin x - \\cos x = \\sqrt{2} \\sin (x - \\frac{\\pi}{4})$ 关于 $\\left(\\frac{\\pi}{4}, 0\\right)$ 中心对称，且 $\\frac{1}{1 + x^2}$ 单减。因此令 $u = x - \\frac{\\pi}{4}$ : ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":69,"id":"bdd6b0e7-a2d4-4634-9e75-3b8a93c64b89","page_size":[595,841],"block_position":"69-3"},{"bbox":[130,257,463,351],"type":"interline_equation","angle":0,"index":4,"text":"$$\n\\begin{array}{r l} & {\\int_ {0} ^ {\\frac {\\pi}{2}} \\frac {\\sin x - \\cos x}{1 + x ^ {2}} \\mathrm{d} x = \\int_ {0} ^ {\\frac {\\pi}{2}} \\frac {\\sqrt {2} \\sin (x - \\frac {\\pi}{4})}{1 + x ^ {2}} \\mathrm{d} x = \\sqrt {2} \\int_ {- \\frac {\\pi}{4}} ^ {\\frac {\\pi}{4}} \\int_ {- \\frac {\\pi}{4}} ^ {\\frac {\\pi}{4}} \\frac {\\sin u}{1 + (\\frac {\\pi}{4} + u) ^ {2}} \\mathrm{d} u} \\\\ & {\\qquad = \\sqrt {2} \\left(\\int_ {0} ^ {\\frac {\\pi}{4}} \\frac {\\sin x}{1 + (\\frac {\\pi}{4} + x) ^ {2}} \\mathrm{d} x - \\int_ {0} ^ {\\frac {\\pi}{4}} \\frac {\\sin x}{1 + (\\frac {\\pi}{4} - x) ^ {2}} \\mathrm{d} x\\right)} \\\\ & {\\qquad = \\sqrt {2} \\int_ {0} ^ {\\frac {\\pi}{4}} \\left(\\frac {1}{(\\frac {\\pi}{4} + x) ^ {2}} - \\frac {1}{(\\frac {\\pi}{4} - x) ^ {2}}\\right) \\sin x \\mathrm{d} x < 0} \\end{array}\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":69,"id":"737f4f38-59f5-499e-94af-ac7610fef6dd","page_size":[595,841],"block_position":"69-4"},{"bbox":[35,358,336,387],"type":"text","angle":0,"index":5,"merge_prev":false,"text":"练习6.23一道特殊题(1)证明 $\\int_0^1\\frac{\\cos x}{\\sqrt{1 - x^2}}\\mathrm{d}x > \\int_0^1\\frac{\\sin x}{\\sqrt{1 - x^2}}\\mathrm{d}x$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":69,"id":"8a645b33-ee24-4fef-8f30-b9faef4deb72","page_size":[595,841],"block_position":"69-5"},{"bbox":[55,402,226,417],"type":"title","angle":0,"index":6,"level":2,"text":"## 6.3.3 利用 Newton-Leibniz 公式","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":69,"id":"41ae11b3-2aec-483a-9ac0-3e750a9470d3","page_size":[595,841],"block_position":"69-6"},{"bbox":[72,429,217,442],"type":"title","angle":0,"index":7,"level":2,"text":"## 定理 6.18 (Newton-Leibniz 放缩)","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":69,"id":"c4781247-d717-4892-ac35-db70e587611b","page_size":[595,841],"block_position":"69-7"},{"bbox":[68,446,477,473],"type":"text","angle":0,"index":8,"merge_prev":false,"text":"若 $f(x)$ 为 $[a,b]$ 上连续可微函数，则对 $\\forall x_1,x_2\\in [a,b]$ 有 $f(x_{2}) = f(x_{1}) + \\int_{x_{1}}^{x_{2}}f^{\\prime}(x)\\mathrm{d}x$ ，因此 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":69,"id":"c257793b-fdfc-433e-9e63-e13ef7f2e455","page_size":[595,841],"block_position":"69-8"},{"bbox":[224,476,368,504],"type":"interline_equation","angle":0,"index":9,"text":"$$\n| f (x _ {2}) | \\leq | f (x _ {1}) | + \\int_ {a} ^ {b} | f ^ {\\prime} (x) | \\mathrm{d} x\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":69,"id":"91837eea-7a4c-49c8-a4f6-b81300d51dbd","page_size":[595,841],"block_position":"69-9"},{"bbox":[35,535,308,547],"type":"text","angle":0,"index":10,"merge_prev":false,"text":"练习 6.24 几道经典题 下面均假设 $f(x)$ 在区间上连续可微 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":69,"id":"816be819-4ef4-44cb-824b-90139a79ae98","page_size":[595,841],"block_position":"69-10"},{"bbox":[73,548,282,572],"type":"text","angle":0,"index":11,"merge_prev":false,"text":"(1) 证明 $|f(0)| \\leq \\frac{1}{a} \\int_{0}^{a} |f(x)| \\, \\mathrm{d}x + \\int_{0}^{a} |f'(x)| \\, \\mathrm{d}x$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":69,"id":"b8d87842-d789-472d-93ec-81501055d0b9","page_size":[595,841],"block_position":"69-11"},{"bbox":[73,572,328,602],"type":"text","angle":0,"index":12,"merge_prev":false,"text":"(2) 经典：证明 $|f(x)| \\leq \\left|\\frac{1}{b - a}\\int_{a}^{b} f(x)\\mathrm{d}x\\right| + \\int_{a}^{b}|f'(x)|\\mathrm{d}x$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":69,"id":"3d6667a0-da46-40e2-afed-dbfccda56b41","page_size":[595,841],"block_position":"69-12"},{"bbox":[73,602,251,628],"type":"text","angle":0,"index":13,"merge_prev":false,"text":"(3) 证明 $|f(x)| \\leq \\int_{0}^{1} (|f(x)| + |f'(x)|) \\, \\mathrm{d}x$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":69,"id":"0a560049-9aee-44d0-8553-a37f6ae81b4a","page_size":[595,841],"block_position":"69-13"},{"bbox":[73,627,323,655],"type":"text","angle":0,"index":14,"merge_prev":false,"text":"(4) 重点：证明 $\\left|f\\left(\\frac{1}{2}\\right)\\right| \\leq \\int_{0}^{1}|f(x)|\\mathrm{d}x + \\frac{1}{2}\\int_{0}^{1}|f'(x)|\\mathrm{d}x$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":69,"id":"1c2f4106-177c-4bfa-a8a0-0664b9198f59","page_size":[595,841],"block_position":"69-14"},{"bbox":[54,653,358,680],"type":"text","angle":0,"index":15,"merge_prev":false,"text":"证明 (1) 用积分第一中值定理可知 $\\frac{1}{a} \\int_{0}^{a} |f(x)| \\mathrm{d}x = |f(\\xi)|$ ，因此得证 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":69,"id":"5c19b0a3-ad3e-4015-a5a2-e6b5f5f2f6fe","page_size":[595,841],"block_position":"69-15"},{"bbox":[73,677,123,690],"type":"text","angle":0,"index":16,"merge_prev":false,"text":"(2)(3) 同理 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":69,"id":"5bb51731-43ee-4581-b1c1-69dc9df53b08","page_size":[595,841],"block_position":"69-16"},{"bbox":[73,687,434,719],"type":"text","angle":0,"index":17,"merge_prev":false,"text":"(4) 设 $|f(x_0)| = \\min_{x \\in [0, \\frac{1}{2}]} |f(x)|$ ，则 $|f(x_0)| = 2\\int_0^{\\frac{1}{2}} |f(x_0)| \\, \\mathrm{d}x \\leq 2\\int_0^{\\frac{1}{2}} |f(x)| \\, \\mathrm{d}x$ ，因此 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":69,"id":"5896cf81-1900-4eb8-b6b1-a611645e9ccf","page_size":[595,841],"block_position":"69-17"},{"bbox":[151,721,442,752],"type":"interline_equation","angle":0,"index":18,"text":"$$\n\\left| f \\left(\\frac {1}{2}\\right) \\right| = \\left| f (x _ {0}) + \\int_ {x _ {0}} ^ {\\frac {1}{2}} f ^ {\\prime} (x) \\mathrm{d} x \\right| \\leq 2 \\int_ {0} ^ {\\frac {1}{2}} | f (x) | \\mathrm{d} x + \\int_ {0} ^ {\\frac {1}{2}} | f ^ {\\prime} (x) | \\mathrm{d} x\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":69,"id":"24e70865-03c0-410c-8c81-9d9a6cf651c6","page_size":[595,841],"block_position":"69-18"},{"bbox":[291,802,302,812],"type":"page_number","angle":0,"index":1000018,"text":"62 ","color":{"line":"rgba(164, 164, 164, 1)","fill":"rgba(164, 164, 164, 1)"},"page_idx":69,"is_discarded":true,"id":"f6c1e84a-2e52-4279-b91a-3f7ef588fc8e","page_size":[595,841],"block_position":"69-19"}],[{"bbox":[458,47,539,58],"type":"header","angle":0,"index":-99999,"text":"6.3 积分 (不) 等式 ","color":{"line":"rgba(164, 164, 164, 1)","fill":"rgba(164, 164, 164, 1)"},"page_idx":70,"is_discarded":true,"id":"4bb77691-6eef-435a-949c-b1ca5f4cf45d","page_size":[595,841],"block_position":"70-0"},{"bbox":[54,68,174,93],"type":"text","angle":0,"index":1,"merge_prev":false,"text":"同理取 $[\\frac{1}{2}, 1]$ 的部分可得到 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":70,"id":"4aa74a8c-1a88-4398-a85a-dd58fd2cd533","page_size":[595,841],"block_position":"70-1"},{"bbox":[205,89,387,120],"type":"interline_equation","angle":0,"index":2,"text":"$$\n\\left| f \\left(\\frac {1}{2}\\right) \\right| \\leq 2 \\int_ {\\frac {1}{2}} ^ {1} | f (x) | \\mathrm{d} x + \\int_ {\\frac {1}{2}} ^ {1} | f ^ {\\prime} (x) | \\mathrm{d} x\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":70,"id":"4b819f63-bb8b-4e34-bb44-8a50b6520185","page_size":[595,841],"block_position":"70-2"},{"bbox":[54,126,202,139],"type":"text","angle":0,"index":3,"merge_prev":false,"text":"因此两式相加除以 2 即得到结果。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":70,"id":"0b8cd0a3-9265-4b91-a01a-b424820af280","page_size":[595,841],"block_position":"70-3"},{"bbox":[55,158,242,172],"type":"title","angle":0,"index":4,"level":2,"text":"## 6.3.4 利用 Cauchy-Schwarz 不等式","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":70,"id":"2e2a0fcf-aa15-4f1b-ab4c-cefdf984819a","page_size":[595,841],"block_position":"70-4"},{"bbox":[72,185,230,198],"type":"title","angle":0,"index":5,"level":2,"text":"## 定理 6.19 (Cauchy-Schwarz 不等式)","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":70,"id":"de069b94-3060-445e-a319-c07d7f0e3186","page_size":[595,841],"block_position":"70-5"},{"bbox":[68,202,214,216],"type":"text","angle":0,"index":6,"merge_prev":false,"text":"设 $f(x), g(x)$ 在 $[a, b]$ 上可积，则 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":70,"id":"d8de35e6-b430-43b9-bbaf-7525de3193ef","page_size":[595,841],"block_position":"70-6"},{"bbox":[196,218,399,252],"type":"interline_equation","angle":0,"index":7,"text":"$$\n\\left(\\int_ {a} ^ {b} f (x) g (x) \\mathrm{d} x\\right) ^ {2} \\leq \\int_ {a} ^ {b} f ^ {2} (x) \\mathrm{d} x \\int_ {a} ^ {b} g ^ {2} (x) \\mathrm{d} x\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":70,"id":"a0a3a7a6-2a09-4ddb-a034-5ee4bf761328","page_size":[595,841],"block_position":"70-7"},{"bbox":[74,273,112,285],"type":"title","angle":0,"index":8,"level":2,"text":"## 推论6.2","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":70,"id":"739272cf-30cd-43c1-9a4a-478e77bc44eb","page_size":[595,841],"block_position":"70-8"},{"bbox":[68,289,189,302],"type":"text","angle":0,"index":9,"merge_prev":false,"text":"若 $f(x), g(x)$ 非负可积，则 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":70,"id":"cfd5df84-6375-4ae1-90ad-eaf81d316b0c","page_size":[595,841],"block_position":"70-9"},{"bbox":[193,305,399,338],"type":"interline_equation","angle":0,"index":10,"text":"$$\n\\int_ {a} ^ {b} f (x) \\mathrm{d} x \\int_ {a} ^ {b} g (x) \\mathrm{d} x \\geq \\left(\\int_ {a} ^ {b} \\sqrt {f (x) g (x)} \\mathrm{d} x\\right) ^ {2}\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":70,"id":"ccc32711-d5b0-45f5-a347-5d6cafdcf2cd","page_size":[595,841],"block_position":"70-10"},{"bbox":[35,367,459,421],"type":"text","angle":0,"index":11,"merge_prev":false,"text":"练习 6.25 一些经典题目 (1) $f(x)$ 在 [0,1] 连续可微， $f(1)=f(0)=1$ ，证明 $\\int_{0}^{1}[f'(x)]^{2}\\mathrm{d}x\\geq1$ 证明 (1) 直接根据 Cauchy-Schwarz 不等式得到 $\\int_{0}^{1}[f'(x)]^{2}\\mathrm{d}x\\geq\\left[\\int_{0}^{1}f'(x)\\mathrm{d}x\\right]^{2}\\geq\\frac{1}{4}$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":70,"id":"9d822852-d4a4-46b6-b3ce-bf84bd59552d","page_size":[595,841],"block_position":"70-11"},{"bbox":[55,436,187,450],"type":"title","angle":0,"index":12,"level":2,"text":"## 6.3.5 利用微分中值定理","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":70,"id":"bb6540bf-a2db-4941-b915-3255ac6b0b3c","page_size":[595,841],"block_position":"70-12"},{"bbox":[35,461,539,497],"type":"text","angle":0,"index":13,"merge_prev":false,"text":"练习 6.26 几道具体分段题 (1) 一阶导: $f(x)$ 在 $[0,2]$ 上可导, $f(0)=f(2)=1$ , $|f'(x)|\\leq1$ , 证明 $1<\\int_{0}^{2}f(x)\\mathrm{d}x<3$ (2) 二阶导: ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":70,"id":"f02784c6-98ca-47af-a292-2be458409fff","page_size":[595,841],"block_position":"70-13"},{"bbox":[52,500,295,513],"type":"text","angle":0,"index":15,"merge_prev":false,"text":"证明 (1) 左侧：进行分段处理，对于 $[0, 1], [1, 2]$ 分别有 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":70,"id":"5b6f7430-365c-4b1c-99a1-aeea5dd67c43","page_size":[595,841],"block_position":"70-14"},{"bbox":[189,517,404,554],"type":"interline_equation","angle":0,"index":16,"text":"$$\n\\left\\{ \\begin{array}{l l} f (x) = f (0) + f ^ {\\prime} (\\xi) x \\geq 1 - x, & x \\in [ 0, 1 ] \\\\ f (x) = f (2) + f ^ {\\prime} (\\eta) (x - 2) \\geq x - 1, & x \\in [ 1, 2 ] \\end{array} \\right.\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":70,"id":"e7bc2221-2924-4bfd-a7c6-a09bf6f82318","page_size":[595,841],"block_position":"70-15"},{"bbox":[53,557,539,596],"type":"text","angle":0,"index":17,"merge_prev":false,"text":"因此 $\\int_0^2 f(x)\\mathrm{d}x\\geq \\int_0^1 (1 - x)\\mathrm{d}x + \\int_1^2 (x - 1)\\mathrm{d}x = 1$ ，且等号取到时在1处不可导，故严格。同理可证明 $f(x)\\geq h(x)$ 其中 $h(x)$ 为 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":70,"id":"d41d9b44-d4c0-47fb-b2f5-0b05496bc080","page_size":[595,841],"block_position":"70-16"},{"bbox":[236,596,356,634],"type":"interline_equation","angle":0,"index":18,"text":"$$\nh (x) = \\left\\{ \\begin{array}{l l} 1 + x, & x \\in [ 0, 1 ] \\\\ 3 - x, & x \\in [ 1, 2 ] \\end{array} \\right.\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":70,"id":"a8e8e356-f16a-45a7-8013-10ab92e88f3b","page_size":[595,841],"block_position":"70-17"},{"bbox":[54,640,146,669],"type":"text","angle":0,"index":19,"merge_prev":false,"text":"因此 $\\int_0^2 f(x)\\mathrm{d}x\\leq 3$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":70,"id":"6221392c-2602-4a2b-b21a-41322df1bfdb","page_size":[595,841],"block_position":"70-18"},{"bbox":[35,679,418,698],"type":"text","angle":0,"index":20,"merge_prev":false,"text":"练习6.27利用一阶导数研究积分 $f(x)$ 在 $[a,b]$ 连续可微， $M = \\max_{x\\in [a,b]}|f'(x)|$ ，证明 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":70,"id":"7361eebc-4dcd-4c55-b59c-e9b5adc9fa53","page_size":[595,841],"block_position":"70-19"},{"bbox":[73,699,393,729],"type":"text","angle":0,"index":21,"merge_prev":false,"text":"(1) 若 $f(a) = 0$ 或 $f(b) = 0$ ，则 $\\left|\\int_{a}^{b} f(x) \\mathrm{d}x\\right| \\leq \\int_{a}^{b} |f(x)| \\mathrm{d}x \\leq \\frac{M}{2}(b - a)^{2}$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":70,"id":"2297d93d-c27b-4db3-aebb-8660bdfdc29c","page_size":[595,841],"block_position":"70-20"},{"bbox":[73,729,374,761],"type":"text","angle":0,"index":22,"merge_prev":false,"text":"(2) 若 $f(a) = f(b) = 0$ ，则 $\\left|\\int_{a}^{b} f(x) \\mathrm{d}x\\right| \\leq \\int_{a}^{b} |f(x)| \\mathrm{d}x \\leq \\frac{M}{4}(b - a)^2$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":70,"id":"3d1502cf-5e34-49e4-b386-6f0d72166c54","page_size":[595,841],"block_position":"70-21"},{"bbox":[291,802,302,812],"type":"page_number","angle":0,"index":1000013,"text":"63 ","color":{"line":"rgba(164, 164, 164, 1)","fill":"rgba(164, 164, 164, 1)"},"page_idx":70,"is_discarded":true,"id":"2f4bff5b-1559-497d-b0cd-f8a60f54ce83","page_size":[595,841],"block_position":"70-22"}],[{"bbox":[458,47,539,58],"type":"header","angle":0,"index":-99999,"text":"6.3 积分 (不) 等式 ","color":{"line":"rgba(164, 164, 164, 1)","fill":"rgba(164, 164, 164, 1)"},"page_idx":71,"is_discarded":true,"id":"20679f2a-c0a3-410b-8e5d-0ddead26e8dc","page_size":[595,841],"block_position":"71-0"},{"bbox":[73,69,170,96],"type":"text","angle":0,"index":1,"merge_prev":false,"text":"(3) 若 $f\\left(\\frac{a + b}{2}\\right) = 0$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":71,"id":"ddcb3d53-7bd0-44cf-8657-7c283a1d4652","page_size":[595,841],"block_position":"71-1"},{"bbox":[55,94,480,133],"type":"text","angle":0,"index":2,"merge_prev":false,"text":"证明 (1) 设 $f(a) = 0$ ，则 $f(x) = f'(\\xi)(x - a)$ ，因此 $\\int_{a}^{b}|f(x)|\\mathrm{d}x \\leq M\\int_{a}^{b}(x - a)\\mathrm{d}x = \\frac{M}{2}(b - a)^2$ (2) 若 $f(a) = f(b) = 0$ ，则分两段在 $a, b$ 展开 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":71,"id":"3ead919a-ac7a-4d3c-856c-ed9d75b5858a","page_size":[595,841],"block_position":"71-2"},{"bbox":[219,136,361,192],"type":"interline_equation","angle":0,"index":3,"text":"$$\n\\left\\{ \\begin{array}{l} \\int_ {a} ^ {\\frac {a + b}{2}} | f (x) | \\mathrm{d} x \\leq \\frac {M}{2} \\cdot \\frac {(b - a) ^ {2}}{4} \\\\ \\int_ {\\frac {a + b}{2}} ^ {b} | f (x) | \\mathrm{d} x \\leq \\frac {M}{2} \\cdot \\frac {(b - a) ^ {2}}{4} \\end{array} \\right.\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":71,"id":"7d0d3846-e071-40cd-af95-f80faee0fa5a","page_size":[595,841],"block_position":"71-3"},{"bbox":[73,194,405,222],"type":"text","angle":0,"index":4,"merge_prev":false,"text":"(3) 若 $f\\left(\\frac{a + b}{2}\\right) = 0$ ，则分两段在 $f\\left(\\frac{a + b}{2}\\right)$ 展开，和 (2) 类似可得到结果 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":71,"id":"097a2038-1151-4961-ad8d-7254dc0b44b8","page_size":[595,841],"block_position":"71-4"},{"bbox":[35,232,394,245],"type":"text","angle":0,"index":5,"merge_prev":false,"text":"练习6.28利用二阶导数研究积分 $f(x)$ 在 $[a,b]$ 二阶可微， $|f''(x)| \\leq M$ ，证明 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":71,"id":"fd51c3a1-ec6a-4f36-b400-0c94d94a67d3","page_size":[595,841],"block_position":"71-5"},{"bbox":[73,245,305,275],"type":"text","angle":0,"index":6,"merge_prev":false,"text":"(1) 若 $f\\left(\\frac{a + b}{2}\\right) = 0$ ，则 $\\left|\\int_{a}^{b}f(x)\\mathrm{d}x\\right| \\leq \\frac{M}{24}(b - a)^3$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":71,"id":"b8b0937d-2fad-40c7-af2c-1ac23a8fba9a","page_size":[595,841],"block_position":"71-6"},{"bbox":[73,276,308,306],"type":"text","angle":0,"index":7,"merge_prev":false,"text":"(2) 若 $f(a) = f(b) = 0$ ，则 $\\left|\\int_{a}^{b} f(x) \\, \\mathrm{d}x\\right| \\leq \\frac{M}{12}(b - a)^3$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":71,"id":"7a281ff7-ea7f-4fda-a327-a0456d3b89a8","page_size":[595,841],"block_position":"71-7"},{"bbox":[52,305,170,328],"type":"text","angle":0,"index":8,"merge_prev":false,"text":"证明 (1) 在 $\\frac{a + b}{2}$ 处展开: ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":71,"id":"79fdefff-b569-4350-9243-9e1dc947b707","page_size":[595,841],"block_position":"71-8"},{"bbox":[144,331,449,359],"type":"interline_equation","angle":0,"index":9,"text":"$$\nf (x) = f \\left(\\frac {a + b}{2}\\right) + f ^ {\\prime} \\left(\\frac {a + b}{2}\\right) \\left(x - \\frac {a + b}{2}\\right) + \\frac {f ^ {\\prime \\prime} (\\xi)}{2} \\left(x - \\frac {a + b}{2}\\right) ^ {2}\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":71,"id":"8c3a2afc-fed0-47e3-81cb-19a634332851","page_size":[595,841],"block_position":"71-9"},{"bbox":[54,362,538,412],"type":"text","angle":0,"index":10,"merge_prev":false,"text":"显然一阶导数项中心对称，故消去，二阶导数项用积分第一中值定理可以得到： $\\int_{a}^{b}f(x)\\mathrm{d}x=(b-a)f\\left(\\frac{a+b}{2}\\right)+\\frac{f''(\\eta)}{24}(b-a)^{3}$ ，经过放缩即结论。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":71,"id":"316eec6e-851f-4970-82a8-2c030772bca5","page_size":[595,841],"block_position":"71-10"},{"bbox":[291,803,303,812],"type":"page_number","angle":0,"index":1000010,"text":"64 ","color":{"line":"rgba(164, 164, 164, 1)","fill":"rgba(164, 164, 164, 1)"},"page_idx":71,"is_discarded":true,"id":"003317bd-4aac-4091-97df-c2978c808213","page_size":[595,841],"block_position":"71-11"}],[{"bbox":[236,74,368,94],"type":"title","angle":0,"index":0,"level":2,"text":"## 第7章 反常积分","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":72,"id":"937eb59d-22c0-4482-a4b1-038e69cbea31","page_size":[595,841],"block_position":"72-0"},{"bbox":[52,118,539,147],"type":"text","angle":0,"index":1,"merge_prev":false,"text":"一般的定积分都定义于闭区间中，但很多实际问题中的积分可能区间无穷或者某点不连续，这就需要引出无穷积分和瑕积分的概念。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":72,"id":"da969457-0191-42c2-9760-3881ac586249","page_size":[595,841],"block_position":"72-1"},{"bbox":[56,169,172,185],"type":"title","angle":0,"index":2,"level":2,"text":"## 7.1 反常积分理论","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":72,"id":"396e919c-5e75-4e34-a602-a23ba4b5552f","page_size":[595,841],"block_position":"72-2"},{"bbox":[56,201,210,216],"type":"title","angle":0,"index":3,"level":2,"text":"## 7.1.1 无穷积分与瑕积分概念","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":72,"id":"d1858cc1-095e-406b-b2ee-34e8cdf160cf","page_size":[595,841],"block_position":"72-3"},{"bbox":[73,229,201,242],"type":"title","angle":0,"index":4,"level":2,"text":"## 定义 7.1 (无穷积分与瑕积分)","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":72,"id":"2adb88ab-c3fe-4080-b289-d83a96d5af2e","page_size":[595,841],"block_position":"72-4"},{"bbox":[83,245,498,273],"type":"text","angle":0,"index":5,"merge_prev":false,"text":"- 无穷积分：设 $f$ 在 $[a, +\\infty)$ 有定义，若极限 $\\lim_{b \\to +\\infty} \\int_{a}^{b} f(x) \\mathrm{d}x$ 存在，则记其为 $\\int_{a}^{+\\infty} f(x) \\mathrm{d}x$ 。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":72,"id":"66980edc-514f-46f5-aaf9-52457ff6829c","page_size":[595,841],"block_position":"72-5"},{"bbox":[83,272,521,302],"type":"text","angle":0,"index":6,"merge_prev":false,"text":"- 瑕积分： $f$ 在 $(a, b]$ 有定义，且 $\\lim_{x \\to a^{+}} f(x) = \\infty$ ，若 $\\lim_{\\epsilon \\to 0^{+}} \\int_{a + \\epsilon}^{b} f(x) \\mathrm{d}x$ 收敛，则记其为 $\\int_{a}^{b} f(x) \\mathrm{d}x$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":72,"id":"c0a742c9-2d7b-4fbc-874f-bc6cb055569f","page_size":[595,841],"block_position":"72-6"},{"bbox":[52,311,533,338],"type":"text","angle":0,"index":7,"merge_prev":false,"text":"笔记 如果 $\\lim_{x\\to a^{+}}f(x)\\neq \\infty$ ，则不能称 $a$ 为 $f(x)$ 的瑕点，比如 $\\int_0^1\\frac{x}{1 - e^x}\\mathrm{d}x$ ，0不是其瑕点，其就是个定积分。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":72,"id":"8d7d4460-41f5-4265-820e-ae6110458652","page_size":[595,841],"block_position":"72-7"},{"bbox":[73,347,218,359],"type":"title","angle":0,"index":8,"level":2,"text":"## 定理 7.1 (反常积分 Leibniz 公式)","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":72,"id":"c0bd59fa-a3c8-4653-b53f-3379e35a3007","page_size":[595,841],"block_position":"72-8"},{"bbox":[83,364,478,391],"type":"text","angle":0,"index":9,"merge_prev":false,"text":"- 无穷积分：设 $\\int_{a}^{+\\infty}f(x)\\mathrm{d}x$ 存在，且 $f$ 有原函数 $F$ ，则 $\\int_{a}^{+\\infty}f(x)\\mathrm{d}x = F(+\\infty) - F(a)$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":72,"id":"6b394ca1-79d4-4660-a8c9-a860dfa86887","page_size":[595,841],"block_position":"72-9"},{"bbox":[83,393,475,418],"type":"text","angle":0,"index":10,"merge_prev":false,"text":"- 瑕积分：若 $F$ 是 $f$ 在 $(a, b]$ 的一个原函数， $a$ 是瑕点，则 $\\int_{a}^{b} f(x) \\mathrm{d}x = F(b) - F(a + 0)$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":72,"id":"e216f775-f9a3-4472-9f33-9ec79c640a5c","page_size":[595,841],"block_position":"72-10"},{"bbox":[74,438,161,450],"type":"title","angle":0,"index":11,"level":2,"text":"## 定理 7.2 (定义判敛)","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":72,"id":"d1af9766-4d2a-4abc-9949-2293a1464a6a","page_size":[595,841],"block_position":"72-11"},{"bbox":[83,454,423,481],"type":"text","angle":0,"index":12,"merge_prev":false,"text":"无穷积分： $\\int_{a}^{+\\infty}f(x)\\mathrm{d}x$ 收敛当且仅当 $\\forall \\epsilon ,\\exists A,\\forall u\\geq A,|\\int_{u}^{+\\infty}f(x)\\mathrm{d}x| < \\epsilon$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":72,"id":"0be99843-f3ad-43c5-b95e-d50ed29be4c2","page_size":[595,841],"block_position":"72-12"},{"bbox":[83,481,426,509],"type":"text","angle":0,"index":13,"merge_prev":false,"text":"- 瑕积分： $\\int_{a}^{b}f(x)\\mathrm{d}x$ 收敛当且仅当 $\\forall \\epsilon ,\\exists \\delta ,\\forall u\\in (a,a + \\delta)$ 有 $|\\int_{a}^{u}f(x)\\mathrm{d}x| < \\epsilon$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":72,"id":"fc9d3255-07dd-4446-bc73-5b7dcb53eeb6","page_size":[595,841],"block_position":"72-13"},{"bbox":[54,523,176,536],"type":"text","angle":0,"index":14,"merge_prev":false,"text":"证明 根据收敛的定义可知。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":72,"id":"c151aef5-f561-4f8c-b2f6-6af5863c7064","page_size":[595,841],"block_position":"72-14"},{"bbox":[73,544,186,568],"type":"title","angle":0,"index":15,"level":2,"text":"## 定理 7.3 ( $\\frac{1}{x^{p}}$ 的反常积分)","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":72,"id":"944cda91-473a-4122-ac1b-6cd0b4115e8c","page_size":[595,841],"block_position":"72-15"},{"bbox":[67,570,424,602],"type":"text","angle":0,"index":16,"merge_prev":false,"text":"$\\int_0^1\\frac{\\mathrm{d}x}{x^p}$ 在 $p < 1$ 时收敛， $p \\geq 1$ 时发散。 $\\int_1^{+\\infty}\\frac{\\mathrm{d}x}{x^p}$ 在 $p > 1$ 时收敛， $p \\leq 1$ 时发散 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":72,"id":"2cbebbc7-06e0-44ee-b9b5-14c41e0fdf84","page_size":[595,841],"block_position":"72-16"},{"bbox":[53,615,224,629],"type":"text","angle":0,"index":17,"merge_prev":false,"text":"证明 只讨论 $[1, +\\infty)$ 上的，直接积分： ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":72,"id":"935911a7-9769-49d0-a230-ab386880be8d","page_size":[595,841],"block_position":"72-17"},{"bbox":[135,632,456,676],"type":"interline_equation","angle":0,"index":18,"text":"$$\n\\int_ {1} ^ {u} \\frac {\\mathrm{d} x}{x ^ {p}} = \\left\\{ \\begin{array}{l l} \\frac {1}{1 - p} (u ^ {1 - p} - 1), & p \\neq 1 \\\\ \\ln u, & p = 1 \\end{array} \\right. \\Rightarrow \\lim _ {u \\to + \\infty} \\int_ {1} ^ {u} \\frac {\\mathrm{d} x}{x ^ {p}} = \\left\\{ \\begin{array}{l l} \\frac {1}{p - 1}, & p > 1 \\\\ + \\infty , & p \\leq 1 \\end{array} \\right.\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":72,"id":"8f091050-79e9-406c-9abd-dc37124d21f5","page_size":[595,841],"block_position":"72-18"}],[{"bbox":[459,46,539,58],"type":"header","angle":0,"index":-99999,"text":"7.1 反常积分理论 ","color":{"line":"rgba(164, 164, 164, 1)","fill":"rgba(164, 164, 164, 1)"},"page_idx":73,"is_discarded":true,"id":"3c59de5e-d1e5-4cd8-ad5d-13d07ae12f7e","page_size":[595,841],"block_position":"73-0"},{"bbox":[56,72,198,87],"type":"title","angle":0,"index":1,"level":2,"text":"## 7.1.2 反常积分比较判别法","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":73,"id":"beaa3f07-8a25-49f1-a82f-54c492299b4a","page_size":[595,841],"block_position":"73-1"},{"bbox":[73,100,196,112],"type":"title","angle":0,"index":2,"level":2,"text":"## 定理 7.4 (Cauchy 收敛准则)","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":73,"id":"cd05858f-fddc-46b1-9111-fcedda9bc917","page_size":[595,841],"block_position":"73-2"},{"bbox":[83,116,394,144],"type":"text","angle":0,"index":3,"merge_prev":false,"text":"无穷积分： $\\int_{a}^{\\infty}f(x)\\mathrm{d}x$ 收敛当且仅当 $\\forall \\epsilon ,\\exists G\\geq a,\\forall u_1,u_2 > G$ ，满足 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":73,"id":"bb7fec67-c498-4629-8560-c5ee1709f6f6","page_size":[595,841],"block_position":"73-3"},{"bbox":[268,145,350,172],"type":"interline_equation","angle":0,"index":4,"text":"$$\n\\left| \\int_ {u _ {1}} ^ {u _ {2}} f (x) \\mathrm{d} x \\right| < \\epsilon\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":73,"id":"dc091771-155b-49b0-b902-1f00f22f4207","page_size":[595,841],"block_position":"73-4"},{"bbox":[83,174,458,202],"type":"text","angle":0,"index":5,"merge_prev":false,"text":"- 瑕积分： $\\int_{a}^{b}f(x)\\mathrm{d}x$ 的瑕点为 $a$ ，则其收敛当且仅当 $\\forall \\epsilon ,\\exists \\delta ,\\forall u_1,u_2\\in (a,a + \\delta)$ 满足 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":73,"id":"855b9d62-0cc1-4888-bd88-3ea91d19e8d1","page_size":[595,841],"block_position":"73-5"},{"bbox":[268,203,349,231],"type":"interline_equation","angle":0,"index":6,"text":"$$\n\\left| \\int_ {u _ {1}} ^ {u _ {2}} f (x) \\mathrm{d} x \\right| < \\epsilon\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":73,"id":"40622785-2621-4a02-ba2f-50969495d37e","page_size":[595,841],"block_position":"73-6"},{"bbox":[73,251,210,264],"type":"title","angle":0,"index":7,"level":2,"text":"## 定义 7.2 (绝对收敛与条件收敛)","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":73,"id":"4cc47989-04b9-4015-83f4-7cb3a166d8b5","page_size":[595,841],"block_position":"73-7"},{"bbox":[69,266,524,307],"type":"text","angle":0,"index":8,"merge_prev":false,"text":"若 $\\int_{a}^{b}|f(x)|\\mathrm{d}x$ 收敛（ $b$ 可以为 $\\infty$ ），则称 $\\int_{a}^{b}f(x)\\mathrm{d}x$ 绝对收敛；若 $\\int_{a}^{b}f(x)\\mathrm{d}x$ 收敛但不绝对收敛，则称其条件收敛。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":73,"id":"edb7f836-e160-4bc9-a733-ab036c94d7e2","page_size":[595,841],"block_position":"73-8"},{"bbox":[73,327,161,339],"type":"title","angle":0,"index":9,"level":2,"text":"## 定理 7.5 (绝对收敛)","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":73,"id":"3b2bd140-f6f6-4df2-8a1c-ceac19a2ae94","page_size":[595,841],"block_position":"73-9"},{"bbox":[83,342,426,372],"type":"text","angle":0,"index":10,"merge_prev":false,"text":"- 无穷积分：若 $\\int_{a}^{+\\infty}|f(x)|\\mathrm{d}x$ 收敛，则 $\\int_{a}^{+\\infty}f(x)\\mathrm{d}x$ 也收敛，且有下述等式 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":73,"id":"967a3218-91e8-4f75-a801-c13c11288be4","page_size":[595,841],"block_position":"73-10"},{"bbox":[236,373,381,401],"type":"interline_equation","angle":0,"index":11,"text":"$$\n| \\int_ {a} ^ {+ \\infty} f (x) \\mathrm{d} x | \\leq \\int_ {a} ^ {+ \\infty} | f (x) | \\mathrm{d} x\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":73,"id":"497053b7-3eab-42cd-b1c9-5b288b21f1d7","page_size":[595,841],"block_position":"73-11"},{"bbox":[83,402,524,440],"type":"text","angle":0,"index":12,"merge_prev":false,"text":"- 瑕积分：设 $f$ 瑕点为 $a$ ， $f$ 在 $(a, b]$ 任一内闭区间上可积，则当 $\\int_{a}^{b} |f(x)| \\mathrm{d}x$ 收敛时， $\\int_{a}^{b} f(x) \\mathrm{d}x$ 也收敛，且 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":73,"id":"7a07d45f-7ec7-4a57-b596-aaa3363bce7d","page_size":[595,841],"block_position":"73-12"},{"bbox":[248,439,370,468],"type":"interline_equation","angle":0,"index":13,"text":"$$\n| \\int_ {a} ^ {b} f (x) \\mathrm{d} x | \\leq \\int_ {a} ^ {b} | f (x) | \\mathrm{d} x\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":73,"id":"9b3092ad-1ef9-4339-8b91-a1ebb3f0fc3b","page_size":[595,841],"block_position":"73-13"},{"bbox":[54,482,401,496],"type":"text","angle":0,"index":14,"merge_prev":false,"text":"证明 用 Cauchy 收敛准则转换至有限区间，再根据积分绝对值放缩可得到结论。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":73,"id":"0a36eccb-e9a5-40a9-b157-db4c907e81b6","page_size":[595,841],"block_position":"73-14"},{"bbox":[73,505,171,518],"type":"title","angle":0,"index":15,"level":2,"text":"## 定理 7.6 (比较判别法)","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":73,"id":"b690a5aa-a63e-4e58-a7da-e9745df77439","page_size":[595,841],"block_position":"73-15"},{"bbox":[69,521,186,534],"type":"text","angle":0,"index":16,"merge_prev":false,"text":"下面考虑 $f(x)$ 非负，则有 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":73,"id":"838dd557-4b5f-47fe-b05e-1cea129ace0d","page_size":[595,841],"block_position":"73-16"},{"bbox":[83,532,525,570],"type":"text","angle":0,"index":17,"merge_prev":false,"text":"- 无穷积分： $f, g$ 在任意有限区间可积，且 $0 \\leq f(x) \\leq g(x)$ ，当 $\\int_{a}^{+\\infty} g(x) \\mathrm{d}x$ 收敛时， $\\int_{a}^{+\\infty} f(x) \\mathrm{d}x$ 必收敛 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":73,"id":"24a17d3f-6dcd-46f7-b55e-7e0af523bc5b","page_size":[595,841],"block_position":"73-17"},{"bbox":[83,568,525,605],"type":"text","angle":0,"index":18,"merge_prev":false,"text":"- 瑕积分： $f, g$ 在内包有限闭区间可积，且 $0 \\leq f(x) \\leq g(x)$ ，当 $\\int_{a}^{b} g(x) \\mathrm{d}x$ 收敛时， $\\int_{a}^{b} f(x) \\mathrm{d}x$ 必收敛 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":73,"id":"2a008a68-c2b0-489c-9891-57a9f50fa11d","page_size":[595,841],"block_position":"73-18"},{"bbox":[73,624,191,637],"type":"title","angle":0,"index":19,"level":2,"text":"## 推论 7.1 (等价无穷小判别)","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":73,"id":"67886f7a-f23f-443a-af05-be27ba039166","page_size":[595,841],"block_position":"73-19"},{"bbox":[82,641,512,693],"type":"text","angle":0,"index":20,"merge_prev":false,"text":"- 无穷积分： $f$ 在任意有限闭区间可积，且 $x \\to \\infty$ 时 $f(x) \\sim \\frac{\\lambda}{x^p}$ ，若 $p > 1$ 则收敛， $p \\leq 1$ 则发散\n- 瑕积分： $f$ 在任意闭区间可积， $x \\to a$ 时 $f(x) \\sim \\frac{\\lambda}{(x - a)^p}$ ，若 $p < 1$ 则收敛， $p \\geq 1$ 则发散 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":73,"id":"6006dc44-386b-47cc-920c-5f7ad175e556","page_size":[595,841],"block_position":"73-20"},{"bbox":[35,698,539,759],"type":"text","angle":0,"index":21,"merge_prev":false,"text":"练习7.1基础训练 判断(1) $\\int_{0}^{+\\infty}\\frac{x\\arctan x}{1+x^{p}}\\mathrm{d}x$ (2) $\\int_{0}^{+\\infty}\\frac{1}{x^{p}+x^{q}}\\mathrm{d}x$ (3) $\\int_{0}^{+\\infty}\\frac{x^{p}}{1+x^{q}}\\mathrm{d}x$ 解(1)0不是瑕点，只需考虑 $\\infty$ 。若p>0， $\\frac{x\\arctan x}{1+x^{p}}\\sim\\frac{\\pi}{2}\\frac{1}{x^{p-1}}$ ，因此要求p>2，此时x不是瑕点。若p=0，显然发散。若p<0， $\\infty$ 显然发散。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":73,"id":"d727f419-37c6-449c-b08c-63827068e2df","page_size":[595,841],"block_position":"73-21"},{"bbox":[291,803,303,812],"type":"page_number","angle":0,"index":1000021,"text":"66 ","color":{"line":"rgba(164, 164, 164, 1)","fill":"rgba(164, 164, 164, 1)"},"page_idx":73,"is_discarded":true,"id":"d95ed40f-aadd-4467-a89a-e490aef942ef","page_size":[595,841],"block_position":"73-22"}],[{"bbox":[459,47,539,58],"type":"header","angle":0,"index":-99998,"text":"7.1 反常积分理论 ","color":{"line":"rgba(164, 164, 164, 1)","fill":"rgba(164, 164, 164, 1)"},"page_idx":74,"is_discarded":true,"id":"a5bfc2b6-ea8e-4c9e-ae34-04f37f560f37","page_size":[595,841],"block_position":"74-0"},{"bbox":[54,68,539,114],"type":"text","angle":0,"index":2,"merge_prev":false,"text":"(2) $0, +\\infty$ 均有可能是瑕点。先考虑 $p = q$ ，此时 $\\frac{1}{x^p + x^q} = \\frac{1}{2x^p}$ ， $\\infty$ 处要求 $p > 1$ ，但是 0 处要求 $p < 1$ ，因此不可能成立。若 $p < q$ ，则 $x \\to 0$ 时 $\\frac{1}{x^p + x^q} \\sim \\frac{1}{x^p}$ ，要求 $p < 1$ ，同理 $x \\to \\infty$ 时要求 $q > 1$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":74,"id":"579a7c52-0cc2-4e03-a771-5403e85d3a24","page_size":[595,841],"block_position":"74-1"},{"bbox":[73,114,470,137],"type":"text","angle":0,"index":3,"merge_prev":false,"text":"(3) $\\frac{x^p}{1 + x^q} = \\frac{1}{x^{-p} + x^{q - p}}$ ，当 $-p < 1 < q - p$ 或者 $q - p < 1 < -p$ 时收敛，其余情况发散。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":74,"id":"82f51c43-565f-4128-9e7c-9f83dc1c8318","page_size":[595,841],"block_position":"74-2"},{"bbox":[72,146,221,158],"type":"title","angle":0,"index":4,"level":2,"text":"## 推论 7.2 (对数、指数比较判别法)","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":74,"id":"644fe36f-5d86-4732-b207-941ab288ffe5","page_size":[595,841],"block_position":"74-3"},{"bbox":[81,165,338,185],"type":"interline_equation","angle":0,"index":5,"text":"$$\n\\forall \\alpha > 0, \\lim _ {x \\rightarrow 0 ^ {+}} x ^ {\\alpha} \\ln x = 0, \\lim _ {x \\rightarrow \\infty} \\frac {\\ln x}{x ^ {\\alpha}} = 0, \\lim _ {x \\rightarrow + \\infty} x ^ {\\alpha} e ^ {- x} = 0:\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":74,"id":"be7889b8-933b-43ca-a7d3-cdd46dbe1434","page_size":[595,841],"block_position":"74-4"},{"bbox":[83,185,279,198],"type":"text","angle":0,"index":6,"merge_prev":false,"text":"- $x \\to \\infty$ 时，等价视为 $\\ln x \\sim x^{\\epsilon}, e^{x} \\sim x^{+\\infty}$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":74,"id":"cd7e3065-e7b5-4704-afa4-d43bdf2c4db8","page_size":[595,841],"block_position":"74-5"},{"bbox":[83,198,331,219],"type":"text","angle":0,"index":7,"merge_prev":false,"text":"- $x \\to 0$ 时， $\\ln (1 + x) \\sim x, e^x - 1 \\sim x, \\ln x \\sim \\frac{1}{x^\\epsilon}, e^x \\sim 1$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":74,"id":"57af3111-bed6-4efb-bf77-2183d395138d","page_size":[595,841],"block_position":"74-6"},{"bbox":[35,243,540,280],"type":"text","angle":0,"index":8,"merge_prev":false,"text":"练习7.2对数、指数比较判别训练 判断收敛性：(1) $\\int_0^{+\\infty}\\frac{x}{1 - e^x}\\mathrm{d}x$ (2) $\\int_0^{+\\infty}\\frac{\\ln(1 + x)}{x^p}\\mathrm{d}x$ (3) $\\int_0^{+\\infty}\\frac{\\ln x}{e^x}\\mathrm{d}x$ （重点！） ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":74,"id":"431d1d5b-1240-42c5-8fff-6ec8d9b198e6","page_size":[595,841],"block_position":"74-7"},{"bbox":[52,281,514,299],"type":"text","angle":0,"index":9,"merge_prev":false,"text":"解(1)由于0处极限为1，不是瑕点。 $\\infty$ 处 $e^x\\sim x^\\infty$ ，因此 $\\lim_{x\\to \\infty}x^2\\frac{x}{1 - e^x} = 0$ ，根据比较判别法可知收敛 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":74,"id":"1ce1f993-32b6-4460-a65a-3f5da38b1f9a","page_size":[595,841],"block_position":"74-8"},{"bbox":[72,300,464,322],"type":"text","angle":0,"index":10,"merge_prev":false,"text":"(2)0处等价于 $\\frac{1}{x^{p - 1}}$ ，要求 $p < 2$ ， $+\\infty$ 处等价于 $\\frac{1}{x^p}$ ，要求 $p > 1$ ，因此 $1 < p < 2$ 时收敛 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":74,"id":"9232dca4-2ca2-407e-99b5-9c759f6cea08","page_size":[595,841],"block_position":"74-9"},{"bbox":[73,322,431,347],"type":"text","angle":0,"index":11,"merge_prev":false,"text":"(3) $\\lim_{x\\to 0}\\frac{\\ln x\\cdot x^{1 / 2}}{e^x} = 0$ ，因此0处收敛。 $\\infty$ 处 $\\lim_{x\\to \\infty}x^2\\frac{\\ln x}{e^x} = 0$ ，因此 $\\infty$ 处也收敛。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":74,"id":"50a126b9-3060-4ce5-84fa-0d0559ffd7cf","page_size":[595,841],"block_position":"74-10"},{"bbox":[35,348,537,409],"type":"text","angle":0,"index":12,"merge_prev":false,"text":"练习7.3对数比较判别进阶 判断 $\\int_{3}^{+\\infty}\\frac{\\mathrm{d}x}{x^p(\\ln x)^q(\\ln\\ln x)^r}$ 的敛散性解 $p > 1$ 肯定收敛， $p < 1$ 肯定发散。若 $p = 1$ ，则积分变为 $\\int_{3}^{+\\infty}\\frac{\\mathrm{d}(\\ln x)}{(\\ln x)^q(\\ln\\ln x)^r}$ ，要求 $q > 1$ 。若 $p = q = 1$ 则同理要求 $r > 1$ 。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":74,"id":"618aa472-2cac-4ad3-954c-82bc97b0bd64","page_size":[595,841],"block_position":"74-11"},{"bbox":[35,407,405,433],"type":"text","angle":0,"index":13,"merge_prev":false,"text":"练习7.4变号情况：绝对收敛 $+$ 比较判别法 判断敛散性：(1) $\\int_{1}^{+\\infty}\\frac{\\cos x\\sin\\frac{1}{x}}{x}\\mathrm{d}x$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":74,"id":"39690907-80dd-46df-b074-7c125dc9a3ff","page_size":[595,841],"block_position":"74-12"},{"bbox":[52,431,229,458],"type":"text","angle":0,"index":14,"merge_prev":false,"text":"解 (1) $\\left|\\frac{\\cos x\\sin\\frac{1}{x}}{x}\\right|\\sim\\frac{1}{x^{2}}$ ，因此绝对收敛 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":74,"id":"d8af5ebf-06a0-4266-be55-dd5ec1a26381","page_size":[595,841],"block_position":"74-13"},{"bbox":[35,460,539,518],"type":"text","angle":0,"index":15,"merge_prev":false,"text":"练习7.5三角双瑕点训练重点： $|q| < 1$ ，判断 $\\int_0^\\pi \\frac{\\sin^{p - 1}x}{|1 + q\\cos x|^p}\\mathrm{d}x$ 的敛散性解0处 $\\frac{\\sin^{p - 1}x}{|1 + q\\cos x|^p}\\sim \\frac{x^{p - 1}}{|1 + q|^p}$ ，要求 $p - 1 < 1$ 。π处 $\\frac{\\sin^{p - 1}x}{|1 + q\\cos x|^p} = \\frac{\\sin^{p - 1}(\\pi -x)}{|1 + q\\cos x|^p}\\sim \\frac{(\\pi -x)^{p - 1}}{|1 + q|^p}$ ，要求 $p - 1 < 1$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":74,"id":"0bf2ea43-583c-4170-a0d1-1e2e150ee9d2","page_size":[595,841],"block_position":"74-14"},{"bbox":[55,536,216,550],"type":"title","angle":0,"index":16,"level":2,"text":"## 7.1.3 Dirichlet 与 Abel 判别法","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":74,"id":"d42c546e-788c-4f5e-beba-8e3550ae27ca","page_size":[595,841],"block_position":"74-15"},{"bbox":[73,564,226,576],"type":"title","angle":0,"index":17,"level":2,"text":"## 定理 7.7 (Dirichlet 与 Abel 判别法)","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":74,"id":"eb6e1826-788a-4342-912b-23046ef1feb9","page_size":[595,841],"block_position":"74-16"},{"bbox":[83,581,525,630],"type":"text","angle":0,"index":18,"merge_prev":false,"text":"- Dirichlet 判别法：若 $(1)F(x) = \\int_{a}^{x}f(t)\\mathrm{d}t$ 在 $[a, + \\infty)$ 有界(2) $g(x)$ 在 $[a, + \\infty)$ 上当 $x\\to \\infty$ 时单调趋于0，则 $\\int_{a}^{+\\infty}f(x)g(x)\\mathrm{d}x$ 收敛。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":74,"id":"9c6a340e-88c0-4f0f-a227-235c88581d2d","page_size":[595,841],"block_position":"74-17"},{"bbox":[83,629,456,660],"type":"text","angle":0,"index":19,"merge_prev":false,"text":"- Abel 判别法：若 (1) $\\int_{a}^{+\\infty} f(x) \\mathrm{d}x$ 收敛 (2) $g(x)$ 单调有界，则 $\\int_{a}^{+\\infty} f(x) g(x) \\mathrm{d}x$ 收敛 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":74,"id":"64fdb53c-76e0-4eb0-be52-3270adbf800f","page_size":[595,841],"block_position":"74-18"},{"bbox":[35,689,486,702],"type":"text","angle":0,"index":20,"merge_prev":false,"text":"练习7.6在 $\\sin x$ 反常积分判敛的应用判断以下带三角函数的反常积分的绝对收敛性与条件收敛性： ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":74,"id":"5bf09938-d505-433a-bbd5-864001be291a","page_size":[595,841],"block_position":"74-19"},{"bbox":[66,702,443,726],"type":"text","angle":0,"index":21,"merge_prev":false,"text":"1. $\\int_{1}^{+\\infty}\\frac{\\sin x}{x^p}\\mathrm{d}x,\\int_{1}^{+\\infty}\\frac{\\cos x}{x^p}\\mathrm{d}x$ ： $p > 0$ 就收敛， $p > 1$ 绝对收敛， $0 < p\\leq 1$ 条件收敛 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":74,"id":"15d4ad74-00fc-4041-a9a1-d17416337d80","page_size":[595,841],"block_position":"74-20"},{"bbox":[65,727,399,751],"type":"text","angle":0,"index":22,"merge_prev":false,"text":"2. $\\int_{0}^{+\\infty}\\frac{\\sin x}{x^{p}}dx:1 1$ 时 $\\left|\\frac{\\sin x}{x^p}\\right| \\leq \\frac{1}{x^p}$ 因此绝对收敛。 $0 < p \\leq 1$ 时根据Dirichlet判别法可知收敛，下证不绝对收敛，根据 $|\\sin x| \\geq \\sin^2 x$ ，得到 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":75,"id":"66ea10d9-c81c-4323-ac7e-45678b03cbc8","page_size":[595,841],"block_position":"75-5"},{"bbox":[215,204,377,231],"type":"interline_equation","angle":0,"index":6,"text":"$$\n\\left| \\frac {\\sin x}{x ^ {p}} \\right| \\geq \\frac {\\sin^ {2} x}{x ^ {p}} = \\frac {1}{2} \\left(\\frac {1}{x ^ {p}} - \\frac {\\cos 2 x}{x ^ {p}}\\right)\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":75,"id":"9873b1a9-7a93-443c-aaa7-83a6a2fbb777","page_size":[595,841],"block_position":"75-6"},{"bbox":[54,237,521,265],"type":"text","angle":0,"index":7,"merge_prev":false,"text":"而 $\\int_{1}^{+\\infty}\\frac{1}{x^p}\\mathrm{d}x$ 不收敛，故 $\\int_1^{+\\infty}\\left|\\frac{\\sin x}{x^p}\\right|$ 在 $0 < p\\leq 1$ 时不收敛。 $p\\leq 0$ 时，取 $u_{1} = 2k\\pi ,u_{2} = 2k\\pi +\\frac{\\pi}{2}$ ，则 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":75,"id":"66fc03f3-402b-4a30-8293-342935b03538","page_size":[595,841],"block_position":"75-7"},{"bbox":[223,267,370,295],"type":"interline_equation","angle":0,"index":8,"text":"$$\n\\left| \\int_ {u _ {1}} ^ {u _ {2}} \\frac {\\sin x}{x ^ {p}} \\mathrm{d} x \\right| \\geq \\int_ {u _ {1}} ^ {u _ {2}} \\sin x \\mathrm{d} x = 1\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":75,"id":"321380ea-8234-4e3b-b1a6-d3111fe9e520","page_size":[595,841],"block_position":"75-8"},{"bbox":[55,297,102,309],"type":"text","angle":0,"index":9,"merge_prev":false,"text":"显然发散。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":75,"id":"58616c75-88d5-4d5a-95da-dfe86f05549c","page_size":[595,841],"block_position":"75-9"},{"bbox":[54,303,539,346],"type":"text","angle":0,"index":10,"merge_prev":false,"text":"(2) 分两段，这里考虑 $\\int_0^1 \\frac{\\sin x}{x^p} \\mathrm{d}x$ ，当 $x \\to 0$ 时 $\\frac{\\sin x}{x^p} \\sim \\frac{1}{x^{p-1}}$ ，因此 $p < 2$ 时收敛且绝对收敛， $p \\geq 2$ 时发散。结合 (1) 得到 $1 < p < 2$ 时绝对收敛， $0 < p \\leq 1$ 时条件收敛，其余发散 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":75,"id":"721526e7-22d0-4390-a981-114d78a51cfd","page_size":[595,841],"block_position":"75-10"},{"bbox":[72,345,355,366],"type":"text","angle":0,"index":11,"merge_prev":false,"text":"(3) $\\frac{\\cos x}{x^{p}}\\sim\\frac{1}{x^{p}}$ ，结合(1)得到0 1$ 时绝对收敛， $0 < 2 - p \\leq 1$ 条件收敛。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":75,"id":"5719c650-f48f-489d-9548-4b058b81a536","page_size":[595,841],"block_position":"75-12"},{"bbox":[52,441,539,470],"type":"text","angle":0,"index":13,"merge_prev":false,"text":"(5) $q = 0$ 时显然无法保证 $0, +\\infty$ 同时收敛，原因在于0要求 $p < 1$ ， $\\infty$ 要求 $p > 1$ 。 $q > 0$ 时，令 $t = x^q$ ，得到 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":75,"id":"1ea40c0a-b7d5-466e-b96d-c17033ec1498","page_size":[595,841],"block_position":"75-13"},{"bbox":[213,467,380,496],"type":"interline_equation","angle":0,"index":15,"text":"$$\n\\int_ {0} ^ {+ \\infty} x ^ {p} \\sin x ^ {q} \\mathrm{d} x = \\frac {1}{q} \\int_ {0} ^ {+ \\infty} \\frac {\\sin t}{t ^ {1 - \\frac {p + 1}{q}}} \\mathrm{d} t\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":75,"id":"29ff06bf-3232-47a1-a7b9-9c251e521aac","page_size":[595,841],"block_position":"75-14"},{"bbox":[54,502,141,513],"type":"text","angle":0,"index":16,"merge_prev":false,"text":"转换为前面的问题。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":75,"id":"4891de26-7717-4dd0-a46e-27600f0d6920","page_size":[595,841],"block_position":"75-15"},{"bbox":[33,511,340,539],"type":"text","angle":0,"index":17,"merge_prev":false,"text":"笔记 $\\int_0^{+\\infty}\\frac{\\sin x}{x^p}\\mathrm{d}x,\\int_0^{+\\infty}\\frac{\\cos x}{x^p}\\mathrm{d}x$ 的收敛情况要当作结论背下来。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":75,"id":"85f47954-a196-4a87-8fbf-932bbac8bba6","page_size":[595,841],"block_position":"75-16"},{"bbox":[35,550,421,563],"type":"text","angle":0,"index":18,"merge_prev":false,"text":"练习 7.7 三角有理式敛散性判别 判断如下几道三角函数有理式的敛散性 (难度较高): ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":75,"id":"753efb9c-026e-481f-a48b-d790e364d76b","page_size":[595,841],"block_position":"75-17"},{"bbox":[72,563,246,587],"type":"text","angle":0,"index":19,"merge_prev":false,"text":"(1) 判断 $\\int_0^{+\\infty} \\frac{\\sin x}{2x + 3\\sin x} \\mathrm{d}x$ 的敛散性 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":75,"id":"657e6c26-b5be-4856-bd6f-965a160bf277","page_size":[595,841],"block_position":"75-18"},{"bbox":[73,588,239,613],"type":"text","angle":0,"index":20,"merge_prev":false,"text":"(2) 判断 $\\int_{a}^{+\\infty} \\frac{\\sin x}{x^p + \\sin x} \\, \\mathrm{d}x$ 的敛散性 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":75,"id":"e3c84335-a521-4076-b0ef-e4af3e198f10","page_size":[595,841],"block_position":"75-19"},{"bbox":[73,613,230,640],"type":"text","angle":0,"index":21,"merge_prev":false,"text":"(3) 证明 $\\int_{0}^{+\\infty} \\frac{\\mathrm{d}x}{1 + x^2 \\sin x^2} \\mathrm{d}x$ 发散 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":75,"id":"4a9e8724-8f73-431f-ba55-703cb96c0efd","page_size":[595,841],"block_position":"75-20"},{"bbox":[52,637,539,674],"type":"text","angle":0,"index":22,"merge_prev":false,"text":"解(1)该积分没有瑕点，只需要考虑 $\\infty$ ，其敛散性等价于 $\\int_2^\\infty \\frac{\\sin x}{2x + 3\\sin x}\\mathrm{d}x$ 。这里必须拆项(不能直接放缩原因在于分母阶数不够) ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":75,"id":"0b3a7693-fe30-4a28-bb40-795e12586c02","page_size":[595,841],"block_position":"75-21"},{"bbox":[100,677,492,704],"type":"interline_equation","angle":0,"index":23,"text":"$$\n{\\frac {\\sin x}{2 x + 3 \\sin x}} = {\\frac {2 x \\sin x}{2 x (2 x + 3 \\sin x)}} = {\\frac {(2 x + 3 \\sin x) \\sin x - 3 \\sin^ {2} x}{2 x (2 x + 3 \\sin x)}} = {\\frac {\\sin x}{2 x}} - {\\frac {3 \\sin^ {2} x}{2 x (2 x + 3 \\sin x)}}\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":75,"id":"eebfc9be-7257-4859-a80c-4025cccd30fd","page_size":[595,841],"block_position":"75-22"},{"bbox":[54,706,232,729],"type":"text","angle":0,"index":24,"merge_prev":false,"text":"考虑前者，显然 $\\frac{1}{2x}$ 单调递减趋于0，而 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":75,"id":"160ac28b-f139-4a5b-baf9-9c3a1354c528","page_size":[595,841],"block_position":"75-23"},{"bbox":[224,729,369,758],"type":"interline_equation","angle":0,"index":25,"text":"$$\n\\left| \\int_ {2} ^ {u} \\sin \\mathrm{d} x \\right| = | \\cos 2 - \\cos u | \\leq 2\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":75,"id":"ad2603f6-1e3b-46d9-b411-cd7043c6479f","page_size":[595,841],"block_position":"75-24"},{"bbox":[291,803,302,812],"type":"page_number","angle":0,"index":1000013,"text":"68 ","color":{"line":"rgba(164, 164, 164, 1)","fill":"rgba(164, 164, 164, 1)"},"page_idx":75,"is_discarded":true,"id":"5aef6fd8-5abe-4a59-b43b-d665d01d27f3","page_size":[595,841],"block_position":"75-25"}],[{"bbox":[459,47,539,58],"type":"header","angle":0,"index":-99999,"text":"7.1 反常积分理论 ","color":{"line":"rgba(164, 164, 164, 1)","fill":"rgba(164, 164, 164, 1)"},"page_idx":76,"is_discarded":true,"id":"fda55aea-858c-4254-bace-158854a7e5a9","page_size":[595,841],"block_position":"76-0"},{"bbox":[54,72,278,84],"type":"text","angle":0,"index":1,"merge_prev":false,"text":"根据 Dirichlet 判别法知前者收敛。后者先进行放缩 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":76,"id":"97f39089-521e-459f-b3c8-069ede0bc410","page_size":[595,841],"block_position":"76-1"},{"bbox":[218,86,373,113],"type":"interline_equation","angle":0,"index":2,"text":"$$\n0 \\leq \\frac {3 \\sin^ {2} x}{2 x (2 x + 3 \\sin x)} \\leq \\frac {3}{2 x (2 x - 3)}\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":76,"id":"ee2dd4ff-63c7-46ca-8f97-8db8e852b6c3","page_size":[595,841],"block_position":"76-2"},{"bbox":[54,114,301,138],"type":"text","angle":0,"index":3,"merge_prev":false,"text":"在无穷处等价于 $\\frac{1}{x^2}$ ，根据比较原则得到收敛。综上收敛 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":76,"id":"90c64b02-2644-4de9-b951-6d34f2860787","page_size":[595,841],"block_position":"76-3"},{"bbox":[55,153,211,167],"type":"title","angle":0,"index":4,"level":2,"text":"## 7.1.4 函数在无穷远处的性质","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":76,"id":"ccc95ee8-1792-4154-a04a-45701b2aa205","page_size":[595,841],"block_position":"76-4"},{"bbox":[54,177,176,190],"type":"title","angle":0,"index":5,"level":2,"text":"## 一、若极限存在，则必为0","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":76,"id":"f97c0ac4-ea0e-420f-9c0c-f69df85e3ecf","page_size":[595,841],"block_position":"76-5"},{"bbox":[35,201,539,280],"type":"text","angle":0,"index":6,"merge_prev":false,"text":"练习7.8 $f(x)$ 在 $[a, u]$ 可积， $\\int_{a}^{+\\infty} f(x) \\mathrm{d}x$ 收敛，若 $\\lim_{x \\to +\\infty} f(x) = A$ ，则 $A = 0$ 。证明 若 $A \\neq 0$ ，不妨设 $A > 0$ ，则 $\\exists M \\geq a, \\forall x \\geq M, f(x) \\geq \\frac{A}{2}$ ，而 $\\int_{a}^{+\\infty} \\frac{A}{2} \\mathrm{d}x$ 发散，根据比较判别法可知 $\\int_{a}^{+\\infty} f(x) \\mathrm{d}x$ 发散 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":76,"id":"f75d1dba-a9d5-416f-9c90-64621307a399","page_size":[595,841],"block_position":"76-6"},{"bbox":[35,283,539,348],"type":"text","angle":0,"index":7,"merge_prev":false,"text":"练习7.9 $f(x)$ 在 $[a, +\\infty)$ 上连续可微，且 $\\int_{a}^{+\\infty} f(x) \\mathrm{d}x, \\int_{a}^{+\\infty} f'(x) \\mathrm{d}x$ 收敛，则 $\\lim_{x \\to +\\infty} f(x) = 0$ 证明由于 $\\int_{a}^{+\\infty} f'(x) \\mathrm{d}x$ 收敛，因此 $\\lim_{u \\to +\\infty} \\int_{a}^{u} f'(x) \\mathrm{d}x = \\lim_{u \\to +\\infty} f(u) - f(a)$ 存在，根据极限存在可知极限为0(前一题结论)。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":76,"id":"abe3c041-c53b-4643-b753-2f3472b8c832","page_size":[595,841],"block_position":"76-7"},{"bbox":[35,357,539,473],"type":"text","angle":0,"index":8,"merge_prev":false,"text":"练习7.10 $f(x)$ 在 $[a, +\\infty)$ 上连续， $\\int_{a}^{+\\infty} f(x) \\mathrm{d}x$ 收敛，则存在递增趋于 $+\\infty$ 的数列 $x_{n}$ ，使得 $\\lim_{n \\to \\infty} f(x_{n}) = 0$ 。证明根据Cauchy收敛准则，取 $\\epsilon_{1} = 1, \\exists A_{1}', A_{1}'' = A_{1}' + 1$ ，使得 $\\left| \\int_{A_1'}^{A_1''} f(x) \\mathrm{d}x \\right| < \\epsilon_{1} = 1$ 。同理对 $\\epsilon_{2} = \\frac{1}{2}, \\exists A_{2}' \\geq A_{1}', A_{2}'' = A_{2}' + 1$ ，使得 $\\left| \\int_{A_2'}^{A_2''} f(x) \\mathrm{d}x \\right| < \\epsilon_{2} = \\frac{1}{2}$ 。以此类推，根据积分第一中值定理 $\\exists x_{i} \\in [A_{i}', A_{i}'']$ 使得 $|f(x_{n})| = \\left| \\int_{A_{n}'}^{A_{n}''} f(x) \\mathrm{d}x \\right|$ ，显然 $x_{n}$ 单增， $|f(x_{n})| < \\frac{1}{n} \\to 0$ 。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":76,"id":"57b03d4f-9a0d-4bf4-85c8-a3711d7f1bbd","page_size":[595,841],"block_position":"76-8"},{"bbox":[52,481,446,494],"type":"title","angle":0,"index":9,"level":2,"text":"## 二、极限不一定存在：光满足非负、连续、可导等并不能说明极限存在，甚至不一定有界","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":76,"id":"83e282df-5684-41b5-ba5f-2d3070202dd9","page_size":[595,841],"block_position":"76-9"},{"bbox":[35,504,486,569],"type":"text","angle":0,"index":10,"merge_prev":false,"text":"练习7.11举例说明 $f(x)$ 在任意有限区间 $[a, u]$ 非负可积， $\\int_{a}^{+\\infty} f(x) \\mathrm{d}x$ 收敛，则 $f(x)$ 不一定有界。解 $f(x) = \\begin{cases} x, & x \\text{为整数} \\\\ 0, & x \\text{不为整数} \\end{cases}$ ，如果要求恒正则取 $g(x) = f(x) + e^{-x}$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":76,"id":"75ee5dc6-43d4-498f-951a-195099d85bd6","page_size":[595,841],"block_position":"76-10"},{"bbox":[35,571,470,622],"type":"text","angle":0,"index":11,"merge_prev":false,"text":"练习7.12举例说明 $f(x)$ 为 $[a, +\\infty)$ 上连续函数， $\\int_{a}^{+\\infty}f(x)\\mathrm{d}x$ 收敛，不一定有 $\\lim_{x\\to +\\infty}f(x) = 0$ 解例如 $\\int_1^{+\\infty}\\sin x^2\\mathrm{d}x$ ，令 $t = x^{2}$ 以及Dirichlet判别法可知收敛，但是极限不为0。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":76,"id":"9bf3fc1a-9860-49ff-8ef7-13de68b46912","page_size":[595,841],"block_position":"76-11"},{"bbox":[35,624,452,679],"type":"text","angle":0,"index":13,"merge_prev":false,"text":"练习7.13举例说明 $[a, +\\infty)$ 恒正的连续可微函数， $\\int_{a}^{+\\infty}f(x)\\mathrm{d}x$ 收敛，则 $f(x)$ 不一定有界解 $\\int_1^{+\\infty}\\frac{x}{1 + x^6\\sin^2x}\\mathrm{d}x$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":76,"id":"dc92ef87-1d66-450c-a9da-1d39225908fb","page_size":[595,841],"block_position":"76-12"},{"bbox":[52,688,518,704],"type":"title","angle":0,"index":14,"level":2,"text":"## 三、获得极限存在：虽然非负、连续、可导不能推出 $\\lim_{x\\to +\\infty}f(x) = 0$ ，但一致连续和单调可以推出该性质。","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":76,"id":"5f6f059d-f2e0-4fd3-a82a-2ad77e197535","page_size":[595,841],"block_position":"76-13"},{"bbox":[35,712,493,740],"type":"text","angle":0,"index":15,"merge_prev":false,"text":"练习7.14经典老题：一致连续 $f(x)$ 在 $[a, +\\infty)$ 上一致连续， $\\int_{a}^{+\\infty}f(x)\\mathrm{d}x$ 收敛，则 $\\lim_{x\\to +\\infty}f(x) = 0$ 证明 反设 $\\lim_{x\\to +\\infty}f(x)\\neq 0$ ，则 $\\exists \\epsilon_0$ 以及趋于无穷的 $x_{n}$ 使得 $|f(x_n)|\\geq \\epsilon_0$ ，对上述 $\\epsilon_0$ ，根据一致连续 $\\exists \\delta ,\\forall x^{\\prime},x^{\\prime \\prime}\\in$ $[a, + \\infty)$ 只要 $|x^{\\prime} - x^{\\prime \\prime}|\\leq \\delta$ ，就有 $|f(x^{\\prime}) - f(x^{\\prime \\prime})| < \\frac{\\epsilon_0}{2}$ 。对每个 $x_{n}$ ，当 $x\\in [x_n,x_n + \\delta ]$ 时， $|f(x) - f(x_n)| < \\frac{\\epsilon_0}{2}$ 从而 $f(x)$ 在 $[x_n,x_n + \\delta ]$ 不变号，且 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":76,"id":"4077de99-c60e-4387-9f21-c22b220629b4","page_size":[595,841],"block_position":"76-14"},{"bbox":[291,802,302,812],"type":"page_number","angle":0,"index":1000011,"text":"69 ","color":{"line":"rgba(164, 164, 164, 1)","fill":"rgba(164, 164, 164, 1)"},"page_idx":76,"is_discarded":true,"id":"ab175d79-caea-45fa-8ce7-782f44a13f9d","page_size":[595,841],"block_position":"76-15"}],[{"bbox":[449,46,539,58],"type":"header","angle":0,"index":-99999,"text":"7.2 反常积分的计算 ","color":{"line":"rgba(164, 164, 164, 1)","fill":"rgba(164, 164, 164, 1)"},"page_idx":77,"is_discarded":true,"id":"88a44875-17df-47d2-8e1e-fa9e5573f108","page_size":[595,841],"block_position":"77-0"},{"bbox":[52,70,540,121],"type":"text","angle":0,"index":1,"merge_prev":false,"lines_deleted":true,"text":"","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":77,"id":"4077de99-c60e-4387-9f21-c22b220629b4","page_size":[595,841],"block_position":"77-1"},{"bbox":[210,123,382,143],"type":"interline_equation","angle":0,"index":2,"text":"$$\n| f (x) | \\geq | f (x _ {n}) | - | f (x) - f (x _ {n}) | > \\frac {\\epsilon_ {0}}{2}\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":77,"id":"3b44bffe-b4e3-4b7e-946b-7a0bd06d6d43","page_size":[595,841],"block_position":"77-2"},{"bbox":[53,143,539,201],"type":"text","angle":0,"index":3,"merge_prev":false,"text":"对 $\\forall M > a, \\exists N, x_N > M$ 满足 $|\\int_{x_N}^{x_N + \\delta} f(x) \\mathrm{d}x| = \\int_{x_N}^{x_N + \\delta} |f(x)| \\mathrm{d}x \\geq \\int_{x_N}^{x_N + \\delta} \\frac{\\epsilon_0}{2} \\mathrm{d}x = \\frac{\\epsilon_0 \\delta}{2}$ 不趋于 0，根据 Cauchy 收敛准则逆否命题可知 $\\int_{a}^{+\\infty} f(x) \\mathrm{d}x$ 发散，矛盾。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":77,"id":"cac75589-19fc-4c17-b5e3-61e108034739","page_size":[595,841],"block_position":"77-3"},{"bbox":[33,204,503,258],"type":"text","angle":0,"index":4,"merge_prev":false,"text":"练习7.15单调 $f(x)$ 在 $[a, +\\infty)$ 单调，且 $\\int_{a}^{+\\infty} f(x) \\mathrm{d}x$ 收敛，则 $\\lim_{x \\to +\\infty} f(x) = 0$ ，且 $\\lim_{x \\to +\\infty} x f(x) = 0$ 。笔记反之不一定，即使有 $\\lim_{x \\to +\\infty} x f(x) = 0$ ，也不一定有 $\\int_{a}^{+\\infty} f(x) \\mathrm{d}x$ 收敛，例如 $f(x) = \\frac{1}{x \\ln x}$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":77,"id":"45d5fe9e-12b9-40ca-b6b0-6fb2dc9e5910","page_size":[595,841],"block_position":"77-4"},{"bbox":[55,287,187,304],"type":"title","angle":0,"index":5,"level":2,"text":"## 7.2 反常积分的计算","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":77,"id":"9c108d61-6056-4a9f-9231-c8c37f16bfee","page_size":[595,841],"block_position":"77-5"},{"bbox":[55,320,223,334],"type":"title","angle":0,"index":6,"level":2,"text":"## 7.2.1 两个常用的基础反常积分","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":77,"id":"e38d93d5-9b27-479e-8117-3d2a2bea803d","page_size":[595,841],"block_position":"77-6"},{"bbox":[35,342,539,420],"type":"text","angle":0,"index":7,"merge_prev":false,"text":"练习7.16计算 $\\int_0^{+\\infty}e^{-ax}\\cos bx\\mathrm{d}x,\\int_0^{+\\infty}e^{-ax}\\sin bx\\mathrm{d}x$ ，其中 $a > 0$ 解直接写不定积分即可(具体计算见配对积分法一节)， $I_{1} = \\frac{e^{-ax}}{a^{2} + b^{2}} (a\\cos bx + b\\sin bx)\\bigg|_{0}^{+\\infty} = \\frac{a}{a^2 + b^2}$ ，同理得到 $I_{2} = \\frac{b}{a^{2} + b^{2}}$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":77,"id":"d1dde64d-74bd-4ac9-bac9-6d1ed20a382e","page_size":[595,841],"block_position":"77-7"},{"bbox":[35,425,539,518],"type":"text","angle":0,"index":8,"merge_prev":false,"text":"练习7.17对数配对积分(1)计算 $I_{n} = \\int_{0}^{1}(\\ln x)^{n}\\mathrm{d}x$ (2)计算 $I(m,n) = \\int_0^1 x^m (\\ln x)^n\\mathrm{d}x$ 解(1)做一次分部积分得到 $I_{n} = x(\\ln x)^{n}\\Big|_{0}^{1} - \\int_{0}^{1}n(\\ln x)^{n - 1}\\mathrm{d}x = -nI_{n - 1}$ ，而 $I_{1} = \\int_{0}^{1}\\ln x\\mathrm{d}x = x\\ln x\\Big|_{0}^{1} - x\\Big|_{0}^{1} = -1$ ，综上得到 $I_{n} = (-1)^{n}n!$ (2) $I(m,n) = -\\frac{n}{m + 1} I(m,n - 1)$ ，结合 $I(m,0) = \\frac{1}{m + 1}$ 可得 $I(m,n) = \\frac{(-1)^{n}n!}{(m + 1)^{n + 1}}$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":77,"id":"9e8ad3a5-8a8b-492f-974c-871b72bbcabd","page_size":[595,841],"block_position":"77-8"},{"bbox":[56,533,163,547],"type":"title","angle":0,"index":9,"level":2,"text":"## 7.2.2 富如兰尼积分","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":77,"id":"172cc48a-79f6-496f-9f54-9fdffb6fa40e","page_size":[595,841],"block_position":"77-9"},{"bbox":[72,560,251,572],"type":"title","angle":0,"index":10,"level":2,"text":"## 定理 7.8 (两侧极限存在则无穷积分收敛)","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":77,"id":"b4b425f0-0f38-4691-84d9-f26d6940aba0","page_size":[595,841],"block_position":"77-10"},{"bbox":[67,576,525,606],"type":"text","angle":0,"index":12,"merge_prev":false,"text":"重点： $f(x)$ 在任意有限区间可积， $\\lim_{x\\to +\\infty}f(x) = A,\\lim_{x\\to -\\infty}f(x) = B$ ，给定 $a\\in \\mathbb{R}$ ，则下面积分收敛，且值为 $a(A - B)$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":77,"id":"1d72bd6f-481d-4361-b50e-7d6d8173481b","page_size":[595,841],"block_position":"77-11"},{"bbox":[239,605,353,634],"type":"interline_equation","angle":0,"index":13,"text":"$$\n\\int_ {- \\infty} ^ {+ \\infty} [ f (a + x) - f (x) ] \\mathrm{d} x\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":77,"id":"f67197ba-dbf3-4b85-a296-97ea99a20448","page_size":[595,841],"block_position":"77-12"},{"bbox":[54,642,503,671],"type":"text","angle":0,"index":14,"merge_prev":false,"text":"证明 根据定义， $\\int_{-\\infty}^{+\\infty}f(x)\\mathrm{d}x = \\lim_{M\\to +\\infty}\\lim_{m\\to -\\infty}\\int_{m}^{M}[f(a + x) - f(x)]\\mathrm{d}x$ ，对有限区域积分做如下变换： ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":77,"id":"062feb1f-b147-4725-bc72-9f09ed70e09b","page_size":[595,841],"block_position":"77-13"},{"bbox":[91,673,502,703],"type":"interline_equation","angle":0,"index":15,"text":"$$\nI = \\int_ {m} ^ {M} [ f (a + x) - f (x) ] \\mathrm{d} x = \\int_ {m + a} ^ {M + a} f (x) \\mathrm{d} x - \\int_ {m} ^ {M} f (x) \\mathrm{d} x = \\int_ {M} ^ {M + a} f (x) \\mathrm{d} x - \\int_ {m} ^ {m + a} f (x) \\mathrm{d} x\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":77,"id":"ac44d8eb-f039-41bc-878c-82a08fe83a3b","page_size":[595,841],"block_position":"77-14"},{"bbox":[54,704,539,749],"type":"text","angle":0,"index":16,"merge_prev":false,"text":"代入 $A, B$ ，得到 $I = \\int_{M}^{M + a}[f(x) - A]\\mathrm{d}x - \\int_{m}^{m + a}[f(x) - B]\\mathrm{d}x + a(A - B)$ ，根据极限 $\\lim_{M\\to +\\infty}[f(x) - A]\\mathrm{d}x = 0$ ，同理 $\\lim_{N\\to +\\infty}[f(x) - B]\\mathrm{d}x = 0$ ，因此取极限得到收敛，且值为 $a(A - B)$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":77,"id":"35e2f76b-d877-42ec-bd71-3ddda27fdb5c","page_size":[595,841],"block_position":"77-15"},{"bbox":[291,802,303,812],"type":"page_number","angle":0,"index":1000010,"text":"70 ","color":{"line":"rgba(164, 164, 164, 1)","fill":"rgba(164, 164, 164, 1)"},"page_idx":77,"is_discarded":true,"id":"10846464-433a-4674-be9a-797fc295816d","page_size":[595,841],"block_position":"77-16"}],[{"bbox":[449,46,539,58],"type":"header","angle":0,"index":-99999,"text":"7.2 反常积分的计算 ","color":{"line":"rgba(164, 164, 164, 1)","fill":"rgba(164, 164, 164, 1)"},"page_idx":78,"is_discarded":true,"id":"92e4a769-22f1-4f06-824f-0e85eb563836","page_size":[595,841],"block_position":"78-0"},{"bbox":[73,73,181,86],"type":"title","angle":0,"index":1,"level":2,"text":"## 定理 7.9 (富如兰尼积分)","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":78,"id":"b4ef1bf6-4fd0-4501-a54a-f6098a90edb1","page_size":[595,841],"block_position":"78-1"},{"bbox":[69,89,245,102],"type":"text","angle":0,"index":2,"merge_prev":false,"text":"$f(x)$ 为 $[0, +\\infty)$ 连续函数， $a, b > 0$ ，则 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":78,"id":"9fdbce69-e97c-454a-b13e-ee1c3438eea3","page_size":[595,841],"block_position":"78-2"},{"bbox":[81,105,199,123],"type":"text","angle":0,"index":3,"merge_prev":false,"text":"1. 若 $\\lim_{x\\to +\\infty}f(x)$ 存在，则 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":78,"id":"7b65b046-bf0a-4109-ae8a-522b3fd75807","page_size":[595,841],"block_position":"78-3"},{"bbox":[207,126,411,153],"type":"interline_equation","angle":0,"index":4,"text":"$$\n\\int_ {0} ^ {+ \\infty} \\frac {f (a x) - f (b x)}{x} \\mathrm{d} x = [ f (0) - f (+ \\infty) ] \\ln \\frac {b}{a}\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":78,"id":"d1bfadc3-3efc-412e-ba60-3aceefce64d4","page_size":[595,841],"block_position":"78-4"},{"bbox":[80,155,275,182],"type":"text","angle":0,"index":5,"merge_prev":false,"text":"2. 若 $\\exists A > 0$ ，使得 $\\int_{A}^{+\\infty} \\frac{f(x)}{x} \\mathrm{d}x$ 收敛，则 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":78,"id":"182d4de3-11fe-4455-a571-766b85edf6a9","page_size":[595,841],"block_position":"78-5"},{"bbox":[232,185,387,212],"type":"interline_equation","angle":0,"index":6,"text":"$$\n\\int_ {0} ^ {+ \\infty} \\frac {f (a x) - f (b x)}{x} \\mathrm{d} x = f (0) \\ln {\\frac {b}{a}}\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":78,"id":"ab05bfee-25b1-46d5-ac5a-a143e551f411","page_size":[595,841],"block_position":"78-6"},{"bbox":[54,222,274,250],"type":"text","angle":0,"index":7,"merge_prev":false,"text":"证明 下面对 $I = \\int_{m}^{M}\\frac{f(ax) - f(bx)}{x}\\mathrm{d}x$ 进行化简： ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":78,"id":"e2332573-55a6-4f45-bea1-a789521e3a08","page_size":[595,841],"block_position":"78-7"},{"bbox":[81,251,511,280],"type":"interline_equation","angle":0,"index":8,"text":"$$\nI = \\int_ {m} ^ {M} \\frac {f (a x)}{x} \\mathrm{d} x - \\int_ {m} ^ {M} \\frac {f (b x)}{x} \\mathrm{d} x = \\int_ {a m} ^ {a M} \\frac {f (x)}{x} \\mathrm{d} x - \\int_ {b m} ^ {b M} \\frac {f (x)}{x} \\mathrm{d} x = \\int_ {a m} ^ {b m} \\frac {f (x)}{x} \\mathrm{d} x - \\int_ {a M} ^ {b M} \\frac {f (x)}{x} \\mathrm{d} x\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":78,"id":"bfe646c4-3a7f-4ba1-a654-5daa8bcf3391","page_size":[595,841],"block_position":"78-8"},{"bbox":[73,282,211,294],"type":"text","angle":0,"index":9,"merge_prev":false,"text":"(1) 根据积分第一中值定理得到 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":78,"id":"953a5db5-d68f-4f49-b293-c4d87b639336","page_size":[595,841],"block_position":"78-9"},{"bbox":[193,297,390,349],"type":"interline_equation","angle":0,"index":10,"text":"$$\n\\left\\{ \\begin{array}{l} \\int_ {a m} ^ {b m} \\frac {f (x)}{x} \\mathrm{d} x = f (\\xi) \\int_ {a \\eta} ^ {b m} \\frac {1}{x} \\mathrm{d} x = f (\\xi) \\ln \\frac {b}{a} \\\\ \\int_ {a M} ^ {b M} \\frac {f (x)}{x} \\mathrm{d} x = f (\\eta) \\int_ {a M} ^ {b M} \\frac {1}{x} \\mathrm{d} x = f (\\eta) \\ln \\frac {b}{a} \\end{array} \\right.\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":78,"id":"40358a86-29bc-477e-8569-93bcec518286","page_size":[595,841],"block_position":"78-10"},{"bbox":[55,353,366,375],"type":"text","angle":0,"index":12,"merge_prev":false,"text":"当 $m \\to 0^{+}, M \\to +\\infty$ 时可代入极限，因此结果为 $[f(0) - f(+\\infty)] \\ln \\frac{b}{a}$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":78,"id":"04cbb3f7-1553-429b-943a-8f8b499f7704","page_size":[595,841],"block_position":"78-11"},{"bbox":[53,373,539,423],"type":"text","angle":0,"index":13,"merge_prev":false,"text":"(2) 由于 $\\int_{A}^{+\\infty} f(x) \\mathrm{d}x$ 收敛，因此第二项 $\\lim_{M \\to +\\infty} \\frac{f(x)}{x} \\mathrm{d}x = 0$ ，同时第一项用积分中值定理取极限得到 $f(0) \\ln \\frac{b}{a}$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":78,"id":"c0a6e1fc-3712-4ef8-a18a-024df2ba3046","page_size":[595,841],"block_position":"78-12"},{"bbox":[36,435,274,447],"type":"text","angle":0,"index":14,"merge_prev":false,"text":"练习 7.18 富如兰尼积分的应用 计算以下反常积分 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":78,"id":"6ce4f0cf-bf9e-4375-a73f-b98614c9ce12","page_size":[595,841],"block_position":"78-13"},{"bbox":[73,445,443,472],"type":"text","angle":0,"index":15,"merge_prev":false,"text":"(1) $\\int_{0}^{+\\infty}\\frac{\\arctan ax - \\arctan bx}{x}\\mathrm{d}x$ (2) $\\int_0^{+\\infty}\\frac{e^{-ax} - e^{-bx}}{x}\\mathrm{d}x$ (3) $\\int_0^{+\\infty}\\frac{\\cos ax - \\cos bx}{x}\\mathrm{d}x$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":78,"id":"29229d53-ae1f-4b8d-a593-cbc3cba09b04","page_size":[595,841],"block_position":"78-14"},{"bbox":[54,471,311,495],"type":"text","angle":0,"index":16,"merge_prev":false,"text":"解 (1) $\\arctan 0 = 0, \\arctan (+\\infty) = \\frac{\\pi}{2}$ ，因此积分为 $-\\frac{\\pi}{2} \\ln \\frac{b}{a}$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":78,"id":"85f49808-d966-4bdb-8b0d-731387d0a7ec","page_size":[595,841],"block_position":"78-15"},{"bbox":[73,493,239,516],"type":"text","angle":0,"index":17,"merge_prev":false,"text":"(2) $e^{0}=1,e^{-\\infty}=0$ ，因此积分为 $\\ln\\frac{b}{a}$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":78,"id":"c8cf24b8-4e5e-435f-8f86-b5fc510c3c68","page_size":[595,841],"block_position":"78-16"},{"bbox":[73,515,295,540],"type":"text","angle":0,"index":18,"merge_prev":false,"text":"(3) $\\cos 0 = 1, \\int_{A}^{+\\infty} \\frac{\\cos x}{x} \\mathrm{d}x$ 收敛，因此积分为 $\\ln \\frac{b}{a}$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":78,"id":"0ff0357c-dbc9-4536-84e8-2bfb1b08850c","page_size":[595,841],"block_position":"78-17"},{"bbox":[57,568,163,581],"type":"title","angle":0,"index":19,"level":2,"text":"## 7.2.3 Dirichlet 积分","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":78,"id":"5247c2d7-0f72-489c-b45c-6ef02c438548","page_size":[595,841],"block_position":"78-18"},{"bbox":[74,594,113,607],"type":"title","angle":0,"index":20,"level":2,"text":"## 引理7.1","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":78,"id":"79cc77c4-c205-45ff-a340-6325f352687c","page_size":[595,841],"block_position":"78-19"},{"bbox":[69,610,390,638],"type":"text","angle":0,"index":21,"merge_prev":false,"text":"设 $p > 0$ ，则含参积分 $\\int_0^{+\\infty}e^{-px}\\frac{\\sin bx - \\sin ax}{x}\\mathrm{d}x = \\arctan \\frac{b}{p} -\\arctan \\frac{a}{p}$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":78,"id":"ecd4b7c2-b654-461c-9bc0-01176993cd1d","page_size":[595,841],"block_position":"78-20"},{"bbox":[53,649,264,676],"type":"text","angle":0,"index":22,"merge_prev":false,"text":"证明 首先 $\\frac{\\sin bx - \\sin ax}{x} = \\int_{a}^{b} \\cos xy \\, dy$ ，因此 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":78,"id":"94c07c74-618b-46cf-8978-a71c3ab48719","page_size":[595,841],"block_position":"78-21"},{"bbox":[229,678,365,706],"type":"interline_equation","angle":0,"index":23,"text":"$$\nI = \\int_ {0} ^ {+ \\infty} \\mathrm{d} x \\int_ {a} ^ {b} e ^ {- p x} \\cos x y \\mathrm{d} y\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":78,"id":"14769691-2b03-4540-b050-49b6f2d532f4","page_size":[595,841],"block_position":"78-22"},{"bbox":[55,707,539,736],"type":"text","angle":0,"index":24,"merge_prev":false,"text":"由于 $|e^{-px}\\cos xy|\\leq e^{-px}$ ，显然 $\\int_0^{+\\infty}e^{-px}\\mathrm{d}x$ 绝对收敛，因此 $\\int_0^{+\\infty}e^{-px}\\cos xy\\mathrm{d}x,y\\in [a,b]$ 一致收敛，交换积 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":78,"id":"32c24d30-3d97-4cee-940d-b2b7392b3edf","page_size":[595,841],"block_position":"78-23"},{"bbox":[291,802,302,812],"type":"page_number","angle":0,"index":1000010,"text":"71 ","color":{"line":"rgba(164, 164, 164, 1)","fill":"rgba(164, 164, 164, 1)"},"page_idx":78,"is_discarded":true,"id":"ff54a7c2-7899-41d4-ac77-d7ee50e0b176","page_size":[595,841],"block_position":"78-24"}],[{"bbox":[449,46,539,58],"type":"header","angle":0,"index":-99999,"text":"7.2 反常积分的计算 ","color":{"line":"rgba(164, 164, 164, 1)","fill":"rgba(164, 164, 164, 1)"},"page_idx":79,"is_discarded":true,"id":"012b9d6b-7b59-4ed1-ab5c-19a7b42b1ebd","page_size":[595,841],"block_position":"79-0"},{"bbox":[54,72,107,84],"type":"text","angle":0,"index":1,"merge_prev":false,"text":"分顺序得到 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":79,"id":"201e4285-8835-4568-b579-91a1db402a35","page_size":[595,841],"block_position":"79-1"},{"bbox":[229,85,365,114],"type":"interline_equation","angle":0,"index":2,"text":"$$\nI = \\int_ {a} ^ {b} \\mathrm{d} y \\int_ {0} ^ {+ \\infty} e ^ {- p x} \\cos x y \\mathrm{d} x\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":79,"id":"2d52f121-fb7e-4a1d-b34a-110c7a298a86","page_size":[595,841],"block_position":"79-2"},{"bbox":[54,116,131,127],"type":"text","angle":0,"index":3,"merge_prev":false,"text":"根据表格法得到: ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":79,"id":"52dda00c-2fd3-468e-84d6-34a3fe664036","page_size":[595,841],"block_position":"79-3"},{"bbox":[106,129,487,158],"type":"interline_equation","angle":0,"index":4,"text":"$$\n\\int_ {0} ^ {+ \\infty} e ^ {- p x} \\cos x y \\mathrm{d} x = - \\frac {1}{p} e ^ {- p x} \\cos x y \\bigg | _ {0} ^ {+ \\infty} + \\frac {y}{p ^ {2}} e ^ {- p x} \\sin x y \\bigg | _ {0} ^ {+ \\infty} - \\int_ {0} ^ {+ \\infty} \\frac {y ^ {2}}{p ^ {2}} e ^ {- p x} \\cos x y \\mathrm{d} x\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":79,"id":"f5b4f485-f10e-4ace-8b20-0aa45de84c07","page_size":[595,841],"block_position":"79-4"},{"bbox":[54,161,219,189],"type":"text","angle":0,"index":5,"merge_prev":false,"text":"因此得到 $\\frac{y^2 + p^2}{p^2}\\int_0^{+\\infty}\\cos xy\\mathrm{d}y = \\frac{1}{p}$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":79,"id":"f4dbe6ea-0037-475e-8d9b-e7cfbd0c72ad","page_size":[595,841],"block_position":"79-5"},{"bbox":[205,190,387,219],"type":"interline_equation","angle":0,"index":6,"text":"$$\nI = \\int_ {a} ^ {b} {\\frac {p}{p ^ {2} + y ^ {2}}} \\mathrm{d} y = \\arctan {\\frac {b}{p}} - \\arctan {\\frac {a}{p}}\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":79,"id":"e88d0d4d-bc27-4132-b3b2-0efda718fa2a","page_size":[595,841],"block_position":"79-6"},{"bbox":[73,231,186,243],"type":"title","angle":0,"index":7,"level":2,"text":"## 定理 7.10 (Dirichlet 积分)","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":79,"id":"a26c5806-1d0d-4f72-84e5-271c417b2bdb","page_size":[595,841],"block_position":"79-7"},{"bbox":[68,248,348,275],"type":"text","angle":0,"index":8,"merge_prev":false,"text":"积分 $\\int_0^{+\\infty}\\frac{\\sin x}{x}\\mathrm{d}x = \\frac{\\pi}{2}$ ，进一步地， $\\int_0^{+\\infty}\\frac{\\sin px}{x}\\mathrm{d}x = \\frac{\\pi}{2}\\mathrm{sgn}(p)$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":79,"id":"70deccbf-025c-4b9b-96c2-c7cbd66c3b91","page_size":[595,841],"block_position":"79-8"},{"bbox":[53,284,540,322],"type":"text","angle":0,"index":9,"merge_prev":false,"text":"证明 令引理中 $a = 0$ ，将 $p$ 看成参数， $G(p) = \\int_{0}^{+\\infty} e^{-px} \\frac{\\sin bx}{x} \\mathrm{d}x = \\arctan \\frac{b}{p}$ ，由于 $\\int_{0}^{+\\infty} \\frac{\\sin bx}{x} \\mathrm{d}x$ 条件收敛， $e^{-px}$ 在固定 $x$ 时关于 $p$ 单调，且一致有界，根据 Abel 判别法可知 $G(p)$ 在 [0,1] 上一致收敛，从而 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":79,"id":"1af0f95b-c830-4dd2-a47e-95c607a17a29","page_size":[595,841],"block_position":"79-9"},{"bbox":[195,326,398,350],"type":"interline_equation","angle":0,"index":10,"text":"$$\nG (0) = \\lim _ {p \\to 0 ^ {+}} G (p) = \\lim _ {p \\to 0 ^ {+}} \\arctan {\\frac {b}{p}} = \\frac {\\pi}{2} \\mathrm{sgn} (b)\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":79,"id":"a41e1208-3d61-47f0-af68-96732b3c778b","page_size":[595,841],"block_position":"79-10"},{"bbox":[54,354,226,380],"type":"text","angle":0,"index":11,"merge_prev":false,"text":"特别地， $b = 1$ 得到 $\\int_0^{+\\infty}\\frac{\\sin x}{x}\\mathrm{d}x = \\frac{\\pi}{2}$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":79,"id":"56e4ffba-ca0b-4b6d-91a7-8e5e03a64c39","page_size":[595,841],"block_position":"79-11"},{"bbox":[35,383,499,411],"type":"text","angle":0,"index":12,"merge_prev":false,"text":"练习7.19Dirichlet积分的应用 计算以下反常积分：(1)重点： $\\int_0^{+\\infty}\\frac{\\sin x^2}{x}\\mathrm{d}x$ (2)重点： $\\int_0^{+\\infty}\\frac{\\sin^2x}{x^2}\\mathrm{d}x$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":79,"id":"9a82c5da-f9d9-486f-b0b4-2aa62c189b0b","page_size":[595,841],"block_position":"79-12"},{"bbox":[73,409,323,436],"type":"interline_equation","angle":0,"index":13,"text":"$$\n(3) \\int_ {0} ^ {+ \\infty} \\frac {\\sin^ {4} x}{x ^ {2}} \\mathrm{d} x (4) \\int_ {0} ^ {+ \\infty} \\frac {\\sin^ {4} x}{x ^ {4}} \\mathrm{d} x (5) \\int_ {0} ^ {+ \\infty} \\frac {x - \\sin x}{x ^ {3}} \\mathrm{d} x\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":79,"id":"d97c474a-3adb-483b-bbb1-3f877827cd56","page_size":[595,841],"block_position":"79-13"},{"bbox":[53,437,242,460],"type":"text","angle":0,"index":14,"merge_prev":false,"text":"解(1)做 $t = x^2$ ，得到 $\\frac{1}{2}\\int_0^{+\\infty}\\frac{\\sin t}{t}\\mathrm{d}t = \\frac{\\pi}{4}$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":79,"id":"461522e1-2df4-4af5-b6ea-5f0d540ac3c3","page_size":[595,841],"block_position":"79-14"},{"bbox":[73,460,336,488],"type":"text","angle":0,"index":15,"merge_prev":false,"text":"(2) $\\int_{0}^{+\\infty}\\frac{\\sin^{2}x}{x^{2}}\\mathrm{d}x=-\\int_{0}^{+\\infty}\\sin^{2}x\\mathrm{d}\\left(\\frac{1}{x}\\right)$ ，用分部积分得到 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":79,"id":"124e5467-a3b7-4721-b376-a44ae2540add","page_size":[595,841],"block_position":"79-15"},{"bbox":[207,491,386,518],"type":"interline_equation","angle":0,"index":16,"text":"$$\nI = - \\frac {\\sin^ {2} x}{x} \\bigg | _ {0} ^ {+ \\infty} + \\int_ {0} ^ {+ \\infty} \\frac {\\sin 2 x}{2 x} \\mathrm{d} 2 x = \\frac {\\pi}{2}\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":79,"id":"13fd4edf-35f1-4bdd-874d-6fa786b911db","page_size":[595,841],"block_position":"79-16"},{"bbox":[73,521,161,533],"type":"text","angle":0,"index":17,"merge_prev":false,"text":"(3) 先分部积分得到 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":79,"id":"17c5705c-4be9-4501-950a-cdb774621918","page_size":[595,841],"block_position":"79-17"},{"bbox":[158,535,435,563],"type":"interline_equation","angle":0,"index":18,"text":"$$\n\\int_ {0} ^ {+ \\infty} \\frac {\\sin^ {4} x}{x ^ {2}} \\mathrm{d} x = - \\int_ {0} ^ {+ \\infty} \\sin^ {4} x \\mathrm{d} \\left(\\frac {1}{x}\\right) = \\int_ {0} ^ {+ \\infty} \\frac {4 \\sin^ {3} x \\cos x}{x} \\mathrm{d} x\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":79,"id":"03caec4f-819d-4d01-88a3-a1226a4ee2fd","page_size":[595,841],"block_position":"79-18"},{"bbox":[54,567,327,588],"type":"text","angle":0,"index":19,"merge_prev":false,"text":"根据 $4\\sin^3 x\\cos x = \\sin 2x(1 - \\cos 2x) = \\sin 2x - \\frac{1}{2}\\sin 4x$ 得到 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":79,"id":"2826cf47-366f-4f39-bfe1-689a9611d274","page_size":[595,841],"block_position":"79-19"},{"bbox":[186,591,406,618],"type":"interline_equation","angle":0,"index":20,"text":"$$\nI = \\int_ {0} ^ {+ \\infty} \\frac {\\sin 2 x}{2 x} \\mathrm{d} (2 x) - \\frac {1}{2} \\int_ {0} ^ {+ \\infty} \\frac {\\sin 4 x}{4 x} \\mathrm{d} (4 x) = \\frac {\\pi}{4}\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":79,"id":"40ffdea2-d0a8-4753-b92c-1ac5bfb0d17d","page_size":[595,841],"block_position":"79-20"},{"bbox":[291,802,302,812],"type":"page_number","angle":0,"index":1000020,"text":"72 ","color":{"line":"rgba(164, 164, 164, 1)","fill":"rgba(164, 164, 164, 1)"},"page_idx":79,"is_discarded":true,"id":"271adfad-329a-4c50-b467-8bec49fdb024","page_size":[595,841],"block_position":"79-21"}],[{"bbox":[236,74,368,94],"type":"title","angle":0,"index":0,"level":2,"text":"## 第8章 数项级数","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":80,"id":"6d9d50c3-b315-446f-ad87-d7301e7d351b","page_size":[595,841],"block_position":"80-0"},{"bbox":[55,122,201,138],"type":"title","angle":0,"index":1,"level":2,"text":"## 8.1 数项级数基本理论","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":80,"id":"12186877-795e-44ca-8d5a-49bbeb7f561e","page_size":[595,841],"block_position":"80-1"},{"bbox":[73,157,180,169],"type":"title","angle":0,"index":2,"level":2,"text":"## 定义 8.1 (数项级数收敛)","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":80,"id":"5b3ecee3-10d8-4d94-88ed-175483e56763","page_size":[595,841],"block_position":"80-2"},{"bbox":[68,173,525,217],"type":"text","angle":0,"index":3,"merge_prev":false,"text":"给定数项级数 $\\sum_{n=1}^{\\infty} u_n$ ，定义 $S_n = \\sum_{k=1}^{n} u_n$ 为部分和函数，若 $\\lim_{n \\to \\infty} S_n = S$ 收敛，则称 $\\sum_{n=1}^{\\infty} u_n$ 收敛，否则称数项级数发散。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":80,"id":"a761226c-0442-4e3d-b972-855e0fffb533","page_size":[595,841],"block_position":"80-3"},{"bbox":[73,237,231,249],"type":"title","angle":0,"index":4,"level":2,"text":"## 定理 8.1 (数项级数收敛的各种条件)","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":80,"id":"6355f100-52c9-49cb-8733-592bd0ce4371","page_size":[595,841],"block_position":"80-4"},{"bbox":[68,253,237,285],"type":"text","angle":0,"index":5,"merge_prev":false,"text":"给定数项级数 $\\sum_{n=1}^{\\infty} u_n$ ，其收敛的条件： ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":80,"id":"68643af9-932b-455c-8d46-052041b20571","page_size":[595,841],"block_position":"80-5"},{"bbox":[83,284,258,298],"type":"text","angle":0,"index":6,"merge_prev":false,"text":"- 必要条件 (一定能推出): $\\lim_{n\\to \\infty}u_n = 0$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":80,"id":"36c5fc83-701f-41e8-836b-7ca02370bf59","page_size":[595,841],"block_position":"80-6"},{"bbox":[83,299,429,313],"type":"text","angle":0,"index":7,"merge_prev":false,"text":"- 充要条件 (Cauchy 准则): $\\forall \\epsilon, \\exists N, \\forall m, m + p > N$ 有 $|u_{m+1} + \\cdots + u_{m+p}| < \\epsilon$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":80,"id":"82d4911c-b125-420b-90c2-1d475e6926b8","page_size":[595,841],"block_position":"80-7"},{"bbox":[52,328,364,342],"type":"title","angle":0,"index":8,"level":2,"text":"## 笔记 一般证明发散要么不满足必要条件，要么用 Cauchy 准则逆否命题","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":80,"id":"c42e39f7-fa45-4805-8fe6-47743daa7156","page_size":[595,841],"block_position":"80-8"},{"bbox":[35,352,539,439],"type":"text","angle":0,"index":9,"merge_prev":false,"text":"练习 8.1Cauchy 收敛准则 (1) 证明 $\\sum_{n=1}^{\\infty}\\frac{1}{n}$ 发散 (2) 用 Cauchy 收敛准则证明 $\\sum_{n=1}^{\\infty}\\frac{1}{n^{2}}$ 收敛\n\n证明 (1) $\\exists\\epsilon_{0}=0,\\forall N,m=N+1,m+p=2N$ ，则 $\\frac{1}{m+1}+\\frac{1}{m+2}+\\cdots+\\frac{1}{2m}\\geq\\frac{1}{2}=\\epsilon_{0}$ ，根据 Cauchy 逆否命题可证 (也可以用积分判别法证明)\n\n(2) 用 $\\frac{1}{n^{2}}<\\frac{1}{n(n-1)}$ ，直接求和或者用 Cauchy 收敛准则都行 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":80,"id":"41f1610f-fd69-419a-a14c-82efcd919041","page_size":[595,841],"block_position":"80-9"},{"bbox":[73,456,161,469],"type":"title","angle":0,"index":10,"level":2,"text":"## 定理 8.2 (子列问题)","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":80,"id":"bfdb4730-ff33-4aed-8c60-f0281c7bbda9","page_size":[595,841],"block_position":"80-10"},{"bbox":[67,470,535,508],"type":"text","angle":0,"index":11,"merge_prev":false,"text":"对于数项级数 $\\sum_{n=1}^{\\infty} u_n$ ，若 $\\lim_{n \\to \\infty} u_n = 0$ ， $S_n = \\sum_{k=1}^{n} u_k$ 有一个子列 $S_{np}$ ( $p$ 为固定整数) 收敛，则 $\\sum_{n=1}^{\\infty} u_n$ 收敛 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":80,"id":"9f91a915-930a-495e-a4b7-11afb4ef2921","page_size":[595,841],"block_position":"80-11"},{"bbox":[36,531,370,555],"type":"text","angle":0,"index":12,"merge_prev":false,"text":"练习 8.2 交错子列 证明 $1 + \\frac{1}{2} + (\\frac{1}{3} - 1) + \\frac{1}{4} + \\frac{1}{5} + (\\frac{1}{6} - \\frac{1}{2}) + \\cdots = \\ln 3$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":80,"id":"5ff98efb-13e2-4f60-a0b7-86bc47425c25","page_size":[595,841],"block_position":"80-12"},{"bbox":[39,550,537,588],"type":"text","angle":0,"index":13,"merge_prev":false,"text":"证明 $S_{3n}=\\sum_{k=1}^{3n}u_{k}=(1+\\frac{1}{2}+\\cdots+\\frac{1}{3n})-(1+\\frac{1}{2}+\\cdots+\\frac{1}{n})$ ，配合 Euler 常数可知 $S_{3n}=\\ln3n+\\gamma-\\ln n-\\gamma=\\ln3$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":80,"id":"50716fee-3855-4b88-9cdf-482ffa2b21af","page_size":[595,841],"block_position":"80-13"},{"bbox":[73,598,170,611],"type":"title","angle":0,"index":14,"level":2,"text":"## 定理 8.3 (去除有限项)","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":80,"id":"7e8ac2be-ff99-4946-9981-ac451db1d1c8","page_size":[595,841],"block_position":"80-14"},{"bbox":[70,615,336,628],"type":"text","angle":0,"index":15,"merge_prev":false,"text":"去掉、增加、改变级数的有限项并不改变数项级数的收敛性。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":80,"id":"d30df763-bde3-40cf-b1e6-1db28a976059","page_size":[595,841],"block_position":"80-15"}],[{"bbox":[479,47,539,58],"type":"header","angle":0,"index":-99999,"text":"8.2 正项级数 ","color":{"line":"rgba(164, 164, 164, 1)","fill":"rgba(164, 164, 164, 1)"},"page_idx":81,"is_discarded":true,"id":"144996ea-5eb5-4dc4-913b-2ecb16074467","page_size":[595,841],"block_position":"81-0"},{"bbox":[55,74,143,90],"type":"title","angle":0,"index":1,"level":2,"text":"## 8.2 正项级数","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":81,"id":"24abd075-e58c-4846-b4bb-844f8e1afa3e","page_size":[595,841],"block_position":"81-1"},{"bbox":[55,106,139,121],"type":"title","angle":0,"index":2,"level":2,"text":"## 8.2.1 比较原则","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":81,"id":"35d84eaf-5205-4327-95f8-3ad4a7f40ab2","page_size":[595,841],"block_position":"81-2"},{"bbox":[73,135,161,147],"type":"title","angle":0,"index":3,"level":2,"text":"## 定理 8.4 (比较原则)","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":81,"id":"531003ab-b4a3-405d-a845-6a48149cf1cc","page_size":[595,841],"block_position":"81-3"},{"bbox":[67,151,207,184],"type":"text","angle":0,"index":4,"merge_prev":false,"text":"$\\sum_{n = 1}^{\\infty}u_n,\\sum_{n = 1}^{\\infty}v_n$ 是两个正项级数， ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":81,"id":"340ad7dc-0957-4231-ada0-0451ff785ae3","page_size":[595,841],"block_position":"81-4"},{"bbox":[82,180,525,243],"type":"text","angle":0,"index":5,"merge_prev":false,"text":"- 非极限形式：若 $\\exists N > 0$ 使得 $\\forall n > N$ 满足 $u_{n} \\leq v_{n}$ ，则根据 $\\sum_{n=1}^{\\infty} v_{n}$ 收敛推出 $\\sum_{n=1}^{\\infty} u_{n}$ 收敛；根据 $\\sum_{n=1}^{\\infty} u_{n}$ 发散，则 $\\sum_{n=1}^{\\infty} v_{n}$ 也发散 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":81,"id":"6684665a-7578-4f36-b602-1d27925c7fa1","page_size":[595,841],"block_position":"81-5"},{"bbox":[83,241,528,307],"type":"text","angle":0,"index":6,"merge_prev":false,"text":"- 极限形式： $\\lim_{n\\to \\infty}\\frac{u_n}{v_n} = \\lambda$ 。当 $0 < \\lambda < \\infty$ ，同敛散。 $\\lambda = 0$ 则 $\\sum_{n = 1}^{\\infty}v_n$ 敛推 $\\sum_{n = 1}^{\\infty}u_n$ 敛。 $\\lambda = \\infty$ 时 $\\sum_{n = 1}^{\\infty}v_n$ 发散推 $\\sum_{n = 1}^{\\infty}v_n$ 发散 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":81,"id":"da426a92-f578-4bd1-9a8f-532437b21f8e","page_size":[595,841],"block_position":"81-6"},{"bbox":[52,319,539,348],"type":"text","angle":0,"index":7,"merge_prev":false,"text":"笔记 比较原则极限形式的 $0 < \\lambda < \\infty$ 情况即可以使用等价无穷小。但在一些特殊情况下也需要用 $\\lambda = 0, \\lambda = \\infty$ 的形式。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":81,"id":"c422cab8-1b74-473d-9ed0-153ff1efb386","page_size":[595,841],"block_position":"81-7"},{"bbox":[52,359,339,402],"type":"text","angle":0,"index":8,"merge_prev":false,"text":"例题8.1常用比较对象 $\\sum_{n = 0}^{\\infty}\\frac{1}{n^p}$ 在 $p > 1$ 时收敛，在 $p\\leq 1$ 时发散证明用积分判别法 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":81,"id":"0ebbd6e6-4272-45e8-8ce9-65c84cf2704f","page_size":[595,841],"block_position":"81-8"},{"bbox":[35,413,364,446],"type":"text","angle":0,"index":9,"merge_prev":false,"text":"练习8.3使用等价无穷小 判断敛散性：(1) $\\sum_{n=1}^{\\infty} \\frac{1}{\\sqrt{n^2 + 1}}$ (2) $\\sum_{n=1}^{\\infty} 2^n \\sin \\frac{\\pi}{3^n}$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":81,"id":"acc10d7b-a418-46b9-835a-fc925d044980","page_size":[595,841],"block_position":"81-9"},{"bbox":[72,444,337,476],"type":"text","angle":0,"index":10,"merge_prev":false,"text":"(3) 重点: $\\sum_{n=1}^{\\infty} (\\sqrt[n]{2} - 1)$ (4) 重点: $\\sum_{n=1}^{\\infty} (a^{\\frac{1}{n}} + a^{-\\frac{1}{n}} - 2)(a > 1)$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":81,"id":"6b308db8-bbfc-45ed-8a9e-38f1cebfc733","page_size":[595,841],"block_position":"81-10"},{"bbox":[54,473,171,499],"type":"text","angle":0,"index":11,"merge_prev":false,"text":"解 (1) $\\frac{1}{\\sqrt{n^2 + 1}} \\sim \\frac{1}{n}$ ，发散 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":81,"id":"7c139910-a119-4e36-9a51-876ef736237b","page_size":[595,841],"block_position":"81-11"},{"bbox":[73,498,251,520],"type":"text","angle":0,"index":12,"merge_prev":false,"text":"(2)和等比级数比，等价于 $(\\frac{2}{3})^n\\pi$ ，收敛 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":81,"id":"acef6d63-d112-46eb-8a9b-2e4b44f3ad49","page_size":[595,841],"block_position":"81-12"},{"bbox":[73,520,357,541],"type":"text","angle":0,"index":13,"merge_prev":false,"text":"(3) 根据 $2^{\\frac{1}{n}} - 1 \\sim \\frac{1}{n} \\ln 2$ ，发散 (注意这里用的是 $a^x \\sim x \\ln a - 1$ ) ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":81,"id":"76990d5f-da43-46b3-b223-2d757ae36cd8","page_size":[595,841],"block_position":"81-13"},{"bbox":[73,540,381,566],"type":"text","angle":0,"index":14,"merge_prev":false,"text":"(4) 根据 $a^{\\frac{1}{n}} + a^{-\\frac{1}{n}} - 2 = (a^{\\frac{1}{2n}} - a^{-\\frac{1}{2n}})^2 = a^{-\\frac{1}{n}}(a^{\\frac{1}{n}} - 1)^2 \\sim \\frac{\\ln^2 a}{n^2}$ 收敛 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":81,"id":"aa34d343-50f3-4b6b-bf5a-3f766d78e42f","page_size":[595,841],"block_position":"81-14"},{"bbox":[35,571,516,604],"type":"text","angle":0,"index":15,"merge_prev":false,"text":"练习8.4含对数与指数(1)重点： $\\sum_{n = 1}^{\\infty}\\frac{1}{3^{\\ln n}}$ (2)重点： $\\sum_{n = 2}^{\\infty}\\frac{1}{(\\ln n)^{\\ln n}}$ (3) $\\sum_{n = 3}^{\\infty}\\frac{1}{(\\ln\\ln n)^{\\ln n}}$ (4)特殊： $\\sum_{n = 2}^{\\infty}\\frac{1}{(\\ln n)^n}$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":81,"id":"55f175f8-5f58-4c63-bff4-de6331483ab9","page_size":[595,841],"block_position":"81-15"},{"bbox":[70,602,368,634],"type":"text","angle":0,"index":16,"merge_prev":false,"text":"(5) 重点: $\\sum_{n=1}^{\\infty} \\frac{1}{n^{2n \\sin \\frac{1}{n}}}$ (6) 重点: $\\sum_{n=1}^{\\infty} \\frac{\\ln(n+1)}{e^n}$ (7) $\\sum_{n=1}^{\\infty} (e^{\\frac{1}{n^2}} - \\cos \\frac{\\pi}{n})$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":81,"id":"86897839-8e6f-4b59-b58c-d67b39c0ae89","page_size":[595,841],"block_position":"81-16"},{"bbox":[54,633,307,645],"type":"text","angle":0,"index":17,"merge_prev":false,"text":"解 对数需要用到恒等式 $a^{\\ln b} = b^{\\ln a}$ (两侧取对数即可证明) ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":81,"id":"9b6c113c-debe-44c6-b1fe-94c511285005","page_size":[595,841],"block_position":"81-17"},{"bbox":[73,645,166,666],"type":"text","angle":0,"index":18,"merge_prev":false,"text":"(1) $\\frac{1}{3^{\\ln n}} = \\frac{1}{n^{\\ln 3}}$ 收敛 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":81,"id":"e570096b-d677-4521-850c-58a67f2f5272","page_size":[595,841],"block_position":"81-18"},{"bbox":[73,665,224,690],"type":"text","angle":0,"index":19,"merge_prev":false,"text":"(2) $\\frac{1}{(\\ln n)^{\\ln n}}=\\frac{1}{n^{\\ln(\\ln n)}}<\\frac{1}{n^{2}}$ 收敛 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":81,"id":"e4e5b91e-2ae4-41fc-b030-c45afbc8a799","page_size":[595,841],"block_position":"81-19"},{"bbox":[73,689,246,713],"type":"text","angle":0,"index":20,"merge_prev":false,"text":"(3) $\\frac{1}{(\\ln \\ln n)^{\\ln n}} = \\frac{1}{n^{\\ln \\ln \\ln n}} < \\frac{1}{n^2}$ ，收敛 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":81,"id":"a2708fba-f186-4ee4-86c5-8b7f798ff385","page_size":[595,841],"block_position":"81-20"},{"bbox":[74,712,177,736],"type":"text","angle":0,"index":21,"merge_prev":false,"text":"(4) $\\frac{1}{(\\ln n)^n} < \\frac{1}{2^n}$ ，收敛 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":81,"id":"6b449b1e-fc78-44fc-9e7b-c8e6b2d79d19","page_size":[595,841],"block_position":"81-21"},{"bbox":[73,736,209,760],"type":"text","angle":0,"index":22,"merge_prev":false,"text":"(5) $\\frac{1}{n^{2n\\sin\\frac{1}{n}}} < \\frac{1}{n^{\\frac{3}{2}}}$ ，因此收敛。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":81,"id":"73d05572-cfb6-48fc-944d-5f1e9a3afdd5","page_size":[595,841],"block_position":"81-22"},{"bbox":[291,802,303,811],"type":"page_number","angle":0,"index":1000022,"text":"74 ","color":{"line":"rgba(164, 164, 164, 1)","fill":"rgba(164, 164, 164, 1)"},"page_idx":81,"is_discarded":true,"id":"07f22601-c278-4980-bfaa-12379d0f7d62","page_size":[595,841],"block_position":"81-23"}],[{"bbox":[480,47,539,58],"type":"header","angle":0,"index":-99999,"text":"8.2 正项级数 ","color":{"line":"rgba(164, 164, 164, 1)","fill":"rgba(164, 164, 164, 1)"},"page_idx":82,"is_discarded":true,"id":"445ce5f2-b13b-4f8e-942c-66c27b008eaa","page_size":[595,841],"block_position":"82-0"},{"bbox":[73,69,185,94],"type":"text","angle":0,"index":1,"merge_prev":false,"text":"(6) $\\lim_{n\\to \\infty}n^2\\frac{\\ln(n + 1)}{e^n}\\rightarrow 0$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":82,"id":"3c5011d8-9e05-4b42-be84-43ad36d1a357","page_size":[595,841],"block_position":"82-1"},{"bbox":[52,91,540,148],"type":"text","angle":0,"index":2,"merge_prev":false,"text":"(7)Taylor 展开: $e^{\\frac{1}{n^2}} - \\cos \\frac{\\pi}{n} = 1 + \\frac{1}{n^2} - 1 + \\frac{\\pi^2}{2n^2} + o\\left(\\frac{1}{n^2}\\right) = \\left(1 + \\frac{\\pi^2}{2}\\right)\\frac{1}{n^2}$ 收敛笔记要灵活使用比较判别法。有时用等价无穷小，有时直接放缩，有时乘上 $n^2$ (一般在分母极大)，有时Taylor展开 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":82,"id":"61b78520-28bf-4ac5-a19b-d61c1ca7c71e","page_size":[595,841],"block_position":"82-2"},{"bbox":[35,160,496,219],"type":"text","angle":0,"index":3,"merge_prev":false,"text":"练习 8.5 等价无穷小的进一步应用 $(1)a_{n}>0$ ，证明 $(1+a_{1})(1+a_{2})\\cdots(1+a_{n})$ 与 $\\sum_{n=1}^{\\infty}a_{n}$ 敛散性相同 $(2)a_{n}=(1-\\frac{p\\ln n}{n})^{n}$ ，讨论 $\\sum_{n=1}^{\\infty}b_{n}$ 的敛散性 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":82,"id":"f443beb5-dde0-4204-ba79-9d059b0edfa8","page_size":[595,841],"block_position":"82-3"},{"bbox":[54,219,539,296],"type":"text","angle":0,"index":4,"merge_prev":false,"text":"(3) $f(x)$ 在 $[-1, 1]$ 二阶连续可微，且 $\\lim_{x \\to 0} \\frac{f(x)}{x} = 0$ ，证明 $\\sum_{n=1}^{\\infty} f\\left(\\frac{1}{n}\\right)$ 绝对收敛证明 (1) 取对数即 $\\sum_{n=1}^{\\infty} \\ln(1 + a_n)$ ， $\\lim_{n \\to \\infty} a_n = 0$ 时， $\\lim_{n \\to \\infty} \\frac{\\ln(1 + a_n)}{a_n} = 1$ ，故同敛散。若 $\\lim_{n \\to \\infty} a_n \\neq 0$ 时， $\\sum_{n=1}^{\\infty} a_n$ 发散，显然 $\\lim_{n \\to \\infty} \\ln(1 + a_n) \\neq 0$ ，故也发散。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":82,"id":"dbaecbe6-521f-411a-9005-b13f6272ac5f","page_size":[595,841],"block_position":"82-4"},{"bbox":[73,296,459,317],"type":"text","angle":0,"index":5,"merge_prev":false,"text":"(2) $a_{n}=e^{n\\ln(1-\\frac{p\\ln n}{n})}=e^{n[-\\frac{p\\ln n}{n}+o(\\frac{1}{n})]}=e^{-p\\ln n}\\sim\\frac{1}{p^{p}}$ ，当p>1时收敛， $p\\leq1$ 时发散。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":82,"id":"f9c58c3b-9473-4b41-a034-378158f5a1ec","page_size":[595,841],"block_position":"82-5"},{"bbox":[73,317,471,342],"type":"text","angle":0,"index":6,"merge_prev":false,"text":"(3) 根据 $\\lim_{x\\to0}\\frac{f(x)}{x}=0$ 得到 $f'(0)=0$ ，根据 Lagrange 余项得到 $f(\\frac{1}{n})=f'(\\xi)\\frac{1}{n^{2}}$ ，因此收敛 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":82,"id":"221038a2-68cd-4093-b7c0-4c251d7b003e","page_size":[595,841],"block_position":"82-6"},{"bbox":[73,358,190,370],"type":"title","angle":0,"index":7,"level":2,"text":"## 定理 8.5 (比式比较判别法)","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":82,"id":"a6d2b3df-797d-40b5-80a5-b60f975530c9","page_size":[595,841],"block_position":"82-7"},{"bbox":[67,372,535,440],"type":"text","angle":0,"index":8,"merge_prev":false,"text":"$\\sum_{n=1}^{\\infty} u_n, \\sum_{n=1}^{\\infty} v_n$ 为正项级数， $\\exists N, \\forall n \\geq N$ 有 $\\frac{u_{n+1}}{u_n} \\leq \\frac{v_{n+1}}{v_n}$ ，则 $\\sum_{n=1}^{\\infty} v_n$ 收敛可推出 $\\sum_{n=1}^{\\infty} u_n$ 收敛， $\\sum_{n=1}^{\\infty} u_n$ 发散可推出 $\\sum_{n=1}^{\\infty} v_n$ 发散 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":82,"id":"b038f093-3fc6-410e-80fa-ed6187ff7265","page_size":[595,841],"block_position":"82-8"},{"bbox":[53,449,539,484],"type":"text","angle":0,"index":9,"merge_prev":false,"text":"证明 根据 $\\frac{u_{n}}{u_{N}} = \\frac{u_{N+1}}{u_{N}} \\frac{u_{N+2}}{u_{N+1}} \\cdots \\frac{u_{n}}{u_{n-1}} \\leq \\frac{v_{N+1}}{v_{N}} \\frac{v_{N+2}}{v_{N+1}} \\cdots \\frac{v_{n}}{v_{n+1}} = \\frac{v_{n}}{v_{N}}$ ，由此推出 $u_{n} \\leq \\frac{u_{N}}{v_{N}} v_{n}$ 只相差一个常数，因此根据一般的比较判别法可得到结论。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":82,"id":"b48bce2e-b5f4-4d86-ae11-d10b6fb99518","page_size":[595,841],"block_position":"82-9"},{"bbox":[36,497,539,555],"type":"text","angle":0,"index":10,"merge_prev":false,"text":"练习8.6比式比较判别法基本应用 $a_{n} > 0,0 < \\alpha \\leq 1,\\beta >1$ ，证明：(1)若 $\\frac{a_{n + 1}}{a_n}\\geq (1 - \\frac{1}{n})^\\alpha$ ，则 $\\sum_{n = 1}^{\\infty}a_n$ 发散(2)若 $\\frac{a_{n + 1}}{a_n}\\leq (1 - \\frac{1}{n})^\\beta$ ，则 $\\sum_{n = 1}^{\\infty}a_n$ 收敛 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":82,"id":"315b174e-32f6-4044-8ce0-8f681928f894","page_size":[595,841],"block_position":"82-10"},{"bbox":[53,555,365,587],"type":"text","angle":0,"index":11,"merge_prev":false,"text":"证明 (1) 根据 $\\frac{a_{n+1}}{a_n} \\geq \\left(\\frac{n-1}{n}\\right)^\\alpha = \\frac{\\frac{1}{n^\\alpha}}{\\frac{1}{(n-1)^\\alpha}}$ ，因此根据结论发散 (2) 同理 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":82,"id":"fda79687-769e-4b07-8a80-2a22d0303b5d","page_size":[595,841],"block_position":"82-11"},{"bbox":[56,602,223,617],"type":"title","angle":0,"index":13,"level":2,"text":"## 8.2.2 比式判别法与根式判别法","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":82,"id":"9ed9d924-0b27-4d2e-89dc-10bff0071cda","page_size":[595,841],"block_position":"82-12"},{"bbox":[73,624,476,650],"type":"text","angle":0,"index":14,"merge_prev":false,"text":"前面比较判别法常用的比较对象是 $\\frac{1}{n^p}$ ，而根式判别法与比式判别法的比较对象是等比级数。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":82,"id":"04699956-2bb5-4bd9-bb34-cec9b311b7aa","page_size":[595,841],"block_position":"82-13"},{"bbox":[73,657,231,669],"type":"title","angle":0,"index":15,"level":2,"text":"## 定理 8.6 (比式判别法与根式判别法)","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":82,"id":"8664b1e6-b65a-499d-9c6f-16bb55957393","page_size":[595,841],"block_position":"82-14"},{"bbox":[68,674,172,705],"type":"text","angle":0,"index":16,"merge_prev":false,"text":"$\\sum_{n = 1}^{\\infty}u_n$ 是正项级数，则 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":82,"id":"876c87d4-58bd-44cb-a929-a88c3cbecb2f","page_size":[595,841],"block_position":"82-15"},{"bbox":[83,704,448,736],"type":"text","angle":0,"index":17,"merge_prev":false,"text":"- 比式判别法：若 $\\varlimsup_{n\\to \\infty}\\frac{u_{n + 1}}{u_n} = q < 1$ ，则 $\\sum_{n = 1}^{\\infty}u_n$ 收敛，若 $\\varliminf_{n\\to \\infty}\\frac{u_{n + 1}}{u_n} >1$ 则发散。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":82,"id":"aae1ac00-2f5c-4060-ba2a-65b4bad2ae52","page_size":[595,841],"block_position":"82-16"},{"bbox":[291,802,302,812],"type":"page_number","angle":0,"index":1000011,"text":"75 ","color":{"line":"rgba(164, 164, 164, 1)","fill":"rgba(164, 164, 164, 1)"},"page_idx":82,"is_discarded":true,"id":"1875a2d2-877c-458a-b179-2e3d8b6b6004","page_size":[595,841],"block_position":"82-17"}],[{"bbox":[479,47,539,58],"type":"header","angle":0,"index":-99999,"text":"8.2 正项级数 ","color":{"line":"rgba(164, 164, 164, 1)","fill":"rgba(164, 164, 164, 1)"},"page_idx":83,"is_discarded":true,"id":"d75a0cc9-f446-47a2-8447-878f54c1d886","page_size":[595,841],"block_position":"83-0"},{"bbox":[83,79,474,112],"type":"text","angle":0,"index":1,"merge_prev":false,"text":"- 根式判别法：若 $\\varlimsup_{n\\to \\infty}\\sqrt[n]{u_n} = l < 1$ 则 $\\sum_{n = 1}^{\\infty}u_n$ 收敛， $\\varlimsup_{n\\to \\infty}\\sqrt[n]{u_n} >1$ 则发散。(双上极限) ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":83,"id":"c398b253-7cbb-46ad-9f2f-51535ccb43c0","page_size":[595,841],"block_position":"83-1"},{"bbox":[53,123,539,166],"type":"text","angle":0,"index":3,"merge_prev":false,"text":"证明 (1) 若 $\\varlimsup_{n\\to \\infty}\\frac{u_{n + 1}}{u_n} = q < 1$ ，则对 $\\forall r\\in (q,1)$ ， $\\exists N,\\forall n > N$ 有 $\\frac{u_{n + 1}}{u_n} < r = \\frac{r^{n + 1}}{r^n}$ ，而 $\\sum_{n = 1}^{\\infty}r^n$ 收敛，根据比式比较判别法可知。另一侧同理。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":83,"id":"b1b09091-6346-46e8-97b1-e224c4993078","page_size":[595,841],"block_position":"83-2"},{"bbox":[52,168,539,225],"type":"text","angle":0,"index":4,"merge_prev":false,"text":"(2.1) 若 $\\varlimsup_{n\\to \\infty}\\sqrt[n]{u_n} = l < 1$ ，则 $\\forall r\\in (l,1)$ ， $\\exists N,\\forall n > N$ 有 $\\sqrt[n]{u_n} < r$ ，即 $u_{n} < r^{n}$ ，由于 $\\sum_{n = 1}^{\\infty}r^n$ 收敛，因此 $\\sum_{n = 1}^{\\infty}u_n$ 收敛。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":83,"id":"6bdabde2-579a-41e9-9dd7-9f79695a02fd","page_size":[595,841],"block_position":"83-3"},{"bbox":[53,226,539,265],"type":"text","angle":0,"index":5,"merge_prev":false,"text":"(2.2) 另一侧若 $\\varlimsup_{n\\to \\infty}\\sqrt[n]{u_n} = l > 1$ ，则 $\\forall r\\in (1,l)$ ，存在子列 $\\sqrt[n]{u_{n_k}}\\geq r$ ， $u_{n_k}\\geq r^{n_k}\\to \\infty$ ，因此 $\\sum_{n = 1}^{\\infty}u_n$ 发散。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":83,"id":"3e682888-4504-4831-ba20-8a504303a4a9","page_size":[595,841],"block_position":"83-4"},{"bbox":[52,267,553,310],"type":"text","angle":0,"index":6,"merge_prev":false,"text":"笔记理论上根式判别法证明发散只需要证上极限, 比比式条件更弱, 因此更好用。而且由于 $\\lim_{n\\to\\infty}\\frac{u_{n+1}}{u_n}\\leq\\lim_{n\\to\\infty}\\sqrt[n]{u_n}\\leq\\overline{\\lim}_{n\\to\\infty}\\sqrt[n]{u_n}\\leq\\overline{\\lim}_{n\\to\\infty}\\frac{u_{n+1}}{u_n}$ 可知能用比式的一定能用根式, 一定要记住 $\\sqrt[n]{n!}\\sim\\frac{n}{e}$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":83,"id":"96b727f8-2303-4e7b-b8f6-b81d67c776be","page_size":[595,841],"block_position":"83-5"},{"bbox":[35,315,426,346],"type":"text","angle":0,"index":7,"merge_prev":false,"text":"练习8.7使用根式判别法用根式判别法法证明(1) $\\sum_{n = 1}^{\\infty}\\frac{n^2}{(2 + \\frac{1}{n})^n}$ (2) $\\sum_{n = 1}^{\\infty}\\frac{n!}{n^n}$ (3) $\\sum_{n = 1}^{\\infty}\\frac{n!3^n}{n^n}$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":83,"id":"cf1d629e-9737-4a6b-b10a-107317f995ea","page_size":[595,841],"block_position":"83-6"},{"bbox":[72,344,424,376],"type":"text","angle":0,"index":8,"merge_prev":false,"text":"(4) 重点: $\\sum_{n=1}^{\\infty} \\frac{(n!)^{2}}{(2n)!}$ (5) 重点: $\\sum_{n=1}^{\\infty} \\frac{(2n-1)!!}{n!}$ (6) 讨论 $\\sum_{n=1}^{\\infty} \\frac{x^{n} n!}{n^{n}}$ 的收敛性, $x \\geq 0$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":83,"id":"2531850d-fbaa-45a4-8df0-fc3a7f4283a6","page_size":[595,841],"block_position":"83-7"},{"bbox":[54,375,154,397],"type":"text","angle":0,"index":9,"merge_prev":false,"text":"解(1)极限为 $\\frac{1}{2}$ , 收敛 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":83,"id":"c2d46f57-6d5b-4b61-87de-addef49ced55","page_size":[595,841],"block_position":"83-8"},{"bbox":[73,396,192,420],"type":"text","angle":0,"index":10,"merge_prev":false,"text":"(2) 极限为 $\\frac{n / e}{n} = \\frac{1}{e}$ , 收敛 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":83,"id":"d3306226-36cd-45fe-b7f1-21a37a391198","page_size":[595,841],"block_position":"83-9"},{"bbox":[73,418,180,441],"type":"text","angle":0,"index":11,"merge_prev":false,"text":"(3) 极限为 $\\frac{3}{e} > 1$ ，发散 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":83,"id":"99f7dc7d-e33e-46a7-aa12-158f511edcdf","page_size":[595,841],"block_position":"83-10"},{"bbox":[73,440,235,474],"type":"text","angle":0,"index":12,"merge_prev":false,"text":"(4) 极限为 $\\sqrt[n]{\\frac{(n!)^2}{(2n)!}} = \\frac{\\left(\\frac{n}{e}\\right)^2}{\\sqrt[2n]{\\left(\\frac{2n}{e}\\right)^2}} \\sim \\frac{1}{4}$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":83,"id":"a0ea3509-2d6e-47ab-9b8e-cd11280538f6","page_size":[595,841],"block_position":"83-11"},{"bbox":[73,475,434,507],"type":"text","angle":0,"index":13,"merge_prev":false,"text":"(5) $\\sum_{n=1}^{\\infty}\\frac{(2n-1)!!}{n!}=\\sum_{n=1}^{\\infty}\\frac{(2n)!}{n!(2n)!!}=\\sum_{n=1}^{\\infty}\\frac{(2n)!}{n!2^{n}n!}$ ，因此根据根式判别法等价于2>1。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":83,"id":"1721748c-f994-4058-baef-5be54d76a682","page_size":[595,841],"block_position":"83-12"},{"bbox":[52,509,539,544],"type":"text","angle":0,"index":14,"merge_prev":false,"text":"(7) $\\lim_{n\\to \\infty}\\sqrt[n]{\\frac{x^n n!}{n^n}} = \\frac{x}{e}, x\\in (0,e)$ 收敛， $x\\in (e, + \\infty)$ 发散。当 $x = e$ 则 $\\frac{a_{n + 1}}{a_n} = \\frac{e}{(1 + \\frac{1}{n})^n} >1$ ，极限不趋于0。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":83,"id":"2158b2cd-10ab-4aca-ae04-47e0a59a27be","page_size":[595,841],"block_position":"83-13"},{"bbox":[35,556,296,587],"type":"text","angle":0,"index":15,"merge_prev":false,"text":"练习8.8仅能用根式判别法(1)证明 $\\sum_{n=1}^{\\infty} \\frac{1}{2^{n+(-1)^n}}$ 收敛 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":83,"id":"7ecff538-e97e-4026-ae26-8985c53f0a54","page_size":[595,841],"block_position":"83-14"},{"bbox":[73,587,208,618],"type":"text","angle":0,"index":16,"merge_prev":false,"text":"(2) 讨论 $\\sum_{n=1}^{\\infty} \\frac{x^n}{1 + x^{2n}}$ 的敛散性 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":83,"id":"d708edfa-87c2-4396-8743-af5f637efd39","page_size":[595,841],"block_position":"83-15"},{"bbox":[53,618,249,643],"type":"text","angle":0,"index":17,"merge_prev":false,"text":"证明 (1) $\\lim_{n\\to \\infty}\\sqrt[n]{\\frac{1}{2^{n + (-1)^n}}} = \\frac{1}{2} < 1$ 因此收敛 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":83,"id":"40b8a97b-f8e3-4777-9ee1-cb9eae69a1e5","page_size":[595,841],"block_position":"83-16"},{"bbox":[73,645,271,662],"type":"text","angle":0,"index":18,"merge_prev":false,"text":"(2) 分母 $\\lim_{n\\to \\infty}\\sqrt[n]{1 + x^{2n}} = \\max \\{1,x^2\\}$ ，因此 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":83,"id":"5b7998ca-78bb-4d88-9d51-5d2f1a166d2b","page_size":[595,841],"block_position":"83-17"},{"bbox":[202,666,389,703],"type":"interline_equation","angle":0,"index":19,"text":"$$\n\\lim _ {n \\to \\infty} \\sqrt [ n ]{u _ {n}} = \\frac {x}{\\max \\{1 , x ^ {2} \\}} = \\left\\{ \\begin{array}{l l} < 1, & x \\neq 1 \\\\ = 1, & x = 1 \\end{array} \\right.\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":83,"id":"90e7a47c-2ecc-4c51-8176-799252c9eff4","page_size":[595,841],"block_position":"83-18"},{"bbox":[54,707,207,719],"type":"text","angle":0,"index":20,"merge_prev":false,"text":"因此 $x \\neq 1$ 时收敛， $x = 1$ 时发散。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":83,"id":"5f4809d5-eb96-4c69-b310-2b5598efea21","page_size":[595,841],"block_position":"83-19"},{"bbox":[35,737,514,766],"type":"text","angle":0,"index":21,"merge_prev":false,"text":"练习 8.9 根式判别法上下极限 (1) 判断 $b + c + b^{2} + c^{2} + \\cdots + b^{n} + c^{n} + \\cdots$ 的敛散性，这里 0 < b < c < 1 解 (1) $\\sqrt[n]{u_{n}}$ 在偶数项为 $\\sqrt{c}$ ，奇数项为 $\\sqrt{b}$ ，因此上极限为 $\\sqrt{c}$ ，故收敛 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":83,"id":"e1c40aac-d49b-4fc0-bbe8-fbc8680ddbe3","page_size":[595,841],"block_position":"83-20"},{"bbox":[291,803,303,812],"type":"page_number","angle":0,"index":1000001,"text":"76 ","color":{"line":"rgba(164, 164, 164, 1)","fill":"rgba(164, 164, 164, 1)"},"page_idx":83,"is_discarded":true,"id":"616b7e53-de25-4659-a554-beda57802668","page_size":[595,841],"block_position":"83-21"}],[{"bbox":[469,47,539,58],"type":"header","angle":0,"index":-99999,"text":"8.3 任意项级数 ","color":{"line":"rgba(164, 164, 164, 1)","fill":"rgba(164, 164, 164, 1)"},"page_idx":84,"is_discarded":true,"id":"459da39d-d01d-4a23-8632-baa37add3721","page_size":[595,841],"block_position":"84-0"},{"bbox":[55,72,223,86],"type":"title","angle":0,"index":1,"level":2,"text":"## 8.2.3 积分判别法和拉贝判别法","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":84,"id":"cee736be-007e-4b8c-97ff-d6b82f04558a","page_size":[595,841],"block_position":"84-1"},{"bbox":[73,100,232,113],"type":"title","angle":0,"index":2,"level":2,"text":"## 定理 8.7 (积分判别法与拉贝判别法)","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":84,"id":"637f1150-990e-4a06-9909-eaee5c3b6fff","page_size":[595,841],"block_position":"84-2"},{"bbox":[68,116,348,148],"type":"text","angle":0,"index":3,"merge_prev":false,"text":"$\\sum_{n = 1}^{\\infty}u_n$ 收敛最本质的充要条件为部分和函数 $S_{n}$ 有界，其推论为 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":84,"id":"c7cf8eb3-991f-43de-8d26-0d43fd35be6c","page_size":[595,841],"block_position":"84-3"},{"bbox":[83,148,530,222],"type":"text","angle":0,"index":4,"merge_prev":false,"text":"- 拉贝判别法(常用于 $\\frac{a_{n+1}}{a_n} \\to 1$ ): $\\lim_{n \\to \\infty} n(1 - \\frac{u_{n+1}}{u_n}) = r > 1$ , 则 $\\sum_{n=1}^{\\infty} u_n$ 收敛。若 $r < 1$ 则发散, $r = 1$ 需要具体判定\n- 积分判别法: $f$ 是 $[1, +\\infty)$ 上非负减函数, 则 $\\sum_{n=1}^{\\infty} f(n)$ 与 $\\int_1^{+\\infty} f(x) \\mathrm{d}x$ 同时收敛与发散 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":84,"id":"8eac490e-a2d6-434f-a644-b2fe33206736","page_size":[595,841],"block_position":"84-4"},{"bbox":[54,235,151,248],"type":"text","angle":0,"index":5,"merge_prev":false,"text":"证明 (1) 拉贝判别法: ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":84,"id":"1ccd14e1-a628-4040-89fe-dedf9d5e904e","page_size":[595,841],"block_position":"84-5"},{"bbox":[73,246,517,272],"type":"text","angle":0,"index":6,"merge_prev":false,"text":"收敛情况：根据条件， $\\exists N, \\forall n > N$ 使得 $n\\left(1 - \\frac{u_{n+1}}{u_n}\\right) > r$ 得到 $\\frac{u_{n+1}}{u_n} < 1 - \\frac{r}{n}$ ，根据伯努利不等式： ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":84,"id":"ae78a29e-1175-4b23-a909-2dcab7165806","page_size":[595,841],"block_position":"84-6"},{"bbox":[175,275,418,301],"type":"interline_equation","angle":0,"index":7,"text":"$$\n(1 + x) ^ {r} \\geq 1 + r x, x > - 1, r > 1 \\Rightarrow \\left(1 - \\frac {1}{n}\\right) ^ {r} \\geq 1 - \\frac {r}{n}\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":84,"id":"be0e582c-e6e7-475f-bfc1-204ac95496dc","page_size":[595,841],"block_position":"84-7"},{"bbox":[54,303,429,335],"type":"text","angle":0,"index":8,"merge_prev":false,"text":"因此 $\\frac{u_{n + 1}}{u_n}\\leq \\left(1 - \\frac{1}{n}\\right)^r = \\left(\\frac{n - 1}{n}\\right)^r = \\frac{\\frac{1}{n^r}}{\\frac{1}{(n - 1)^r}}$ ，显然 $\\sum_{n = 1}^{\\infty}\\frac{1}{n^r}$ 收敛，因此 $\\sum_{n = 1}^{\\infty}u_n$ 收敛 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":84,"id":"f0855b9e-25f6-4bd3-b6f7-a95fefd1f63a","page_size":[595,841],"block_position":"84-8"},{"bbox":[73,335,386,365],"type":"text","angle":0,"index":9,"merge_prev":false,"text":"发散情况： $\\frac{u_{n + 1}}{u_n}\\geq 1 - \\frac{1}{n} = \\frac{n - 1}{n} = \\frac{\\frac{1}{n}}{\\frac{1}{n - 1}}$ ，而 $\\sum_{n = 1}^{\\infty}\\frac{1}{n}$ 发散，因此发散。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":84,"id":"90e5e1b2-dba0-4395-8d00-fb8713baf70f","page_size":[595,841],"block_position":"84-9"},{"bbox":[33,367,333,381],"type":"title","angle":0,"index":10,"level":2,"text":"## 笔记 拉贝判别法更多地用于比式和根式判别法都失效的情况下。","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":84,"id":"fc2e14be-362a-4a47-8080-2c1e03155f60","page_size":[595,841],"block_position":"84-10"},{"bbox":[35,394,255,423],"type":"text","angle":0,"index":11,"merge_prev":false,"text":"练习 8.10 拉贝判别法 判断敛散性 (1) $\\sum_{n=1}^{\\infty}\\frac{n^{n}}{e^{n}n!}$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":84,"id":"22100951-c5e2-4d65-9716-3a3b6ef0ca7e","page_size":[595,841],"block_position":"84-11"},{"bbox":[52,421,538,457],"type":"text","angle":0,"index":12,"merge_prev":false,"text":"解 (1) 显然根式判别法失效。因此设 $u_{n} = \\frac{n^{n}}{e^{n}n!}$ ，使用拉贝判别法以及 $(1 + \\frac{1}{n})^{n} = e - \\frac{e}{2n}$ (用 $(1 + x)^{\\frac{1}{x}}$ 在 0 处 Taylor 展开) 得到 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":84,"id":"82786f4d-48df-4670-a044-572d773c8d0d","page_size":[595,841],"block_position":"84-12"},{"bbox":[126,460,467,488],"type":"interline_equation","angle":0,"index":13,"text":"$$\n\\lim _ {n \\to \\infty} n \\left(1 - \\frac {u _ {n + 1}}{u _ {n}}\\right) = \\lim _ {n \\to \\infty} n \\left[ 1 - \\frac {(1 + \\frac {1}{n}) ^ {n}}{e} \\right] = \\lim _ {n \\to \\infty} n \\left[ \\frac {1}{2 n} + o (\\frac {1}{n}) \\right] = \\frac {1}{2} < 1\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":84,"id":"6db826e4-c8b0-4683-98a3-05a88f557371","page_size":[595,841],"block_position":"84-13"},{"bbox":[54,491,172,502],"type":"text","angle":0,"index":14,"merge_prev":false,"text":"根据拉贝判别法可知发散。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":84,"id":"ed01907a-8093-47df-855a-d740a319bad5","page_size":[595,841],"block_position":"84-14"},{"bbox":[55,521,174,534],"type":"title","angle":0,"index":15,"level":2,"text":"## 8.2.4 放缩证明敛散性","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":84,"id":"4856d4d0-5230-49e4-9620-ea5ded7c60ce","page_size":[595,841],"block_position":"84-15"},{"bbox":[35,543,398,574],"type":"text","angle":0,"index":16,"merge_prev":false,"text":"练习8.11两个简单放缩 $(1)a_{n}\\geq 0$ ，讨论 $\\sum_{n = 1}^{\\infty}a_n$ 与 $\\sum_{n = 1}^{\\infty}\\sqrt{a_na_{n + 1}}$ 收敛性的关系 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":84,"id":"d6d421e4-8b7c-4718-9b93-289f4e233350","page_size":[595,841],"block_position":"84-16"},{"bbox":[73,572,351,606],"type":"text","angle":0,"index":17,"merge_prev":false,"text":"(2) 设 $\\sum_{n=1}^{\\infty} a_n^2, \\sum_{n=1}^{\\infty} b_n^2$ 收敛，证明 $\\sum_{n=1}^{\\infty} a_n b_n, \\sum_{n=1}^{\\infty} (a_n + b_n)^2$ 均收敛 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":84,"id":"34f196c8-cb56-4252-9ece-4725628a7399","page_size":[595,841],"block_position":"84-17"},{"bbox":[54,605,287,627],"type":"text","angle":0,"index":18,"merge_prev":false,"text":"证明 $(1)\\sqrt{a_n + b_n} \\leq \\frac{1}{2}(a_n + b_n)$ ，根据比较定理可知 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":84,"id":"c63c5189-848e-4422-b4ce-9f09a166c455","page_size":[595,841],"block_position":"84-18"},{"bbox":[54,624,539,667],"type":"text","angle":0,"index":19,"merge_prev":false,"text":"(2) 根据 Cauchy 不等式， $\\sum_{n=1}^{N}a_{n}b_{n}\\leq\\left(\\sum_{n=1}^{N}a_{n}^{2}\\right)\\left(\\sum_{n=1}^{N}b_{n}^{2}\\right)$ ，因此收敛。 $\\sum_{n=1}^{N}(a_{n}+b_{n})^{2}=\\sum_{n=1}^{N}a_{n}^{2}+2a_{n}b_{n}+b_{n}^{2}$ 即可 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":84,"id":"faa07cff-c6a9-4212-a0d0-d58b4f5038cf","page_size":[595,841],"block_position":"84-19"},{"bbox":[291,802,302,812],"type":"page_number","angle":0,"index":1000019,"text":"77 ","color":{"line":"rgba(164, 164, 164, 1)","fill":"rgba(164, 164, 164, 1)"},"page_idx":84,"is_discarded":true,"id":"2744f84a-b44a-4a06-a2dd-a24bb067de5b","page_size":[595,841],"block_position":"84-20"}],[{"bbox":[469,47,539,58],"type":"header","angle":0,"index":-99998,"text":"8.3 任意项级数 ","color":{"line":"rgba(164, 164, 164, 1)","fill":"rgba(164, 164, 164, 1)"},"page_idx":85,"is_discarded":true,"id":"bf380594-c462-48fa-9d65-0cdcd21a7678","page_size":[595,841],"block_position":"85-0"},{"bbox":[35,632,47,648],"type":"image","angle":0,"index":0,"img_path":"/6a2f4b6cc6f1a88a4b7ff3a2c0646790a46f3c0d06e9f32f3b45405572f20b4a.jpg","color":{"line":"rgba(89, 92, 220, 1)","fill":"rgba(89, 92, 220, 1)"},"img_ratio":0.2,"page_idx":85,"id":"2004542a-d428-4892-8d6d-c61da66cdc3a","page_size":[595,841],"block_position":"85-1"},{"bbox":[55,74,158,90],"type":"title","angle":0,"index":2,"level":2,"text":"## 8.3 任意项级数","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":85,"id":"6e11f424-9c9b-4467-8c56-3cc231b41df6","page_size":[595,841],"block_position":"85-2"},{"bbox":[55,107,151,121],"type":"title","angle":0,"index":3,"level":2,"text":"## 8.3.1 交错项级数","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":85,"id":"8fe4bae7-cba7-4cf3-a6c3-985b0d3b41ca","page_size":[595,841],"block_position":"85-3"},{"bbox":[73,135,186,147],"type":"title","angle":0,"index":4,"level":2,"text":"## 定理 8.8 (Leibniz 判别法)","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":85,"id":"805b52b4-bd4c-40de-af08-dd76a111d733","page_size":[595,841],"block_position":"85-4"},{"bbox":[67,151,312,184],"type":"text","angle":0,"index":5,"merge_prev":false,"text":"若 $a_{n}$ 单减趋于0，那么交替项级数 $\\sum_{n = 1}^{\\infty}(-1)^{n - 1}a_{n}$ 收敛 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":85,"id":"3a1ba7ca-67f1-4aad-a0dc-7589a0325c5f","page_size":[595,841],"block_position":"85-5"},{"bbox":[52,193,354,224],"type":"text","angle":0,"index":6,"merge_prev":false,"text":"证明 设 $a_{n}$ 单减趋于 0, $S_{n} = \\sum_{k=1}^{n} (-1)^{k-1} a_{k}$ , 下面考虑奇偶部分和: ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":85,"id":"a64397ac-aff4-4ab5-bf93-1e0b6bf92ad3","page_size":[595,841],"block_position":"85-6"},{"bbox":[158,230,424,266],"type":"interline_equation","angle":0,"index":7,"text":"$$\n\\left\\{ \\begin{array}{l} S _ {2 n} = (a _ {1} - a _ {2}) + (a _ {3} - a _ {4}) + \\dots + (a _ {2 n - 1} - a _ {2 n}) \\\\ S _ {2 n + 1} = (a _ {1}) - (a _ {2} - a _ {3}) - (a _ {4} - a _ {5}) - \\dots - (a _ {2 n} - a _ {2 n + 1}) \\end{array} \\right.\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":85,"id":"e1995f94-4515-4296-ba9f-2da133937186","page_size":[595,841],"block_position":"85-7"},{"bbox":[54,270,537,298],"type":"text","angle":0,"index":8,"merge_prev":false,"text":"显然 $S_{2n}$ 单增， $S_{2n+1}$ 单减， $a_{1}-a_{2}\\leq S_{2n}\\leq S_{2n}+a_{2n+1}=S_{2n+1}\\leq a_{1}$ ，因此 $S_{2n}, S_{2n+1}$ 均有界，根据单调有界定理可知收敛，而奇偶子列均收敛说明 $S_{n}$ 收敛。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":85,"id":"ccc7d54e-738d-4d61-98fb-62ee6d846f7e","page_size":[595,841],"block_position":"85-8"},{"bbox":[35,317,273,327],"type":"text","angle":0,"index":9,"merge_prev":false,"text":"练习 8.12 讨论下面级数的条件收敛与绝对收敛性: ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":85,"id":"d4c3c122-5466-4cdd-a77d-379e2b3d6e0e","page_size":[595,841],"block_position":"85-9"},{"bbox":[73,328,311,358],"type":"text","angle":0,"index":10,"merge_prev":false,"text":"(1) 重点: $\\sum_{n=1}^{\\infty} \\frac{(-1)^n}{n^{p+\\frac{1}{n}}}$ (2) $\\sum_{n=2}^{\\infty} \\frac{(-1)^n}{n^p \\ln^q n}$ (3) $\\sum_{n=1}^{\\infty} (-1)^n \\sin \\frac{x}{n}$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":85,"id":"9e0bf9b6-12fa-4019-9a9b-ae1b4e8ac026","page_size":[595,841],"block_position":"85-10"},{"bbox":[73,358,419,387],"type":"text","angle":0,"index":11,"merge_prev":false,"text":"(4) $\\sum_{n=1}^{\\infty} (-1)^{n} \\frac{\\arctan n}{n^{x}}$ (5) 重点: $\\sum_{n=1}^{\\infty} \\sin(n\\pi + \\frac{1}{n^{p}})$ (6) 重点: $\\sum_{n=1}^{\\infty} \\sin(\\pi \\sqrt{n^{2} + p^{2}})$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":85,"id":"07776c1b-597c-4b48-8201-f3dba9ce6559","page_size":[595,841],"block_position":"85-11"},{"bbox":[53,387,539,454],"type":"text","angle":0,"index":12,"merge_prev":false,"text":"解 (1) $\\frac{1}{n^{p + \\frac{1}{n}}} \\sim \\frac{1}{n^p}$ 因此 $p > 1$ 时绝对收敛。条件收敛性只需要看是否单调，考虑 $1 \\geq p > 0$ ， $f(x) = x^{p + \\frac{1}{x}}$ ， $f'(x) = px^{p - 1}x^{\\frac{1}{x}} + x^p x^{\\frac{1}{x}}\\frac{1 - \\ln x}{x^2} = x^{p - 1}x^{\\frac{1}{x}}(p + \\frac{1 - \\ln x}{x}) > 0$ （ $x$ 足够大时），因此整体单减趋于 0。 $p \\leq 0$ 时 $\\lim_{n \\to \\infty} \\frac{(-1)^n}{n^{p + \\frac{1}{n}}} \\neq 0$ ，因此发散。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":85,"id":"8dd19e35-2006-4c7b-af4a-a74b1f283746","page_size":[595,841],"block_position":"85-12"},{"bbox":[53,454,539,497],"type":"text","angle":0,"index":13,"merge_prev":false,"text":"(2) 加绝对值后可以用积分判别法，在 $p > 1$ 时绝对收敛， $p = 1, q > 1$ 绝对收敛。 $p > 0$ ，构造 $g(x) = x^p \\ln^q x$ ，得到 $g'(x) = x^{p-1} \\ln^{q-1} x (p \\ln x + q) > 0$ （ $x$ 足够大），从而 $p = 1, q \\leq 1$ 时， $0 < p < 1$ 条件收敛 $p = 0$ 时 $q > 0$ 条件收敛， $q \\leq 0$ 发散。 $p < 0$ 发散。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":85,"id":"7d17bca9-edb1-4358-a153-f9c6425cfb7f","page_size":[595,841],"block_position":"85-13"},{"bbox":[73,497,517,521],"type":"text","angle":0,"index":14,"merge_prev":false,"text":"(3) $x = 0$ 则绝对收敛。 $x \\neq 0$ 时 $|(-1)^n \\sin \\frac{x}{n} | \\sim \\frac{|x|}{n}$ 发散。而 $\\sin \\frac{x}{n}$ 在 $n$ 足够大时单调，因此条件收敛 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":85,"id":"6e3b8c12-0c69-4f1e-aa2b-e2636533b982","page_size":[595,841],"block_position":"85-14"},{"bbox":[54,520,539,563],"type":"text","angle":0,"index":15,"merge_prev":false,"text":"(4) 加绝对值时等价于 $\\frac{1}{n^x}$ ，因此 $x > 1$ 时绝对收敛。不加绝对值时， $x < 0$ 一定发散， $x > 0$ 时，取 $g(y) = \\frac{\\arctan y}{y^x}$ ，求导发现小于 0，因此收敛 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":85,"id":"b29ded2d-0f2a-43ce-b50d-03c847d05ab2","page_size":[595,841],"block_position":"85-15"},{"bbox":[72,562,412,596],"type":"text","angle":0,"index":16,"merge_prev":false,"text":"(5) 等于 $\\sum_{n=1}^{\\infty} (-1)^{n} \\sin \\frac{1}{n^{p}}$ , $p > 1$ 绝对收敛, $0 < p \\leq 1$ 条件收敛, $p \\leq 0$ 发散。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":85,"id":"3ae43ffa-2bdb-43e5-a6a0-c12903e7fda9","page_size":[595,841],"block_position":"85-16"},{"bbox":[54,592,537,633],"type":"text","angle":0,"index":17,"merge_prev":false,"text":"(6) 等于 $\\sum_{n=1}^{\\infty} (-1)^n \\sin (\\pi \\sqrt{n^2 + p^2} - n\\pi) = \\sum_{n=1}^{\\infty} (-1)^n \\sin \\frac{p^2\\pi}{\\sqrt{n^2 + p^2} + n}$ ，因此 $p = 0$ 时绝对收敛， $p \\neq 0$ 条件收敛。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":85,"id":"6eaa4782-e430-4d41-9f5d-418db9550385","page_size":[595,841],"block_position":"85-17"},{"bbox":[51,637,193,649],"type":"text","angle":0,"index":18,"merge_prev":false,"text":"笔记 单调性常常使用求导辅助。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":85,"id":"87e4da7f-a35e-4f53-a9ab-66f88329969d","page_size":[595,841],"block_position":"85-18"},{"bbox":[291,802,302,812],"type":"page_number","angle":0,"index":1000018,"text":"78 ","color":{"line":"rgba(164, 164, 164, 1)","fill":"rgba(164, 164, 164, 1)"},"page_idx":85,"is_discarded":true,"id":"1bae0111-5770-4e7d-9da5-87349d3ffe85","page_size":[595,841],"block_position":"85-19"}],[{"bbox":[459,47,539,58],"type":"header","angle":0,"index":-99999,"text":"8.4 数项级数理论 ","color":{"line":"rgba(164, 164, 164, 1)","fill":"rgba(164, 164, 164, 1)"},"page_idx":86,"is_discarded":true,"id":"b57f5889-5b2a-4d77-bdc5-2ea299fe8660","page_size":[595,841],"block_position":"86-0"},{"bbox":[55,72,216,86],"type":"title","angle":0,"index":1,"level":2,"text":"## 8.3.2 Dirichlet 与 Abel 判别法","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":86,"id":"a3b07efb-b6fe-40a0-9f99-5787fa27149d","page_size":[595,841],"block_position":"86-1"},{"bbox":[73,100,191,112],"type":"title","angle":0,"index":2,"level":2,"text":"## 定理 8.9 (Dirichlet 判别法)","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":86,"id":"0bc5123f-dc83-409a-98bc-db35c8a1aed6","page_size":[595,841],"block_position":"86-2"},{"bbox":[68,117,403,151],"type":"text","angle":0,"index":3,"merge_prev":false,"text":"$a_{k}, b_{k}$ 为两个数列，且满足(1) $\\sum_{i=1}^{k} a_{i}$ 有界(2) $b_{k}$ 单调趋于0，则 $\\sum_{k=1}^{\\infty} a_{k} b_{k}$ 收敛 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":86,"id":"f9022a8e-c192-4256-a9bf-57ce6ed7b979","page_size":[595,841],"block_position":"86-3"},{"bbox":[73,171,178,184],"type":"title","angle":0,"index":4,"level":2,"text":"## 定理 8.10 (Abel 判别法)","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":86,"id":"0a3e22f6-ee4e-4f0f-923f-a982d9dbbe1c","page_size":[595,841],"block_position":"86-4"},{"bbox":[67,187,325,222],"type":"text","angle":0,"index":5,"merge_prev":false,"text":"$a_{k}, b_{k}$ 满足 (1) $\\sum_{i=1}^{\\infty} a_{i}$ 收敛 (2) $b_{k}$ 单调有界，则 $\\sum_{k=1}^{\\infty} a_{k} b_{k}$ 收敛 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":86,"id":"87e6cfb7-b680-492b-a2c6-a5cec7d3924d","page_size":[595,841],"block_position":"86-5"},{"bbox":[52,237,336,249],"type":"title","angle":0,"index":6,"level":2,"text":"## 笔记 Leibniz 判别法本质只是 Dirichlet 和 Abel 判别法的一个特例","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":86,"id":"eea6100c-be48-4ea0-ba6f-bffb4fa91e52","page_size":[595,841],"block_position":"86-6"},{"bbox":[36,248,491,280],"type":"text","angle":0,"index":7,"merge_prev":false,"text":"练习8.13常用三角结论(1)重点： $a_{n}$ 单减趋于0， $x\\neq 2k\\pi$ ，证明： $\\sum_{n = 1}^{\\infty}a_n\\cos nx,\\sum_{n = 1}^{\\infty}a_n\\sin nx$ 收敛 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":86,"id":"47e7454e-89ea-430f-b9a3-937b848039b8","page_size":[595,841],"block_position":"86-7"},{"bbox":[73,277,392,308],"type":"text","angle":0,"index":8,"merge_prev":false,"text":"(2) 重点： $x \\neq k\\pi$ ，讨论 $\\sum_{n=1}^{\\infty} \\frac{\\cos nx}{n^{p}}$ ， $\\sum_{n=1}^{\\infty} \\frac{\\sin nx}{n^{p}}$ 的条件收敛和绝对收敛性 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":86,"id":"41171ab4-e547-4e1b-a48b-0f0c011e24a1","page_size":[595,841],"block_position":"86-8"},{"bbox":[74,308,402,338],"type":"text","angle":0,"index":9,"merge_prev":false,"text":"(3) 重点：讨论 $\\sum_{n=1}^{\\infty} (-1)^n \\frac{\\sin n}{n}$ 与 $\\sum_{n=1}^{\\infty} (-1)^n \\frac{\\sin^2 n}{n}$ 的条件收敛与绝对收敛性 证明 (1) 根据 $|\\sum_{k=1}^{n}\\cos kx| = |\\frac{\\sin(n + \\frac{1}{2})x - \\sin\\frac{x}{2}}{2\\sin\\frac{x}{2}}| \\leq \\frac{1}{|\\sin\\frac{x}{2}|}$ ，同理 $|\\sum_{k=1}^{\\infty}\\sin kx| = |\\frac{\\cos\\frac{x}{2} - \\cos(n + \\frac{1}{2})x}{2\\sin\\frac{x}{2}}| \\leq \\frac{1}{|\\sin\\frac{x}{2}|}$ ，根据 Dirichlet 判别法可知。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":86,"id":"7fc6098b-6ff5-4c7a-8adf-ebc349c53ada","page_size":[595,841],"block_position":"86-9"},{"bbox":[54,337,539,392],"type":"text","angle":0,"index":10,"merge_prev":false,"lines_deleted":true,"text":"","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":86,"id":"7fc6098b-6ff5-4c7a-8adf-ebc349c53ada","page_size":[595,841],"block_position":"86-10"},{"bbox":[54,394,539,427],"type":"text","angle":0,"index":11,"merge_prev":false,"text":"(2) $p > 1$ 时 $\\left|\\frac{\\cos nx}{n^p}\\right| \\leq \\frac{1}{n^p}$ ，因此绝对收敛。 $p > 0$ 时根据Dirichlet可知条件收敛， $p \\leq 0$ 时 $\\lim_{n \\to \\infty} \\frac{\\cos nx}{n^p} \\neq 0$ 发散。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":86,"id":"a35681ba-2856-43d2-a2a1-1980ba1e10c3","page_size":[595,841],"block_position":"86-11"},{"bbox":[74,428,227,441],"type":"text","angle":0,"index":12,"merge_prev":false,"text":"下面证明 $0 < p \\leq 1$ 时不绝对收敛： ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":86,"id":"dcd7f54a-0059-4cba-aeda-a3bd659c928c","page_size":[595,841],"block_position":"86-12"},{"bbox":[218,444,374,470],"type":"interline_equation","angle":0,"index":13,"text":"$$\n\\left| \\frac {\\cos n x}{n ^ {p}} \\right| \\geq \\left| \\frac {\\cos^ {2} n x}{n ^ {p}} \\right| \\geq \\left| \\frac {1 - \\sin^ {2} x}{n ^ {p}} \\right|\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":86,"id":"e2fb6195-aea2-4151-b2f2-ae40534222ee","page_size":[595,841],"block_position":"86-13"},{"bbox":[54,473,176,504],"type":"text","angle":0,"index":14,"merge_prev":false,"text":"而 $\\sum_{n=1}^{\\infty} \\frac{1}{n^p}$ 发散，因此发散。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":86,"id":"86d653a7-e7ee-4d49-895e-844d697c89f6","page_size":[595,841],"block_position":"86-14"},{"bbox":[52,507,561,596],"type":"text","angle":0,"index":15,"merge_prev":false,"text":"(3) 这种双交错要把 $(-1)^n$ 吸收进三角。 $\\sin(x + n\\pi) = (-1)^n \\sin x, \\cos(n + n\\pi) = (-1)^n \\cos x$ ，因此 $\\sum_{n=1}^{\\infty} (-1)^n \\frac{\\sin n}{n} = \\sum_{n=1}^{\\infty} \\frac{\\sin(1 + \\pi)n}{n}$ 条件收敛 (根据 (1) 结论)。第二个转换为 $\\frac{1}{2} \\sum_{n=1}^{\\infty} \\frac{(-1)^n}{n} - \\frac{1}{2} \\sum_{n=1}^{\\infty} \\frac{\\cos(2 + \\pi)n}{n}$ 条件收敛。绝对值发散的原因是 $\\sum_{n=1}^{\\infty} \\frac{\\sin^2 n}{n} = \\frac{1}{2} \\sum_{n=1}^{\\infty} \\frac{1}{n} - \\frac{1}{2} \\sum_{n=1}^{\\infty} \\frac{\\cos 2n}{n}$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":86,"id":"7237ac10-bc77-4a2b-b541-457bf41555a0","page_size":[595,841],"block_position":"86-15"},{"bbox":[36,597,293,627],"type":"text","angle":0,"index":16,"merge_prev":false,"text":"练习8.14具体例子讨论 $\\sum_{n = 1}^{\\infty}\\frac{\\cos 3n}{n} (1 + \\frac{1}{n})^n$ 的收敛性 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":86,"id":"1cffb529-9900-41d1-8093-ae7226f83143","page_size":[595,841],"block_position":"86-16"},{"bbox":[52,627,412,657],"type":"text","angle":0,"index":17,"merge_prev":false,"text":"解 $\\sum_{n=1}^{\\infty} \\frac{\\cos 3n}{n}$ 根据前面结论收敛， $(1 + \\frac{1}{n})^n \\to e$ 单调有界，因此由 Abel 可知收敛 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":86,"id":"7df2ea1a-b016-46ba-8056-74e55f75b79b","page_size":[595,841],"block_position":"86-17"},{"bbox":[55,677,172,692],"type":"title","angle":0,"index":18,"level":2,"text":"## 8.4 数项级数理论","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":86,"id":"1ca484dd-250c-4f63-8c9a-f60cfeab9bce","page_size":[595,841],"block_position":"86-18"},{"bbox":[55,709,163,724],"type":"text","angle":0,"index":19,"merge_prev":false,"text":"8.4.1 级数的加括号 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":86,"id":"b8a03720-254a-4367-a975-54e483339443","page_size":[595,841],"block_position":"86-19"},{"bbox":[73,733,296,746],"type":"text","angle":0,"index":20,"merge_prev":false,"text":"级数的加括号本质是数列收敛及其子列收敛的关系 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":86,"id":"f1e9ce2c-71e1-48b6-b000-5c1eedc7a904","page_size":[595,841],"block_position":"86-20"},{"bbox":[291,802,302,812],"type":"page_number","angle":0,"index":1000020,"text":"79 ","color":{"line":"rgba(164, 164, 164, 1)","fill":"rgba(164, 164, 164, 1)"},"page_idx":86,"is_discarded":true,"id":"7db4881a-bae5-4c80-97c1-276f47ba0efc","page_size":[595,841],"block_position":"86-21"}],[{"bbox":[459,47,539,58],"type":"header","angle":0,"index":-99999,"text":"8.4 数项级数理论 ","color":{"line":"rgba(164, 164, 164, 1)","fill":"rgba(164, 164, 164, 1)"},"page_idx":87,"is_discarded":true,"id":"57a91668-ecb9-436a-a2ea-ebf2641997a4","page_size":[595,841],"block_position":"87-0"},{"bbox":[73,73,176,85],"type":"title","angle":0,"index":1,"level":2,"text":"## 定理 8.11 (加括号收敛)","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":87,"id":"c65d59a2-8fdc-4ad8-a6b0-39f08bff9fe4","page_size":[595,841],"block_position":"87-1"},{"bbox":[67,89,525,132],"type":"text","angle":0,"index":2,"merge_prev":false,"text":"$\\sum_{n=1}^{\\infty} u_n$ 收敛，则 $\\sum_{n=1}^{\\infty} u_n$ 中任意加括号得到级数 $\\sum_{n=1}^{\\infty} v_n$ 也收敛，且 $\\sum_{n=1}^{\\infty} u_n = \\sum_{n=1}^{\\infty} v_n$ (即加括号既不改变收敛性，又不改变和) ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":87,"id":"c5b78c91-3825-4314-bb14-7ecb750ea08f","page_size":[595,841],"block_position":"87-2"},{"bbox":[54,142,415,174],"type":"text","angle":0,"index":3,"merge_prev":false,"text":"证明 显然 $S_{n} = \\sum_{k = 1}^{n}u_{k}$ 收敛，从而 $S_{k}^{\\prime} = \\sum_{k = 1}^{m}v_{k}$ 是 $S_{n}$ 子列，从而收敛且极限相同。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":87,"id":"c4407d60-1ed6-47a6-b4f1-26bf158edb46","page_size":[595,841],"block_position":"87-3"},{"bbox":[52,174,539,214],"type":"text","angle":0,"index":4,"merge_prev":false,"text":"笔记 若级数发散，则加括号得到的新级数可能收敛，例如 $\\sum_{n=1}^{\\infty} (-1)^n$ 。若级数加括号以后发散，则原级数一定收敛。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":87,"id":"5e32d0fc-ec98-4bb0-a69c-ea71be901ecc","page_size":[595,841],"block_position":"87-4"},{"bbox":[72,222,236,234],"type":"title","angle":0,"index":5,"level":2,"text":"## 定理 8.12 (加括号收敛反推级数收敛)","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":87,"id":"662fd429-ef2d-43a8-9930-fee36687363b","page_size":[595,841],"block_position":"87-5"},{"bbox":[70,238,210,271],"type":"text","angle":0,"index":6,"merge_prev":false,"text":"若 $\\sum_{n = 1}^{\\infty}u_n$ 加括号后 $\\sum_{n = 1}^{\\infty}v_{n}$ 收敛， ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":87,"id":"3fb1d191-8295-4dc9-9745-2c0d4a74cfac","page_size":[595,841],"block_position":"87-6"},{"bbox":[83,269,386,301],"type":"text","angle":0,"index":7,"merge_prev":false,"text":"若 $\\lim_{n\\to \\infty}u_n = 0$ 且每个括号里项个数小于固定值 $L$ ，则 $\\sum_{n = 1}^{\\infty}u_n$ 收敛 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":87,"id":"eb35b1d0-84f7-45b9-9add-9ce391dddf2a","page_size":[595,841],"block_position":"87-7"},{"bbox":[83,299,269,331],"type":"text","angle":0,"index":8,"merge_prev":false,"text":"- 若每个括号里符号相同，则 $\\sum_{n=1}^{\\infty} u_n$ 收敛 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":87,"id":"610cfdb1-6b43-4865-96c9-c4ad59d90d7d","page_size":[595,841],"block_position":"87-8"},{"bbox":[56,363,151,377],"type":"title","angle":0,"index":9,"level":2,"text":"## 8.4.2 级数的重排","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":87,"id":"d4e42b0a-1678-4b6b-b738-7634695dbf7e","page_size":[595,841],"block_position":"87-9"},{"bbox":[73,390,141,402],"type":"title","angle":0,"index":10,"level":2,"text":"## 定义 8.2 (重排)","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":87,"id":"e09fb707-5640-47bd-9a10-0a03456e1311","page_size":[595,841],"block_position":"87-10"},{"bbox":[69,409,274,439],"type":"text","angle":0,"index":11,"merge_prev":false,"text":"考虑级数 $\\sum_{n=1}^{\\infty} u_n$ , $f$ 是 $[n] \\to [n]$ 的映射, 则称 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":87,"id":"bb212650-30b0-4d63-b98e-571a3aae1ad7","page_size":[595,841],"block_position":"87-11"},{"bbox":[255,441,337,474],"type":"interline_equation","angle":0,"index":12,"text":"$$\n\\sum_ {n = 1} ^ {\\infty} v _ {n} = \\sum_ {n = 1} ^ {\\infty} u _ {f (n)}\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":87,"id":"02cd039e-e2d5-4489-9a86-ae369392dc0a","page_size":[595,841],"block_position":"87-12"},{"bbox":[69,475,165,507],"type":"text","angle":0,"index":13,"merge_prev":false,"text":"为 $\\sum_{n=1}^{\\infty} u_n$ 的一个重排 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":87,"id":"b6f3a422-d2c5-4afb-88ae-d1c47b1ccc62","page_size":[595,841],"block_position":"87-13"},{"bbox":[73,528,196,540],"type":"title","angle":0,"index":14,"level":2,"text":"## 定理 8.13 (正项级数的重排)","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":87,"id":"cbd3802b-ee8e-4fce-bedd-237e610a5317","page_size":[595,841],"block_position":"87-14"},{"bbox":[68,544,329,576],"type":"text","angle":0,"index":15,"merge_prev":false,"text":"正项级数 $\\sum_{n=1}^{\\infty} u_n$ 收敛，则其任意重排 $\\sum_{n=1}^{\\infty} u_{f(n)}$ 也收敛，且 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":87,"id":"4f46bfa0-00ff-40bc-867a-aa53da5f44f7","page_size":[595,841],"block_position":"87-15"},{"bbox":[254,578,337,610],"type":"interline_equation","angle":0,"index":16,"text":"$$\n\\sum_ {n = 1} ^ {\\infty} u _ {n} = \\sum_ {n = 1} ^ {\\infty} u _ {f (n)}\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":87,"id":"5546d470-c638-4b93-83c9-715a3bb45a61","page_size":[595,841],"block_position":"87-16"},{"bbox":[54,621,539,667],"type":"text","angle":0,"index":17,"merge_prev":false,"text":"证明 记 $S_{n}=\\sum_{k=1}^{n}u_{k},S_{n}^{\\prime}=\\sum_{k=1}^{n}u_{f(k)}$ ，记 $M=\\max\\{f(1),\\cdots,f(n)\\}$ ，则显然 $S_{n}^{\\prime}\\leq S_{M}$ ，根据 $\\lim_{n\\to\\infty}S_{n}$ 存在可知 $\\lim_{n\\to\\infty}S_{n}^{\\prime}$ 存在，且 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":87,"id":"42b805b8-c082-4ae0-a2a6-a0962a8fd1f5","page_size":[595,841],"block_position":"87-17"},{"bbox":[254,666,339,684],"type":"interline_equation","angle":0,"index":18,"text":"$$\n\\lim _ {n \\to \\infty} S _ {n} ^ {\\prime} \\leq \\lim _ {n \\to \\infty} S _ {n} ^ {\\prime}\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":87,"id":"3b2831ee-8348-411c-a633-9ae797902e12","page_size":[595,841],"block_position":"87-18"},{"bbox":[74,689,363,707],"type":"text","angle":0,"index":19,"merge_prev":false,"text":"反之也可以将 $S_{n}$ 看作 $S_{n}^{\\prime}$ 的重排，因此也满足 $\\lim_{n\\to \\infty}S_n\\leq \\lim_{n\\to \\infty}S_n'$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":87,"id":"e8bd4930-a9fa-49cc-9ad1-4b0a02e27455","page_size":[595,841],"block_position":"87-19"},{"bbox":[291,803,303,812],"type":"page_number","angle":0,"index":1000019,"text":"80 ","color":{"line":"rgba(164, 164, 164, 1)","fill":"rgba(164, 164, 164, 1)"},"page_idx":87,"is_discarded":true,"id":"44d6f577-ffe4-4eea-93fa-0bd5b193c61a","page_size":[595,841],"block_position":"87-20"}],[{"bbox":[459,47,539,58],"type":"header","angle":0,"index":-99999,"text":"8.4 数项级数理论 ","color":{"line":"rgba(164, 164, 164, 1)","fill":"rgba(164, 164, 164, 1)"},"page_idx":88,"is_discarded":true,"id":"eac82866-c545-46e9-affb-c512f86b34bf","page_size":[595,841],"block_position":"88-0"},{"bbox":[73,73,206,86],"type":"title","angle":0,"index":1,"level":2,"text":"## 定理 8.14 (一般项级数的重排)","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":88,"id":"2c9664c6-a4a4-42d4-ad6a-f75cbd5ca7af","page_size":[595,841],"block_position":"88-1"},{"bbox":[68,89,483,122],"type":"text","angle":0,"index":2,"merge_prev":false,"text":"$f$ 是重排映射，若 $\\exists M\\in \\mathbb{Z}^{+}$ 使得 $|f(n) - n|\\leq M$ ，则 $\\sum_{n = 1}^{\\infty}u_n$ 收敛当且仅当 $\\sum_{n = 1}^{\\infty}u_{f(n)}$ 收敛，且 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":88,"id":"f661509e-afb5-494f-a202-1b9812626a2a","page_size":[595,841],"block_position":"88-2"},{"bbox":[254,123,337,156],"type":"interline_equation","angle":0,"index":3,"text":"$$\n\\sum_ {n = 1} ^ {\\infty} u _ {n} = \\sum_ {n = 1} ^ {\\infty} u _ {f (n)}\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":88,"id":"217ad318-22d7-42f1-a7dd-d8ddccbaa51a","page_size":[595,841],"block_position":"88-3"},{"bbox":[74,176,215,189],"type":"title","angle":0,"index":4,"level":2,"text":"## 定理 8.15 (绝对收敛级数的重排)","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":88,"id":"0e2f2651-9aa3-4ecb-bea0-f5a575babac3","page_size":[595,841],"block_position":"88-4"},{"bbox":[69,193,329,225],"type":"text","angle":0,"index":5,"merge_prev":false,"text":"设 $\\sum_{n = 1}^{\\infty}u_n$ 绝对收敛，则其任意重排 $\\sum_{n = 1}^{\\infty}u_{f(n)}$ 绝对收敛，且 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":88,"id":"0ef733fd-70ad-4a60-b35c-1570b215d32a","page_size":[595,841],"block_position":"88-5"},{"bbox":[254,227,337,259],"type":"interline_equation","angle":0,"index":6,"text":"$$\n\\sum_ {n = 1} ^ {\\infty} u _ {n} = \\sum_ {n = 1} ^ {\\infty} u _ {f (n)}\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":88,"id":"ce7bf0b1-6d57-4a4f-a359-d161f0a464d9","page_size":[595,841],"block_position":"88-6"},{"bbox":[55,275,151,287],"type":"text","angle":0,"index":7,"merge_prev":false,"text":"证明 正负部分开即可 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":88,"id":"f4eb218d-deb0-414b-ba1f-f250a540dd5e","page_size":[595,841],"block_position":"88-7"},{"bbox":[73,296,287,309],"type":"title","angle":0,"index":8,"level":2,"text":"## 定理 8.16 (Riemann 定理：条件收敛级数的重排)","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":88,"id":"8c499b28-1d1c-42ea-8dc0-ec3a0dc4590c","page_size":[595,841],"block_position":"88-8"},{"bbox":[68,313,419,346],"type":"text","angle":0,"index":9,"merge_prev":false,"text":"设数项级数 $\\sum_{n=1}^{\\infty} u_n$ 条件收敛，则对 $\\forall S \\in \\mathbb{R} \\cup \\{\\pm \\infty\\}$ ，存在重排 $\\sum_{n=1}^{\\infty} v_n$ 收敛到 $S$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":88,"id":"f638f690-8980-4de7-84d4-fd07380ded48","page_size":[595,841],"block_position":"88-9"},{"bbox":[56,377,151,391],"type":"title","angle":0,"index":10,"level":2,"text":"## 8.4.3 正部与负部","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":88,"id":"e380fd01-7865-4162-ba4d-53b4023e6d21","page_size":[595,841],"block_position":"88-10"},{"bbox":[54,399,539,434],"type":"text","angle":0,"index":11,"merge_prev":false,"text":"对于任意数列 $u_{n}$ ，考虑非负数列 $p_n = \\frac{|u_n| + u_n}{2}, q_n = \\frac{|u_n| - u_n}{2}$ ，分别为 $u_{n}$ 的正部与负部，下面研究它们之间的关系。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":88,"id":"8bfc371c-d7d0-477a-9c7a-3db5ae02ba6e","page_size":[595,841],"block_position":"88-11"},{"bbox":[74,442,155,454],"type":"title","angle":0,"index":12,"level":2,"text":"## 定理 8.17 (正负部)","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":88,"id":"6c4f84e4-be0a-42de-9b1d-bd136b00c121","page_size":[595,841],"block_position":"88-12"},{"bbox":[66,456,528,545],"type":"text","angle":0,"index":13,"merge_prev":false,"text":"若 $\\sum_{n=1}^{\\infty} u_n$ 绝对收敛，则 $\\sum_{n=1}^{\\infty} p_n, \\sum_{n=1}^{\\infty} q_n$ 均收敛。若 $\\sum_{n=1}^{\\infty} u_n$ 条件收敛，则 $\\sum_{k=1}^{n} p_k, \\sum_{k=1}^{n} q_k$ 发散到正无穷，且 $\\lim_{n \\to \\infty} \\frac{\\sum_{k=1}^{n} p_k}{\\sum_{k=1}^{n} q_k} = 1$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":88,"id":"a64f7dec-2a1e-4c5a-85c9-f359494c3cf2","page_size":[595,841],"block_position":"88-13"},{"bbox":[54,552,432,581],"type":"text","angle":0,"index":14,"merge_prev":false,"text":"证明 (1) $0 \\leq p_{n}, q_{n} \\leq |u_{n}|$ ，因此根据比较原则可发现 $\\sum_{n=1}^{\\infty} p_{n}, q_{n}$ 都收敛，且极限可拆开 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":88,"id":"5aa7e1aa-3431-4733-9bba-a2dfce6f8415","page_size":[595,841],"block_position":"88-14"},{"bbox":[54,580,538,644],"type":"text","angle":0,"index":15,"merge_prev":false,"text":"(2) $\\sum_{n=1}^{\\infty} u_n$ 条件收敛，则 $\\sum_{n=1}^{\\infty} (p_n - q_n)$ 收敛，但 $\\sum_{n=1}^{\\infty} (p_n + q_n) = +\\infty$ ，假设 $\\sum_{n=1}^{\\infty} p_n$ 收敛，则 $\\sum_{n=1}^{\\infty} q_n$ 收敛，这与 $\\sum_{n=1}^{\\infty} (p_n + q_n)$ 发散矛盾。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":88,"id":"df9e4c0e-5361-4923-b945-c4d7fc33f901","page_size":[595,841],"block_position":"88-15"},{"bbox":[36,643,399,671],"type":"text","angle":0,"index":17,"merge_prev":false,"text":"练习8.15 $a_{n}$ 单减， $a_{n} \\geq 0$ ， $\\sum_{n=1}^{\\infty} a_{n}$ 发散，证明 $\\lim_{n \\to \\infty} \\frac{a_{2} + a_{4} + \\cdots + a_{2n}}{a_{1} + a_{3} + \\cdots + a_{2n-1}} = 1$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":88,"id":"c23fbde4-b018-4e27-8e7e-ffba2a9c19b8","page_size":[595,841],"block_position":"88-16"},{"bbox":[48,671,540,709],"type":"text","angle":0,"index":18,"merge_prev":false,"text":"证明 极限小于等于 1 显然。而 $\\frac{a_{2}+\\cdots+a_{2n}}{a_{1}+\\cdots+a_{2n-1}}\\geq\\frac{a_{3}+a_{5}+\\cdots+a_{2n-1}}{a_{1}+a_{3}+\\cdots+a_{2n-1}}=1-\\frac{a_{1}}{a_{1}+\\cdots+a_{2n-1}}$ ，根据单减以及发散可知 $2(a_{1}+a_{3}+\\cdots+a_{2n-1})\\geq a_{1}+a_{2}+\\cdots+a_{2n}\\rightarrow\\infty$ ，因此大于等于 1。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":88,"id":"a913a336-21e1-4232-9e88-5c9edce330ef","page_size":[595,841],"block_position":"88-17"},{"bbox":[291,803,302,812],"type":"page_number","angle":0,"index":1000015,"text":"81 ","color":{"line":"rgba(164, 164, 164, 1)","fill":"rgba(164, 164, 164, 1)"},"page_idx":88,"is_discarded":true,"id":"e02c8670-4e4d-419f-a703-74ee36400984","page_size":[595,841],"block_position":"88-18"}],[{"bbox":[459,47,539,58],"type":"header","angle":0,"index":-99999,"text":"8.4 数项级数理论 ","color":{"line":"rgba(164, 164, 164, 1)","fill":"rgba(164, 164, 164, 1)"},"page_idx":89,"is_discarded":true,"id":"0ca4d8c5-ee41-4fed-94e9-80a345e3f8a8","page_size":[595,841],"block_position":"89-0"},{"bbox":[55,72,151,86],"type":"title","angle":0,"index":1,"level":2,"text":"## 8.4.4 级数的乘积","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":89,"id":"abf5c90c-9b46-4242-89bb-25b12492c259","page_size":[595,841],"block_position":"89-1"},{"bbox":[56,97,174,111],"type":"title","angle":0,"index":2,"level":2,"text":"## 8.4.5 级数的相互控制","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":89,"id":"77b7d6e1-4276-46de-b89b-3de69e3335da","page_size":[595,841],"block_position":"89-2"},{"bbox":[73,125,186,137],"type":"title","angle":0,"index":3,"level":2,"text":"## 定理 8.18 (级数相互控制)","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":89,"id":"ee05ace7-8727-4254-800a-9dcfb697490b","page_size":[595,841],"block_position":"89-3"},{"bbox":[69,142,153,154],"type":"text","angle":0,"index":4,"merge_prev":false,"text":"设 $a_{n}\\leq c_{n}\\leq b_{n}$ 则 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":89,"id":"38a68274-f670-4abe-bfa0-577c18549cbe","page_size":[595,841],"block_position":"89-4"},{"bbox":[83,157,241,169],"type":"text","angle":0,"index":5,"merge_prev":false,"text":"- 数列 $a_{n}, b_{n}$ 收敛不能得出 $c_{n}$ 收敛 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":89,"id":"b251edbf-7046-4964-9066-c4e62ba51e7f","page_size":[595,841],"block_position":"89-5"},{"bbox":[83,169,282,200],"type":"text","angle":0,"index":6,"merge_prev":false,"text":"- 级数 $\\sum_{n=1}^{\\infty} a_n, \\sum_{n=1}^{\\infty} b_n$ 收敛可得出 $\\sum_{n=1}^{\\infty} c_n$ 收敛 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":89,"id":"7c13df1d-2452-4b8f-a87a-c3fc9752c9fa","page_size":[595,841],"block_position":"89-6"},{"bbox":[53,211,290,234],"type":"text","angle":0,"index":7,"merge_prev":false,"text":"证明 (1) 例如: $a_{n} = -2 - \\frac{1}{n}, b_{n} = 2 + \\frac{1}{n}, c_{n} = (-1)^{n}$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":89,"id":"bf11bbef-4ef8-4570-b862-4cf9fa0a8e51","page_size":[595,841],"block_position":"89-7"},{"bbox":[53,232,539,294],"type":"text","angle":0,"index":8,"merge_prev":false,"text":"(2) 此时 $0 \\leq c_{n} - a_{n} \\leq b_{n} - a_{n}$ , 显然 $\\sum_{n=1}^{\\infty} b_{n} - a_{n}$ 收敛, 根据比较原则得到 $\\sum_{n=1}^{\\infty} c_{n} - a_{n}$ 收敛, 因此得到 $\\sum_{n=1}^{\\infty} c_{n}$ 收敛。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":89,"id":"1f6c07ba-9fca-4bf8-8b9e-46b4d4a5128e","page_size":[595,841],"block_position":"89-8"},{"bbox":[73,303,180,316],"type":"title","angle":0,"index":9,"level":2,"text":"## 推论 8.1 (控制是相互的)","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":89,"id":"6ebb7846-5a69-4ef4-b10a-55c4d87816f7","page_size":[595,841],"block_position":"89-9"},{"bbox":[67,319,409,352],"type":"text","angle":0,"index":10,"merge_prev":false,"text":"若 $a_{n} \\leq b_{n} \\leq a_{n + 1}$ ，则显然推出 $b_{n - 1} \\leq a_{n} \\leq b_{n}$ ，因此 $\\sum_{n = 1}^{\\infty} a_{n}, \\sum_{n = 1}^{\\infty} b_{n}$ 同敛散。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":89,"id":"3dff270a-5cc1-429e-ad78-544b02d6c862","page_size":[595,841],"block_position":"89-10"},{"bbox":[35,377,537,418],"type":"text","angle":0,"index":11,"merge_prev":false,"text":"练习8.16相互控制练习(1)重点： $a_{n}$ 每一项都非零， $\\lim_{n\\to \\infty}a_n = a\\neq 0$ ，证明 $\\sum_{n = 1}^{\\infty}|a_{n + 1} - a_n|,\\sum_{n = 1}^{\\infty}\\left|\\frac{1}{a_{n + 1}} -\\frac{1}{a_n}\\right|$ 敛散性相同 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":89,"id":"7153557f-ae77-4592-8f17-c8481a7f4861","page_size":[595,841],"block_position":"89-11"},{"bbox":[73,417,337,449],"type":"text","angle":0,"index":12,"merge_prev":false,"text":"(2) $a_{n}$ 是单调减的正数列，证明 $\\sum_{n = 1}^{\\infty}a_n$ 与 $\\sum_{m = 1}^{\\infty}2^{m}a_{2^{m}}$ 同敛散 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":89,"id":"629dc2f1-f517-4c35-903c-87692c3f03f0","page_size":[595,841],"block_position":"89-12"},{"bbox":[53,447,472,474],"type":"text","angle":0,"index":13,"merge_prev":false,"text":"证明 (1) 根据极限保号性， $\\exists N, \\forall n > N$ 有 $\\frac{1}{2} a \\leq a_n \\leq \\frac{3}{2} a$ ，而 $\\left|\\frac{1}{a_{n+1}} - \\frac{1}{a_n}\\right| = \\left|\\frac{a_n - a_{n+1}}{a_na_{n+1}}\\right|$ 因此 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":89,"id":"3aa131e5-523f-45de-80eb-5af9535be293","page_size":[595,841],"block_position":"89-13"},{"bbox":[192,476,401,502],"type":"interline_equation","angle":0,"index":14,"text":"$$\n\\frac {4}{9 a ^ {2}} | a _ {n + 1} - a _ {n} | \\leq \\frac {| a _ {n + 1} - a _ {n} |}{| a _ {n} a _ {n + 1} |} \\leq \\frac {4}{a ^ {2}} | a _ {n + 1} - a _ {n} |\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":89,"id":"d49cc439-07fa-4bf7-b630-96336cf470fa","page_size":[595,841],"block_position":"89-14"},{"bbox":[54,505,157,517],"type":"text","angle":0,"index":15,"merge_prev":false,"text":"从而相互控制，同敛散 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":89,"id":"0fb6bc2a-0225-4305-9743-e2fc1b083926","page_size":[595,841],"block_position":"89-15"},{"bbox":[73,520,176,533],"type":"text","angle":0,"index":16,"merge_prev":false,"text":"(2) 根据 $a_{n}$ 单调减可得 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":89,"id":"065ac444-11e9-419a-a878-5b32293af06b","page_size":[595,841],"block_position":"89-16"},{"bbox":[178,539,413,553],"type":"interline_equation","angle":0,"index":17,"text":"$$\n2 ^ {k - 1} a _ {2 ^ {k}} \\leq a _ {2 ^ {k - 1} + 1} + a _ {2 ^ {k - 1} + 2} + \\dots + a _ {2 ^ {k}} \\leq 2 ^ {k - 1} a _ {2 ^ {k - 1}}\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":89,"id":"5cc38201-38a0-4eae-b763-f6bd4f470e0a","page_size":[595,841],"block_position":"89-17"},{"bbox":[54,559,118,571],"type":"text","angle":0,"index":18,"merge_prev":false,"text":"对 $k$ 求和得到 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":89,"id":"f4ac7424-5392-4c1b-98a6-4fe187541d40","page_size":[595,841],"block_position":"89-18"},{"bbox":[184,570,408,602],"type":"interline_equation","angle":0,"index":19,"text":"$$\n\\frac {1}{2} \\sum_ {k = 1} ^ {n} 2 ^ {k} a _ {2 ^ {k}} \\leq a _ {1} + a _ {2} + \\dots + a _ {2 ^ {n}} \\leq a _ {1} + \\sum_ {k = 0} ^ {n - 1} 2 ^ {k} a _ {2 ^ {k}}\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":89,"id":"154fa6f7-85dd-4980-8540-5a4a53907f4b","page_size":[595,841],"block_position":"89-19"},{"bbox":[54,609,221,621],"type":"text","angle":0,"index":20,"merge_prev":false,"text":"这说明两者部分和等价，因此同敛散。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":89,"id":"440fda78-a4ee-4fe2-a30c-710b36acd173","page_size":[595,841],"block_position":"89-20"},{"bbox":[36,635,538,694],"type":"text","angle":0,"index":21,"merge_prev":false,"text":"练习 8.17 比较原则对于一般项级数不成立的反例 (1) $a_{n}, b_{n}$ 满足 $\\lim_{n\\to\\infty}\\frac{a_{n}}{b_{n}}=l>0$ ，此时显然 $\\sum_{n=1}^{\\infty}|a_{n}|,\\sum_{n=1}^{\\infty}|b_{n}|$ 敛散性相同，但举例说明 $\\sum_{n=1}^{\\infty}a_{n},\\sum_{n=1}^{\\infty}b_{n}$ 敛散性可以不同。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":89,"id":"9db50f92-1c7f-4f35-a7b9-9bebf732ea00","page_size":[595,841],"block_position":"89-21"},{"bbox":[73,692,509,726],"type":"text","angle":0,"index":22,"merge_prev":false,"text":"(2) $\\lim_{n\\to \\infty}\\frac{a_n}{b_n} = 0$ ，根据 $\\sum_{n = 1}^{\\infty}b_n$ 绝对收敛可推出 $\\sum_{n = 1}^{\\infty}a_n$ 绝对收敛，举例 $\\sum_{n = 1}^{\\infty}b_n$ 收敛推不出 $\\sum_{n = 1}^{\\infty}a_n$ 收敛 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":89,"id":"735c8590-3381-4a33-a457-b28ec56730af","page_size":[595,841],"block_position":"89-22"},{"bbox":[73,724,274,756],"type":"text","angle":0,"index":23,"merge_prev":false,"text":"(3)举例说明 $\\sum_{n = 1}^{\\infty}a_n$ 收敛， $\\sum_{n = 1}^{\\infty}a_n^2$ 不一定收敛 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":89,"id":"8cb9581a-33ab-41f3-8201-37d74b65b28a","page_size":[595,841],"block_position":"89-23"},{"bbox":[291,802,302,812],"type":"page_number","angle":0,"index":1000023,"text":"82 ","color":{"line":"rgba(164, 164, 164, 1)","fill":"rgba(164, 164, 164, 1)"},"page_idx":89,"is_discarded":true,"id":"75815e5c-bf53-4394-b241-7b26ef9b1875","page_size":[595,841],"block_position":"89-24"}],[{"bbox":[459,47,539,58],"type":"header","angle":0,"index":-99999,"text":"8.4 数项级数理论 ","color":{"line":"rgba(164, 164, 164, 1)","fill":"rgba(164, 164, 164, 1)"},"page_idx":90,"is_discarded":true,"id":"857dd7d4-9553-4995-8c9e-5e3912ef69a0","page_size":[595,841],"block_position":"90-0"},{"bbox":[54,68,418,95],"type":"text","angle":0,"index":1,"merge_prev":false,"text":"解 (1) 例如 $a_{n} = \\frac{(-1)^{n}}{\\sqrt{n}}, b_{n} = \\frac{(-1)^{n}}{\\sqrt{n}} + \\frac{1}{n}$ , 显然 $\\lim_{n\\to \\infty}\\frac{a_n}{b_n} = 1$ , 但 $a_{n}$ 收敛, $b_{n}$ 发散 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":90,"id":"52184ba1-822b-43b0-be25-d70dc3c52518","page_size":[595,841],"block_position":"90-1"},{"bbox":[73,95,201,119],"type":"text","angle":0,"index":2,"merge_prev":false,"text":"(2) 比如 $a_{n} = \\frac{1}{n}, b_{n} = \\frac{(-1)^{n}}{\\sqrt{n}}$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":90,"id":"eedd5f9c-5723-41c2-979d-0c9cc9b9233a","page_size":[595,841],"block_position":"90-2"},{"bbox":[74,119,165,144],"type":"text","angle":0,"index":3,"merge_prev":false,"text":"(3) 例如 $a_{n} = \\frac{(-1)^{n}}{\\sqrt{n}}$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":90,"id":"8df92457-b904-49b0-a186-2d501d849a6d","page_size":[595,841],"block_position":"90-3"},{"bbox":[291,803,302,811],"type":"page_number","angle":0,"index":1000003,"text":"83 ","color":{"line":"rgba(164, 164, 164, 1)","fill":"rgba(164, 164, 164, 1)"},"page_idx":90,"is_discarded":true,"id":"0332d17f-10e4-45fb-b2b6-ac4924264b56","page_size":[595,841],"block_position":"90-4"}],[{"bbox":[193,74,411,94],"type":"title","angle":0,"index":0,"level":1,"text":"# 第9章 函数列与函数项级数","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":91,"id":"e1a4c01a-c41d-4742-83c0-d79992d3e666","page_size":[595,841],"block_position":"91-0"},{"bbox":[55,122,201,139],"type":"title","angle":0,"index":1,"level":2,"text":"## 9.1 函数列的一致收敛","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":91,"id":"bb3e98f8-bd62-45d0-a64a-65e05e30ad41","page_size":[595,841],"block_position":"91-1"},{"bbox":[55,156,174,169],"type":"title","angle":0,"index":2,"level":2,"text":"## 9.1.1 一致收敛的概念","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":91,"id":"0a21c8cc-8446-42e7-a6e2-791485424482","page_size":[595,841],"block_position":"91-2"},{"bbox":[71,182,161,195],"type":"title","angle":0,"index":3,"level":2,"text":"## 定义 9.1 (一致收敛)","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":91,"id":"36618efa-116c-47fb-b97c-195cb147e834","page_size":[595,841],"block_position":"91-3"},{"bbox":[67,198,537,220],"type":"text","angle":0,"index":4,"merge_prev":false,"text":"$f_{n}$ 是 $I$ 上的函数列，若对 $\\forall \\epsilon, \\exists N, \\forall x, \\forall n > N$ 有 $|f_{n}(x) - f(x)| < \\epsilon$ ，则称 $f_{n}$ 一致收敛至 $f$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":91,"id":"a092c9ff-23d4-4249-9b84-a3aeabbd6b6b","page_size":[595,841],"block_position":"91-4"},{"bbox":[52,229,539,243],"type":"text","angle":0,"index":5,"merge_prev":false,"text":"笔记一致收敛与点态收敛的区别：一致收敛的 N 不受 x 影响，因此当 n > N 时，所有 x 都会在 $f(x) \\pm \\epsilon$ 带中。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":91,"id":"c21d6298-6522-4b07-9ddb-a9b8521481d6","page_size":[595,841],"block_position":"91-5"},{"bbox":[35,259,539,311],"type":"text","angle":0,"index":6,"merge_prev":false,"text":"练习9.1定义说明不一致收敛 $(1)\\frac{x}{n}$ 在 $(0,a]$ 一致收敛到0，但在 $(0, + \\infty)$ 不一致收敛(2)说明 $x^{n}$ 在 $(0,1)$ 上收敛但不一致收敛证明 $(1)\\exists \\epsilon_0 = \\frac{1}{2},\\forall N,\\exists n_0 = N + 1,\\exists x_0 = n_0$ 使得 $|\\frac{x_0}{n_0} -0| = 1 > \\epsilon_0$ ，因此不一致收敛 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":91,"id":"75cc28d4-4381-478c-9a08-b72cd1a40503","page_size":[595,841],"block_position":"91-6"},{"bbox":[35,309,539,365],"type":"text","angle":0,"index":7,"merge_prev":false,"text":"(2) 点态收敛显然。不一致收敛的原因在 1 处，因此考虑 $x_{n} = 1 - \\frac{1}{n}$ ，则 $\\lim_{n\\to \\infty}|x_n^n -0| = \\lim_{n\\to \\infty}(1 - \\frac{1}{n})^n = e^{-1}\\neq 0$ 笔记 $\\frac{x}{n}, x^n$ 是极度重要的不一致收敛反例，在很多反例里都会用到它 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":91,"id":"d9f8c90e-4c40-44ed-846e-ba637cd4ec31","page_size":[595,841],"block_position":"91-7"},{"bbox":[73,381,196,394],"type":"title","angle":0,"index":8,"level":2,"text":"## 定理 9.1 (Cauchy 收敛准则)","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":91,"id":"e939b661-594c-4c76-8e0e-0cd64f339b42","page_size":[595,841],"block_position":"91-8"},{"bbox":[61,396,536,427],"type":"text","angle":0,"index":9,"merge_prev":false,"text":"$f_{n}$ 是 $I$ 上的函数列， $f_{n}$ 在 $I$ 上一致收敛于 $f$ 的充要条件为 $\\forall \\epsilon, \\exists N, \\forall n, m > N$ 满足 $\\sup_{x \\in I} |f_{m}(x) - f_{n}(x)| < \\epsilon$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":91,"id":"89d3a939-3d0e-4e74-86ff-5b93ce05025f","page_size":[595,841],"block_position":"91-9"},{"bbox":[36,434,539,464],"type":"text","angle":0,"index":10,"merge_prev":false,"text":"笔记 Cauchy 判别法一般用于证明其他判别法，本身对于证明具体一致收敛用处较少 (因为具体函数列的极限一般容易算出) ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":91,"id":"da1fb404-1327-49c1-add2-094f2eefb383","page_size":[595,841],"block_position":"91-10"},{"bbox":[73,472,181,485],"type":"title","angle":0,"index":11,"level":2,"text":"## 定义 9.2 (内闭一致收敛)","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":91,"id":"00259738-8c8b-430f-87b1-162239af2f06","page_size":[595,841],"block_position":"91-11"},{"bbox":[66,488,536,510],"type":"text","angle":0,"index":12,"merge_prev":false,"text":"若函数列 $f_{n}(x)$ 在区间 $D$ 任意一个闭区间都一致收敛，则称 $f_{n}(x)$ 在 $D$ 上内闭一致收敛。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":91,"id":"0ca11396-fa02-4eba-956d-f08633197d71","page_size":[595,841],"block_position":"91-12"},{"bbox":[52,519,538,548],"type":"text","angle":0,"index":13,"merge_prev":false,"text":"笔记 内闭一致收敛的意义在于大部分不一致收敛都是在区间端点出现了问题，因此内闭一致收敛反应了区间整体的性质，不受个别端点影响。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":91,"id":"2c254c05-db73-42e7-8f37-68a84610f2ba","page_size":[595,841],"block_position":"91-13"},{"bbox":[55,566,210,580],"type":"title","angle":0,"index":14,"level":2,"text":"## 9.1.2 一致收敛的基本判别法","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":91,"id":"cbe60672-7b43-4822-9d00-422e11d35fc5","page_size":[595,841],"block_position":"91-14"},{"bbox":[73,593,199,606],"type":"title","angle":0,"index":15,"level":2,"text":"## 定理 9.2 (Weiestrass 判别法)","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":91,"id":"225a2dc5-0ef1-4ac8-843d-cdcdcc7fb0cf","page_size":[595,841],"block_position":"91-15"},{"bbox":[67,608,536,636],"type":"text","angle":0,"index":16,"merge_prev":false,"text":"若存在 $\\lim_{n\\to \\infty}a_n = 0$ 的正项数列 $a_{n}$ ，使得 $I$ 上有 $|f(x) - f_n(x)|\\leq a_n$ ，则 $f_{n}(x)$ 一致收敛到 $f(x)$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":91,"id":"e15dcfc8-9384-4509-b109-e14941fc5bf5","page_size":[595,841],"block_position":"91-16"},{"bbox":[54,645,352,657],"type":"text","angle":0,"index":17,"merge_prev":false,"text":"证明 根本原因是 $a_{n}$ 收敛的 N 与 x 无关，用 Cauchy 收敛准则易证。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":91,"id":"cc948fa0-8d6f-4396-82ee-d0d078f9ed1d","page_size":[595,841],"block_position":"91-17"},{"bbox":[35,668,393,696],"type":"text","angle":0,"index":18,"merge_prev":false,"text":"练习9.2M判别法的使用(1)重点：证明 $\\sqrt{x^2 + \\frac{1}{n^2}}$ 在任意区间 $I$ 上一致收敛 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":91,"id":"357961f1-7b16-402a-a914-d36c18d5cb15","page_size":[595,841],"block_position":"91-18"},{"bbox":[52,697,536,756],"type":"text","angle":0,"index":19,"merge_prev":false,"text":"(2) 重点： $f(x)$ 为 $[a, b]$ 上的连续函数， $f_{0}(x) = f(x)$ ， $f_{n+1}(x) = \\int_{a}^{x} f_{n}(t) \\, dt$ ，证明 $f_{n}(x)$ 在 $[a, b]$ 一致收敛\n\n证明 (1) $\\left|\\sqrt{x^{2} + \\frac{1}{n^{2}}} - |x|\\right| = \\frac{\\frac{1}{n^{2}}}{\\sqrt{x^{2} + \\frac{1}{n^{2}}} + |x|} \\leq \\frac{1}{n} \\to 0$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":91,"id":"b2843cc5-4040-42e6-b4ce-62cb997deb11","page_size":[595,841],"block_position":"91-19"}],[{"bbox":[439,47,539,58],"type":"header","angle":0,"index":-99998,"text":"9.1 函数列的一致收敛 ","color":{"line":"rgba(164, 164, 164, 1)","fill":"rgba(164, 164, 164, 1)"},"page_idx":92,"is_discarded":true,"id":"d246f942-1a87-41ae-96b3-9104794a10dc","page_size":[595,841],"block_position":"92-0"},{"bbox":[54,69,539,121],"type":"text","angle":0,"index":2,"merge_prev":false,"text":"(2) 由于连续，因此 $|f(x)| \\leq M$ ，因此 $|f_1(x)| \\leq \\int_0^x |f_1(t)|\\mathrm{d}t \\leq M(x - a)$ (暂时不要进一步放缩)，同理得到 $|f_2(x)| \\leq \\int_a^x M(t - a)\\mathrm{d}t = \\frac{(x - a)^2}{2!} M$ ，因此 $|f_n(x)| \\leq \\frac{(x - a)^n}{n!} M \\leq \\frac{(b - a)^n}{n!} M \\to 0$ (最后再放缩) ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":92,"id":"f4e782c3-7d91-47cf-9b7c-32fc5751dae4","page_size":[595,841],"block_position":"92-1"},{"bbox":[72,136,171,148],"type":"title","angle":0,"index":3,"level":2,"text":"## 定理 9.3 (确界判别法)","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":92,"id":"3072285b-0637-428d-81e1-be5abbf005fc","page_size":[595,841],"block_position":"92-2"},{"bbox":[68,153,165,164],"type":"text","angle":0,"index":4,"merge_prev":false,"text":"$f_{n}$ 为 $I$ 上函数列，则 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":92,"id":"258d48e0-cf4d-4f01-ac2b-ef6b1c9569bf","page_size":[595,841],"block_position":"92-3"},{"bbox":[83,167,382,184],"type":"text","angle":0,"index":5,"merge_prev":false,"text":"- $f_{n}(x)$ 一致收敛到 $f(x)$ 的充要条件为 $\\lim_{n\\to \\infty}\\sup_{x\\in I}|f_n(x) - f(x)| = 0$ 。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":92,"id":"df36b09a-9815-46a2-8327-a8023f341552","page_size":[595,841],"block_position":"92-4"},{"bbox":[83,187,495,200],"type":"text","angle":0,"index":6,"merge_prev":false,"text":"- $f_{n}(x)$ 不一致收敛于 $f(x)$ 的充要条件是存在 $x_{n}$ ，使得 $f_{n}(x_{n}) - f(x_{n})$ 不收敛到0(极度重要) ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":92,"id":"b14a7c01-1abe-4eda-992e-337831a95225","page_size":[595,841],"block_position":"92-5"},{"bbox":[53,216,508,232],"type":"text","angle":0,"index":7,"merge_prev":false,"text":"证明 左推右：对 $\\forall \\epsilon, \\exists N, \\forall x, \\forall n > N$ 有 $|f_n(x) - f(x)| < \\epsilon$ ，根据上确界定义可知 $\\sup_{\\epsilon \\in I} |f_n(x) - f(x)| \\leq \\epsilon$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":92,"id":"f518c7cc-236b-4ab7-80c8-a060b2d58fa3","page_size":[595,841],"block_position":"92-6"},{"bbox":[54,235,537,268],"type":"text","angle":0,"index":8,"merge_prev":false,"text":"右推左： $\\forall \\epsilon, \\exists N, \\forall n > N$ 有 $\\sup_{x \\in I} |f_n(x) - f(x)| < \\epsilon$ ，而对 $\\forall x \\in I$ 有 $|f_n(x) - f(x)| \\leq \\sup_{x \\in I} |f_n(x) - f(x)|$ ，因此 $\\forall x, |f_n(x) - f(x)| < \\epsilon$ ，即一致收敛。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":92,"id":"ee93369a-393a-4b1f-9c12-27daa8ebdf41","page_size":[595,841],"block_position":"92-7"},{"bbox":[52,270,415,282],"type":"text","angle":0,"index":9,"merge_prev":false,"text":"笔记 一般一致收敛用 Weiestrass 判别法判定，不一致收敛用确界判别法第二条判定 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":92,"id":"e4fee76a-2738-49ec-a40e-112457e54987","page_size":[595,841],"block_position":"92-8"},{"bbox":[55,301,186,315],"type":"title","angle":0,"index":10,"level":2,"text":"## 9.1.3 一致收敛判别练习","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":92,"id":"d659fa0e-5c61-4792-9480-cf829a32f5c8","page_size":[595,841],"block_position":"92-9"},{"bbox":[35,325,373,337],"type":"text","angle":0,"index":11,"merge_prev":false,"text":"练习 9.3 三角相关 求下列函数列的收敛区间、极限函数与一致收敛区间： ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":92,"id":"21030c3e-cae7-4e87-b659-4fb6f2a005e7","page_size":[595,841],"block_position":"92-10"},{"bbox":[73,338,386,359],"type":"interline_equation","angle":0,"index":12,"text":"$$\n(1) f _ {n} (x) = \\frac {\\sin x}{n} (2) f _ {n} (x) = \\sin \\frac {x}{n} (3) f _ {n} (x) = \\frac {x}{n} \\sin \\frac {x}{n} (4) f _ {n} (x) = n \\sin \\frac {x}{n}\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":92,"id":"b093c1fd-e1f7-479d-8542-a984aaf78f8c","page_size":[595,841],"block_position":"92-11"},{"bbox":[53,359,207,379],"type":"text","angle":0,"index":13,"merge_prev":false,"text":"解(1)根据 $|f_n(x)|\\leq \\frac{1}{n}\\to 0$ ，显然 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":92,"id":"451e5779-ace9-4ba9-a762-91a9e86ef4a8","page_size":[595,841],"block_position":"92-12"},{"bbox":[52,379,539,429],"type":"text","angle":0,"index":14,"merge_prev":false,"text":"(2) 先求极限函数， $\\forall x \\in (-\\infty, +\\infty)$ ， $\\lim_{n \\to \\infty} f_n(x) = 0$ ，因此点态收敛到 0。取 $x_n = n$ ，则 $\\sin \\frac{x_n}{n} = \\sin 1$ 不趋于 0，因此在 $(- \\infty, +\\infty)$ 不一致收敛。但是任意 $[a, b] \\subseteq (-\\infty, +\\infty)$ ，显然 $|x| \\leq M$ ， $|\\sin \\frac{x}{n}| \\leq |\\frac{x}{n}| \\leq \\frac{M}{n} \\to 0$ ，因此内闭一致收敛 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":92,"id":"0ce97640-f000-4932-ad0c-520ea488a777","page_size":[595,841],"block_position":"92-13"},{"bbox":[54,433,539,470],"type":"text","angle":0,"index":15,"merge_prev":false,"text":"(3) 点态收敛到 $f(x) = 0$ ，取 $x_{n} = n$ ，则 $\\lim_{n\\to \\infty}f_n(x_n) = \\sin 1\\neq 0$ 。取 $[a,b]\\subseteq (-\\infty , + \\infty)$ ，则 $|f_n(x) - 0|\\leq |\\frac{x}{n} |\\rightarrow 0$ 因此内闭一致收敛。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":92,"id":"d7d53334-bd0a-4046-a508-b33b7f3ca3ac","page_size":[595,841],"block_position":"92-14"},{"bbox":[54,468,540,519],"type":"text","angle":0,"index":16,"merge_prev":false,"text":"(4) 极限函数为 1，考虑 $x_{n} = n$ ，则 $\\lim_{n\\to \\infty}|f_n(x_n) - x_n| = |n\\sin 1 - n| \\neq 0$ ，因此在 $(- \\infty, + \\infty)$ 不一致收敛。取 $[a,b] \\subseteq (-\\infty, + \\infty)$ ，则 $|f_n(x) - x| \\leq |n(\\sin \\frac{x}{n} - \\frac{x}{n})| \\leq n(\\frac{x}{n})^2 \\leq \\frac{M^2}{n} \\to 0$ ，中间的不等式使用了 Taylor 展开，因此内闭一致收敛。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":92,"id":"ab298fee-d139-4109-a475-bf2a6443338d","page_size":[595,841],"block_position":"92-15"},{"bbox":[35,537,268,549],"type":"text","angle":0,"index":17,"merge_prev":false,"text":"练习 9.4 分式相关 判断在 $[0, +\\infty)$ 的一致收敛性 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":92,"id":"7b3d2e41-9e22-4244-b4ae-3fe346a0eee1","page_size":[595,841],"block_position":"92-16"},{"bbox":[73,549,422,572],"type":"interline_equation","angle":0,"index":18,"text":"$$\n(1) f _ {n} (x) = \\frac {x}{1 + n ^ {2} x ^ {2}} (2) f _ {n} (x) = \\frac {x}{1 + n x} (3) f _ {n} (x) = \\frac {1}{1 + n x} (4) f _ {n} (x) = \\frac {n x}{1 + n ^ {2} x ^ {2}}\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":92,"id":"c4a74bc9-20ba-494b-bc6b-ca50528924b9","page_size":[595,841],"block_position":"92-17"},{"bbox":[52,572,389,593],"type":"text","angle":0,"index":19,"merge_prev":false,"text":"解 $(1)f_{n}(x)\\leq \\frac{x}{2nx} = \\frac{1}{2n} (x\\neq 0)\\to 0$ ，而 $x = 0$ 时 $f_{n}(0) = 0$ ，因此一致收敛 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":92,"id":"d893c41c-6971-45bb-acda-af74accf5f08","page_size":[595,841],"block_position":"92-18"},{"bbox":[72,592,299,615],"type":"text","angle":0,"index":20,"merge_prev":false,"text":"(2) $x = 0$ 显然， $x \\neq 0$ 时 $f_{n}(x) \\leq \\frac{1}{n} \\to 0$ ，一致收敛 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":92,"id":"d4c1821a-6188-44c9-96b2-370d462e4e5f","page_size":[595,841],"block_position":"92-19"},{"bbox":[52,614,538,656],"type":"text","angle":0,"index":21,"merge_prev":false,"text":"(3) 先考虑极限函数 $x = 0$ 时, 极限函数为 $1, x \\neq 0$ 时, 极限函数为 0。考虑 $(0, +\\infty)$ , 取 $x_{n} = \\frac{1}{n}, f_{n}(x) - f(x) = \\frac{1}{2} \\neq 0$ , 因此不一致收敛。在 $[a, +\\infty)$ 有 $f_{n}(x) - 0 = \\frac{1}{1 + nx} \\leq \\frac{1}{1 + na} \\to 0$ 一致。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":92,"id":"fd80282b-b8bf-4e34-954a-c66b8891b4b0","page_size":[595,841],"block_position":"92-20"},{"bbox":[54,658,539,701],"type":"text","angle":0,"index":22,"merge_prev":false,"text":"(4) 极限函数为 0。取 $x_{n} = \\frac{1}{n}$ ，显然 $f_{n}(x) - 0 = \\frac{1}{2}$ 不趋于 0，因此不一致收敛。在 $[a, +\\infty)$ 时 $f_{n}(x) \\leq \\frac{nx}{n^{2}x^{2}} \\leq \\frac{1}{nx} \\leq \\frac{1}{na} \\to 0$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":92,"id":"a46db3d8-5ae9-499f-80ab-04b628c6b0aa","page_size":[595,841],"block_position":"92-21"},{"bbox":[35,712,539,751],"type":"text","angle":0,"index":23,"merge_prev":false,"text":"练习 9.5 导数与积分一致收敛 (1) $f(x)$ 在 R 上有连续导函数， $a_{n}$ 为单调趋于 $+\\infty$ 的正数列，证明 $f_{n}(x) = a_{n}\\left[f\\left(x + \\frac{1}{a_{n}}\\right) - f(x)\\right]$ 在任意有限闭区间 $[a, b]$ 上一致收敛到 $f'(x)$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":92,"id":"68dc6710-5792-4803-bc3a-ff211f2fe365","page_size":[595,841],"block_position":"92-22"},{"bbox":[291,803,302,812],"type":"page_number","angle":0,"index":999999,"text":"85 ","color":{"line":"rgba(164, 164, 164, 1)","fill":"rgba(164, 164, 164, 1)"},"page_idx":92,"is_discarded":true,"id":"d2f8abf3-339a-4fb4-b0f2-3bef11878402","page_size":[595,841],"block_position":"92-23"}],[{"bbox":[439,47,539,58],"type":"header","angle":0,"index":-99999,"text":"9.1 函数列的一致收敛 ","color":{"line":"rgba(164, 164, 164, 1)","fill":"rgba(164, 164, 164, 1)"},"page_idx":93,"is_discarded":true,"id":"24b44f1d-56e6-469a-bc51-81a5ae2a9674","page_size":[595,841],"block_position":"93-0"},{"bbox":[72,69,456,100],"type":"text","angle":0,"index":1,"merge_prev":false,"text":"(2) 特例：证明 $f_{n}(x) = n\\left(\\sqrt{x + \\frac{1}{n}} - \\sqrt{x}\\right)$ 在任意有限闭区间 $[a, b]$ 上一致收敛到 $\\frac{1}{2\\sqrt{x}}$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":93,"id":"4f3b7db6-8763-4d41-83ed-748d9c7e87f9","page_size":[595,841],"block_position":"93-1"},{"bbox":[54,101,539,158],"type":"text","angle":0,"index":2,"merge_prev":false,"text":"(3) $f(x)$ 为 $\\mathbb{R}$ 上连续函数，记 $f_{n}(x) = \\frac{1}{n}\\sum_{k = 1}^{n}f\\left(x + \\frac{k}{n}\\right)$ ，证明 $f_{n}(x)$ 在任意有限区间 $[a,b]$ 上一致收敛于 $F(x) = \\int_0^1 f(x + t)\\mathrm{d}t$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":93,"id":"6d76094d-48d4-4bd9-bde7-9a0388f98eca","page_size":[595,841],"block_position":"93-2"},{"bbox":[54,157,539,232],"type":"text","angle":0,"index":4,"merge_prev":false,"text":"(4) 特例：证明 $f_{n}(x) = \\sum_{k=1}^{n} \\frac{1}{n} \\cos \\left( x + \\frac{k}{n} \\right)$ 在任意有限区域 $[a, b]$ 上一致收敛到 $\\sin (x + 1) - \\sin x$ 证明 (1) 给定 $n$ ，根据 Lagrange 中值定理， $f_{n}(x) = f'(x + \\xi_{n})$ ，这里 $\\xi_{n} < \\frac{1}{a_{n}}$ 。由于 $f'(x)$ 在 $[a, b + 1]$ 连续，因此一致连续， $\\forall \\epsilon, \\exists \\delta \\in (0, 1)$ 当 $|x_{1} - x_{2}| < \\delta$ ， $|f'(x_{1}) - f'(x_{2})| < \\epsilon$ ，而 $\\exists N, \\forall n > N, \\xi_{n} < \\frac{1}{a_{n}} < \\delta$ ，此时 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":93,"id":"dc4c9d96-2a3a-4890-be47-3deeb0d3c4ee","page_size":[595,841],"block_position":"93-3"},{"bbox":[202,235,391,249],"type":"interline_equation","angle":0,"index":5,"text":"$$\n\\left| f _ {n} (x) - f ^ {\\prime} (x) \\right| = \\left| f ^ {\\prime} \\left(x _ {n} + \\xi_ {n}\\right) - f ^ {\\prime} (x) \\right| < \\epsilon\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":93,"id":"bffa7c99-1237-47a1-b588-3584d4894d73","page_size":[595,841],"block_position":"93-4"},{"bbox":[54,255,117,267],"type":"text","angle":0,"index":6,"merge_prev":false,"text":"因此一致收敛 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":93,"id":"582e1a8b-0c22-450a-9865-7f47799c78dc","page_size":[595,841],"block_position":"93-5"},{"bbox":[73,270,148,282],"type":"text","angle":0,"index":7,"merge_prev":false,"text":"(2) 即 (1) 的特例 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":93,"id":"f3a8046d-3286-4a02-88df-e54462f15d4e","page_size":[595,841],"block_position":"93-6"},{"bbox":[73,281,236,314],"type":"text","angle":0,"index":8,"merge_prev":false,"text":"(3) $F(x)=\\sum_{k=1}^{n}\\int_{\\frac{k-1}{n}}^{\\frac{k}{n}}f(x+t)\\mathrm{d}t$ ，因此 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":93,"id":"83c4df67-8038-403c-96e4-197b9551a3cf","page_size":[595,841],"block_position":"93-7"},{"bbox":[79,316,514,349],"type":"interline_equation","angle":0,"index":9,"text":"$$\n\\left| f _ {n} (x) - F (x) \\right| = \\left| \\sum_ {k = 1} ^ {n} \\int_ {\\frac {k - 1}{n}} ^ {\\frac {k}{n}} f \\left(x + \\frac {k}{n}\\right) - \\sum_ {k = 1} ^ {n} \\int_ {\\frac {k - 1}{n}} ^ {\\frac {k}{n}} f (x + t) \\mathrm{d} t \\right| \\leq \\sum_ {k = 1} ^ {n} \\int_ {\\frac {k - 1}{n}} ^ {\\frac {k}{n}} \\left| f \\left(x + \\frac {k}{n}\\right) - f (x + t) \\right| \\mathrm{d} t\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":93,"id":"5f952a12-10ea-4fc4-920b-571e55e78778","page_size":[595,841],"block_position":"93-8"},{"bbox":[53,354,539,398],"type":"text","angle":0,"index":10,"merge_prev":false,"text":"显然 $f(x)$ 在 $[a, b + 1]$ 一致连续， $\\forall \\epsilon, \\exists \\delta, |x_1 - x_2| < \\delta$ 时 $|f(x') - f(x'')| < \\epsilon$ 。而 $\\exists N, \\forall n > N, \\frac{1}{n} < \\delta$ ，对于 $\\forall x \\in [a, b], t \\in \\left[\\frac{k - 1}{n}, \\frac{k}{n}\\right]$ ，有 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":93,"id":"d4455a89-34be-4f74-947c-146c460c11e9","page_size":[595,841],"block_position":"93-9"},{"bbox":[208,396,386,423],"type":"interline_equation","angle":0,"index":11,"text":"$$\n\\left| \\left(x + \\frac {k}{n}\\right) - (x + t) \\right| = \\left| \\frac {k}{n} - t \\right| \\leq \\frac {1}{n} < \\delta\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":93,"id":"61209c74-d92f-410c-b53d-e3bfb7aa403b","page_size":[595,841],"block_position":"93-10"},{"bbox":[54,430,97,443],"type":"text","angle":0,"index":12,"merge_prev":false,"text":"综上得到 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":93,"id":"e65632bf-7440-4cb0-a543-bb272969574b","page_size":[595,841],"block_position":"93-11"},{"bbox":[196,440,396,473],"type":"interline_equation","angle":0,"index":13,"text":"$$\n\\forall x \\in [ a, b ], | f _ {n} (x) - F (x) | < \\sum_ {k = 1} ^ {n} \\int_ {\\frac {k - 1}{n}} ^ {\\frac {k}{n}} \\epsilon \\mathrm{d} t = \\epsilon\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":93,"id":"8d262ded-13ed-4a8d-9f54-6a8a442d8446","page_size":[595,841],"block_position":"93-12"},{"bbox":[55,497,270,511],"type":"title","angle":0,"index":14,"level":2,"text":"## 9.1.4 四则运算、复合极限的一致收敛性","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":93,"id":"e160e2ff-3c83-46fc-8ba6-2f19b3f418fb","page_size":[595,841],"block_position":"93-13"},{"bbox":[73,523,161,536],"type":"title","angle":0,"index":15,"level":2,"text":"## 定义 9.3 (一致有界)","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":93,"id":"24f65869-6d58-4c96-a02f-b44a4fc89eaf","page_size":[595,841],"block_position":"93-14"},{"bbox":[68,541,205,555],"type":"text","angle":0,"index":16,"merge_prev":false,"text":"$\\exists M, \\forall n, \\forall x \\in I$ 有 $|f_n(x)| \\leq M$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":93,"id":"5ba69463-099b-402f-88b9-63b74d0907bc","page_size":[595,841],"block_position":"93-15"},{"bbox":[74,575,181,587],"type":"title","angle":0,"index":17,"level":2,"text":"## 引理 9.1 (极限的有界性)","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":93,"id":"3602a3a5-ff90-4a1d-98ee-99d6c62311b4","page_size":[595,841],"block_position":"93-16"},{"bbox":[69,592,252,604],"type":"text","angle":0,"index":18,"merge_prev":false,"text":"设 $f_{n}(x)$ 在区间 $I$ 上点态收敛到 $f(x)$ ，则 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":93,"id":"7557a4da-f69c-40e7-ae66-9b2e56b4aace","page_size":[595,841],"block_position":"93-17"},{"bbox":[83,608,376,620],"type":"text","angle":0,"index":19,"merge_prev":false,"text":"- 若一致收敛，且每个 $f_{n}(x)$ 有界，则 $f(x)$ 有界， $f_{n}(x)$ 一致有界 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":93,"id":"00832367-7485-4b86-80a7-888d0ccc7686","page_size":[595,841],"block_position":"93-18"},{"bbox":[83,623,378,635],"type":"text","angle":0,"index":20,"merge_prev":false,"text":"若一致收敛，且 $f(x)$ 有界，则 $f_{n}(x)$ 除去至多有限项外一致有界 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":93,"id":"691dbc1a-670b-42d4-8efd-799360c67777","page_size":[595,841],"block_position":"93-19"},{"bbox":[83,638,239,651],"type":"text","angle":0,"index":21,"merge_prev":false,"text":"若 $f_{n}(x)$ 一致有界，则 $f(x)$ 有界 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":93,"id":"cf2a46f6-a0e2-4e0f-b27c-29c3fea1f876","page_size":[595,841],"block_position":"93-20"},{"bbox":[53,668,539,711],"type":"text","angle":0,"index":22,"merge_prev":false,"text":"证明 (1) 设 $|f_n(x)| \\leq M_n$ 。根据一致收敛，取 $\\epsilon = 1, \\exists N, \\forall n > N, \\forall x \\in I$ 有 $|f_n(x) - f(x)| \\leq 1$ 。因此得到 $|f(x)| \\leq M_N + 1 := M$ 。而 $n \\geq N$ 时， $\\forall x \\in I$ 有 $|f_n(x)| \\leq |f(x)| + 1 = M + 1$ ，前面有限项取最大即可，因此一致有界 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":93,"id":"80c13f73-ac20-48dc-91f9-05943272b8d2","page_size":[595,841],"block_position":"93-21"},{"bbox":[73,715,197,727],"type":"text","angle":0,"index":23,"merge_prev":false,"text":"(2) 包含于 (1) 的推导过程中 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":93,"id":"e39e926c-74b7-4972-983d-d69372b819eb","page_size":[595,841],"block_position":"93-22"},{"bbox":[54,730,539,760],"type":"text","angle":0,"index":24,"merge_prev":false,"text":"(3) 若 $\\exists M, \\forall n, |f_n(x)| \\leq M$ ，根据 $\\lim_{n \\to \\infty} f_n(x) = f(x)$ ，则 $\\forall x \\in I, \\exists \\epsilon_0 = 1, \\exists N, n \\geq N$ 有 $|f(x) - f_n(x)| \\leq 1$ ，因此 $|f(x)| \\leq |f_n(x)| + 1 \\leq M + 1$ ，因此有界。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":93,"id":"5b6c2c51-0895-457a-bb57-4630972404af","page_size":[595,841],"block_position":"93-23"},{"bbox":[291,803,303,812],"type":"page_number","angle":0,"index":1000002,"text":"86 ","color":{"line":"rgba(164, 164, 164, 1)","fill":"rgba(164, 164, 164, 1)"},"page_idx":93,"is_discarded":true,"id":"f6d2f7c7-ba81-4aee-889a-b5e82713c7ac","page_size":[595,841],"block_position":"93-24"}],[{"bbox":[439,47,539,58],"type":"header","angle":0,"index":-99999,"text":"9.2 函数列极限的性质 ","color":{"line":"rgba(164, 164, 164, 1)","fill":"rgba(164, 164, 164, 1)"},"page_idx":94,"is_discarded":true,"id":"aa919904-b64c-46af-8d48-7425143a6b10","page_size":[595,841],"block_position":"94-0"},{"bbox":[73,91,161,104],"type":"title","angle":0,"index":1,"level":2,"text":"## 定理 9.4 (四则运算)","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":94,"id":"efe1814c-5f94-46d4-975b-d69c9b401d2c","page_size":[595,841],"block_position":"94-1"},{"bbox":[69,108,264,121],"type":"text","angle":0,"index":2,"merge_prev":false,"text":"$f_{n}, g_{n}$ 在 $I$ 上分别一致收敛到 $f(x), g(x)$ ，则 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":94,"id":"44c30c43-22a2-4653-b4a6-7d44a03ef703","page_size":[595,841],"block_position":"94-2"},{"bbox":[83,124,236,136],"type":"text","angle":0,"index":3,"merge_prev":false,"text":"- 数乘： $kf_{n}(x)$ 一致收敛到 $kf(x)$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":94,"id":"e2126681-ef9b-48c5-bb25-178275553d53","page_size":[595,841],"block_position":"94-3"},{"bbox":[83,139,291,152],"type":"text","angle":0,"index":4,"merge_prev":false,"text":"- 加法： $f_{n}(x) + g_{n}(x)$ 一致收敛到 $f(x) + g(x)$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":94,"id":"823849be-4bb6-487c-ac8e-610b5c400ba6","page_size":[595,841],"block_position":"94-4"},{"bbox":[83,154,406,168],"type":"text","angle":0,"index":5,"merge_prev":false,"text":"- 乘法：若每个 $f_{n}(x), g_{n}(x)$ 均有界，则 $f_{n}(x)g_{n}(x)$ 一致收敛到 $f(x)g(x)$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":94,"id":"a2d0133d-67f4-44fb-81de-1ffb76bb579d","page_size":[595,841],"block_position":"94-5"},{"bbox":[83,170,497,183],"type":"text","angle":0,"index":6,"merge_prev":false,"text":"- 乘法推论：若 $f_{n}(x), g_{n}(x)$ 为连续函数列且 $I$ 为闭区间，则 $f_{n}(x)g_{n}(x)$ 一致收敛到 $f(x)g(x)$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":94,"id":"76607f0d-0c3a-4fd8-a7bd-ab8131a7e916","page_size":[595,841],"block_position":"94-6"},{"bbox":[83,183,536,210],"type":"text","angle":0,"index":7,"merge_prev":false,"text":"- 除法：若 $\\exists M$ 使得 $|f(x)| \\geq M$ ，则 $\\exists N, n > N$ 时 $f_n(x)$ 在 $I$ 上无零点，且 $\\frac{1}{f_n(x)}$ 一致收敛到 $\\frac{1}{f(x)}$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":94,"id":"aada4062-8734-4234-80aa-2ae634344f2e","page_size":[595,841],"block_position":"94-7"},{"bbox":[54,223,124,235],"type":"title","angle":0,"index":8,"level":2,"text":"## 证明 (1,2) 显然","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":94,"id":"68b79229-ca94-47a3-ac36-71df3f5a9136","page_size":[595,841],"block_position":"94-8"},{"bbox":[53,238,538,300],"type":"text","angle":0,"index":9,"merge_prev":false,"text":"(3) 用 $|f_n(x)g_n(x) - f(x)g(x)| \\leq |f_n(x)||g_n(x) - g(x)| + |g(x)||f_n(x) - f(x)|$ (4) 取 $\\epsilon = \\frac{M}{2}, \\exists N, \\forall n > N, \\forall x$ 有 $|f_n(x) - f(x)| \\leq \\frac{M}{2}$ ，因此 $|f_n(x)| \\geq |f(x)| - \\frac{M}{2} \\geq \\frac{M}{2}$ ，因此无零点。第二个结论用 $|\\frac{1}{f_n(x)} - \\frac{1}{f(x)}| = |\\frac{f_n(x) - f(x)}{f_n(x)f(x)}| \\leq \\frac{|f_n(x) - f(x)|}{M \\cdot \\frac{M}{2}} < \\epsilon$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":94,"id":"bd3a949e-a2ca-4b7e-a288-77bebc59ea10","page_size":[595,841],"block_position":"94-9"},{"bbox":[35,308,539,366],"type":"text","angle":0,"index":10,"merge_prev":false,"text":"练习 9.6 反例 (1) 举例说明 $f_{n}(x), g_{n}(x)$ 分别一致收敛到 $f(x), g(x)$ ，但 $f_{n}(x)g_{n}(x)$ 不一致收敛到 $f(x)g(x)$ 解 (1) 根据定理函数列要无界，取 $f_{n}(x)=\\frac{1}{n}, g_{n}(x)=\\frac{1}{x}$ ，显然 $f_{n}(x)$ 一致收敛到 0， $g_{n}(x)$ 一致收敛到 $\\frac{1}{x}$ ，但 $f_{n}(x)g_{n}(x)=\\frac{1}{nx}$ 不一致收敛到 0，因为取 $x_{n}=\\frac{1}{n}$ 即可。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":94,"id":"6b4ac851-efdc-44ec-8d13-df0935e4be02","page_size":[595,841],"block_position":"94-10"},{"bbox":[72,382,142,394],"type":"title","angle":0,"index":11,"level":2,"text":"## 定理 9.5 (复合)","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":94,"id":"9c527977-5a20-4126-a380-9c7d9653832b","page_size":[595,841],"block_position":"94-11"},{"bbox":[66,397,530,429],"type":"text","angle":0,"index":12,"merge_prev":false,"text":"$f_{n}(x)$ 在 $I$ 上一致收敛到 $f(x)$ ，且每个 $f_{n}(x)$ 均有界， $g(x)$ 为 $\\mathbb{R}$ 上连续函数，则 $g(f_{n}(x))$ 一致收敛到 $g(f(x))$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":94,"id":"42ab7da8-297d-4ffc-86ec-7cec4e10f1c9","page_size":[595,841],"block_position":"94-12"},{"bbox":[54,444,399,456],"type":"text","angle":0,"index":13,"merge_prev":false,"text":"证明 根据一致收敛以及个个有界得到 $f_{n}(x)$ 一致有界，不妨设对 $\\forall n, \\forall x \\in I$ 有 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":94,"id":"848f4c33-e467-411e-877f-702e05d03810","page_size":[595,841],"block_position":"94-13"},{"bbox":[268,464,325,476],"type":"interline_equation","angle":0,"index":14,"text":"$$\n\\left| f _ {n} (x) \\right| \\leq M\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":94,"id":"7cb2f36d-e6e4-49f0-9c5d-8e486ff4ff5a","page_size":[595,841],"block_position":"94-14"},{"bbox":[54,482,539,511],"type":"text","angle":0,"index":15,"merge_prev":false,"text":"由于 $g(x)$ 在 $[-M, M]$ 上一致连续， $\\forall \\epsilon, \\exists \\delta, |x_1 - x_2| < \\delta$ 时， $|g(x_1) - g(x_2)| < \\epsilon$ ，对上述 $\\delta$ ，由于 $f_n(x)$ 一致收敛， $\\exists N, \\forall n > N$ 有 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":94,"id":"fbae5558-df38-4856-b541-409b46ac5dfe","page_size":[595,841],"block_position":"94-15"},{"bbox":[239,514,355,527],"type":"interline_equation","angle":0,"index":16,"text":"$$\n\\left| f _ {n} (x) - f (x) \\right| < \\delta , \\forall x \\in I\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":94,"id":"72f4b1c8-ed85-4345-ac47-c1187035786d","page_size":[595,841],"block_position":"94-16"},{"bbox":[54,537,288,550],"type":"text","angle":0,"index":17,"merge_prev":false,"text":"从而 $n > N$ 时， $\\forall x\\in I$ 满足 $|g(f_n(x)) - g(f(x))| < \\epsilon$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":94,"id":"9eca9772-d7b3-4131-9d16-46bc056c5475","page_size":[595,841],"block_position":"94-17"},{"bbox":[55,572,201,588],"type":"title","angle":0,"index":18,"level":2,"text":"## 9.2 函数列极限的性质","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":94,"id":"9ec41a3d-ae56-41e3-b2c7-f05ed9633bb9","page_size":[595,841],"block_position":"94-18"},{"bbox":[55,605,186,619],"type":"title","angle":0,"index":19,"level":2,"text":"## 9.2.1 极限函数的连续性","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":94,"id":"4646a6ab-1359-4ddf-aade-f9c6d334ce42","page_size":[595,841],"block_position":"94-19"},{"bbox":[72,633,181,645],"type":"title","angle":0,"index":20,"level":2,"text":"## 定理 9.6 (极限顺序交换)","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":94,"id":"c26a1813-4556-4ecc-b3ce-9ecc4441d3db","page_size":[595,841],"block_position":"94-20"},{"bbox":[68,650,439,665],"type":"text","angle":0,"index":21,"merge_prev":false,"text":"$f_{n}(x)$ 在 $(a,x_0)\\cup (x_0,b)$ 上一致收敛于 $f(x)$ ，且 $\\forall n$ ，极限 $\\lim_{x\\to x_0}f_n(x) = a_n$ 存在，则 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":94,"id":"a2d95944-ee39-4f90-aab5-385d659643cc","page_size":[595,841],"block_position":"94-21"},{"bbox":[219,669,374,687],"type":"interline_equation","angle":0,"index":22,"text":"$$\n\\lim _ {n \\to \\infty} \\lim _ {x \\to x _ {0}} f _ {n} (x) = \\lim _ {x \\to x _ {0}} \\lim _ {n \\to \\infty} f _ {n} (x)\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":94,"id":"6bec58c0-05e4-40f2-84f5-92fe065d9c82","page_size":[595,841],"block_position":"94-22"},{"bbox":[53,703,539,735],"type":"text","angle":0,"index":23,"merge_prev":false,"text":"证明 由于 $f_{n}(x)$ 一致收敛到 $f(x)$ ，则对 $\\forall \\epsilon, \\exists N, \\forall m, n > N, \\forall x \\in (a, b)$ 有 $|f_{n}(x) - f_{m}(x)| < \\epsilon$ ，让 $x \\to x_{0}$ ，则 $|a_{n} - a_{m}| < \\epsilon$ ，因此 $a_{n}$ 收敛，设 $\\lim_{n \\to \\infty} a_{n} = A$ 。下证右侧 $\\lim_{x \\to x_{0}} f(x) = A$ ，根据 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":94,"id":"7f022885-2c64-4a38-ae6d-6193c9cdad5e","page_size":[595,841],"block_position":"94-23"},{"bbox":[167,740,425,752],"type":"interline_equation","angle":0,"index":24,"text":"$$\n| f (x) - A | \\leq | f (x) - f _ {n} (x) | + | f _ {n} (x) - a _ {n} | + | a _ {n} - A | < 3 \\epsilon\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":94,"id":"df0bb358-7b0b-4c19-8b62-088975cfae94","page_size":[595,841],"block_position":"94-24"},{"bbox":[54,759,359,771],"type":"text","angle":0,"index":26,"merge_prev":false,"text":"第一个要求 N 足够大，第二个要求 $\\delta$ 足够小，第三个要求 N 足够大。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":94,"id":"c5bcda04-de9a-420c-9861-17e85a914408","page_size":[595,841],"block_position":"94-25"},{"bbox":[291,802,302,811],"type":"page_number","angle":0,"index":1000024,"text":"87 ","color":{"line":"rgba(164, 164, 164, 1)","fill":"rgba(164, 164, 164, 1)"},"page_idx":94,"is_discarded":true,"id":"0d1f98f7-04c7-4499-9c1a-739e889d8e51","page_size":[595,841],"block_position":"94-26"}],[{"bbox":[439,47,538,58],"type":"header","angle":0,"index":-99998,"text":"9.2 函数列极限的性质 ","color":{"line":"rgba(164, 164, 164, 1)","fill":"rgba(164, 164, 164, 1)"},"page_idx":95,"is_discarded":true,"id":"befe8bcb-7d89-4b37-bc64-065580228aab","page_size":[595,841],"block_position":"95-0"},{"bbox":[73,73,171,86],"type":"title","angle":0,"index":2,"level":2,"text":"## 推论 9.1 (局部连续性)","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":95,"id":"a3010b05-0cac-44ff-aa51-d0eb646225bd","page_size":[595,841],"block_position":"95-1"},{"bbox":[68,89,486,104],"type":"text","angle":0,"index":3,"merge_prev":false,"text":"若 $f_{n}(x)$ 在 $x_0$ 邻域 $U(x_0,\\delta)$ 内一致收敛， $f_{n}(x)$ 在 $x_0$ 处连续，则极限函数 $f(x)$ 也在 $x_0$ 处连续 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":95,"id":"4e633e12-73ce-4cca-ace9-7c8afd872d64","page_size":[595,841],"block_position":"95-2"},{"bbox":[54,119,412,138],"type":"text","angle":0,"index":4,"merge_prev":false,"text":"证明 根据 $\\lim_{x\\to x_0}f(x) = \\lim_{x\\to x_0}\\lim_{n\\to \\infty}f_n(x) = \\lim_{n\\to \\infty}\\lim_{x\\to x_0}f_n(x) = \\lim_{n\\to \\infty}f_n(x_0) = f(x_0)$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":95,"id":"6c3ff27d-0005-4b72-ad79-7124f7e39bd2","page_size":[595,841],"block_position":"95-3"},{"bbox":[72,148,171,160],"type":"title","angle":0,"index":5,"level":2,"text":"## 推论 9.2 (整体连续性)","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":95,"id":"ba48baae-430d-4064-85b0-a595f6f53ffd","page_size":[595,841],"block_position":"95-4"},{"bbox":[69,164,321,176],"type":"text","angle":0,"index":6,"merge_prev":false,"text":"若连续函数列在 $I$ 上内闭一致收敛于 $f$ ，则 $f$ 在 $I$ 上连续 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":95,"id":"a128af8c-18b2-4551-8d56-14514b696c8b","page_size":[595,841],"block_position":"95-5"},{"bbox":[53,194,539,221],"type":"text","angle":0,"index":7,"merge_prev":false,"text":"证明 原因在于单点的连续性仅与小邻域有关(甚至可以只有半边)，因此内闭的一致收敛即可推出极限函数的连续 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":95,"id":"51a4a8c7-acaa-4ab2-8465-6febc0e5dd1a","page_size":[595,841],"block_position":"95-6"},{"bbox":[54,225,370,238],"type":"text","angle":0,"index":8,"merge_prev":false,"text":"例题 9.1 例如 $x^{n}$ 在 $(0,1)$ 内闭一致收敛，因此极限函数 0 在 $(0,1)$ 连续。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":95,"id":"83b3d88a-f19c-48c7-8308-a818b67b22a9","page_size":[595,841],"block_position":"95-7"},{"bbox":[73,260,171,272],"type":"title","angle":0,"index":9,"level":2,"text":"## 定理 9.7 (一致连续性)","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":95,"id":"f2ffb483-1846-4bbc-bb6c-d6e41ebc6426","page_size":[595,841],"block_position":"95-8"},{"bbox":[69,276,434,290],"type":"text","angle":0,"index":10,"merge_prev":false,"text":"若 $f_{n}(x)$ 在 $I$ 中一致收敛，每一项都一致连续，则极限函数 $f(x)$ 在 $I$ 上也一致连续 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":95,"id":"7856b527-093e-4ca6-af73-a4de4b1704ab","page_size":[595,841],"block_position":"95-9"},{"bbox":[53,306,539,334],"type":"text","angle":0,"index":11,"merge_prev":false,"text":"证明 思路： $|f(x_{1})-f(x_{2})|\\leq|f(x_{1})-f_{n}(x_{1})|+|f_{n}(x_{1})-f_{n}(x_{2})|+|f_{n}(x_{2})-f(x_{2})|<3\\epsilon$ ，中间一项用 $f_{n}(x)$ 的一致连续性，左右两项用一致收敛性。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":95,"id":"459d941f-9e95-41f0-a11f-eccb7fcad919","page_size":[595,841],"block_position":"95-10"},{"bbox":[35,352,539,382],"type":"text","angle":0,"index":12,"merge_prev":false,"text":"练习9.7函数列连续性练习 $(1)f_{n}(x)$ 在 $[a,b]$ 一致收敛到 $f(x)$ ，数列 $x_{n} \\in [a,b]$ ， $\\lim_{n \\to \\infty} x_{n} = x_{0}$ ，若 $f(x)$ 在 $x_{0}$ 处连续，证明 $\\lim_{n \\to \\infty} f_{n}(x_{n}) = f(x_{0})$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":95,"id":"1c50de61-ce04-4318-9f8d-db60bfdab695","page_size":[595,841],"block_position":"95-11"},{"bbox":[54,384,539,414],"type":"text","angle":0,"index":13,"merge_prev":false,"text":"(2) $f_{n}(x)$ 在 $[a,b]$ 收敛到连续 $f(x)$ ，证明 $f_{n}(x)$ 在 $[a,b]$ 一致收敛到 $f(x)$ 的充要条件为任意满足 $\\lim_{n\\to \\infty}x_n = x_0$ 的数列，都有 $\\lim_{n\\to \\infty}f_n(x_n) = f(x_0)$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":95,"id":"ee2bca38-58df-43fe-b278-bf67d616f482","page_size":[595,841],"block_position":"95-12"},{"bbox":[53,415,537,443],"type":"text","angle":0,"index":14,"merge_prev":false,"text":"证明 (1) 注意与定理的区别，这里有双重 $n$ 。思路为 $|f_n(x_n) - f(x_0)| \\leq |f_n(x_n) - f(x_n)| + |f(x_n) - f(x_0)|$ ，前者用一致收敛，后者用连续性。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":95,"id":"7303b421-c2e5-46dd-bff3-83215ca23328","page_size":[595,841],"block_position":"95-13"},{"bbox":[55,446,539,476],"type":"text","angle":0,"index":15,"merge_prev":false,"text":"根据一致收敛性， $\\forall \\epsilon, \\exists N_1, \\forall n > N_1, \\forall x \\in [a, b]$ 有 $|f_n(x) - f(x)| < \\epsilon$ ，根据 $f(x)$ 在 $x_0$ 连续以及 $\\lim_{n \\to \\infty} x_n = x_0$ 可知 $\\exists N_2, \\forall n > N_2, |f(x_n) - f(x)| < \\epsilon$ ，取 $N = \\max\\{N_1, N_2\\}$ ， $\\forall n > N$ 有 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":95,"id":"1309175b-9b46-4adc-8617-085c33b0e4d6","page_size":[595,841],"block_position":"95-14"},{"bbox":[166,481,427,495],"type":"interline_equation","angle":0,"index":16,"text":"$$\n\\left| f _ {n} \\left(x _ {n}\\right) - f \\left(x _ {0}\\right) \\right| \\leq \\left| f _ {n} \\left(x _ {n}\\right) - f \\left(x _ {n}\\right) \\right| + \\left| f \\left(x _ {n}\\right) - f \\left(x _ {0}\\right) \\right| < 2 \\epsilon\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":95,"id":"4a5e9b2a-45e8-40bf-8353-8c565bbd5f44","page_size":[595,841],"block_position":"95-15"},{"bbox":[53,501,537,529],"type":"text","angle":0,"index":17,"merge_prev":false,"text":"(2) 左推右：思路： $|f_{n}(x_{n}) - f(x_{0})| \\leq |f_{n}(x_{n}) - f(x_{n})| + |f(x_{n}) - f(x_{0})|$ ，注意不要写成 $|f_{n}(x_{n}) - f_{n}(x_{0})| + |f_{n}(x_{0}) - f(x_{0})|$ ！原因在于 $f_{n}(x)$ 不一定连续。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":95,"id":"a7b3731a-ddd3-48d9-b0a5-7afd387a9fb2","page_size":[595,841],"block_position":"95-16"},{"bbox":[53,532,539,575],"type":"text","angle":0,"index":18,"merge_prev":false,"text":"右推左：反设不一致收敛，则 $\\exists \\epsilon_0, \\forall N, \\exists n > N, \\exists x_n \\in [a, b]$ 满足 $|f_n(x_n) - f(x_n)| \\geq \\epsilon_0$ ，由于 $x_n$ 有界，因此存在收敛子列，不妨就设 $x_n \\to x_0 \\in [a, b]$ ，则 $\\lim_{n \\to \\infty} |f_n(x_n) - f(x_n)| = \\lim_{n \\to \\infty} |f_n(x_n) - f(x_0)| \\geq \\epsilon_0$ ，与所给条件矛盾。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":95,"id":"87627013-7aaa-4383-a29c-cf26585a0dea","page_size":[595,841],"block_position":"95-17"},{"bbox":[52,578,389,592],"type":"text","angle":0,"index":19,"merge_prev":false,"text":"笔记 反设法下不一致收敛常用的讨论即存在 $x_{n}$ ，使得 $\\left|f_n(x_n) - f(x_n)\\right| \\geq \\epsilon_0$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":95,"id":"52594fae-a90c-4a49-990d-7922cfb9da4c","page_size":[595,841],"block_position":"95-18"},{"bbox":[55,609,186,623],"type":"title","angle":0,"index":20,"level":2,"text":"## 9.2.2 极限函数的可积性","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":95,"id":"d8386989-6a2c-46f9-8842-92706e494b8d","page_size":[595,841],"block_position":"95-19"},{"bbox":[73,637,151,650],"type":"title","angle":0,"index":21,"level":2,"text":"## 定理 9.8 (可积性)","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":95,"id":"f324dbc8-b18b-401b-b66f-59edc576c2ae","page_size":[595,841],"block_position":"95-20"},{"bbox":[68,653,443,667],"type":"text","angle":0,"index":22,"merge_prev":false,"text":"$f_{n}(x)$ 在 $[a,b]$ 上一致收敛到 $f(x)$ ， $f_{n}(x)$ 均在 $[a,b]$ 上可积，则 $f(x)$ 在 $[a,b]$ 可积，且 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":95,"id":"e5cfcf12-2c88-4c26-92f4-046c1a31df89","page_size":[595,841],"block_position":"95-21"},{"bbox":[185,669,408,698],"type":"interline_equation","angle":0,"index":23,"text":"$$\n\\lim _ {n \\rightarrow \\infty} \\int_ {a} ^ {b} f _ {n} (x) \\mathrm{d} x = \\int_ {a} ^ {b} f (x) \\mathrm{d} x = \\int_ {a} ^ {b} \\lim _ {n \\rightarrow \\infty} f _ {n} (x) \\mathrm{d} x\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":95,"id":"ba807d0b-07fb-423b-bd89-31e25bfbdb1e","page_size":[595,841],"block_position":"95-22"},{"bbox":[54,711,528,731],"type":"text","angle":0,"index":24,"merge_prev":false,"text":"证明 (1) 先说明可积: $\\forall \\epsilon, \\exists N$ 使得 $\\forall x, |f(x) - f_N(x)| < \\frac{\\epsilon}{b - a}$ , 做划分 $T$ , 取 $x_1, x_2$ 于同一区间, 考虑振幅: ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":95,"id":"2260e354-cdd0-4d9d-bc5d-7566b9a817c3","page_size":[595,841],"block_position":"95-23"},{"bbox":[79,734,512,756],"type":"interline_equation","angle":0,"index":25,"text":"$$\n| f (x _ {1}) - f (x _ {2}) | \\leq | f (x _ {1}) - f _ {N} (x _ {1}) | + | f _ {N} (x _ {1}) - f _ {N} (x _ {2}) | + | f _ {N} (x _ {2}) - f (x _ {2}) | \\leq \\frac {\\epsilon}{b - a} + \\omega_ {i} ^ {f _ {N}} + \\frac {\\epsilon}{b - a}\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":95,"id":"3ad4843e-523d-4b92-a411-af034c1d8795","page_size":[595,841],"block_position":"95-24"},{"bbox":[291,802,302,812],"type":"page_number","angle":0,"index":999999,"text":"88 ","color":{"line":"rgba(164, 164, 164, 1)","fill":"rgba(164, 164, 164, 1)"},"page_idx":95,"is_discarded":true,"id":"d1a9a0ae-b7f6-4210-a58b-64dad201f7f6","page_size":[595,841],"block_position":"95-25"}],[{"bbox":[439,46,538,58],"type":"header","angle":0,"index":-99999,"text":"9.2 函数列极限的性质 ","color":{"line":"rgba(164, 164, 164, 1)","fill":"rgba(164, 164, 164, 1)"},"page_idx":96,"is_discarded":true,"id":"b6eb32a6-d08e-419e-be70-0cf1aa40f3d2","page_size":[595,841],"block_position":"96-0"},{"bbox":[54,68,243,94],"type":"text","angle":0,"index":1,"merge_prev":false,"text":"得到 $f(x)$ 的振幅 $\\omega_{i}^{f}\\leq \\frac{2\\epsilon}{b - a} +\\omega_{i}^{f_{N}}$ ，因此 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":96,"id":"08389fc6-f7f5-49cd-8da9-357b48e7a2ea","page_size":[595,841],"block_position":"96-1"},{"bbox":[193,96,400,123],"type":"interline_equation","angle":0,"index":2,"text":"$$\n\\sum_ {T} \\omega_ {i} ^ {f} \\Delta x _ {i} \\leq \\sum_ {T} \\frac {2 \\epsilon}{b - a} \\Delta x _ {i} + \\sum_ {T} \\omega_ {i} ^ {f _ {N}} \\Delta x _ {i} < 3 \\epsilon\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":96,"id":"2bb1e2d0-bce0-4461-bef9-2169a0d062e4","page_size":[595,841],"block_position":"96-2"},{"bbox":[73,126,463,155],"type":"text","angle":0,"index":3,"merge_prev":false,"text":"(2) 求积分值：考虑 $|\\int_{a}^{b} f_{n}(x) \\mathrm{d}x - \\int_{a}^{b} f(x) \\mathrm{d}x| \\leq \\int_{a}^{b} |f_{n}(x) - f(x)| \\mathrm{d}x < \\int_{a}^{b} \\frac{\\epsilon}{b - a} \\mathrm{d}x = \\epsilon$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":96,"id":"6c4f9e1a-26a0-40f4-bf99-ebf0f3c3b86a","page_size":[595,841],"block_position":"96-3"},{"bbox":[52,157,205,169],"type":"text","angle":0,"index":4,"merge_prev":false,"text":"笔记 该定理在 ZJU2020 中出现过。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":96,"id":"495ad2af-a06f-4323-88d3-f6c81e342fba","page_size":[595,841],"block_position":"96-4"},{"bbox":[35,182,537,258],"type":"text","angle":0,"index":5,"merge_prev":false,"text":"练习9.8不可积反例举例连续函数列 $f_{n}(x)$ 在 $[a,b]$ 收敛到连续函数 $f(x)$ ，但无 $\\lim_{n\\to \\infty}\\int_a^b f_n(x)\\mathrm{d}x = \\int_a^b f(x)\\mathrm{d}x$ 解 $f_{n}(x) = nx^{n}(1 - x^{n}),x\\in [0,1]$ ，极限函数为 $f(x) = 0$ ，此时 $\\int_0^1 f(x)\\mathrm{d}x = 0$ ，但是 $\\int_0^1 f_n(x)\\mathrm{d}x = \\frac{n}{n + 1} -$ $\\frac{n}{2n + 1} = \\frac{1}{2}$ 。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":96,"id":"9dc01ac7-4b94-4a2b-b7cc-b64c946bbb60","page_size":[595,841],"block_position":"96-5"},{"bbox":[35,268,540,344],"type":"text","angle":0,"index":6,"merge_prev":false,"text":"练习9.9可积性的应用 $f_{n}(x)$ 每项都在 $[a,b]$ 连续可微， $g(x)$ 为 $[a,b]$ 连续函数，若存在 $M > 0$ ，使得 $\\forall n \\in \\mathbb{N}^{+}$ ， $x', x'' \\in [a,b]$ 有 $|f_{n}(x') - f_{n}(x'')| \\leq \\frac{M}{n} |x' - x''|$ ，证明： $\\lim_{n \\to \\infty} \\int_{a}^{b} f_{n}'(x) g(x) \\, \\mathrm{d}x = 0$ 证明根据条件得到 $|\\frac{f_n(x) - f_n(y)}{x - y}| \\leq \\frac{M}{n}$ ，令 $y \\to x$ 得到 $|f_n'(x)| \\leq \\frac{M}{n} \\to 0$ 。由于 $g(x)$ 有界，因此 $f_{n}'(x) g(x)$ 一致收敛到0，根据一致收敛性质可将极限和积分互换位置，从而结论成立。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":96,"id":"ea2e6c1b-59b4-4939-8fff-250a8fb7f8ef","page_size":[595,841],"block_position":"96-6"},{"bbox":[73,366,171,378],"type":"title","angle":0,"index":7,"level":2,"text":"## 推论 9.3 (可积性推广)","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":96,"id":"08a56c23-a05e-489e-aeb8-9feca85163a4","page_size":[595,841],"block_position":"96-7"},{"bbox":[67,381,532,423],"type":"text","angle":0,"index":9,"merge_prev":false,"text":"$f_{n}(x)$ 在 $[a,b]$ 一致收敛到 $f(x)$ ，且 $f_{n}(x)$ 均可积，则 $f(x)$ 在 $[a,b]$ 可积，且对任意可积 $g(x)$ 和 $x_0 \\in [a,b]$ ， $\\int_{x_0}^{x} f_n(t) g(t) \\mathrm{d}t$ 在 $[a,b]$ 上一致收敛到 $\\int_{x_0}^{x} f(t) g(t) \\mathrm{d}t$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":96,"id":"53c72dbb-c61e-45c5-bc09-b3d67a4c5312","page_size":[595,841],"block_position":"96-8"},{"bbox":[36,431,503,486],"type":"text","angle":0,"index":10,"merge_prev":false,"text":"证明 考虑 $\\left|\\int_{x_{0}}^{x}f_{n}(t)g(t)\\mathrm{d}t-\\int_{x_{0}}^{x}f(t)g(t)\\mathrm{d}t\\right|\\leq\\left|\\int_{x_{0}}^{x}|f_{n}(t)-f(t)||g(t)|\\mathrm{d}t\\right|\\leq M\\int_{a}^{b}|f_{n}(x)-f(x)|\\mathrm{d}x<\\epsilon$ 笔记 上述推论常用的特例： $\\int_{x_{0}}^{x}f_{n}(t)\\mathrm{d}t$ 在 $[a,b]$ 上一致收敛到 $\\int_{x_{0}}^{x}f(t)\\mathrm{d}t$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":96,"id":"8fedaa9f-ce59-4e96-b00d-99941f4c24f7","page_size":[595,841],"block_position":"96-9"},{"bbox":[55,501,186,516],"type":"title","angle":0,"index":11,"level":2,"text":"## 9.2.3 极限函数的可微性","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":96,"id":"e00b7146-7de6-43e8-b6a5-c2b7e7c7fa10","page_size":[595,841],"block_position":"96-10"},{"bbox":[73,528,151,541],"type":"title","angle":0,"index":12,"level":2,"text":"## 定理 9.9 (可微性)","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":96,"id":"b7b9dd70-76bd-40f3-a313-93dafc6464bc","page_size":[595,841],"block_position":"96-11"},{"bbox":[67,544,524,575],"type":"text","angle":0,"index":13,"merge_prev":false,"text":"$f_{n}(x)$ 的每项在 $[a,b]$ 有连续导函数， $f_{n}^{\\prime}(x)$ 在 $[a,b]$ 一致收敛到 $g(x)$ ，若 $\\exists x_0\\in [a,b]$ 使得 $f_{n}(x_{0})$ 收敛，则 $f_{n}(x)$ 在 $[a,b]$ 一致收敛于一个连续可微函数 $f(x)$ ，且 $f^{\\prime}(x) = g(x)$ ，即 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":96,"id":"2f9ab44d-dc80-46a1-96d7-377d9e5dd5c7","page_size":[595,841],"block_position":"96-12"},{"bbox":[236,580,357,598],"type":"interline_equation","angle":0,"index":14,"text":"$$\n(\\lim _ {n \\to \\infty} f _ {n} (x)) ^ {\\prime} = \\lim _ {n \\to \\infty} f _ {n} ^ {\\prime} (x)\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":96,"id":"d9a97c29-16e3-4949-aeda-e9d82fe00142","page_size":[595,841],"block_position":"96-13"},{"bbox":[53,610,539,684],"type":"text","angle":0,"index":15,"merge_prev":false,"text":"证明 (1) 极限函数存在：根据 Newton-Leibniz 公式可知 $f_{n}(x) = f_{n}(x_{0}) + \\int_{x_{0}}^{x} f_{n}'(t) \\mathrm{d}t$ ，根据条件 $f_{n}(x_{0})$ 收敛且与 $x$ 无关。根据前面可积性推论可知 $\\int_{x_{0}}^{x} f_{n}'(t) \\mathrm{d}t$ 一致收敛到 $\\int_{x_{0}}^{x} g(t) \\mathrm{d}t$ ，根据一致收敛的四则运算可知 $f_{n}(x)$ 本身也一致收敛至 $f(x) = \\lim_{n \\to \\infty} f_{n}(x_{0}) + \\int_{x_{0}}^{x} g(t) \\mathrm{d}t$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":96,"id":"21d85fdf-fbae-4910-a38f-211c32cc5406","page_size":[595,841],"block_position":"96-14"},{"bbox":[58,684,420,698],"type":"text","angle":0,"index":16,"merge_prev":false,"text":"(2) 导数：由于 $f_n'(x)$ 在 $[a, b]$ 连续， $g(x)$ 连续，从而 $f(x)$ 可导且 $f'(x) = g(x)$ 。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":96,"id":"dee8d962-48c4-42dc-ba13-5a8a04f58cc9","page_size":[595,841],"block_position":"96-15"},{"bbox":[52,698,539,733],"type":"text","angle":0,"index":17,"merge_prev":false,"text":"笔记 定理中 $f_{n}(x)$ 存在 $x_{0}$ 这个收敛点的条件必不可少，这为使用 Leibniz 公式奠定了基础。在实际使用中，往往可以直接看出 $f_{n}(x), f_{n}^{\\prime}(x)$ 都一致收敛，则可直接得出 $(\\lim_{n\\to\\infty}f_{n}(x))^{\\prime}=\\lim_{n\\to\\infty}f_{n}^{\\prime}(x)$ 的结论，不用管 $x_{0}$ 。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":96,"id":"14316168-6b45-4393-9512-3e4443f2594a","page_size":[595,841],"block_position":"96-16"},{"bbox":[35,741,495,767],"type":"text","angle":0,"index":18,"merge_prev":false,"text":"练习 9.10 具体计算 (1) 已知 $u_{n}(x)=\\frac{\\cos nx}{n\\sqrt{n}}$ ，判定其在 $(0,2\\pi)$ 上是否逐项可导，若可导则求出 $S'(x)$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":96,"id":"df0ccebe-000b-44b2-8311-e223082b3c05","page_size":[595,841],"block_position":"96-17"},{"bbox":[291,802,302,811],"type":"page_number","angle":0,"index":1000007,"text":"89 ","color":{"line":"rgba(164, 164, 164, 1)","fill":"rgba(164, 164, 164, 1)"},"page_idx":96,"is_discarded":true,"id":"dc9a19e3-dcb2-466f-8905-102d8060fd75","page_size":[595,841],"block_position":"96-18"}],[{"bbox":[409,47,538,58],"type":"header","angle":0,"index":-99999,"text":"9.3 函数项级数的一致收敛性 ","color":{"line":"rgba(164, 164, 164, 1)","fill":"rgba(164, 164, 164, 1)"},"page_idx":97,"is_discarded":true,"id":"722748e4-1dc4-4535-ba82-c2cd4c169d59","page_size":[595,841],"block_position":"97-0"},{"bbox":[53,70,509,100],"type":"text","angle":0,"index":1,"merge_prev":false,"text":"解(1)考虑 $u_{n}^{\\prime}(x) = -\\frac{\\sin nx}{\\sqrt{n}}$ ，根据Dirichlet判别法可知一致收敛，因此逐项可导， $S^{\\prime}(x) = -\\sum_{n = 1}^{\\infty}\\frac{\\sin nx}{\\sqrt{n}}$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":97,"id":"9b905ebe-a3e8-448f-a6c5-c3116b03ea75","page_size":[595,841],"block_position":"97-1"},{"bbox":[55,117,235,131],"type":"title","angle":0,"index":2,"level":2,"text":"## 9.2.4 连续 + 单调：推出一致收敛","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":97,"id":"e37cbbde-1377-47fa-ad1e-33137d6a8334","page_size":[595,841],"block_position":"97-2"},{"bbox":[72,144,177,157],"type":"title","angle":0,"index":3,"level":2,"text":"## 定理 9.10 (关于 x 单调)","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":97,"id":"a52006c0-0466-47f9-a84b-3ce782fa30d7","page_size":[595,841],"block_position":"97-3"},{"bbox":[67,161,529,193],"type":"text","angle":0,"index":4,"merge_prev":false,"text":"函数列 $f_{n}(x)$ 在 $[a,b]$ 收敛到连续函数 $f(x)$ ，且每个 $f_{n}(x)$ 在 $[a,b]$ 上关于 $x$ 单调增，则 $f_{n}(x)$ 一致收敛到 $f(x)$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":97,"id":"05005bdd-11b5-4dcf-8f1e-4b69441761f1","page_size":[595,841],"block_position":"97-4"},{"bbox":[54,207,78,219],"type":"text","angle":0,"index":5,"merge_prev":false,"text":"证明 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":97,"id":"0a79732d-2a56-4e35-a5fc-0708f934ea23","page_size":[595,841],"block_position":"97-5"},{"bbox":[72,242,227,255],"type":"title","angle":0,"index":6,"level":2,"text":"## 定理 9.11 (Dini 定理：关于 n 单调)","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":97,"id":"a4f1754a-6ed1-45f1-a779-acc1cb3a8049","page_size":[595,841],"block_position":"97-6"},{"bbox":[67,259,535,291],"type":"text","angle":0,"index":7,"merge_prev":false,"text":"连续函数列 $f_{n}(x)$ 在 $[a,b]$ 收敛于连续函数 $f(x)$ ，若固定 $x$ ，每个 $f_{n}(x)$ 关于 $n$ 单调，则 $f_{n}(x)$ 在 $[a,b]$ 上一致收敛到 $f(x)$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":97,"id":"04c00950-312e-4ebb-ace6-5ee01ce996c9","page_size":[595,841],"block_position":"97-7"},{"bbox":[52,305,539,349],"type":"text","angle":0,"index":8,"merge_prev":false,"text":"证明 (1) 先证明在小区间成立：固定 $x_0 \\in [a, b]$ ， $f_n(x_0) - f(x_0)$ 单调趋于 0， $\\forall \\epsilon, \\exists N$ 有 $|f_N(x_0) - f(x_0)| < \\epsilon$ 。固定 $N$ ，由于 $f_N(x) - f(x)$ 连续，因此根据保号性 $\\exists \\delta, \\forall x \\in U(x_0, \\delta)$ 有 $|f_N(x) - f(x)| < \\epsilon$ 。重新根据单调性得到 $\\forall n > N, \\forall x \\in U(x_0, \\delta)$ 有 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":97,"id":"aa35d094-e0de-49bc-8e1c-f221d503a024","page_size":[595,841],"block_position":"97-8"},{"bbox":[217,352,377,365],"type":"interline_equation","angle":0,"index":9,"text":"$$\n\\left| f _ {n} (x) - f (x) \\right| \\leq \\left| f _ {N} (x) - f (x) \\right| < \\epsilon\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":97,"id":"b74e1daf-2b92-490f-8377-afdba18e8d46","page_size":[595,841],"block_position":"97-9"},{"bbox":[54,375,539,404],"type":"text","angle":0,"index":10,"merge_prev":false,"text":"(2) 再证明在整个区间成立：显然区间集 $\\{U(x_{0},\\delta):x_{0}\\in[a,b]\\}$ 是 $[a,b]$ 的开覆盖，根据有限覆盖定理得到存在有限个点 $x_{1},\\cdots,x_{k}$ 的邻域覆盖 $[a,b]$ ，每个对应着 $N_{i},\\delta_{i}$ 使得 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":97,"id":"70f54e97-8afe-4b0d-96e9-afcd5f9f7384","page_size":[595,841],"block_position":"97-10"},{"bbox":[202,410,390,423],"type":"interline_equation","angle":0,"index":12,"text":"$$\n\\forall n > N _ {i}, \\forall x \\in U (x _ {i}, \\delta_ {i}), | f _ {n} (x) - f (x) | < \\epsilon\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":97,"id":"92cc722d-1074-4c1e-bd04-b2ae21dceb08","page_size":[595,841],"block_position":"97-11"},{"bbox":[52,428,537,457],"type":"text","angle":0,"index":13,"merge_prev":false,"text":"因此取 $N = \\max\\{N_{1}, \\cdots, N_{k}\\}$ ， $\\forall n > N$ 时，对 $\\forall x \\in [a, b]$ ，总能找到 $x_{i}$ 使得 $x \\in U(x_{i}, \\delta_{i})$ ，且 $|f_{n}(x) - f(x)| < \\epsilon$ 笔记 ZJU2020 默写了 Dini 定理的证明 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":97,"id":"5384c5b5-64d0-471a-8c74-87de4297d948","page_size":[595,841],"block_position":"97-12"},{"bbox":[55,480,243,496],"type":"title","angle":0,"index":14,"level":2,"text":"## 9.3 函数项级数的一致收敛性","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":97,"id":"34ea0446-0a31-4abf-833e-a5ca3c82f447","page_size":[595,841],"block_position":"97-13"},{"bbox":[55,513,234,527],"type":"title","angle":0,"index":15,"level":2,"text":"## 9.3.1 一致收敛的基本性质与判别","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":97,"id":"e7c14bf6-3124-4f3b-bbcb-e41cec233158","page_size":[595,841],"block_position":"97-14"},{"bbox":[73,540,186,553],"type":"title","angle":0,"index":16,"level":2,"text":"## 定理 9.12 (一致收敛性质)","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":97,"id":"b0cc87b7-0bae-401d-9177-932d9c1aba44","page_size":[595,841],"block_position":"97-15"},{"bbox":[67,559,333,589],"type":"text","angle":0,"index":17,"merge_prev":false,"text":"$\\sum_{n=1}^{\\infty}u_{n}(x)$ 在 D 上一致收敛，则 $u_{n}(x)$ 在 D 上一致收敛到 0。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":97,"id":"b6af011e-8911-44ba-91a3-5a89eba28d91","page_size":[595,841],"block_position":"97-16"},{"bbox":[35,614,408,645],"type":"text","angle":0,"index":18,"merge_prev":false,"text":"练习9.11收敛但不一致收敛的级数说明 $\\sum_{n = 1}^{\\infty}\\frac{n}{1 + n^3x}$ 在(0,1)收敛但不一致收敛 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":97,"id":"451e294a-7b5a-4716-99ee-a5a47d8f6294","page_size":[595,841],"block_position":"97-17"},{"bbox":[52,645,465,668],"type":"text","angle":0,"index":19,"merge_prev":false,"text":"证明 (1) 收敛: 根据 $\\frac{n}{1 + n^3 x} \\sim \\frac{1}{n^2 x}$ , 若 $x$ 固定, 则作为数项级数收敛, 因此函数项级数收敛 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":97,"id":"589d2ed3-ea96-47f7-9d89-ffac0eac678e","page_size":[595,841],"block_position":"97-18"},{"bbox":[73,668,393,690],"type":"text","angle":0,"index":20,"merge_prev":false,"text":"(2) 不一致收敛: $x_{n} = \\frac{1}{n^{3}}$ , 则 $\\frac{n}{1 + n^{3}x_{n}} = \\frac{n}{2}$ 不趋于 0, 因此不一致收敛 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":97,"id":"e5162abd-60b3-4af7-9343-432b407afc36","page_size":[595,841],"block_position":"97-19"},{"bbox":[73,707,205,719],"type":"title","angle":0,"index":21,"level":2,"text":"## 定理 9.13 (一致收敛充要条件)","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":97,"id":"2e7cfbfa-1768-4cc0-9d76-e8e0ebdc7980","page_size":[595,841],"block_position":"97-20"},{"bbox":[68,724,288,756],"type":"text","angle":0,"index":22,"merge_prev":false,"text":"$\\sum_{n = 1}^{\\infty}u_n(x)$ 在 $D$ 上一致收敛到 $S(x)$ 的充要条件为： ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":97,"id":"f74dae8e-094b-4931-ae7b-fa56710a4774","page_size":[595,841],"block_position":"97-21"},{"bbox":[290,802,303,812],"type":"page_number","angle":0,"index":1000010,"text":"90 ","color":{"line":"rgba(164, 164, 164, 1)","fill":"rgba(164, 164, 164, 1)"},"page_idx":97,"is_discarded":true,"id":"1bc29b10-7d7f-46ed-835e-f95fe70532db","page_size":[595,841],"block_position":"97-22"}],[{"bbox":[409,47,538,58],"type":"header","angle":0,"index":-99999,"text":"9.3 函数项级数的一致收敛性 ","color":{"line":"rgba(164, 164, 164, 1)","fill":"rgba(164, 164, 164, 1)"},"page_idx":98,"is_discarded":true,"id":"a261158f-4536-436f-845e-6b92f01bcd5f","page_size":[595,841],"block_position":"98-0"},{"bbox":[83,80,396,112],"type":"text","angle":0,"index":1,"merge_prev":false,"text":"- Cauchy 收敛准则： $\\forall \\epsilon, \\exists N, \\forall n > m > N, \\forall x \\in D$ 有 $|\\sum_{k=m+1}^{n} u_k(x)| < \\epsilon$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":98,"id":"892eff3a-bbda-4f94-aa8c-5ac330942dc0","page_size":[595,841],"block_position":"98-1"},{"bbox":[83,111,245,144],"type":"text","angle":0,"index":2,"merge_prev":false,"text":"- 确界： $\\lim_{n\\to \\infty}\\sup_{x\\in D}|\\sum_{k = n + 1}^{\\infty}u_k(x)| = 0$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":98,"id":"84cc0d1d-8dff-4dd8-9ba5-28c2a6e96daf","page_size":[595,841],"block_position":"98-2"},{"bbox":[36,176,539,224],"type":"text","angle":0,"index":3,"merge_prev":false,"text":"练习9.12 $I$ 上函数列 $u_{n}(x)$ 满足 $u_{i}(x)u_{j}(x) = 0, i \\neq j, \\forall x \\in I$ ，若存在正数列 $a_{n}$ 满足 $|u_{n}(x)| \\leq a_{n}$ ， $\\lim_{n \\to \\infty} a_{n} = 0$ ，证明： $\\sum_{n=1}^{\\infty} u_{n}(x)$ 在 $I$ 一致收敛 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":98,"id":"86d78bda-3559-46be-8d30-dc5fe25cf43f","page_size":[595,841],"block_position":"98-3"},{"bbox":[52,225,539,274],"type":"text","angle":0,"index":4,"merge_prev":false,"text":"证明 根据条件 $\\forall x \\in I, u_{1}(x), \\cdots, u_{n}(x), \\cdots$ 中至多只有一个非零，因此根据确界条件 $\\sup_{x \\in I} |\\sum_{k=n+1}^{\\infty} u_{k}(x)| \\leq \\sup_{n} \\{a_{n+1}, a_{n+2}, \\cdots\\} \\to 0$ 。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":98,"id":"bb3a4213-b94d-478b-831e-f8a1473eb766","page_size":[595,841],"block_position":"98-4"},{"bbox":[73,289,168,301],"type":"title","angle":0,"index":5,"level":2,"text":"## 定理 9.14 (M 判别法)","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":98,"id":"5afcbc93-c9a4-40ba-a3a6-6ef21cdea0c5","page_size":[595,841],"block_position":"98-5"},{"bbox":[67,304,533,371],"type":"text","angle":0,"index":6,"merge_prev":false,"text":"$\\sum_{n=1}^{\\infty} u_n(x)$ 为 $D$ 上的函数项级数，若存在收敛的数项级数 $\\sum_{n=1}^{\\infty} M_n$ 满足 $\\forall x \\in D, |u_n(x)| \\leq M_n$ ，则得到 $\\sum_{n=1}^{\\infty} u_n(x)$ 在 $D$ 上一致收敛 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":98,"id":"2485bbab-84a1-4514-b895-6634640ddcb5","page_size":[595,841],"block_position":"98-6"},{"bbox":[35,394,449,423],"type":"text","angle":0,"index":7,"merge_prev":false,"text":"练习9.13 讨论一致收敛性：(1) $\\sum_{n=1}^{\\infty} \\arctan \\frac{2x}{x^2 + n^3}$ 在 $(- \\infty, + \\infty)$ (2) $\\sum_{n=1}^{\\infty} \\frac{x^{2n}}{1 + x^{2n+1}}$ 在 $x \\geq 0$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":98,"id":"4869694d-a803-4dca-a6e2-115ae2c6d8f3","page_size":[595,841],"block_position":"98-7"},{"bbox":[53,423,361,449],"type":"text","angle":0,"index":8,"merge_prev":false,"text":"解 (1) $|\\arctan \\frac{2x}{x^2 + n^3} |\\leq |\\frac{2x}{x^2 + n^3} |\\leq \\frac{|2x|}{2|x|n^{3/2}} = \\frac{1}{n^{3/2}}$ ，因此一致收敛 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":98,"id":"32b23a65-79d8-4842-a55b-9f9e41357d6c","page_size":[595,841],"block_position":"98-8"},{"bbox":[54,450,538,499],"type":"text","angle":0,"index":9,"merge_prev":false,"text":"(2) $x \\geq 1$ 时， $\\lim_{n \\to \\infty} \\frac{x^{2n}}{1 + x^{2n+1}} = \\frac{1}{x}$ ，发散。 $0 \\leq 1 < 1$ 时根据 $\\frac{x^{2n}}{1 + x^{2n+1}} \\leq x^{2n}$ 点态收敛，但是感觉出 1 处会导致一致收敛出问题，因此取 $x_n = 1 - \\frac{1}{n}$ ， $\\frac{x_n^{2n}}{1 + x_n^{2n+1}} \\to \\frac{e^{-2}}{1 + e^{-2}} \\neq 0$ ，因此不一致收敛。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":98,"id":"c1ea0bff-4a5a-46a3-99a5-bfe8c391403b","page_size":[595,841],"block_position":"98-9"},{"bbox":[35,505,539,568],"type":"text","angle":0,"index":10,"merge_prev":false,"text":"练习9.14积分格式 $f(x)$ 在 $[a,b]$ 上连续函数， $f_0(x) = f(x)$ ， $f_{n + 1}(x) = \\int_a^x f_n(t)\\mathrm{d}t$ ，证明 $\\sum_{n = 1}^{\\infty}f_n(x)$ 在 $[a,b]$ 一致收敛。证明同前面， $|f_n(x)|\\leq \\frac{(b - a)^n}{n!} M$ ，根据根式或者比式判别法易知。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":98,"id":"e082cfdc-1226-475a-a821-4f8ac40ddc38","page_size":[595,841],"block_position":"98-10"},{"bbox":[55,582,216,596],"type":"title","angle":0,"index":11,"level":2,"text":"## 9.3.2 Dirichlet 与 Abel 判别法","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":98,"id":"5d31a9fd-69e0-46dc-83cf-d8a7cd3d04ac","page_size":[595,841],"block_position":"98-11"},{"bbox":[73,610,231,622],"type":"title","angle":0,"index":12,"level":2,"text":"## 定理 9.15 (Dirichlet 与 Abel 判别法)","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":98,"id":"967ae174-1a14-4db2-bbec-6ee0b1af5945","page_size":[595,841],"block_position":"98-12"},{"bbox":[69,624,520,723],"type":"text","angle":0,"index":13,"merge_prev":false,"text":"对于 $I$ 上的函数项级数 $\\sum_{n=1}^{\\infty} u_n(x) v_n(x)$ ，若满足下面任意一条，则一致收敛\n- Dirichlet: (1) $\\sum_{k=1}^{n} u_k(x)$ 在 $I$ 上一致有界 (2) $v_n(x)$ 固定 $x \\in I$ 关于 $n$ 单调，且在 $I$ 上一致收敛到 0\n- Abel: (1) $\\sum_{n=1}^{\\infty} u_n(x)$ 一致收敛 (2) $v_n(x)$ 固定 $x \\in I$ 关于 $n$ 单调，且在 $I$ 上一致有界 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":98,"id":"6bd53f22-34eb-4a9f-b965-3adae5868f7d","page_size":[595,841],"block_position":"98-13"},{"bbox":[52,735,227,747],"type":"text","angle":0,"index":14,"merge_prev":false,"text":"笔记 单调都是关于 $n$ ，一致都是关于 $x$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":98,"id":"23d0ee3a-375f-45d6-a71f-61db7daf2e99","page_size":[595,841],"block_position":"98-14"},{"bbox":[290,802,302,811],"type":"page_number","angle":0,"index":1000014,"text":"91 ","color":{"line":"rgba(164, 164, 164, 1)","fill":"rgba(164, 164, 164, 1)"},"page_idx":98,"is_discarded":true,"id":"5075a8d0-5c3a-42d2-9dc2-7e6dc0f71f54","page_size":[595,841],"block_position":"98-15"}],[{"bbox":[399,47,538,58],"type":"header","angle":0,"index":-99998,"text":"9.4 函数项级数极限函数的性质 ","color":{"line":"rgba(164, 164, 164, 1)","fill":"rgba(164, 164, 164, 1)"},"page_idx":99,"is_discarded":true,"id":"2bbf3877-bd1c-46c8-a756-e24bb957475a","page_size":[595,841],"block_position":"99-0"},{"bbox":[36,70,465,112],"type":"text","angle":0,"index":2,"merge_prev":false,"text":"练习9.15经典一致收敛题(1) $a_{n}$ 单调收敛于0，证明： $\\sum_{n = 1}^{\\infty}a_n\\cos nx$ 在 $[\\alpha ,2\\pi -\\alpha ]$ 上一致收敛证明(1)首先判断一致有界性： ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":99,"id":"08eb1661-a53c-4b4e-acc4-6f859230ef90","page_size":[595,841],"block_position":"99-1"},{"bbox":[157,114,437,147],"type":"interline_equation","angle":0,"index":3,"text":"$$\n\\left| \\sum_ {k = 1} ^ {n} \\cos k x \\right| = \\left| \\frac {\\sin \\left(n + \\frac {1}{2}\\right) x}{2 \\sin \\frac {x}{2}} - \\frac {1}{2} \\right| \\leq \\frac {1}{2 \\left| \\sin \\frac {x}{2} \\right|} + \\frac {1}{2} \\leq \\frac {1}{2 \\sin \\frac {\\alpha}{2}} + 1\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":99,"id":"f2781b75-53d2-4f79-b946-cb3efe87772b","page_size":[595,841],"block_position":"99-2"},{"bbox":[54,151,472,167],"type":"text","angle":0,"index":4,"merge_prev":false,"text":"因此 $\\sum \\cos nx$ 一致有界，而 $a_{n}$ 固定 $x$ 时单调趋于0，因此根据Dirichlet判别法可知一致收敛。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":99,"id":"54778f9f-1dcb-4ca3-ba7a-a1d633ed519f","page_size":[595,841],"block_position":"99-3"},{"bbox":[55,187,258,204],"type":"title","angle":0,"index":5,"level":2,"text":"## 9.4 函数项级数极限函数的性质","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":99,"id":"2d4a880c-6708-4428-b517-5bfcb215c319","page_size":[595,841],"block_position":"99-4"},{"bbox":[73,222,155,235],"type":"title","angle":0,"index":6,"level":2,"text":"## 定理 9.16 (端点法)","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":99,"id":"eed1256e-d38f-40cb-8446-e12fc91afcf5","page_size":[595,841],"block_position":"99-5"},{"bbox":[67,237,527,280],"type":"text","angle":0,"index":7,"merge_prev":false,"text":"$\\sum_{n=1}^{\\infty} u_n(x)$ 在 $(a, b)$ 一致收敛，且每个 $u_n(x)$ 在 $[a, b)$ 连续，则 $\\sum_{n=1}^{\\infty} u_n(x)$ 在 $[a, b)$ 一致收敛，特别的， $\\sum_{n=1}^{\\infty} u_n(a)$ 收敛 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":99,"id":"2e8a7ac6-7de7-4b51-9947-24ad368ab3e3","page_size":[595,841],"block_position":"99-6"},{"bbox":[35,291,47,307],"type":"image","angle":0,"index":8,"img_path":"/c161fa1ac200508a16f4bc9d4e777e4dbc7aa39cf920488e81a4031f73dfdf45.jpg","color":{"line":"rgba(89, 92, 220, 1)","fill":"rgba(89, 92, 220, 1)"},"img_ratio":0.2,"page_idx":99,"id":"608f16f0-71d4-4ecc-9aed-d45bb0c2e86a","page_size":[595,841],"block_position":"99-7"},{"bbox":[54,296,539,324],"type":"text","angle":0,"index":9,"merge_prev":false,"text":"笔记 开区间一致收敛的函数列往往能延拓到闭区间，因此题目中开区间的函数列往往是不一致收敛的 (除非无定义)。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":99,"id":"70d3cfb7-1f92-4e3e-86fb-e7c6d8609c0c","page_size":[595,841],"block_position":"99-8"},{"bbox":[73,333,186,345],"type":"title","angle":0,"index":10,"level":2,"text":"## 定理 9.17 (逐项单向极限)","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":99,"id":"269c649e-c47c-4c87-a6e2-7248a39c8290","page_size":[595,841],"block_position":"99-9"},{"bbox":[67,350,349,381],"type":"text","angle":0,"index":11,"merge_prev":false,"text":"$\\sum_{n=1}^{\\infty} u_n(x)$ 在 $(a, b)$ 一致收敛， $\\lim_{x \\to a^+} u_n(x) = a_n, \\sum_{n=1}^{\\infty} a_n$ 收敛，则 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":99,"id":"d1ee5d06-dbf2-4515-ad76-0ecac03c7116","page_size":[595,841],"block_position":"99-10"},{"bbox":[202,384,390,417],"type":"interline_equation","angle":0,"index":12,"text":"$$\n\\lim _ {x \\to a ^ {+}} \\sum_ {n = 1} ^ {\\infty} u _ {n} (x) = \\sum_ {n = 1} ^ {\\infty} \\lim _ {x \\to a ^ {+}} u _ {n} (x) = \\sum_ {n = 1} ^ {\\infty} a _ {n}\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":99,"id":"6442cf4a-cc69-4381-a7b8-af4e66425795","page_size":[595,841],"block_position":"99-11"},{"bbox":[73,437,155,449],"type":"title","angle":0,"index":13,"level":2,"text":"## 定理 9.18 (连续性)","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":99,"id":"8bb02139-982f-44df-a226-48779b43ee27","page_size":[595,841],"block_position":"99-12"},{"bbox":[68,454,512,486],"type":"text","angle":0,"index":14,"merge_prev":false,"text":"$\\sum_{n=1}^{\\infty} u_n(x)$ 在 $U(x_0, \\delta)$ 内一致收敛到 $S(x)$ ，且每个 $u_n(x)$ 均在 $x_0$ 处连续，则 $S(x)$ 也在 $x_0$ 处连续，即 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":99,"id":"019ace3f-d989-4563-b5fe-1811ff126a9a","page_size":[595,841],"block_position":"99-13"},{"bbox":[220,486,371,520],"type":"interline_equation","angle":0,"index":15,"text":"$$\n\\sum_ {n = 1} ^ {\\infty} (\\lim _ {x \\to x _ {0}} u _ {n} (x)) = \\lim _ {x \\to x _ {0}} \\sum_ {n = 1} ^ {\\infty} u _ {n} (x)\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":99,"id":"94bdeb30-140f-4985-ab9a-af43262c7573","page_size":[595,841],"block_position":"99-14"},{"bbox":[73,539,166,552],"type":"title","angle":0,"index":16,"level":2,"text":"## 定理 9.19 (逐项积分)","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":99,"id":"65f8c2a0-62fa-4d66-b3e2-ff53a08889e9","page_size":[595,841],"block_position":"99-15"},{"bbox":[69,559,487,589],"type":"text","angle":0,"index":17,"merge_prev":false,"text":"若 $\\sum_{n=1}^{\\infty} u_n(x)$ 在 $[a, b]$ 一致收敛到 $S(x)$ ，且每个 $u_n(x)$ 在 $[a, b]$ 可积，则 $S(x)$ 也在 $[a, b]$ 可积，且 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":99,"id":"7867c527-04a2-4392-b343-ed3392491bbd","page_size":[595,841],"block_position":"99-16"},{"bbox":[219,590,374,622],"type":"interline_equation","angle":0,"index":18,"text":"$$\n\\sum_ {n = 1} ^ {\\infty} \\int_ {a} ^ {b} u _ {n} (x) \\mathrm{d} x = \\int_ {a} ^ {b} \\sum_ {n = 1} ^ {\\infty} u _ {n} (x) \\mathrm{d} x\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":99,"id":"23bdbbef-6fcc-4a4a-b313-83be5c9f51d8","page_size":[595,841],"block_position":"99-17"},{"bbox":[74,643,166,655],"type":"title","angle":0,"index":19,"level":2,"text":"## 定理 9.20 (逐项可微)","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":99,"id":"0b15c320-5784-458b-a583-1ab49329f096","page_size":[595,841],"block_position":"99-18"},{"bbox":[67,660,525,723],"type":"text","angle":0,"index":20,"merge_prev":false,"text":"$\\sum_{n=1}^{\\infty} u_n(x)$ 的每个 $u_n(x)$ 在 $[a, b]$ 有连续偏导数， $\\sum_{n=1}^{\\infty} u_n'(x)$ 在 $[a, b]$ 一致收敛到 $g(x)$ ，若存在 $x_0 \\in [a, b]$ 使得 $\\sum_{n=1}^{\\infty} u_n(x_0)$ 收敛，则 $\\sum_{n=1}^{\\infty} u_n(x)$ 在 $[a, b]$ 上一致收敛，和函数 $S(x)$ 可导， $S'(x) = g(x)$ ，即 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":99,"id":"43647b37-e89d-4251-baaf-b2fd971b78be","page_size":[595,841],"block_position":"99-19"},{"bbox":[248,724,345,756],"type":"interline_equation","angle":0,"index":21,"text":"$$\n(\\sum_ {n = 1} ^ {\\infty} u _ {n} (x)) ^ {\\prime} = \\sum_ {n = 1} ^ {\\infty} u _ {n} ^ {\\prime}\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":99,"id":"9a6a5031-88dd-49e8-b5bf-5c97a31a6fef","page_size":[595,841],"block_position":"99-20"},{"bbox":[290,802,302,811],"type":"page_number","angle":0,"index":999999,"text":"92 ","color":{"line":"rgba(164, 164, 164, 1)","fill":"rgba(164, 164, 164, 1)"},"page_idx":99,"is_discarded":true,"id":"fc43d11d-8255-4752-81e1-02bf9e25a139","page_size":[595,841],"block_position":"99-21"}],[{"bbox":[240,74,364,94],"type":"title","angle":0,"index":0,"level":2,"text":"## 第 10 章 幂级数","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":100,"id":"38b9cbff-2a34-4e28-b4cb-00e05a958925","page_size":[595,841],"block_position":"100-0"},{"bbox":[57,122,251,138],"type":"title","angle":0,"index":1,"level":2,"text":"## 10.1 幂级数概念及其收敛区间","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":100,"id":"adbf1a20-13c6-41c7-b543-4469b4f1fb2d","page_size":[595,841],"block_position":"100-1"},{"bbox":[66,160,539,211],"type":"table_body","angle":0,"index":2,"img_path":"/","id":"85b9005a-15fd-431e-a98c-61a9d2913034","color":{"line":"rgba(103, 194, 63, 1)","fill":"rgba(103, 194, 63, 1)"},"table_body":"
定义10.1(幂级数)
\\sum_{n=0}^{\\infty}a_n(x-x_0)^n称为幂级数,但一般只研究\\sum_{n=0}^{\\infty}a_nx^n
","page_idx":100,"page_size":[595,841],"block_position":"100-2"},{"bbox":[57,238,204,252],"type":"title","angle":0,"index":3,"level":2,"text":"## 10.1.1 收敛区间与收敛半径","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":100,"id":"4dbfb013-dfa1-4e67-8f10-b41c2a39fd6f","page_size":[595,841],"block_position":"100-3"},{"bbox":[73,265,168,277],"type":"title","angle":0,"index":4,"level":2,"text":"## 定理 10.1 (Abel 定理)","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":100,"id":"8fe6f990-6623-473e-8f39-84a00aafe2c8","page_size":[595,841],"block_position":"100-4"},{"bbox":[67,280,527,327],"type":"text","angle":0,"index":5,"merge_prev":false,"text":"若幂级数 $\\sum_{n=1}^{\\infty} a_n x^n$ 在 $x_0 \\neq 0$ 收敛，则对 $\\forall x, |x| < |x_0|$ ， $\\sum_{n=1}^{\\infty} a_n x^n$ 绝对收敛；若 $x_0$ 处发散，则 $\\forall x, |x| > |x_0|$ 均发散。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":100,"id":"acce78f3-bd51-4ea1-8a86-4c12156a0144","page_size":[595,841],"block_position":"100-5"},{"bbox":[54,335,473,366],"type":"text","angle":0,"index":6,"merge_prev":false,"text":"证明 若 $\\sum_{n=0}^{\\infty} a_n x_0^n$ 收敛，则 $|a_n x^n| = |a_n x_0^n| \\cdot |\\frac{x}{x_0}|^n \\leq M r^n$ ，这里 $r = |\\frac{x}{x_0}| \\in [0,1)$ 。另一侧同理 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":100,"id":"c0f315fd-ec31-42eb-91bd-3df3da227648","page_size":[595,841],"block_position":"100-6"},{"bbox":[73,374,166,386],"type":"title","angle":0,"index":7,"level":2,"text":"## 定义 10.2 (收敛半径)","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":100,"id":"d6465524-3c92-4592-b1b0-c1b3db52702b","page_size":[595,841],"block_position":"100-7"},{"bbox":[68,391,524,420],"type":"text","angle":0,"index":8,"merge_prev":false,"text":"根据Abel定理可知 $\\exists R\\in \\mathbb{R}$ ，使得 $|x| < R$ 时幂级数绝对收敛(若 $R = 0$ 则仅在 $x = 0$ 处收敛)，对一切 $|x| > R$ 均发散，称 $R$ 为幂级数收敛半径， $(-R,R)$ 为收敛区间。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":100,"id":"3a352d70-a0fe-422b-8b2a-ca2f0c91fafd","page_size":[595,841],"block_position":"100-8"},{"bbox":[73,441,196,454],"type":"title","angle":0,"index":9,"level":2,"text":"## 定理 10.2 (收敛半径的计算)","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":100,"id":"b17a76c5-2638-4b37-be4d-fc4c0af3edf4","page_size":[595,841],"block_position":"100-9"},{"bbox":[69,458,317,470],"type":"text","angle":0,"index":10,"merge_prev":false,"text":"幂级数收敛半径 $R$ (可以为 $0, \\infty$ ) 的计算有以下几种方式： ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":100,"id":"35cc1661-8043-47c9-b2bb-b6ebf3422c67","page_size":[595,841],"block_position":"100-10"},{"bbox":[83,470,321,490],"type":"text","angle":0,"index":11,"merge_prev":false,"text":"- 根式 (柯西-阿达马): $\\lim_{n\\to \\infty}\\sqrt[n]{|a_n|} = \\frac{1}{R}$ (可以为 $0,\\infty$ ) ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":100,"id":"7dea7b9f-be21-40ab-86cd-507d7a17ecc7","page_size":[595,841],"block_position":"100-11"},{"bbox":[83,490,391,516],"type":"text","angle":0,"index":12,"merge_prev":false,"text":"- 比式 (一般用得少): 若 $\\lim_{n\\to \\infty}\\frac{|a_{n + 1}|}{|a_n|} = \\frac{1}{R}$ , 也能推出 $\\lim_{n\\to \\infty}\\sqrt[n]{a_n} = \\frac{1}{R}$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":100,"id":"7f152fdc-9c89-433f-9f9c-8c188d88eec1","page_size":[595,841],"block_position":"100-12"},{"bbox":[53,526,537,581],"type":"text","angle":0,"index":13,"merge_prev":false,"text":"证明 (1) 根式：只需要考虑 $\\sum_{n=0}^{\\infty}|a_n x^n|$ ，用数项级数的根式判别法为 $\\varlimsup_{n \\to \\infty} \\sqrt[n]{|a_n x^n|} = \\varliminf_{n \\to \\infty} \\sqrt[n]{|a_n|} \\cdot |x|$ ，因此若记 $\\varlimsup_{n \\to \\infty} \\sqrt[n]{|a_n|} = \\rho$ ，则 $|x| < \\frac{1}{\\rho}$ 则收敛， $\\frac{1}{\\rho}$ 为收敛半径。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":100,"id":"b3ae18fb-1147-4a5e-a0b9-b8f02fee5925","page_size":[595,841],"block_position":"100-13"},{"bbox":[36,585,437,618],"type":"text","angle":0,"index":14,"merge_prev":false,"text":"练习10.1计算收敛半径 计算收敛半径与收敛域(1) $\\sum_{n = 1}^{\\infty}\\frac{x^n}{n^2}$ (2) $\\sum_{n = 1}^{\\infty}\\frac{x^n}{n}$ (3) $\\sum_{n = 1}^{\\infty}\\frac{(2n - 1)!!}{(2n)!!} x^n$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":100,"id":"c08c280c-9ac2-4160-85ca-035c9e175d19","page_size":[595,841],"block_position":"100-14"},{"bbox":[53,615,275,643],"type":"text","angle":0,"index":15,"merge_prev":false,"text":"解 (1) $\\lim_{n\\to \\infty}\\sqrt[n]{\\frac{1}{n^2}} = 1$ ，收敛半径1，收敛域[-1,1] ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":100,"id":"6ae5c165-81ee-4d3c-8d1b-0be5959e233d","page_size":[595,841],"block_position":"100-15"},{"bbox":[73,644,278,667],"type":"text","angle":0,"index":16,"merge_prev":false,"text":"(2) $\\lim_{n\\to \\infty}\\sqrt[n]{\\frac{1}{n}} = 1$ ，收敛半径1，收敛域[-1,1) ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":100,"id":"6aba6cad-3837-4d15-83de-2d7c287a243e","page_size":[595,841],"block_position":"100-16"},{"bbox":[55,665,545,721],"type":"text","angle":0,"index":17,"merge_prev":false,"text":"(3) $a_{n}=\\frac{(2n-1)!!}{(2n)!!}$ ，用Wallis+根式或者比式 $\\lim_{n\\to\\infty}\\frac{a_{n+1}}{a_{n}}=\\lim_{n\\to\\infty}\\frac{2n+1}{2n+2}=1$ 得到收敛半径为1。而 $\\frac{(2n-1)!!}{(2n)!!}\\sim\\frac{1}{\\sqrt{n\\pi}}$ ，因此发散。 $\\sum_{n=1}^{\\infty}\\frac{(2n-1)!!}{(2n)!!}(-1)^{n}$ 根据Leibniz可知收敛。故收敛域为 $[-1,1)$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":100,"id":"1c6a62ba-e3f4-45f6-88ca-881c32f47c5c","page_size":[595,841],"block_position":"100-17"},{"bbox":[52,725,339,738],"type":"text","angle":0,"index":18,"merge_prev":false,"text":"笔记 上述题目也说明在逐项求导后，端点处收敛域可能发生变换 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":100,"id":"b1af84a0-ae47-4fb9-813b-1c06b75da90d","page_size":[595,841],"block_position":"100-18"}],[{"bbox":[405,47,539,58],"type":"header","angle":0,"index":-99999,"text":"10.1 幂级数概念及其收敛区间 ","color":{"line":"rgba(164, 164, 164, 1)","fill":"rgba(164, 164, 164, 1)"},"page_idx":101,"is_discarded":true,"id":"39934723-2edf-4d6c-b532-3ccd11541d53","page_size":[595,841],"block_position":"101-0"},{"bbox":[36,69,428,100],"type":"text","angle":0,"index":1,"merge_prev":false,"text":"练习 10.2 常规格式幂级数 (1) 求收敛域: $\\sum_{n=1}^{\\infty}(1+\\frac{1}{2}+\\cdots+\\frac{1}{n})x^{n}\\sum_{n=1}^{\\infty}\\frac{x^{n}}{1+\\frac{1}{2}+\\cdots+\\frac{1}{n}}$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":101,"id":"fd28177a-2023-47af-8861-7754071fbe21","page_size":[595,841],"block_position":"101-1"},{"bbox":[73,100,308,131],"type":"text","angle":0,"index":2,"merge_prev":false,"text":"(2) $0 < a < b$ ，求 $\\sum_{n=1}^{\\infty} \\frac{x^n}{a^n + b^n}, \\sum_{n=1}^{\\infty} (a^n + b^n)x^n$ 收敛域 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":101,"id":"7b713b89-2594-4b5f-9176-b10cea434b82","page_size":[595,841],"block_position":"101-2"},{"bbox":[73,130,212,163],"type":"text","angle":0,"index":3,"merge_prev":false,"text":"(3) $\\sum_{n=1}^{\\infty} \\frac{x^{2n}}{n-3^{2n}}, \\sum_{n=1}^{\\infty} \\frac{x^{n^2}}{2^n}$ 收敛域 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":101,"id":"6f96df0a-955c-458f-aeb7-790363013880","page_size":[595,841],"block_position":"101-3"},{"bbox":[73,162,193,190],"type":"text","angle":0,"index":4,"merge_prev":false,"text":"(4) $\\sum_{n=1}^{\\infty}\\frac{3^{n}+(-2)^{n}}{n}(x+1)^{n}$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":101,"id":"c9834b2e-96e6-437a-96af-b29c8ecd83ba","page_size":[595,841],"block_position":"101-4"},{"bbox":[53,190,539,227],"type":"text","angle":0,"index":5,"merge_prev":false,"text":"解 (1) 要么根据 Euler 常数得到 $\\sqrt[n]{1+\\frac{1}{2}+\\cdots+\\frac{1}{n}}\\sim\\sqrt[n]{\\ln n}=1$ ，要么用 $1\\leq1+\\frac{1}{2}+\\cdots+\\frac{1}{n}\\leq n$ 得到收敛半径为 R=1，显然收敛域为 $(-1,1)$ 第二个同理得到收敛域为 $[-1,1]$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":101,"id":"1a8985d7-18a4-495e-a01c-89d4ec8e8800","page_size":[595,841],"block_position":"101-5"},{"bbox":[53,230,539,264],"type":"text","angle":0,"index":6,"merge_prev":false,"text":"(2) 第一个 $\\lim_{n\\to \\infty}\\sqrt[n]{\\frac{1}{a^n + b^n}} = \\frac{1}{b}$ ，收敛半径为 $b$ ， $x = \\pm b$ 都震荡发散。第二个 $\\lim_{n\\to \\infty}\\sqrt[n]{a^n + b^n} = b$ ，收敛半径 $b$ ， $x = \\pm b$ 震荡发散 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":101,"id":"3c5cd419-1210-4103-98fd-469586d169eb","page_size":[595,841],"block_position":"101-6"},{"bbox":[54,265,540,315],"type":"text","angle":0,"index":7,"merge_prev":false,"text":"(3) 由于不是 $x^n$ 因此开 $2n$ 次根号， $\\lim_{n \\to \\infty} \\sqrt[2n]{|\\frac{1}{n - 3^{2n}}|} = \\frac{1}{3}$ ，半径为 $3$ ， $x = \\pm 3$ 时通项不趋于 0，因此发散。 $\\lim_{n \\to \\infty} \\sqrt[n^2]{\\frac{1}{2^n}} = 1$ ，而 $x = \\pm 1$ 均收敛 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":101,"id":"89f958eb-a602-4669-a1c1-183ff2eaca35","page_size":[595,841],"block_position":"101-7"},{"bbox":[52,314,539,398],"type":"text","angle":0,"index":8,"merge_prev":false,"text":"(4) 将 $x + 1$ 视为 $y$ 即可， $\\lim_{n \\to \\infty} \\sqrt[n]{\\frac{3^n + (-2)^n}{n}} = 3$ ，收敛半径为 $\\frac{1}{3}$ ， $y = \\frac{1}{3}$ 时为 $\\sum_{n=1}^{\\infty} \\left(\\frac{1}{n} + \\frac{1}{n} \\left(-\\frac{2}{3}\\right)^n\\right)$ 前者发散，后者收敛，因此整体发散。 $y = -\\frac{1}{3}$ 时为 $\\sum_{n=1}^{\\infty} \\left(\\frac{(-1)^n}{n} + \\frac{1}{n} \\left(\\frac{2}{3}\\right)^n\\right)$ 收敛，因此收敛域为 $[- \\frac{1}{3} - 1, \\frac{1}{3} - 1)$ 即 $x \\in [-\\frac{4}{3}, -\\frac{2}{3})$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":101,"id":"96f310e9-6648-4a2e-ae12-8d6e6fc42d5e","page_size":[595,841],"block_position":"101-8"},{"bbox":[36,403,581,476],"type":"text","angle":0,"index":9,"merge_prev":false,"text":"练习10.3非常规格式幂级数(1) $\\sum_{n=1}^{\\infty}\\frac{(x^{2}+x+1)^{n}}{3n}$ 的收敛域(2) $\\sum_{n=1}^{\\infty}\\frac{1}{2^{n}}\\left(\\frac{x-1}{2x+1}\\right)^{n^{2}}$ (3) $\\sum_{n=1}^{\\infty}\\frac{1^{n}+2^{n}+\\cdots+2022^{n}}{n^{2022}}\\left(\\frac{1-x}{1+x}\\right)^{n}$ 解(1)设 $y=x^{2}+x+1$ ，考虑 $\\lim_{n\\to\\infty}\\sqrt[n]{\\frac{1}{3n}}=1$ ，R=1，而y=1发散，y=-1收敛，计算 $x^{2}+x+1\\in[-1,1)$ ，即 $x^{2}+x\\in[-2,0)$ ，注意最终得到 $x\\in(-1,0)$ 两侧都是开的！ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":101,"id":"6369b31d-a6d9-41ac-b828-8a6d1ab57d85","page_size":[595,841],"block_position":"101-9"},{"bbox":[52,475,540,607],"type":"text","angle":0,"index":11,"merge_prev":false,"text":"(2) 考虑 $y = \\frac{x - 1}{2x + 1}$ ，计算得到 $y$ 的收敛域为 $[-1, 1]$ ，计算 $-1 \\leq \\frac{x - 1}{2x + 1} \\leq 1$ ，直接讨论分母容易出错，因此这里看成两个不等式。 $\\frac{x - 1}{2x + 1} + 1 = \\frac{3x}{2x + 1} \\geq 0$ 以及 $\\frac{x - 1}{2x + 1} - 1 = \\frac{-x - 2}{2x + 1} \\leq 0$ ，第一个要求 $x \\geq 0$ 或者 $x < -\\frac{1}{2}$ ，第二个要求 $x \\leq -2$ 或者 $x \\geq -\\frac{1}{2}$ ，求交得到 $x \\in (-\\infty, -2] \\cup [0, +\\infty)$ (3) $y = \\frac{1 - x}{1 + x}$ 得到收敛半径为 $\\frac{1}{2022}$ 且端点收敛，考虑 $-\\frac{1}{2022} \\leq \\frac{1 - x}{1 + x} \\leq \\frac{1}{2022}$ ，分为两部分，得到 $\\frac{2023 - 2021x}{1 + x} \\geq 0$ 以及 $\\frac{2021 - 2023x}{1 + x} \\leq 0$ ，前者解出 $-1 < x \\leq \\frac{2023}{2021}$ ，后者解出 $x < -1$ 或者 $x \\geq \\frac{2021}{2023}$ ，得出结果为 $x \\in [\\frac{2021}{2023}, \\frac{2023}{2021}]$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":101,"id":"74921200-f338-4e14-b031-9bf91466c513","page_size":[595,841],"block_position":"101-10"},{"bbox":[57,622,204,636],"type":"title","angle":0,"index":12,"level":2,"text":"## 10.1.2 幂级数的一致收敛性","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":101,"id":"3dc0ed0f-96b5-4791-b206-7a767e608711","page_size":[595,841],"block_position":"101-11"},{"bbox":[74,644,539,677],"type":"text","angle":0,"index":13,"merge_prev":false,"text":"下面考虑幂级数 $\\sum_{n=1}^{\\infty} a_n x^n$ 的一致收敛性，假设其收敛半径为 $R$ ，收敛域在端点处需要讨论，下面开始分析。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":101,"id":"5def3082-eae8-4e8c-9c71-f714fd50c161","page_size":[595,841],"block_position":"101-12"},{"bbox":[73,686,188,698],"type":"title","angle":0,"index":14,"level":2,"text":"## 定理 10.3 (Abel 第二定理)","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":101,"id":"fc54e68d-682d-46b6-80a1-d663511c7fee","page_size":[595,841],"block_position":"101-13"},{"bbox":[69,703,392,736],"type":"text","angle":0,"index":15,"merge_prev":false,"text":"若幂级数收敛半径为 $R$ ， $\\sum_{n = 1}^{\\infty}a_nx^n$ 在 $\\langle -R,R\\rangle$ 内闭一致收敛(端点未确定) ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":101,"id":"c83fd2d1-a5f2-4a1f-bfde-3da1013621fa","page_size":[595,841],"block_position":"101-14"},{"bbox":[291,803,303,811],"type":"page_number","angle":0,"index":1000009,"text":"94 ","color":{"line":"rgba(164, 164, 164, 1)","fill":"rgba(164, 164, 164, 1)"},"page_idx":101,"is_discarded":true,"id":"029d2bb0-ba67-4abd-96c5-7960686a5929","page_size":[595,841],"block_position":"101-15"}],[{"bbox":[436,47,539,58],"type":"header","angle":0,"index":-99999,"text":"10.2 幂级数展开式求和 ","color":{"line":"rgba(164, 164, 164, 1)","fill":"rgba(164, 164, 164, 1)"},"page_idx":102,"is_discarded":true,"id":"9158945d-53a7-4045-b8ef-948db38ea917","page_size":[595,841],"block_position":"102-0"},{"bbox":[53,68,539,132],"type":"text","angle":0,"index":1,"merge_prev":false,"text":"证明 设 $\\sum_{n=0}^{\\infty} a_n x_0^n$ 收敛，(1) 若 $x_0 > 0$ ，则 $\\forall x \\in [0, x_0]$ 有 $\\sum_{n=0}^{\\infty} a_n x^n = \\sum_{n=0}^{\\infty} a_n x_0^n \\left(\\frac{x}{x_0}\\right)^n$ ，根据 Abel 判别法可证。(2) 若 $x_0 < 0$ ，则 $\\forall x \\in [x_0, 0]$ 有 $\\sum_{n=0}^{\\infty} a_n x^n = \\sum_{n=0}^{\\infty} a_n x_0^n \\left(\\frac{x}{x_0}\\right)^n$ ，根据 Abel 可证。进而任选闭区间都一致收敛。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":102,"id":"99cbbdd9-bdae-4faf-b552-56ddde74daf7","page_size":[595,841],"block_position":"102-1"},{"bbox":[73,140,155,152],"type":"title","angle":0,"index":2,"level":2,"text":"## 定理 10.4 (连续性)","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":102,"id":"1eec1701-521e-47be-a311-11e0eaa7db8a","page_size":[595,841],"block_position":"102-2"},{"bbox":[69,158,309,189],"type":"text","angle":0,"index":3,"merge_prev":false,"text":"幂级数的和函数 $\\sum_{n=0}^{\\infty} a_n x^n$ 在收敛域 $\\langle -R, R \\rangle$ 是连续的。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":102,"id":"9ea90e1e-60c1-4b08-8f95-accd587be094","page_size":[595,841],"block_position":"102-3"},{"bbox":[73,210,176,222],"type":"title","angle":0,"index":4,"level":2,"text":"## 定理 10.5 (区间不变性)","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":102,"id":"d9be8dae-29a3-4c61-a7c5-4e87d47809f2","page_size":[595,841],"block_position":"102-4"},{"bbox":[68,227,526,269],"type":"text","angle":0,"index":5,"merge_prev":false,"text":"幂级数 $\\sum_{n=0}^{\\infty} a_n x^n$ 的收敛半径为 $R$ ，则其逐项求导和逐项积分后得到的幂级数的收敛半径仍为 $R$ ，不过端点处的一致收敛情况可能改变！ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":102,"id":"546a6d83-b57e-456a-aad9-1d30ab236728","page_size":[595,841],"block_position":"102-5"},{"bbox":[53,277,543,334],"type":"text","angle":0,"index":6,"merge_prev":false,"text":"证明 考虑 $\\sum_{n=0}^{\\infty} a_n x^n$ 的逐项求导与逐项积分, 得到 $\\sum_{n=1}^{\\infty} na_n$ 和 $\\sum_{n=0}^{\\infty} \\frac{a_n}{n+1} x^{n+1}$ , 根据 $\\varlimsup_{n \\to \\infty} \\sqrt[n]{|a_n|} = \\varliminf_{n \\to \\infty} \\sqrt[n-1]{|na_n|} = \\varlimsup_{n \\to \\infty} \\sqrt[n-1]{|\\frac{a_n}{n+1}|}$ 可知收敛半径相同 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":102,"id":"51575e13-0d26-4550-8cad-2188904bcb3a","page_size":[595,841],"block_position":"102-6"},{"bbox":[52,333,460,367],"type":"text","angle":0,"index":7,"merge_prev":false,"text":"笔记 端点处一致收敛性不保证，例如 $\\sum_{n=1}^{\\infty} \\frac{x^n}{n}$ 收敛域为 $[-1, 1)$ ，而 $\\sum_{n=1}^{\\infty} x^{n-1}$ 收敛域为 $(-1, 1)$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":102,"id":"bb0321ea-a452-47f4-b27e-e970240293a3","page_size":[595,841],"block_position":"102-7"},{"bbox":[73,375,176,386],"type":"title","angle":0,"index":8,"level":2,"text":"## 定理 10.6 (求导与求积)","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":102,"id":"d8f63946-26c4-4828-9fe6-abf9036a116e","page_size":[595,841],"block_position":"102-8"},{"bbox":[69,391,314,423],"type":"text","angle":0,"index":10,"merge_prev":false,"text":"设 $(-R, R)$ 上 $f(x) = \\sum_{n=0}^{\\infty} a_n x^n$ ，则对 $\\forall x \\in (-R, R)$ 有： ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":102,"id":"1f26ca44-a981-4dcf-b747-9e1a696a510e","page_size":[595,841],"block_position":"102-9"},{"bbox":[186,424,404,456],"type":"interline_equation","angle":0,"index":11,"text":"$$\nf ^ {\\prime} (x) = \\sum_ {n = 1} ^ {\\infty} n a _ {n} x ^ {n - 1}, \\int_ {0} ^ {x} f (t) \\mathrm{d} t = \\sum_ {n = 0} ^ {\\infty} \\frac {a _ {n}}{n + 1} x ^ {n + 1}\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":102,"id":"4e3fcc44-bbb9-4090-9593-c4a4122d84c4","page_size":[595,841],"block_position":"102-10"},{"bbox":[52,473,483,486],"type":"text","angle":0,"index":12,"merge_prev":false,"text":"笔记 幂级数的逐项求导和逐项积分要注意其只能保证在收敛区间成立，但不保证在收敛域上成立！ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":102,"id":"dd59a6be-3f38-4417-a550-1471ad5fc7dd","page_size":[595,841],"block_position":"102-11"},{"bbox":[57,505,217,518],"type":"title","angle":0,"index":13,"level":2,"text":"## 10.1.3 缺项幂级数的收敛区域","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":102,"id":"8176e78b-19a7-4866-b6a0-73d013058910","page_size":[595,841],"block_position":"102-12"},{"bbox":[35,528,330,559],"type":"text","angle":0,"index":14,"merge_prev":false,"text":"练习10.4缺项幂级数的收敛范围(1)求 $\\sum_{n = 1}^{\\infty}n^{n^2}x^{n^3}$ 的收敛范围 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":102,"id":"fa8d40c7-e5ba-48cf-bae1-8aea77682321","page_size":[595,841],"block_position":"102-13"},{"bbox":[52,560,322,595],"type":"text","angle":0,"index":15,"merge_prev":false,"text":"解令 $a_{k} = \\left\\{ \\begin{array}{ll}n^{n^{2}}, & k = n^{3}\\\\ 0, & k\\neq n^{3} \\end{array} \\right.$ ，则 $\\sum_{n = 1}^{\\infty}n^{n^2}x^{n^3} = \\sum_{k = 1}^{\\infty}a_kx^k$ ，此时 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":102,"id":"f242a3c4-6197-4fd6-8f7b-51788e3b7bfa","page_size":[595,841],"block_position":"102-14"},{"bbox":[183,599,409,624],"type":"interline_equation","angle":0,"index":16,"text":"$$\n\\frac {1}{R} = \\varlimsup_ {k \\to \\infty} \\sqrt [ k ]{| a _ {k} |} = \\lim _ {n \\to \\infty} \\sqrt [ n ^ {3} ]{n ^ {n ^ {2}}} = \\lim _ {n \\to \\infty} (n ^ {n ^ {2}}) ^ {\\frac {1}{n ^ {3}}} = 1\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":102,"id":"93ffde8a-7367-42ef-8e58-596bdc86592b","page_size":[595,841],"block_position":"102-15"},{"bbox":[57,641,207,657],"type":"title","angle":0,"index":17,"level":2,"text":"## 10.2 幂级数展开式求和","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":102,"id":"cd077d87-a12b-4902-a855-d11ab259cc48","page_size":[595,841],"block_position":"102-16"},{"bbox":[73,675,196,687],"type":"title","angle":0,"index":18,"level":2,"text":"## 定理 10.7 (必备幂级数展开)","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":102,"id":"9064f8bf-ad0b-4d9a-a554-7d768881fab7","page_size":[595,841],"block_position":"102-17"},{"bbox":[69,692,381,704],"type":"text","angle":0,"index":19,"merge_prev":false,"text":"以下的幂级数展开会经常使用，基本上是分母较为简单的 Taylor 展开式 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":102,"id":"084d498c-801a-4f98-b2cf-fde99088d207","page_size":[595,841],"block_position":"102-18"},{"bbox":[81,704,164,735],"type":"text","angle":0,"index":20,"merge_prev":false,"text":"1. $\\frac{1}{1 - x} = \\sum_{n=0}^{\\infty} x^n$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":102,"id":"2882b210-9856-4d20-b937-ebd6b23d8a0e","page_size":[595,841],"block_position":"102-19"},{"bbox":[291,803,302,811],"type":"page_number","angle":0,"index":1000008,"text":"95 ","color":{"line":"rgba(164, 164, 164, 1)","fill":"rgba(164, 164, 164, 1)"},"page_idx":102,"is_discarded":true,"id":"a824e63d-f754-42ae-82e6-0bbf51c2980e","page_size":[595,841],"block_position":"102-20"}],[{"bbox":[436,47,539,58],"type":"header","angle":0,"index":-99999,"text":"10.2 幂级数展开式求和 ","color":{"line":"rgba(164, 164, 164, 1)","fill":"rgba(164, 164, 164, 1)"},"page_idx":103,"is_discarded":true,"id":"d5eb2496-fdfe-4746-b3d3-893c36b909c5","page_size":[595,841],"block_position":"103-0"},{"bbox":[81,81,150,112],"type":"text","angle":0,"index":1,"merge_prev":false,"text":"2. $e^x = \\sum_{n=0}^{\\infty} \\frac{x^n}{n!}$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":103,"id":"468f8124-598e-4e80-9b63-bc36ed05e6a9","page_size":[595,841],"block_position":"103-1"},{"bbox":[81,112,213,141],"type":"text","angle":0,"index":2,"merge_prev":false,"text":"3. $\\ln (1 + x) = \\sum_{n = 1}^{\\infty}(-1)^{n - 1}\\frac{x^n}{n}$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":103,"id":"37db7115-b9d6-4e98-af8a-3e7281fc8bc9","page_size":[595,841],"block_position":"103-2"},{"bbox":[80,142,214,173],"type":"text","angle":0,"index":3,"merge_prev":false,"text":"4. $\\arctan x = \\sum_{n=0}^{\\infty} \\frac{(-1)^n x^{2n+1}}{2n+1}$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":103,"id":"073b2f7a-7b58-4718-9758-46c9a9145024","page_size":[595,841],"block_position":"103-3"},{"bbox":[35,185,47,200],"type":"image","angle":0,"index":4,"img_path":"/f2fe059e28c6b6c8debade1d0b18e4ac23151c56b894b7eeec42bd44c39050ed.jpg","color":{"line":"rgba(89, 92, 220, 1)","fill":"rgba(89, 92, 220, 1)"},"img_ratio":0.2,"page_idx":103,"id":"391ace66-96b3-4cba-a444-fee3191c5fae","page_size":[595,841],"block_position":"103-4"},{"bbox":[52,190,289,202],"type":"text","angle":0,"index":5,"merge_prev":false,"text":"笔记 具体用哪个主要看 $(-1)$ 的指数次和分母的关系。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":103,"id":"ae3d687a-44b6-4e91-ba64-4472caca6aed","page_size":[595,841],"block_position":"103-5"},{"bbox":[57,221,145,234],"type":"title","angle":0,"index":6,"level":2,"text":"## 10.2.1 等比级数","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":103,"id":"3ab483b0-431e-4191-b06a-52c108565be5","page_size":[595,841],"block_position":"103-6"},{"bbox":[73,248,166,260],"type":"title","angle":0,"index":7,"level":2,"text":"## 定理 10.8 (等比级数)","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":103,"id":"16ed780e-950b-4755-8714-3e59bd607d7a","page_size":[595,841],"block_position":"103-7"},{"bbox":[69,265,240,296],"type":"text","angle":0,"index":8,"merge_prev":false,"text":"根据 $f(x) = \\sum_{n=0}^{\\infty} x^n = \\frac{1}{1 - x}$ 可以推出： ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":103,"id":"e39931ef-52ad-4a2c-a652-bf26ba1b3c47","page_size":[595,841],"block_position":"103-8"},{"bbox":[81,295,333,327],"type":"text","angle":0,"index":9,"merge_prev":false,"text":"1. $\\sum_{n=1}^{\\infty} nx^{n} = \\frac{x}{(1 - x)^{2}}$ ，方法为对 $\\sum_{n=0}^{\\infty} x^{n}$ 求导，再同乘 $x$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":103,"id":"9c85297e-30a3-407a-80da-3f6c37973193","page_size":[595,841],"block_position":"103-9"},{"bbox":[80,326,334,357],"type":"text","angle":0,"index":10,"merge_prev":false,"text":"2. $\\sum_{n=1}^{\\infty} n^{2} x^{n} = \\frac{x(1 + x)}{(1 - x)^{3}}$ ，方法为对 $\\sum_{n=1}^{\\infty} nx^{n}$ 求导，再乘 $x$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":103,"id":"ce8cfcd6-a795-4203-b94c-bcd32147af0c","page_size":[595,841],"block_position":"103-10"},{"bbox":[36,367,278,398],"type":"text","angle":0,"index":11,"merge_prev":false,"text":"练习10.5等比级数 (1) $\\sum_{n=1}^{\\infty} \\frac{n^2 + 3n}{2^n}$ (2) $\\sum_{n=1}^{\\infty} \\frac{x^n}{n(n+1)}$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":103,"id":"0b412c43-802e-4efc-afc7-6286f4ea818e","page_size":[595,841],"block_position":"103-11"},{"bbox":[53,396,218,409],"type":"text","angle":0,"index":12,"merge_prev":false,"text":"解(1)这里展示正推, 进行公式推导: ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":103,"id":"53ca331e-be59-4f9d-a127-6c741631244b","page_size":[595,841],"block_position":"103-12"},{"bbox":[150,412,432,501],"type":"interline_equation","angle":0,"index":13,"text":"$$\n\\left\\{ \\begin{array}{l} f _ {1} (x) = \\sum_ {n = 0} ^ {\\infty} x ^ {n} = \\frac {1}{1 - x} \\Rightarrow f _ {1} ^ {\\prime} (x) = \\sum_ {n = 1} ^ {\\infty} n x ^ {n - 1} = \\frac {1}{(1 - x) ^ {2}} \\\\ f _ {2} (x) = \\sum_ {n = 1} ^ {\\infty} n x ^ {n} = \\frac {x}{(1 - x) ^ {2}} \\Rightarrow f _ {2} ^ {\\prime} (x) = \\sum_ {n = 1} ^ {\\infty} n ^ {2} x ^ {n - 1} = \\frac {1 + x}{(1 - x) ^ {3}} \\\\ f _ {3} (x) = \\sum_ {n = 1} ^ {\\infty} n ^ {2} x ^ {n} = \\frac {x (1 + x)}{(1 - x) ^ {3}} \\end{array} \\right.\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":103,"id":"6afeead7-c86a-4102-b8fd-83281a5f13b6","page_size":[595,841],"block_position":"103-13"},{"bbox":[55,503,224,523],"type":"text","angle":0,"index":14,"merge_prev":false,"text":"因此 $I = f_{3}\\left(\\frac{1}{2}\\right) + 3f_{2}\\left(\\frac{1}{2}\\right) = 6 + 6 = 12$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":103,"id":"d0fd8fcd-7260-4e9e-8212-564f65547611","page_size":[595,841],"block_position":"103-14"},{"bbox":[73,524,155,536],"type":"text","angle":0,"index":15,"merge_prev":false,"text":"(2) 这里展示反推: ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":103,"id":"a176417b-951f-437f-99ad-273bd088377c","page_size":[595,841],"block_position":"103-15"},{"bbox":[113,540,468,601],"type":"interline_equation","angle":0,"index":16,"text":"$$\n\\left\\{ \\begin{array}{l} f _ {1} (x) = \\sum_ {n = 1} ^ {\\infty} \\frac {x ^ {n}}{n (n + 1)} \\\\ f _ {2} (x) = \\sum_ {n = 1} ^ {\\infty} \\frac {x ^ {n + 1}}{n (n + 1)} \\Rightarrow f _ {2} ^ {\\prime} (x) = \\sum_ {n = 1} ^ {\\infty} \\frac {x ^ {n}}{n} \\Rightarrow f _ {2} ^ {\\prime \\prime} (x) = \\sum_ {n = 1} ^ {\\infty} x ^ {n - 1} = \\sum_ {n = 0} ^ {\\infty} x ^ {n} = \\frac {1}{1 - x} \\end{array} \\right.\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":103,"id":"fc7995bb-8fc4-4d76-a148-a483f4484902","page_size":[595,841],"block_position":"103-16"},{"bbox":[54,603,465,626],"type":"text","angle":0,"index":17,"merge_prev":false,"text":"因此推出 $f_{2}^{\\prime}(x) = -\\ln (1 - x),f_{2}(x) = (1 - x)\\ln (1 - x) + x$ ，因此 $f_{1}(x) = \\frac{1 - x}{x}\\ln (1 - x) + 1$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":103,"id":"8707b236-5b82-415a-b24f-ba5e2e925918","page_size":[595,841],"block_position":"103-17"},{"bbox":[57,641,165,654],"type":"title","angle":0,"index":18,"level":2,"text":"## 10.2.2 $\\arctan x$ 级数","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":103,"id":"899ee5e4-1630-469a-9a99-21a0d0ee7692","page_size":[595,841],"block_position":"103-18"},{"bbox":[35,663,253,695],"type":"text","angle":0,"index":20,"merge_prev":false,"text":"练习 10.6arctan x (1) 重点：求 $\\sum_{n=1}^{\\infty}\\frac{(-1)^{n-1}}{n(2n+1)}$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":103,"id":"a0e3ab5d-e698-4093-a1db-a2bbfee69e0c","page_size":[595,841],"block_position":"103-19"},{"bbox":[291,803,303,812],"type":"page_number","angle":0,"index":1000018,"text":"96 ","color":{"line":"rgba(164, 164, 164, 1)","fill":"rgba(164, 164, 164, 1)"},"page_idx":103,"is_discarded":true,"id":"700cdf72-eceb-4cf5-ae3a-955c61106953","page_size":[595,841],"block_position":"103-20"}],[{"bbox":[436,47,539,58],"type":"header","angle":0,"index":-99999,"text":"10.3 幂级数求和式展开 ","color":{"line":"rgba(164, 164, 164, 1)","fill":"rgba(164, 164, 164, 1)"},"page_idx":104,"is_discarded":true,"id":"59157455-8804-4a9b-98e2-481aa286bab4","page_size":[595,841],"block_position":"104-0"},{"bbox":[53,68,465,101],"type":"text","angle":0,"index":1,"merge_prev":false,"text":"解(1)相差较大，选择裂项： $\\sum_{n = 1}^{\\infty}\\frac{(-1)^{n - 1}}{n(2n + 1)} = 2\\sum_{n = 1}^{\\infty}\\left(\\frac{(-1)^{n - 1}}{2n} -\\frac{(-1)^{n - 1}}{2n + 1}\\right)$ ，两者均收敛，且 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":104,"id":"2d237526-7ea6-44b0-81a0-e9a75ca5f545","page_size":[595,841],"block_position":"104-1"},{"bbox":[226,104,356,163],"type":"interline_equation","angle":0,"index":2,"text":"$$\n\\left\\{ \\begin{array}{l} \\ln (1 + x) = \\sum_ {n = 1} ^ {\\infty} \\frac {(- 1) ^ {n - 1} x ^ {n}}{n} \\\\ \\arctan x = \\sum_ {n = 0} ^ {\\infty} \\frac {(- 1) ^ {n} x ^ {2 n + 1}}{2 n + 1} \\end{array} \\right.\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":104,"id":"f5559177-6424-4ed0-826e-59ebc419f18d","page_size":[595,841],"block_position":"104-2"},{"bbox":[54,165,163,188],"type":"text","angle":0,"index":3,"merge_prev":false,"text":"因此结果为 $\\ln 2 + \\frac{\\pi}{2} - 2$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":104,"id":"1fe0aa80-c3a6-457f-aecb-71c275429fa9","page_size":[595,841],"block_position":"104-3"},{"bbox":[57,203,134,216],"type":"title","angle":0,"index":4,"level":2,"text":"## 10.2.3 $e^x$ 级数","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":104,"id":"9b81f7c5-83c7-4b11-9086-6ba0873036d6","page_size":[595,841],"block_position":"104-4"},{"bbox":[36,225,289,251],"type":"text","angle":0,"index":5,"merge_prev":false,"text":"练习 $10.7e^{x}$ (1) 计算 $\\lim_{n\\to\\infty}\\left[1+\\frac{1}{2!}+\\frac{1}{4!}+\\cdots+\\frac{1}{(2n)!}\\right]$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":104,"id":"d40cfa11-7926-47d6-b634-ac06475eebdf","page_size":[595,841],"block_position":"104-5"},{"bbox":[53,252,296,282],"type":"text","angle":0,"index":6,"merge_prev":false,"text":"解 (1) 由于 $e^x = \\sum_{n=0}^{\\infty} \\frac{x^n}{n!}$ ，因此 $e^{-x} = \\sum_{n=0}^{\\infty} \\frac{(-1)^n x^n}{n!}$ ，故 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":104,"id":"7a9a876b-0521-4091-8af1-186c53b291a3","page_size":[595,841],"block_position":"104-6"},{"bbox":[248,285,344,316],"type":"interline_equation","angle":0,"index":7,"text":"$$\n\\frac {e ^ {x} + e ^ {- x}}{2} = \\sum_ {n = 0} ^ {\\infty} \\frac {x ^ {2 n}}{(2 n) !}\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":104,"id":"34c5f759-8f85-4677-b6eb-ad0f30c147b2","page_size":[595,841],"block_position":"104-7"},{"bbox":[54,318,214,349],"type":"text","angle":0,"index":9,"merge_prev":false,"text":"取 $x = 1$ ，得到 $\\sum_{n=0}^{\\infty}\\frac{1}{(2n)!} = \\frac{e + e^{-1}}{2}$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":104,"id":"299bc7e4-e983-4900-a709-4f99727c20ab","page_size":[595,841],"block_position":"104-8"},{"bbox":[57,370,207,385],"type":"title","angle":0,"index":10,"level":2,"text":"## 10.3 幂级数求和式展开","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":104,"id":"1d4280e0-bc53-406f-8951-64c1c6b49bc4","page_size":[595,841],"block_position":"104-9"},{"bbox":[57,401,204,416],"type":"title","angle":0,"index":11,"level":2,"text":"## 10.3.1 基本理论与直接展开","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":104,"id":"b34ec183-5a2b-451e-8ed7-5dbc8f2943ae","page_size":[595,841],"block_position":"104-10"},{"bbox":[73,429,176,441],"type":"title","angle":0,"index":12,"level":2,"text":"## 定理 10.9 (幂级数展开)","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":104,"id":"7770d26d-8afd-4266-80f7-bec4414a4b75","page_size":[595,841],"block_position":"104-11"},{"bbox":[67,445,524,475],"type":"text","angle":0,"index":13,"merge_prev":false,"text":"若 $f(x)$ 在 $(x_0 - R, x_0 + R)$ 上存在任意阶导数，且 $\\exists M, N > 0$ ，当 $n > N$ 时， $\\forall x \\in (x_0 - R, x_0 + R)$ ，均有 $|f^{(n)}(x_0)| \\leq M$ ，则 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":104,"id":"ab36b59b-dbf1-4b5e-8e45-622afeeecf53","page_size":[595,841],"block_position":"104-12"},{"bbox":[179,477,412,508],"type":"interline_equation","angle":0,"index":14,"text":"$$\nf (x) = \\sum_ {n = 0} ^ {\\infty} \\frac {f ^ {(n)} (x _ {0})}{n !} (x - x _ {0}) ^ {n}, \\forall x \\in (x _ {0} - R, x _ {0} + R)\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":104,"id":"2d6588ff-8e25-4189-93ce-1bbc9ddcf241","page_size":[595,841],"block_position":"104-13"},{"bbox":[35,534,407,560],"type":"text","angle":0,"index":15,"merge_prev":false,"text":"练习10.8幂级数直接展开(1) $\\frac{ax + b}{cx + d}$ 型： $f(x) = \\frac{x - 3}{x + 1}$ 在 $x = 1$ 处的幂级数展开 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":104,"id":"40f1eb35-7145-49d1-b76f-77ce0b79d7c5","page_size":[595,841],"block_position":"104-14"},{"bbox":[73,558,261,585],"type":"text","angle":0,"index":16,"merge_prev":false,"text":"(2) $f(x)=\\frac{1}{x(x+1)}$ 在 x=1 处幂级数展开 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":104,"id":"fdf70894-e028-4753-acdd-d4764d845d23","page_size":[595,841],"block_position":"104-15"},{"bbox":[73,584,251,607],"type":"text","angle":0,"index":17,"merge_prev":false,"text":"(3) $f(x)=\\frac{1}{1-3x+2x^{2}}$ 在 x=0 处展开 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":104,"id":"348da962-85e9-43ff-a812-3f9db9c22759","page_size":[595,841],"block_position":"104-16"},{"bbox":[73,605,246,624],"type":"text","angle":0,"index":18,"merge_prev":false,"text":"(4) $f(x) = \\frac{x}{1 + x - 2x^2}$ 在 $x = 0$ 处展开 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":104,"id":"6479a35d-6ac7-4e1e-8731-49ca61870f21","page_size":[595,841],"block_position":"104-17"},{"bbox":[54,624,539,664],"type":"text","angle":0,"index":19,"merge_prev":false,"text":"解(1)首先 $f(x)=1-\\frac{4}{1+x}$ ，设t=x-1，则 $f(x)=g(t)=1-\\frac{4}{2+t}$ ，因此 $f(x)=1-2\\cdot\\frac{1}{1+\\frac{t}{2}}=1-2\\sum_{n=0}^{\\infty}\\left(\\frac{-t}{2}\\right)^{n}$ ，\n因此得到 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":104,"id":"c307c503-a8fe-4b7e-bc85-a6db46855082","page_size":[595,841],"block_position":"104-18"},{"bbox":[244,662,348,694],"type":"interline_equation","angle":0,"index":20,"text":"$$\nf (x) = 1 - \\sum_ {n = 0} ^ {\\infty} \\left(\\frac {- t}{2}\\right) ^ {n}\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":104,"id":"214ee78c-7c5f-4a61-b46d-5952268f9ab6","page_size":[595,841],"block_position":"104-19"},{"bbox":[73,700,364,732],"type":"text","angle":0,"index":21,"merge_prev":false,"text":"(2) 设 $t = x - 1$ ，可以得到 $f(x) = \\sum_{n=0}^{\\infty} (-1)^n \\left(1 - \\frac{1}{2^{n+1}}\\right)(x - 1)^n$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":104,"id":"69cd9fb6-02d5-4b90-aa88-6d4a31fe26a9","page_size":[595,841],"block_position":"104-20"},{"bbox":[290,802,302,811],"type":"page_number","angle":0,"index":1000007,"text":"97 ","color":{"line":"rgba(164, 164, 164, 1)","fill":"rgba(164, 164, 164, 1)"},"page_idx":104,"is_discarded":true,"id":"97b832bb-a899-4b00-a77a-b2108241e96b","page_size":[595,841],"block_position":"104-21"}],[{"bbox":[436,47,539,58],"type":"header","angle":0,"index":-99999,"text":"10.3 幂级数求和式展开 ","color":{"line":"rgba(164, 164, 164, 1)","fill":"rgba(164, 164, 164, 1)"},"page_idx":105,"is_discarded":true,"id":"ba66a0c2-c717-40d6-b10e-ae9f0ce1b3e7","page_size":[595,841],"block_position":"105-0"},{"bbox":[57,72,157,86],"type":"title","angle":0,"index":1,"level":2,"text":"## 10.3.2 求导与求积","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":105,"id":"b33b2401-8fdb-4a5c-b657-f3d58dc6523c","page_size":[595,841],"block_position":"105-1"},{"bbox":[35,94,432,119],"type":"text","angle":0,"index":2,"merge_prev":false,"text":"练习10.9几个经典的积分与求导法(1)计算 $f(x) = \\frac{1}{(1 + x)^2}$ 在 $x = 0$ 处的幂级数展开 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":105,"id":"0fc8862b-71d7-451c-b3e4-00127f19511b","page_size":[595,841],"block_position":"105-2"},{"bbox":[73,119,292,132],"type":"text","angle":0,"index":3,"merge_prev":false,"text":"(2) 计算 $f(x) = \\arcsin x$ 在 $x = 0$ 处的幂级数展开 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":105,"id":"c572c3e4-923e-499c-9511-351e60c6c9fe","page_size":[595,841],"block_position":"105-3"},{"bbox":[73,131,298,155],"type":"text","angle":0,"index":4,"merge_prev":false,"text":"(3) 计算 $f(x) = \\int_{0}^{x} \\frac{\\sin t}{t} t$ 在 $x = 0$ 处的幂级数展开 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":105,"id":"c5679f3a-1f2e-4e7e-b6a5-3355e3a0168a","page_size":[595,841],"block_position":"105-4"},{"bbox":[53,153,547,185],"type":"text","angle":0,"index":5,"merge_prev":false,"text":"解(1)积分得到 $\\int_0^x f(t)\\mathrm{d}t = -\\frac{1}{1 + t}\\bigg|_0^x = 1 - \\frac{1}{1 + x} = 1 - \\sum_{n = 0}^{\\infty}(-x)^n$ ，因此 $f(x) = \\left(\\int_0^x f(t)\\mathrm{d}t\\right)' = \\sum_{n = 1}^{\\infty}(-1)^{n + 1}nx^{n - 1}$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":105,"id":"76e0eb4f-5dec-4cf6-9213-68ab96f8f406","page_size":[595,841],"block_position":"105-5"},{"bbox":[73,185,183,210],"type":"interline_equation","angle":0,"index":6,"text":"$$\n(2) f ^ {\\prime} (x) = \\frac {1}{\\sqrt {1 - x ^ {2}}}, \\text { 而 }\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":105,"id":"e39cacd7-b1bf-4c0d-b40e-3650e7dc59b9","page_size":[595,841],"block_position":"105-6"},{"bbox":[119,211,473,243],"type":"interline_equation","angle":0,"index":7,"text":"$$\n(1 - x ^ {2}) ^ {- \\frac {1}{2}} = 1 + \\sum_ {n = 1} ^ {\\infty} \\frac {(- \\frac {1}{2}) (- \\frac {3}{2}) \\cdots (- \\frac {1}{2} - n + 1)}{n !} (- x ^ {2}) ^ {n} = 1 + \\sum_ {n = 1} ^ {\\infty} \\frac {(2 n - 1) ! !}{(2 n) ! !} x ^ {2 n}\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":105,"id":"6ad4dd0e-d5da-43f7-aecd-8ce8d604bb1f","page_size":[595,841],"block_position":"105-7"},{"bbox":[54,247,325,277],"type":"text","angle":0,"index":8,"merge_prev":false,"text":"因此 $f(x) = f(0) + \\int_0^x f'(t)\\mathrm{d}t = x + \\sum_{n = 1}^{\\infty}\\frac{(2n - 1)!!}{(2n + 1)(2n)!!} x^{2n + 1}$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":105,"id":"ac8dd419-bf73-4e55-8f1b-605ca9127a73","page_size":[595,841],"block_position":"105-8"},{"bbox":[35,285,241,297],"type":"text","angle":0,"index":9,"merge_prev":false,"text":"练习 10.10arctan 经典题目 求幂级数展开: ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":105,"id":"eb04ae7f-0c6d-4ab6-ace4-3e35a1db4835","page_size":[595,841],"block_position":"105-9"},{"bbox":[73,301,158,314],"type":"text","angle":0,"index":10,"merge_prev":false,"text":"(1) $f(x)=\\arctan x$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":105,"id":"aa7d1a7a-6bf9-43f0-bd98-7fd092552fe7","page_size":[595,841],"block_position":"105-10"},{"bbox":[73,312,259,335],"type":"text","angle":0,"index":11,"merge_prev":false,"text":"(2) 重点 (ZJU2021): $g(x) = \\arctan \\frac{1 - kx}{1 + kx}$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":105,"id":"c46a070a-0bb2-4077-b470-27f6f52f24e9","page_size":[595,841],"block_position":"105-11"},{"bbox":[73,334,293,358],"type":"text","angle":0,"index":12,"merge_prev":false,"text":"(3) 重点: $h(x) = \\arctan \\frac{2x}{1 - x^2}$ , 求导结果很简单 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":105,"id":"12f936e6-71b5-4de0-a414-36a107801510","page_size":[595,841],"block_position":"105-12"},{"bbox":[54,359,236,387],"type":"text","angle":0,"index":13,"merge_prev":false,"text":"解 (1) $f'(x)=\\frac{1}{1+x^{2}}=\\sum_{n=0}^{\\infty}(-x^{2})^{n}$ ，因此 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":105,"id":"55af5293-5bb3-4c2d-b1cc-16c5c32d966d","page_size":[595,841],"block_position":"105-13"},{"bbox":[180,391,412,422],"type":"interline_equation","angle":0,"index":14,"text":"$$\nf (x) = f (0) + \\sum_ {n = 0} ^ {\\infty} \\int_ {0} ^ {x} (- 1) ^ {n} t ^ {2 n} \\mathrm{d} t = \\sum_ {n = 0} ^ {\\infty} (- 1) ^ {n} \\frac {x ^ {2 n + 1}}{2 n + 1}\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":105,"id":"ac1b113d-1b2f-4e10-8613-04a981edbbd3","page_size":[595,841],"block_position":"105-14"},{"bbox":[73,423,279,449],"type":"text","angle":0,"index":15,"merge_prev":false,"text":"(2) $\\arctan \\frac{1 - kx}{1 + kx} = \\arctan 1 - \\arctan kx$ ，因此 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":105,"id":"a390336a-12b3-42fb-a304-efa3c462accc","page_size":[595,841],"block_position":"105-15"},{"bbox":[220,451,372,482],"type":"interline_equation","angle":0,"index":16,"text":"$$\ng (x) = \\frac {\\pi}{4} - \\sum_ {n = 0} ^ {\\infty} \\frac {(- 1) ^ {n} k ^ {2 n + 1}}{2 n + 1} x ^ {2 n + 1}\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":105,"id":"74a599ae-dc3e-43be-9854-fbf1ca4ea6da","page_size":[595,841],"block_position":"105-16"},{"bbox":[73,485,380,516],"type":"text","angle":0,"index":17,"merge_prev":false,"text":"(3) $h'(x)=\\frac{2}{1+x^{2}}$ (别怕求导！)，因此 $h'(x)=2\\sum_{n=0}^{\\infty}(-x^{2})^{n}$ ，积分得到 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":105,"id":"9151cc38-1cb5-48b5-b35a-c8749cea2bbb","page_size":[595,841],"block_position":"105-17"},{"bbox":[238,518,353,550],"type":"interline_equation","angle":0,"index":18,"text":"$$\nh (x) = 2 \\sum_ {n = 0} ^ {\\infty} \\frac {(- 1) ^ {n}}{2 n + 1} x ^ {2 n + 1}\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":105,"id":"cf6d3e6d-69cc-4728-8840-05f5255bb4bc","page_size":[595,841],"block_position":"105-18"},{"bbox":[290,802,302,812],"type":"page_number","angle":0,"index":1000018,"text":"98 ","color":{"line":"rgba(164, 164, 164, 1)","fill":"rgba(164, 164, 164, 1)"},"page_idx":105,"is_discarded":true,"id":"56d1f440-233d-4474-8fb5-19a2ab036904","page_size":[595,841],"block_position":"105-19"}],[{"bbox":[223,74,381,94],"type":"title","angle":0,"index":0,"level":2,"text":"## 第 11 章 傅里叶级数","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":106,"id":"7fbea3fb-f5b3-4931-a599-6e158694662f","page_size":[595,841],"block_position":"106-0"},{"bbox":[57,122,260,139],"type":"title","angle":0,"index":1,"level":2,"text":"## 11.1 以 $2\\pi$ 为周期的傅里叶级数","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":106,"id":"0ac775d2-6d17-4704-a064-0574f60580a8","page_size":[595,841],"block_position":"106-1"},{"bbox":[72,156,210,169],"type":"title","angle":0,"index":2,"level":2,"text":"## 定义 11.1 (2π 周期傅里叶展开)","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":106,"id":"501b4acb-ea13-42d8-bfeb-708b5b136a28","page_size":[595,841],"block_position":"106-2"},{"bbox":[68,174,470,205],"type":"text","angle":0,"index":3,"merge_prev":false,"text":"对于 $2\\pi$ 周期函数 $f(x)$ ，定义其傅里叶展开为 $f(x) \\sim \\frac{a_0}{2} + \\sum_{n=1}^{\\infty} (a_n \\cos nx + b_n \\sin nx)$ ，其中 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":106,"id":"0a5565ed-e8f1-4792-a1a8-688f9a8c1ac1","page_size":[595,841],"block_position":"106-3"},{"bbox":[201,207,392,234],"type":"interline_equation","angle":0,"index":4,"text":"$$\n\\int a _ {n} = \\frac {1}{\\pi} \\int_ {- \\pi} ^ {\\pi} f (x) \\cos n x \\mathrm{d} x \\quad n = 0, 1, 2, \\dots\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":106,"id":"73e0695b-95df-4503-9116-209a4922afae","page_size":[595,841],"block_position":"106-4"},{"bbox":[202,233,384,259],"type":"interline_equation","angle":0,"index":5,"text":"$$\n\\left\\{b _ {n} = \\frac {1}{\\pi} \\int_ {- \\pi} ^ {\\pi} f (x) \\sin n x \\mathrm{d} x \\quad n = 1, 2, \\dots \\right.\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":106,"id":"cc2c1c25-0967-4d58-ac09-3334ca03254d","page_size":[595,841],"block_position":"106-5"},{"bbox":[35,269,47,286],"type":"image","angle":0,"index":6,"img_path":"/94b85b028233997a3678393537cecdea4088e96528b29e922edcbe8d2e7f4793.jpg","color":{"line":"rgba(89, 92, 220, 1)","fill":"rgba(89, 92, 220, 1)"},"img_ratio":0.2,"page_idx":106,"id":"a48e225d-a474-4768-b66d-4196ba052a45","page_size":[595,841],"block_position":"106-6"},{"bbox":[36,271,220,293],"type":"text","angle":0,"index":7,"merge_prev":false,"text":"笔记 计算时一定要注意第一项是 $\\frac{a_0}{2}$ ! ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":106,"id":"485ebdc1-808d-43c5-aafd-29a477489bdf","page_size":[595,841],"block_position":"106-7"},{"bbox":[73,301,166,314],"type":"title","angle":0,"index":8,"level":2,"text":"## 定理 11.1 (收敛定理)","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":106,"id":"5f507eb5-448d-4a7f-96b8-902d835580fb","page_size":[595,841],"block_position":"106-8"},{"bbox":[68,317,525,347],"type":"text","angle":0,"index":9,"merge_prev":false,"text":"若以 $2\\pi$ 为周期的函数 $f$ 在 $[- \\pi, \\pi]$ 上按段光滑，则对 $\\forall x \\in [-\\pi, \\pi]$ ，傅里叶级数收敛于 $f$ 在 $x$ 左右极限的算术平均值。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":106,"id":"e5242524-9bb0-4830-88c7-f4d6f3ca5c09","page_size":[595,841],"block_position":"106-9"},{"bbox":[73,366,186,379],"type":"title","angle":0,"index":10,"level":2,"text":"## 推论 11.1 (连续特殊情况)","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":106,"id":"5fe84d7d-e034-4d90-ba36-713b815c9000","page_size":[595,841],"block_position":"106-10"},{"bbox":[68,381,537,401],"type":"text","angle":0,"index":11,"merge_prev":false,"text":"若 $f$ 是以 $2\\pi$ 为周期的连续函数，且在 $[- \\pi, \\pi]$ 上按段光滑，则 $f$ 的傅里叶级数在 $(- \\infty, +\\infty)$ 上收敛于 $f$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":106,"id":"43a5e834-5a5c-4e45-9c79-3e3c37bce8db","page_size":[595,841],"block_position":"106-11"},{"bbox":[35,423,414,453],"type":"text","angle":0,"index":12,"merge_prev":false,"text":"练习11.1几个经典展开(1)求 $f(x) = x^{2}$ 在 $(- \\pi, \\pi)$ 上的傅里叶级数，并求 $\\sum_{n=1}^{\\infty} \\frac{1}{n^{2}}$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":106,"id":"4ce25e4b-cd87-499c-971e-4b65a0861ad2","page_size":[595,841],"block_position":"106-12"},{"bbox":[52,451,382,478],"type":"text","angle":0,"index":13,"merge_prev":false,"text":"解(1)计算得到 $a_0 = \\frac{2}{3}\\pi^2$ ， $a_{n} = \\frac{1}{\\pi}\\int_{-\\pi}^{\\pi}x^{2}\\cos nx\\mathrm{d}x = (-1)^{n}\\frac{4}{n^{2}}$ ，因此得到 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":106,"id":"1a5a35e1-f1d6-4f59-88d0-0ecef8b5ee6a","page_size":[595,841],"block_position":"106-13"},{"bbox":[240,481,352,512],"type":"interline_equation","angle":0,"index":14,"text":"$$\nx ^ {2} = \\frac {1}{3} \\pi^ {2} + \\sum_ {n = 1} ^ {\\infty} (- 1) ^ {n} \\frac {4}{n ^ {2}}\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":106,"id":"baeedbb2-a82e-42d7-bb00-d9f033eade4a","page_size":[595,841],"block_position":"106-14"},{"bbox":[54,514,117,527],"type":"text","angle":0,"index":15,"merge_prev":false,"text":"令 $x = \\pi$ 得到 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":106,"id":"5a99eb2a-931d-4202-931d-59788b0422ab","page_size":[595,841],"block_position":"106-15"},{"bbox":[216,524,376,556],"type":"interline_equation","angle":0,"index":16,"text":"$$\n\\pi^ {2} = \\frac {1}{3} \\pi^ {2} + \\sum_ {n = 1} ^ {\\infty} \\frac {4}{n ^ {2}} \\Rightarrow \\sum_ {n = 1} ^ {\\infty} \\frac {1}{n ^ {2}} = \\frac {\\pi^ {2}}{6}\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":106,"id":"c8ecc067-8a09-4bce-a03e-cddeea3ee53e","page_size":[595,841],"block_position":"106-16"},{"bbox":[34,562,48,579],"type":"image","angle":0,"index":17,"img_path":"/d54523c5e53334e427bb9e9822136c9ad2e44ba8407c1ddcc6e3a92d9c8e6f2d.jpg","color":{"line":"rgba(89, 92, 220, 1)","fill":"rgba(89, 92, 220, 1)"},"img_ratio":0.2,"page_idx":106,"id":"fccdbad2-4f0d-4e1b-b2b5-8db329919249","page_size":[595,841],"block_position":"106-17"},{"bbox":[52,566,172,580],"type":"text","angle":0,"index":18,"merge_prev":false,"text":"笔记 注意 $\\cos n\\pi = (-1)^n$ 。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":106,"id":"2ac4e4fc-a844-45fa-a7dd-bd31587264a7","page_size":[595,841],"block_position":"106-18"},{"bbox":[57,618,255,634],"type":"title","angle":0,"index":19,"level":2,"text":"## 11.2 以 $2l$ 为周期的傅里叶级数","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":106,"id":"2a8d33d1-b710-4551-874a-27a3e48e04a3","page_size":[595,841],"block_position":"106-19"},{"bbox":[73,652,217,665],"type":"title","angle":0,"index":20,"level":2,"text":"## 定义 11.2 (2l 周期的傅里叶级数)","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":106,"id":"e0862c25-bfac-48e1-9f2a-8b4720b498be","page_size":[595,841],"block_position":"106-20"},{"bbox":[68,669,453,699],"type":"text","angle":0,"index":21,"merge_prev":false,"text":"$f(x)$ 以 $2l$ 为周期，则其傅里叶级数为 $f(x) \\sim \\frac{a_0}{2} + \\sum_{n=1}^{\\infty} (a_n \\cos \\frac{n\\pi x}{l} + b_n \\sin \\frac{n\\pi x}{l})$ ，其中 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":106,"id":"92e0a7f9-9b0c-457f-9a0d-a30c680b4672","page_size":[595,841],"block_position":"106-21"},{"bbox":[197,703,396,731],"type":"interline_equation","angle":0,"index":22,"text":"$$\n\\int a _ {n} = \\frac {1}{l} \\int_ {- l} ^ {l} f (x) \\cos \\frac {n \\pi x}{l} \\mathrm{d} x, \\quad n = 0, 1, 2, \\dots\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":106,"id":"439584ed-aeaf-43e2-821a-b91384214308","page_size":[595,841],"block_position":"106-22"},{"bbox":[198,731,387,757],"type":"interline_equation","angle":0,"index":23,"text":"$$\n\\left\\{b _ {n} = \\frac {1}{l} \\int_ {- l} ^ {l} f (x) \\sin {\\frac {n \\pi x}{l}} \\mathrm{d} x, n = 1, 2, \\dots \\right.\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":106,"id":"21dc2a1a-3191-4713-9a7e-32148d1868d6","page_size":[595,841],"block_position":"106-23"}],[{"bbox":[459,47,539,58],"type":"header","angle":0,"index":-99999,"text":"11.3 Parseval 等式 ","color":{"line":"rgba(164, 164, 164, 1)","fill":"rgba(164, 164, 164, 1)"},"page_idx":107,"is_discarded":true,"id":"8ae5e58c-6051-4a58-9573-fb3bece6ca94","page_size":[595,841],"block_position":"107-0"},{"bbox":[57,74,177,90],"type":"title","angle":0,"index":1,"level":2,"text":"## 11.3 Parseval 等式","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":107,"id":"01a9170d-87f4-4874-8fb3-76261ba50a04","page_size":[595,841],"block_position":"107-1"},{"bbox":[73,109,185,121],"type":"text","angle":0,"index":2,"merge_prev":false,"text":"定理 11.2 (Parseval 等式) ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":107,"id":"c65b1f77-192b-43d8-a256-ed5c56274223","page_size":[595,841],"block_position":"107-2"},{"bbox":[69,126,233,156],"type":"interline_equation","angle":0,"index":3,"text":"$$\n\\frac {1}{\\pi} \\int_ {- \\pi} ^ {\\pi} [ f (x) ] ^ {2} \\mathrm{d} x = \\frac {a _ {0} ^ {2}}{2} + \\sum_ {n = 1} ^ {\\infty} (a _ {n} ^ {2} + b _ {n} ^ {2})\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":107,"id":"76b63678-38d0-4f15-b953-3a21c64bc71c","page_size":[595,841],"block_position":"107-3"},{"bbox":[36,169,353,199],"type":"text","angle":0,"index":4,"merge_prev":false,"text":"练习11.2计算数项级数(1)对 $f(x) = x^{2}$ 展开，并证明 $\\sum_{n=1}^{\\infty}\\frac{1}{n^4} = \\frac{\\pi^4}{90}$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":107,"id":"c2caee07-0046-42c6-9d49-5e90e69bf454","page_size":[595,841],"block_position":"107-4"},{"bbox":[289,802,305,812],"type":"page_number","angle":0,"index":1000004,"text":"100 ","color":{"line":"rgba(164, 164, 164, 1)","fill":"rgba(164, 164, 164, 1)"},"page_idx":107,"is_discarded":true,"id":"409b60a2-7bf4-4230-9b69-38a46a675743","page_size":[595,841],"block_position":"107-5"}],[{"bbox":[35,245,47,261],"type":"image","angle":0,"index":0,"img_path":"/cc6eab2313c8b054f7a9294640558a696f8d7f28aec6df75223441dd840f7835.jpg","color":{"line":"rgba(89, 92, 220, 1)","fill":"rgba(89, 92, 220, 1)"},"img_ratio":0.2,"page_idx":108,"id":"1a0afdae-1f59-4103-ad8e-af0cdedfdf22","page_size":[595,841],"block_position":"108-0"},{"bbox":[35,574,47,590],"type":"image","angle":0,"index":1,"img_path":"/ef8b8195cbadf4e004099909588a2b27bf2e5ea56b2eedc28d61fe25a11d8d12.jpg","color":{"line":"rgba(89, 92, 220, 1)","fill":"rgba(89, 92, 220, 1)"},"img_ratio":0.2,"page_idx":108,"id":"b5d0417b-b8ea-4d29-bf5a-81162aecdce7","page_size":[595,841],"block_position":"108-1"},{"bbox":[205,74,398,94],"type":"title","angle":0,"index":2,"level":2,"text":"## 第 12 章 多元极限与连续","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":108,"id":"99556bcf-cf28-48cc-87a9-1f67febe3b53","page_size":[595,841],"block_position":"108-2"},{"bbox":[57,122,208,139],"type":"title","angle":0,"index":3,"level":2,"text":"## 12.1 累次极限与重极限","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":108,"id":"97eab85e-1356-4909-a97a-86bf2cc95c14","page_size":[595,841],"block_position":"108-3"},{"bbox":[57,155,145,169],"type":"title","angle":0,"index":4,"level":2,"text":"## 12.1.1 基本概念","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":108,"id":"666fd57e-c39b-4c89-9d19-68638129685b","page_size":[595,841],"block_position":"108-4"},{"bbox":[72,183,156,195],"type":"title","angle":0,"index":5,"level":2,"text":"## 定义 12.1 (重极限)","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":108,"id":"f2c3cecb-8811-4e65-887c-deee59aeb7a8","page_size":[595,841],"block_position":"108-5"},{"bbox":[67,198,525,231],"type":"text","angle":0,"index":6,"merge_prev":false,"text":"$a \\in \\mathbb{R}^n$ , $f: \\mathbb{R}^n \\to \\mathbb{R}$ 在 $a$ 的去心邻域中有定义, 若 $\\exists A, \\forall \\epsilon, \\exists \\delta$ 当 $\\mathbf{x} \\in \\mathbb{R}^n$ 满足 $|\\mathbf{x} - a| < \\delta$ 时, 有 $|f(\\mathbf{x}) - A| < \\epsilon$ , 则称 $\\lim_{\\mathbf{x} \\to a} f(\\mathbf{x}) = A$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":108,"id":"6a90a8dd-f40a-4e1f-bd2e-8de34509a7b6","page_size":[595,841],"block_position":"108-6"},{"bbox":[35,247,164,261],"type":"text","angle":0,"index":7,"merge_prev":false,"text":"笔记 计算重极限的方法: ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":108,"id":"7791ffa0-bd41-49ef-bdca-cca21e12099d","page_size":[595,841],"block_position":"108-7"},{"bbox":[68,259,311,283],"type":"text","angle":0,"index":8,"merge_prev":false,"text":"- 放缩 + 迫敛：例如 $\\left|(x + y)\\sin \\frac{1}{x}\\sin \\frac{1}{y}\\right| \\leq |x + y| = 0$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":108,"id":"55202dea-d8bd-47b8-a409-e9b6cbab87ec","page_size":[595,841],"block_position":"108-8"},{"bbox":[69,283,439,312],"type":"text","angle":0,"index":9,"merge_prev":false,"text":"- 极坐标变换：例如 $\\left|\\frac{x^3 + y^3}{x^2 + y^2}\\right| = \\left|\\frac{r^3\\cos^3\\theta + r^3\\sin^3\\theta}{r^2}\\right| = r|\\cos^3\\theta +\\sin^3\\theta |\\leq 2r\\to 0$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":108,"id":"4eb76b19-988a-41c7-8f29-c5979c4104db","page_size":[595,841],"block_position":"108-9"},{"bbox":[52,313,214,327],"type":"text","angle":0,"index":10,"merge_prev":false,"text":"笔记 证明重极限不存在的常用方法: ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":108,"id":"883e4299-136a-45ee-9450-5252d5e5ef1f","page_size":[595,841],"block_position":"108-10"},{"bbox":[69,329,202,342],"type":"text","angle":0,"index":11,"merge_prev":false,"text":"- 两个累次极限存在但不相等 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":108,"id":"90657174-3520-4996-ba62-48ba32a010d7","page_size":[595,841],"block_position":"108-11"},{"bbox":[69,345,324,358],"type":"text","angle":0,"index":12,"merge_prev":false,"text":"- 找特殊的趋近方式，如 $y = kx$ ，使得获得的重极限不同 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":108,"id":"cd367557-5899-41e8-b748-39d49967f0c6","page_size":[595,841],"block_position":"108-12"},{"bbox":[35,376,357,395],"type":"text","angle":0,"index":13,"merge_prev":false,"text":"练习12.1重极限的计算(1)已知 $f(x,y)$ 如下，计算 $\\lim_{(x,y)\\to (0,0)}f(x,y)$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":108,"id":"fd0a8f04-3b09-487d-9c5c-d2fcb478e312","page_size":[595,841],"block_position":"108-13"},{"bbox":[218,399,375,442],"type":"interline_equation","angle":0,"index":14,"text":"$$\nf (x, y) = \\left\\{ \\begin{array}{l l} x y \\frac {x ^ {2} - y ^ {2}}{x ^ {2} + y ^ {2}}, & (x, y) \\neq 0 \\\\ 0, & (x, y) = 0 \\end{array} \\right.\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":108,"id":"9e76bf25-e9d7-4eef-9ea4-b6d5390f98c1","page_size":[595,841],"block_position":"108-14"},{"bbox":[53,446,210,459],"type":"text","angle":0,"index":15,"merge_prev":false,"text":"解 (1) 设 $x = r\\cos \\theta, y = r\\sin \\theta$ ，则 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":108,"id":"b69e68c2-d290-4b29-a8f8-95359dc32869","page_size":[595,841],"block_position":"108-15"},{"bbox":[139,461,453,488],"type":"interline_equation","angle":0,"index":16,"text":"$$\n| f (x, y) | = \\left| \\frac {r ^ {2} \\sin \\theta \\cos \\theta (r ^ {2} \\cos^ {2} \\theta - r ^ {2} \\sin^ {2} \\theta)}{r ^ {2}} \\right| = \\frac {1}{4} r ^ {2} | \\sin 4 \\theta | \\leq \\frac {1}{4} r ^ {2} \\rightarrow 0\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":108,"id":"7b8c711f-554e-4b68-8016-1d85f571d2e1","page_size":[595,841],"block_position":"108-16"},{"bbox":[55,491,116,503],"type":"text","angle":0,"index":17,"merge_prev":false,"text":"因此极限为0 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":108,"id":"b58096cb-d237-4fc6-8d9e-31f5c083e707","page_size":[595,841],"block_position":"108-17"},{"bbox":[73,525,166,538],"type":"title","angle":0,"index":18,"level":2,"text":"## 定义 12.2 (累次极限)","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":108,"id":"27fcc0d8-32de-4658-b873-110441e0731a","page_size":[595,841],"block_position":"108-18"},{"bbox":[68,542,505,563],"type":"text","angle":0,"index":19,"merge_prev":false,"text":"设 $f(x,y)$ 在 $x_0,y_0$ 的某去心邻域中有定义，则称 $\\lim_{y\\to y_0}\\lim_{x\\to x_0}f(x,y),\\lim_{x\\to x_0}\\lim_{y\\to y_0}f(x,y)$ 称为累次极限。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":108,"id":"026a8606-5c1e-4c03-81a0-96b40de53585","page_size":[595,841],"block_position":"108-19"},{"bbox":[52,577,396,591],"type":"text","angle":0,"index":20,"merge_prev":false,"text":"笔记 重极限是多个分量同时趋近于 $\\mathbf{x}_0$ ，而累次极限相当于依次取极限的极限。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":108,"id":"7293839e-ed00-415e-a853-491468c69eb6","page_size":[595,841],"block_position":"108-20"},{"bbox":[72,599,236,612],"type":"title","angle":0,"index":21,"level":2,"text":"## 定理 12.1 (重极限与累次极限的关系)","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":108,"id":"d090d0e2-73b5-4a27-9118-a4eb4df424ce","page_size":[595,841],"block_position":"108-21"},{"bbox":[68,616,518,640],"type":"text","angle":0,"index":22,"merge_prev":false,"text":"若重极限 $\\lim_{(x,y)\\to (x_0,y_0)}f(x,y) = A$ 存在，且累次极限存在(累次极限可能不存在)，则累次极限的值为 $A$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":108,"id":"f0d5a512-8201-41b8-baa7-b4642da66b90","page_size":[595,841],"block_position":"108-22"},{"bbox":[54,653,539,683],"type":"text","angle":0,"index":23,"merge_prev":false,"text":"证明 $\\forall \\epsilon, \\exists \\delta, |\\mathbf{x} - \\mathbf{x}_0| < \\delta$ 时， $|f(x, y) - A| < \\epsilon$ ，两侧取极限 $x \\to x_0$ ，得到 $|\\lim_{x \\to x_0} f(x, y) - A| < \\epsilon$ ，当 $|y - y_0| < \\delta$ 时显然结论成立。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":108,"id":"66020cf9-fa7a-4851-8922-5cbec1c138b8","page_size":[595,841],"block_position":"108-23"},{"bbox":[73,691,196,703],"type":"title","angle":0,"index":24,"level":2,"text":"## 定理 12.2 (重极限的任意性)","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":108,"id":"ea86385e-c73f-4118-919d-01c6de1e5d52","page_size":[595,841],"block_position":"108-24"},{"bbox":[69,708,318,721],"type":"text","angle":0,"index":25,"merge_prev":false,"text":"设 $(x,y)$ 沿着任意路径趋于 $(x_0,y_0)$ 时， $f(x,y)\\to A$ ，则 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":108,"id":"2068fcd7-4d1c-4d6c-9699-73f57417315a","page_size":[595,841],"block_position":"108-25"},{"bbox":[244,728,349,747],"type":"interline_equation","angle":0,"index":26,"text":"$$\n\\lim _ {(x, y) \\to (x _ {0}, y _ {0})} f (x, y) = A\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":108,"id":"d7eca7ed-57aa-4f48-90a0-92f54351061a","page_size":[595,841],"block_position":"108-26"}],[{"bbox":[445,47,538,58],"type":"header","angle":0,"index":-99999,"text":"12.2 多元函数连续性 ","color":{"line":"rgba(164, 164, 164, 1)","fill":"rgba(164, 164, 164, 1)"},"page_idx":109,"is_discarded":true,"id":"6f79edcd-0473-4462-9041-5a0e3fab56ba","page_size":[595,841],"block_position":"109-0"},{"bbox":[52,71,548,104],"type":"text","angle":0,"index":1,"merge_prev":false,"text":"证明 反设 $\\lim_{(x,y)\\to(x_{0},y_{0})}f(x,y)\\neq A$ ，则 $\\exists\\epsilon_{0}$ 和 $\\{x_{n}\\}$ 使得 $\\lim_{n\\to\\infty}x_{n}=(x_{0},y_{0})$ ，但 $|f(x_{n})-A|\\geq\\epsilon_{0}$ ，将 $x_{1},x_{2},\\cdots,x_{n},\\cdots$ 连接成折线，折线趋近于 $(x_{0},y_{0})$ ，但 $f(x,y)$ 不趋于 A，这与已知矛盾。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":109,"id":"a9dd76cb-2593-4fb0-9f7b-cf9611ef6d37","page_size":[595,841],"block_position":"109-1"},{"bbox":[35,121,292,133],"type":"text","angle":0,"index":2,"merge_prev":false,"text":"练习 12.2 常用反例 讨论 $(0,0)$ 处的重极限与累次极限 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":109,"id":"9a0c35ca-3d11-44f3-95b0-d8f6359faa8b","page_size":[595,841],"block_position":"109-2"},{"bbox":[72,135,198,156],"type":"text","angle":0,"index":3,"merge_prev":false,"text":"(1) 重点: $f(x, y) = \\frac{xy}{x^2 + y^2}$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":109,"id":"e2f905bf-031a-4db4-a964-0eff4e4e4710","page_size":[595,841],"block_position":"109-3"},{"bbox":[73,156,159,179],"type":"text","angle":0,"index":4,"merge_prev":false,"text":"(2) $f(x,y)=\\frac{x-y}{x+y}$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":109,"id":"14f9ba78-521f-4a6a-ae92-b321d75eb2e7","page_size":[595,841],"block_position":"109-4"},{"bbox":[73,179,160,199],"type":"text","angle":0,"index":5,"merge_prev":false,"text":"(3) $f(x,y)=x\\sin\\frac{1}{y}$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":109,"id":"fc7e3aad-fbb3-4a8e-bae2-f7f6033c4ad2","page_size":[595,841],"block_position":"109-5"},{"bbox":[73,197,202,224],"type":"text","angle":0,"index":7,"merge_prev":false,"text":"(4) $f(x,y)=x\\sin\\frac{1}{y}+y\\sin\\frac{1}{x}$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":109,"id":"082390d4-04df-4685-9f28-26e0dc1044ca","page_size":[595,841],"block_position":"109-6"},{"bbox":[72,225,310,261],"type":"text","angle":0,"index":8,"merge_prev":false,"text":"(5) 重点: $f(x, y) = \\begin{cases} 1, & 0 < y < x^2, -\\infty < x < +\\infty \\\\ 0, & \\text{其他} \\end{cases}$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":109,"id":"232380ca-0215-4f46-809b-1153fbd967ff","page_size":[595,841],"block_position":"109-7"},{"bbox":[52,260,483,287],"type":"text","angle":0,"index":9,"merge_prev":false,"text":"解(1)累次都是0。重极限若选择 $y = kx$ ，则得到 $f(x, y) = \\frac{kx^2}{(k^2 + 1)x^2} = \\frac{k}{k^2 + 1}$ ，重极限不存在。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":109,"id":"b83a1276-3a48-48c5-bffc-60ba84a057f0","page_size":[595,841],"block_position":"109-8"},{"bbox":[72,287,404,311],"type":"text","angle":0,"index":10,"merge_prev":false,"text":"(2) 累次为 $\\pm 1$ 。重极限考虑 $y = kx$ ，则 $f(x, y) = \\frac{x - kx}{x + kx} = \\frac{1 - k}{1 + k}$ ，不存在 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":109,"id":"192ef468-b979-440e-866c-776e91ceb28e","page_size":[595,841],"block_position":"109-9"},{"bbox":[72,312,445,327],"type":"text","angle":0,"index":11,"merge_prev":false,"text":"(3) 先 $x$ 后 $y$ 极限为 0，先 $y$ 后 $x$ 不存在。重极限 $x \\sin \\frac{1}{y} < |x| \\to 0$ ，因此重极限为 0 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":109,"id":"73a495f2-a3aa-4be2-afb7-04d6032f8ca5","page_size":[595,841],"block_position":"109-10"},{"bbox":[72,332,348,344],"type":"text","angle":0,"index":12,"merge_prev":false,"text":"(4) 累次极限不存在。而 $|f(x, y)| \\leq |x| + |y| \\to 0$ ，重极限为 0。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":109,"id":"3c5bca9a-1563-41da-a4c4-98b3ede7f3b5","page_size":[595,841],"block_position":"109-11"},{"bbox":[72,347,518,359],"type":"text","angle":0,"index":13,"merge_prev":false,"text":"(5) 画出图像，发现沿着直线逼近都是 0，沿着抛物线 $y = kx^{2} (0 < k < 1)$ 时趋于 1，因此极限不存在。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":109,"id":"7e73b151-3df2-49c0-9dc7-3b9666b8e0c8","page_size":[595,841],"block_position":"109-12"},{"bbox":[35,378,288,390],"type":"text","angle":0,"index":14,"merge_prev":false,"text":"练习 12.3 反例应用 上述反例可作为以下问题的解答: ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":109,"id":"bf504663-90e5-47d3-9397-4342288db227","page_size":[595,841],"block_position":"109-13"},{"bbox":[66,391,218,412],"type":"text","angle":0,"index":15,"merge_prev":false,"text":"1. 累次极限存在但不相等: $\\frac{x - y}{x + y}$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":109,"id":"0775bb74-d476-4cdc-a798-a9b7ef070cb9","page_size":[595,841],"block_position":"109-14"},{"bbox":[64,412,297,432],"type":"text","angle":0,"index":16,"merge_prev":false,"text":"2. 累次极限存在且相等，但重极限不存在： $\\frac{xy}{x^2 + y^2}$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":109,"id":"d0c98158-871c-4985-b719-b30d280fed0a","page_size":[595,841],"block_position":"109-15"},{"bbox":[65,432,302,456],"type":"text","angle":0,"index":17,"merge_prev":false,"text":"3. 累次极限不存在，但重极限存在： $x \\sin \\frac{1}{y} + y \\sin \\frac{1}{x}$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":109,"id":"b6381b64-1000-4091-9783-f58d604a460d","page_size":[595,841],"block_position":"109-16"},{"bbox":[65,454,251,480],"type":"text","angle":0,"index":18,"merge_prev":false,"text":"4. 重极限与双累次极限都不存在: $\\frac{1}{x} + \\frac{1}{y}$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":109,"id":"103f1b24-b54c-4b2a-9ad3-c98bc6216cc4","page_size":[595,841],"block_position":"109-17"},{"bbox":[56,507,169,521],"type":"title","angle":0,"index":19,"level":2,"text":"## 12.1.2 重极限的计算","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":109,"id":"e00ad58a-6bf0-4f25-ac06-1e052fac21f3","page_size":[595,841],"block_position":"109-18"},{"bbox":[35,539,537,592],"type":"text","angle":0,"index":20,"merge_prev":false,"text":"练习 12.4 多元分式 讨论 (0,0) 处的重极限: (1) $f(x,y)=\\frac{xy}{x+y}$ (2) $f(x,y)=\\frac{xy}{x^{2}+y^{2}}$ (3) $f(x,y)=\\frac{x^{2}y^{2}}{(x^{2}+y^{2})^{2}}$ (4) $f(x,y)=\\frac{x^{6}y^{8}}{(x^{2}+y^{4})^{5}}$ (5) $f(x,y)=\\frac{x^{2}y^{2}}{x^{2}y^{2}+(x-y)^{2}}$ (6) $f(x,y)=\\frac{x^{2}y^{2}}{x^{3}+y^{3}}$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":109,"id":"398c2daa-df70-42dd-81ef-e6027808ade3","page_size":[595,841],"block_position":"109-19"},{"bbox":[52,590,539,618],"type":"text","angle":0,"index":21,"merge_prev":false,"text":"解 (1) 不齐次, 一方面将分子次数升高, $y = x$ 得到极限为 0 。另一方面常数消去公因式, 将分母变简单, 例如 $y = x^{2} - x$ 化简为 $x - 1 \\rightarrow -1$ , 因此重极限不存在 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":109,"id":"d695f662-046f-407b-94c5-438495236737","page_size":[595,841],"block_position":"109-20"},{"bbox":[73,621,192,634],"type":"text","angle":0,"index":22,"merge_prev":false,"text":"(2) 齐次， $y = kx$ ，不存在 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":109,"id":"f68675b6-f13c-42b6-90de-cc0b3663d7f4","page_size":[595,841],"block_position":"109-21"},{"bbox":[73,637,192,649],"type":"text","angle":0,"index":23,"merge_prev":false,"text":"(3) 齐次， $y = kx$ ，不存在 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":109,"id":"52ee56cf-5af2-4e2d-9e9f-03dbe6ce8360","page_size":[595,841],"block_position":"109-22"},{"bbox":[72,647,344,672],"type":"text","angle":0,"index":24,"merge_prev":false,"text":"(4) 尝试将分母变齐次， $x = ky^{2}$ ，得到 $\\frac{k^{6}y^{20}}{(k^{2} + 1)^{5}y^{20}}$ ，不存在 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":109,"id":"fe30ee0f-7b35-4dda-be7e-ab35483fb892","page_size":[595,841],"block_position":"109-23"},{"bbox":[72,673,451,685],"type":"text","angle":0,"index":25,"merge_prev":false,"text":"(5) 尝试化简 $x = y$ 得到 1，其他想不到办法，尝试取零， $x = 0$ 得到极限为 0。不存在 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":109,"id":"7d193a8f-d753-44ae-9ac0-d62bd60f0000","page_size":[595,841],"block_position":"109-24"},{"bbox":[72,685,540,709],"type":"text","angle":0,"index":26,"merge_prev":false,"text":"(6) 分子次数高： $y = x$ 显然极限为 0。另一方面化简分母，设 $y^{3} = x^{4} - x^{3}$ ，得到 $\\frac{x^2(x^4x^3)^{\\frac{2}{3}}}{x^4}$ ，极限不是 0。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":109,"id":"db981b2e-58fe-4470-8753-c9323c565f3a","page_size":[595,841],"block_position":"109-25"},{"bbox":[35,723,221,741],"type":"text","angle":0,"index":27,"merge_prev":false,"text":"练习 12.5 指数形式 (1) $\\lim_{(x,y)\\to(0^{+},0^{+})}x^{y}$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":109,"id":"b7082da9-fef7-45c6-9c59-4b3d75eb699e","page_size":[595,841],"block_position":"109-26"},{"bbox":[52,741,308,754],"type":"text","angle":0,"index":28,"merge_prev":false,"text":"解(1)用累次极限看出一个0, 一个1, 显然重极限不存在 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":109,"id":"7fc811cd-3422-45a0-a66f-083310d1b983","page_size":[595,841],"block_position":"109-27"},{"bbox":[289,802,305,812],"type":"page_number","angle":0,"index":1000005,"text":"102 ","color":{"line":"rgba(164, 164, 164, 1)","fill":"rgba(164, 164, 164, 1)"},"page_idx":109,"is_discarded":true,"id":"8a02fa27-1a02-4a8d-89dd-344e750ca3c8","page_size":[595,841],"block_position":"109-28"}],[{"bbox":[445,46,538,58],"type":"header","angle":0,"index":-99999,"text":"12.2 多元函数连续性 ","color":{"line":"rgba(164, 164, 164, 1)","fill":"rgba(164, 164, 164, 1)"},"page_idx":110,"is_discarded":true,"id":"c026199d-64e7-4956-85d8-00d0d7699ff3","page_size":[595,841],"block_position":"110-0"},{"bbox":[57,74,193,90],"type":"title","angle":0,"index":1,"level":2,"text":"## 12.2 多元函数连续性","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":110,"id":"383e04eb-d5d7-4c19-bf2f-e72353f00dfb","page_size":[595,841],"block_position":"110-1"},{"bbox":[57,107,181,121],"type":"title","angle":0,"index":2,"level":2,"text":"## 12.2.1 多元函数连续性","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":110,"id":"3e7043e9-bfc2-4eb3-a1ac-c6f2e87951e3","page_size":[595,841],"block_position":"110-2"},{"bbox":[73,134,186,147],"type":"title","angle":0,"index":3,"level":2,"text":"## 定义 12.3 (多元函数连续)","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":110,"id":"5aa280fc-fca6-4343-8358-aded78fd8504","page_size":[595,841],"block_position":"110-3"},{"bbox":[60,148,475,179],"type":"text","angle":0,"index":4,"merge_prev":false,"text":"若 $f(x,y)$ 在 $(x_0,y_0)$ 处满足 $\\lim_{(x,y)\\to (x_0,y_0)}f(x,y) = f(x_0,y_0)$ ，则称 $f(x,y)$ 在 $(x_0,y_0)$ 处连续。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":110,"id":"6161ae61-b852-43ed-84ac-dc089064b9ad","page_size":[595,841],"block_position":"110-4"},{"bbox":[53,188,443,202],"type":"text","angle":0,"index":5,"merge_prev":false,"text":"例题 12.1 关于 x, y 连续，不多元连续 举例说明 $f(x, y)$ 关于 x, y 均连续，但不多元连续。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":110,"id":"3e2d40ed-03c6-45c1-a9fe-4bb8907bb04c","page_size":[595,841],"block_position":"110-5"},{"bbox":[215,215,377,258],"type":"interline_equation","angle":0,"index":6,"text":"$$\nf (x, y) = \\left\\{ \\begin{array}{l l} \\frac {x y}{x ^ {2} + y ^ {2}}, & (x, y) \\neq (0, 0) \\\\ 0, & (x, y) = (0, 0) \\end{array} \\right.\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":110,"id":"e4297493-ef91-401b-a228-6dd486964d87","page_size":[595,841],"block_position":"110-6"},{"bbox":[53,264,301,285],"type":"text","angle":0,"index":7,"merge_prev":false,"text":"$f(x,y)$ 关于 $x,y$ 均连续，但是 $\\lim_{(x,y)\\to (0,0)}f(x,y)$ 不存在。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":110,"id":"97dbcfa8-5c4c-45a4-a320-99a2d1ae40fe","page_size":[595,841],"block_position":"110-7"},{"bbox":[35,296,449,309],"type":"text","angle":0,"index":8,"merge_prev":false,"text":"练习 12.6 加强条件 若 $f(x,y)$ 在 D 中关于 x,y 连续，且增加任意条件可证明 $f(x,y)$ 连续： ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":110,"id":"8f36faab-275b-41eb-9d44-7a6461c6c70e","page_size":[595,841],"block_position":"110-8"},{"bbox":[66,312,261,323],"type":"text","angle":0,"index":9,"merge_prev":false,"text":"1. 一致连续条件： $y$ 的连续是关于 $x$ 一致的 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":110,"id":"1e0f751f-3fc7-48ce-8766-70aa5af5e283","page_size":[595,841],"block_position":"110-9"},{"bbox":[65,327,324,339],"type":"text","angle":0,"index":10,"merge_prev":false,"text":"2. Lip 条件： $f(x,y)$ 关于 x 满足 Lip 条件，则 $f(x,y)$ 连续 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":110,"id":"2194d2e4-a158-4883-b0d0-e141392dff3d","page_size":[595,841],"block_position":"110-10"},{"bbox":[66,343,204,354],"type":"text","angle":0,"index":11,"merge_prev":false,"text":"3. 单调条件：关于 $x$ 或 $y$ 单调 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":110,"id":"03d6af6b-d3f7-4675-ab95-1db37b4bec96","page_size":[595,841],"block_position":"110-11"},{"bbox":[53,357,538,386],"type":"text","angle":0,"index":12,"merge_prev":false,"text":"证明 (1)(2) 用 $|f(x,y) - f(x_0,y_0)| \\leq |f(x,y) - f(x,y_0)| + |f(x,y_0) - f(x_0,y_0)| < 2\\epsilon$ 即可，注意第一个绝对值放缩要求 $y$ 的连续关于 $x$ 一致！ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":110,"id":"18a5bdd7-8c0d-42dc-8ceb-e279d08a43d7","page_size":[595,841],"block_position":"110-12"},{"bbox":[53,389,539,418],"type":"text","angle":0,"index":13,"merge_prev":false,"text":"(3) 设关于 $y$ 单调增。考虑一点 $(x_0, y_0)$ ，用 $(x, y)$ 逼近，设 $y_0 - \\delta_1 < y < y_0 + \\delta_1$ ，首先根据 $y$ 的单调性可知 $f(x, y) \\leq f(x, y_0 + \\delta_1)$ 根据 $y$ 的连续性可知 $f(x, y_0 + \\delta_1) \\leq f(x, y_0) + \\epsilon$ 。进一步地根据 $x$ 的连续性可知 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":110,"id":"45056bfa-5921-4452-ae51-6f98991ef69c","page_size":[595,841],"block_position":"110-13"},{"bbox":[213,425,381,438],"type":"interline_equation","angle":0,"index":14,"text":"$$\nf (x, y) \\leq f (x, y _ {0}) + \\epsilon \\leq f (x _ {0}, y _ {0}) + 2 \\epsilon\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":110,"id":"b1b23e37-8f8d-4247-809b-ef08dab630a4","page_size":[595,841],"block_position":"110-14"},{"bbox":[54,444,375,457],"type":"text","angle":0,"index":15,"merge_prev":false,"text":"同理可证得 $f(x,y)\\geq f(x_0,y_0) - 2\\epsilon$ ，综上可证 $\\left|f(x,y) - f(x_0,y_0)\\right|\\leq 2\\epsilon$ 。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":110,"id":"754e197d-12f7-4718-93ea-33877efca126","page_size":[595,841],"block_position":"110-15"},{"bbox":[57,475,216,488],"type":"title","angle":0,"index":16,"level":2,"text":"## 12.2.2 多元函数的一致连续性","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":110,"id":"d4b4abfb-4c8a-4adf-aa57-a4b5a383b01b","page_size":[595,841],"block_position":"110-16"},{"bbox":[73,502,196,515],"type":"title","angle":0,"index":17,"level":2,"text":"## 引理 12.1 (多元三角不等式)","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":110,"id":"40f24ce4-d721-4012-8712-f422aa8c56db","page_size":[595,841],"block_position":"110-17"},{"bbox":[67,518,537,539],"type":"text","angle":0,"index":18,"merge_prev":false,"text":"对 $\\forall \\mathbf{x},\\mathbf{y}\\in \\mathbb{R}^n$ ，有 $||\\mathbf{x} - |\\mathbf{y}||\\leq |\\mathbf{x} - \\mathbf{y}|$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":110,"id":"5b95fc99-d9bc-4425-a60c-4c377d39271f","page_size":[595,841],"block_position":"110-18"},{"bbox":[54,549,513,563],"type":"text","angle":0,"index":19,"merge_prev":false,"text":"证明 $|\\mathbf{x}| = |(\\mathbf{x} - \\mathbf{y}) + \\mathbf{y}| \\leq |\\mathbf{x} - \\mathbf{y}| + |\\mathbf{y}|$ ，因此移项得到 $|\\mathbf{x}| - |\\mathbf{y}| \\leq |\\mathbf{x} - \\mathbf{y}|$ 。同理可得 $|\\mathbf{y}| - |\\mathbf{x}| \\leq |\\mathbf{x} - \\mathbf{y}|$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":110,"id":"6804484e-541d-43e7-b6ba-e9cbed26e5a4","page_size":[595,841],"block_position":"110-19"},{"bbox":[36,579,539,674],"type":"text","angle":0,"index":20,"merge_prev":false,"text":"练习 12.7 多元一致连续 (1) $f(\\mathbf{x}) = |\\mathbf{x}|$ ，证明 $f(\\mathbf{x})$ 在 $\\mathbb{R}^n$ 上一致连续 (2) $f(x, y) = \\sin(xy)$ 在 $\\mathbb{R}^2$ 的一致连续性 (3) $f(x, y) = \\frac{1}{1 - xy}$ 在 $(0, 1) \\times (0, 1)$ 的一致连续性\n\n证明 (1) $\\forall \\epsilon, \\forall \\mathbf{x}, \\mathbf{y} \\in \\mathbb{R}^n$ ，若 $|\\mathbf{x} - \\mathbf{y}| < \\epsilon$ ，则根据三角不等式可知 $|f(\\mathbf{x}) - f(\\mathbf{y})| = ||\\mathbf{x}| - |\\mathbf{y}|| \\leq |\\mathbf{x} - \\mathbf{y}| < \\epsilon$ ，因此取 $\\delta = \\epsilon$ 即可。\n\n(2) 选 $a_n = (\\sqrt{2n\\pi}, \\sqrt{2n\\pi}), b_n = (\\sqrt{2n\\pi + \\frac{\\pi}{2}}, \\sqrt{2n\\pi + \\frac{\\pi}{2}})$ ，得到 $|b_n - a_n| \\to 0$ ，但 $|f(b_n) - f(a_n)| \\neq 0$ ，因此不一致连续。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":110,"id":"4b9cd376-734e-4e23-93ca-a60e78ce6a08","page_size":[595,841],"block_position":"110-20"},{"bbox":[53,676,540,724],"type":"text","angle":0,"index":21,"merge_prev":false,"text":"(3) 由于是在 $(1,1)$ 处出问题，可以考虑 $a_{n} = (1 - \\frac{1}{n}, 1 - \\frac{1}{n}), b_{n} = (1 - \\frac{1}{n + 1}, 1 - \\frac{1}{n + 1})$ ，显然 $|a_{n} - b_{n}| \\to 0$ ， $|f(a_{n}) - f(b_{n})| = \\left|\\frac{n^{2} - 3n^{2} + o(n^{2})}{4n^{2} - 1}\\right| \\to \\frac{1}{2}$ ，因此不一致连续。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":110,"id":"098ed437-d7e6-4312-87fd-8f60b959d8b4","page_size":[595,841],"block_position":"110-21"},{"bbox":[289,802,305,812],"type":"page_number","angle":0,"index":1000021,"text":"103 ","color":{"line":"rgba(164, 164, 164, 1)","fill":"rgba(164, 164, 164, 1)"},"page_idx":110,"is_discarded":true,"id":"a423ae74-8710-4c48-abdd-4a8c5b1c9a62","page_size":[595,841],"block_position":"110-22"}],[{"bbox":[446,47,538,58],"type":"header","angle":0,"index":-99999,"text":"12.2 多元函数连续性 ","color":{"line":"rgba(164, 164, 164, 1)","fill":"rgba(164, 164, 164, 1)"},"page_idx":111,"is_discarded":true,"id":"7d34370f-9732-487f-909f-90801cc06a2a","page_size":[595,841],"block_position":"111-0"},{"bbox":[73,74,235,85],"type":"title","angle":0,"index":1,"level":2,"text":"## 定理 12.3 (多元 Lip 连续与一致连续)","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":111,"id":"2ae38bf5-ada2-469c-9722-eac9fe3e9c60","page_size":[595,841],"block_position":"111-1"},{"bbox":[68,91,351,104],"type":"text","angle":0,"index":2,"merge_prev":false,"text":"$f(x,y)$ 是二元函数，若对 $\\forall (x_1,y),(x_2,y),(x,y_1),(x,y_2)\\in \\mathbb{R}^2$ 有 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":111,"id":"8182c209-4aa2-48be-8453-bfb9bd35b897","page_size":[595,841],"block_position":"111-2"},{"bbox":[145,111,448,124],"type":"interline_equation","angle":0,"index":3,"text":"$$\n\\left| f \\left(x _ {1}, y\\right) - f \\left(x _ {2}, y\\right) \\right| \\leq L \\left| x _ {1} - x _ {2} \\right|, \\quad \\left| f \\left(x, y _ {1}\\right) - f \\left(x, y _ {2}\\right) \\right| \\leq L \\left| y _ {1} - y _ {2} \\right|\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":111,"id":"4706f7a0-ffc8-48cb-9bce-69a1138cd402","page_size":[595,841],"block_position":"111-3"},{"bbox":[69,129,193,142],"type":"text","angle":0,"index":4,"merge_prev":false,"text":"则 $f(x,y)$ 在 $\\mathbb{R}^2$ 上一致连续 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":111,"id":"3bebea26-e217-4444-9bbd-9f968767fab6","page_size":[595,841],"block_position":"111-4"},{"bbox":[53,159,508,173],"type":"text","angle":0,"index":5,"merge_prev":false,"text":"证明 考虑 $\\left|f(x',y')-f(x'',y'')\\right|\\leq\\left|f(x',y')-f(x',y'')\\right|+\\left|f(x',y'')-f(x'',y'')\\right|$ ，用 Lip 条件放缩即可。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":111,"id":"c6a4da89-8830-4cee-a48f-952747dc0828","page_size":[595,841],"block_position":"111-5"},{"bbox":[289,803,305,811],"type":"page_number","angle":0,"index":1000005,"text":"104 ","color":{"line":"rgba(164, 164, 164, 1)","fill":"rgba(164, 164, 164, 1)"},"page_idx":111,"is_discarded":true,"id":"bdb777be-f5e7-4211-ae80-5798c00ad0c1","page_size":[595,841],"block_position":"111-6"}],[{"bbox":[223,74,381,94],"type":"title","angle":0,"index":0,"level":1,"text":"# 第 13 章 多元微分学","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":112,"id":"b988241c-286e-4d9d-9fb9-c5b927ff2931","page_size":[595,841],"block_position":"112-0"},{"bbox":[57,122,208,139],"type":"title","angle":0,"index":1,"level":2,"text":"## 13.1 多变量导数与微分","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":112,"id":"53c698c0-61b4-4a1c-8617-875e7a348efe","page_size":[595,841],"block_position":"112-1"},{"bbox":[57,155,180,169],"type":"title","angle":0,"index":2,"level":2,"text":"## 13.1.1 全微分与偏导数","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":112,"id":"bb6c3410-6456-467d-9d0a-102d80d0a989","page_size":[595,841],"block_position":"112-2"},{"bbox":[72,183,157,195],"type":"title","angle":0,"index":3,"level":2,"text":"## 定义 13.1 (偏导数)","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":112,"id":"54b16faa-40f4-47de-8a4a-e5f009c82169","page_size":[595,841],"block_position":"112-3"},{"bbox":[67,198,335,225],"type":"text","angle":0,"index":4,"merge_prev":false,"text":"$f$ 在 $\\mathbf{x}_0\\in D$ 的偏导数定义为 $\\frac{\\partial f}{\\partial x_i} := \\lim_{t\\to 0}\\frac{f(\\mathbf{x}_0 + t\\mathbf{e}_i) - f(\\mathbf{x}_0)}{t}$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":112,"id":"0c43f98f-f7a2-432a-bfa9-78da957e36bc","page_size":[595,841],"block_position":"112-4"},{"bbox":[33,239,439,255],"type":"text","angle":0,"index":5,"merge_prev":false,"text":"笔记 计算偏导 $f_{x}(x_{0},y_{0})$ 时，预先将 $y=y_{0}$ 代入，得到 $f(x,y_{0})$ ，再对 x 求导更加方便。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":112,"id":"14d74e76-3cbe-484e-9ae3-70a4951efb94","page_size":[595,841],"block_position":"112-5"},{"bbox":[72,264,147,276],"type":"title","angle":0,"index":6,"level":2,"text":"## 定义 13.2 (可微)","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":112,"id":"57168308-adda-41b0-9375-996d58eb73cf","page_size":[595,841],"block_position":"112-6"},{"bbox":[67,280,478,294],"type":"text","angle":0,"index":7,"merge_prev":false,"text":"若 $z = f(x,y)$ 在 $P_0(x_0,y_0)$ 邻域 $U(P_0)$ 中有定义，且对于 $U(P_0)$ 中任意点 $(x,y)$ ， $\\exists A, B$ 使得 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":112,"id":"d2e00825-a8fb-4924-9289-19bb989e8172","page_size":[595,841],"block_position":"112-7"},{"bbox":[136,300,456,313],"type":"interline_equation","angle":0,"index":8,"text":"$$\nf (x, y) - f \\left(x _ {0}, y _ {0}\\right) = A \\left(x - x _ {0}\\right) + B \\left(y - y _ {0}\\right) + o \\left(\\sqrt {\\left(x - x _ {0}\\right) ^ {2} + \\left(y - y _ {0}\\right) ^ {2}}\\right)\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":112,"id":"0a8e8492-c319-4e6d-ae0c-b4a657a88fa8","page_size":[595,841],"block_position":"112-8"},{"bbox":[67,314,528,343],"type":"text","angle":0,"index":9,"merge_prev":false,"text":"则称 $f(x,y)$ 在 $P_0$ 可微。可知这里 $A = \\frac{\\partial f}{\\partial x}, B = \\frac{\\partial f}{\\partial y}$ ，因此记 $\\mathrm{d}f(x_0,y_0) = \\frac{\\partial f}{\\partial x}(x_0,y_0)\\mathrm{d}x + \\frac{\\partial f}{\\partial y}(x_0,y_0)\\mathrm{d}y$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":112,"id":"cb180851-4336-479c-8ac2-1b4f3ef1c257","page_size":[595,841],"block_position":"112-9"},{"bbox":[52,356,539,386],"type":"text","angle":0,"index":10,"merge_prev":false,"text":"笔记可微定义中的范数可以更改，例如 $\\Delta z = A\\Delta x + B\\Delta y + \\alpha \\Delta x + \\beta \\Delta y$ ，在 $(\\Delta x,\\Delta y)\\to (0,0)$ 时 $\\alpha ,\\beta \\rightarrow 0$ 该条件也等价于可微。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":112,"id":"3ddc900c-2248-4d73-bfee-050f58ee319e","page_size":[595,841],"block_position":"112-10"},{"bbox":[73,393,176,407],"type":"title","angle":0,"index":11,"level":2,"text":"## 推论 13.1 (偏导与可微)","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":112,"id":"33f5f7e0-3b23-4591-99ff-69353335e54e","page_size":[595,841],"block_position":"112-11"},{"bbox":[69,410,311,424],"type":"text","angle":0,"index":12,"merge_prev":false,"text":"若 $f$ 在 $(x_0, y_0)$ 可微，则 $f_x(x_0, y_0), f_y(x_0, y_0)$ 一定存在 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":112,"id":"081f41ec-35f4-4b1e-8a90-b4afd96fa708","page_size":[595,841],"block_position":"112-12"},{"bbox":[53,440,174,453],"type":"title","angle":0,"index":13,"level":2,"text":"## 笔记 判断是否可微分两步：","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":112,"id":"7cabf35b-eb33-4113-80cd-beabc3992dc8","page_size":[595,841],"block_position":"112-13"},{"bbox":[66,456,261,468],"type":"text","angle":0,"index":14,"merge_prev":false,"text":"1. 看看 $f_{x}, f_{y}$ 是否都存在，不存在则不可微 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":112,"id":"abf61368-81df-410c-9e14-b4d62d799fd3","page_size":[595,841],"block_position":"112-14"},{"bbox":[64,468,410,497],"type":"text","angle":0,"index":15,"merge_prev":false,"text":"2. 计算 $\\frac{f(x + \\Delta x, y + \\Delta y) - f(x, y) - f_x \\Delta x - f_y \\Delta y}{\\sqrt{(\\Delta x)^2 + (\\Delta y)^2}}$ 是否极限为 0，是则可微 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":112,"id":"78b70c49-12d6-4aed-97e1-71e0ae6e70b0","page_size":[595,841],"block_position":"112-15"},{"bbox":[35,505,346,518],"type":"text","angle":0,"index":16,"merge_prev":false,"text":"练习13.1计算全微分(1)求 $u = xe^{yz} + e^{-z} + y$ 在原点处的全微分 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":112,"id":"a7a3ee43-9a24-4bda-944d-5d6c71e11af2","page_size":[595,841],"block_position":"112-16"},{"bbox":[52,520,364,534],"type":"text","angle":0,"index":17,"merge_prev":false,"text":"解(1)原点处 $u_{x} = 1, u_{y} = 1, u_{z} = -1$ ，因此原点处 $\\mathrm{d}u = \\mathrm{d}x + \\mathrm{d}y - \\mathrm{d}z$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":112,"id":"445fb647-88a4-4e70-98cc-27e2950d9ed0","page_size":[595,841],"block_position":"112-17"},{"bbox":[35,545,417,588],"type":"text","angle":0,"index":18,"merge_prev":false,"text":"练习13.2判断可微性 判断 $f(x,y) = \\left\\{ \\begin{array}{ll}\\frac{xy}{\\sqrt{x^2 + y^2}}, & (x,y)\\neq (0,0)\\\\ 0, & (x,y) = (0,0) \\end{array} \\right.$ 在原点的可微性 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":112,"id":"445a27c1-e2c1-445b-b2ac-985fb1420839","page_size":[595,841],"block_position":"112-18"},{"bbox":[52,590,395,602],"type":"text","angle":0,"index":19,"merge_prev":false,"text":"解 首先计算偏导，计算 x 偏导时将 y 固定为 0(而不是固定为一个很小的值！) ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":112,"id":"728de846-5eb5-40b1-a37b-e3130233e0ca","page_size":[595,841],"block_position":"112-19"},{"bbox":[222,604,371,629],"type":"interline_equation","angle":0,"index":20,"text":"$$\nf _ {x} (0, 0) = \\frac {f (\\Delta x , 0) - f (0 , 0)}{\\Delta x} = 0\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":112,"id":"7825125e-f7c8-4e13-b46c-dcfbf07901e5","page_size":[595,841],"block_position":"112-20"},{"bbox":[54,629,209,642],"type":"text","angle":0,"index":21,"merge_prev":false,"text":"同理可知 $f_{y}(0,0) = 0$ ，再考虑极限 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":112,"id":"88c366ed-3542-4feb-8285-14b2673482e2","page_size":[595,841],"block_position":"112-21"},{"bbox":[168,645,423,672],"type":"interline_equation","angle":0,"index":22,"text":"$$\n\\lim _ {\\rho \\rightarrow 0} \\frac {f (\\Delta x , \\Delta y) - f (0 , 0) - 0 \\Delta x - 0 \\Delta y}{\\sqrt {\\Delta x ^ {2} + \\Delta y ^ {2}}} = \\lim _ {\\rho \\rightarrow 0} \\frac {\\Delta x \\Delta y}{\\Delta x ^ {2} + \\Delta y ^ {2}}\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":112,"id":"b49afa4b-eb45-48d6-8acf-0a5e87fb41b4","page_size":[595,841],"block_position":"112-22"},{"bbox":[54,676,172,688],"type":"text","angle":0,"index":23,"merge_prev":false,"text":"极限不存在，因此不可微。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":112,"id":"c8daf126-7f7a-4a3d-8feb-be719c1fc312","page_size":[595,841],"block_position":"112-23"},{"bbox":[56,707,169,721],"type":"title","angle":0,"index":24,"level":2,"text":"## 13.1.2 可微性的证明","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":112,"id":"84feb055-a054-42e6-b6c4-f408aaf5daa1","page_size":[595,841],"block_position":"112-24"},{"bbox":[35,729,442,769],"type":"text","angle":0,"index":25,"merge_prev":false,"text":"练习13.3判断可微性(1)讨论 $f(x,y) = \\left\\{ \\begin{array}{ll}y\\ln (x^2 +y^2), & (x,y)\\neq (0,0)\\\\ 0, & (x,y) = (0,0) \\end{array} \\right.$ 在原点的可微性 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":112,"id":"e9731e79-fe03-4375-b79e-08ab7f698313","page_size":[595,841],"block_position":"112-25"}],[{"bbox":[435,46,538,58],"type":"header","angle":0,"index":-99999,"text":"13.1 多变量导数与微分 ","color":{"line":"rgba(164, 164, 164, 1)","fill":"rgba(164, 164, 164, 1)"},"page_idx":113,"is_discarded":true,"id":"afa4b48b-3bfc-445a-8b7a-87d25873b0aa","page_size":[595,841],"block_position":"113-0"},{"bbox":[73,72,283,84],"type":"text","angle":0,"index":1,"merge_prev":false,"text":"(2) $\\alpha > 0$ ，讨论 $f(x, y) = |xy|^{\\alpha}$ 在原点的可微性 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":113,"id":"180a0a12-969c-4d8b-811a-0ed3afb85317","page_size":[595,841],"block_position":"113-1"},{"bbox":[73,84,387,125],"type":"text","angle":0,"index":2,"merge_prev":false,"text":"(3) $\\alpha > 0$ ，讨论 $f(x, y) = \\begin{cases} \\frac{|x|^{\\alpha}y}{x^2 + y^2}, & (x, y) \\neq (0, 0) \\\\ 0, & (x, y) = (0, 0) \\end{cases}$ 在原点处的可微性 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":113,"id":"60ae8283-8ef4-4994-b46a-04b359483d85","page_size":[595,841],"block_position":"113-2"},{"bbox":[54,126,539,153],"type":"text","angle":0,"index":3,"merge_prev":false,"text":"解(1)先看看偏导数， $f_{x}(0,0)=\\lim_{x\\to0}\\frac{f(x,0)-f(0,0)}{x}=\\frac{0}{x}=0,\\quad f_{y}(0,0)=\\lim_{y\\to0}\\frac{y\\ln(y^{2})-0}{y}=-\\infty$ ，因此不可微 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":113,"id":"2fb5bb46-69d9-4ca7-af80-0bb173dc8ade","page_size":[595,841],"block_position":"113-3"},{"bbox":[72,152,377,174],"type":"text","angle":0,"index":4,"merge_prev":false,"text":"(2) 先看看偏导数 $f_{x}(0,0) = \\lim_{x\\to 0}\\frac{0}{x} = 0,\\quad f_{y}(0,0) = \\lim_{y\\to 0}\\frac{0}{y} = 0$ ，考虑： ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":113,"id":"9fcecf21-254c-4e60-a8ae-462312c7c8aa","page_size":[595,841],"block_position":"113-4"},{"bbox":[181,176,412,205],"type":"interline_equation","angle":0,"index":5,"text":"$$\n\\frac {f (x , y) - f (0 , 0) - f _ {x} (0 , 0) x - f _ {y} (0 , 0) y}{\\sqrt {x ^ {2} + y ^ {2}}} = \\frac {| x y | ^ {\\alpha}}{\\sqrt {x ^ {2} + y ^ {2}}}\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":113,"id":"bfd64907-d264-4628-81ca-633b48f2748d","page_size":[595,841],"block_position":"113-5"},{"bbox":[54,206,471,230],"type":"text","angle":0,"index":6,"merge_prev":false,"text":"考察上述分式是否趋于0，用极坐标变换即可得出若 $\\alpha >\\frac{1}{2}$ ，则可微，若 $0 < \\alpha \\leq \\frac{1}{2}$ ，则不可微 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":113,"id":"da13f404-72ee-4b95-8861-2e6b967fea7f","page_size":[595,841],"block_position":"113-6"},{"bbox":[54,228,540,276],"type":"text","angle":0,"index":7,"merge_prev":false,"text":"(3) $f_{x}(0,0) = f_{y}(0,0) = 0$ ，而 $\\frac{f(x,y)}{\\sqrt{x^2 + y^2}} = \\frac{|x|^{\\alpha}y}{(x^2 + y^2)^{\\frac{3}{2}}}$ ，使用极坐标变换得到 $r^{\\alpha -2}|\\cos \\theta |^{\\alpha}\\sin \\theta$ 。因此 $\\alpha >2$ 时可微， $0 < \\alpha \\leq 2$ 时不可微（例如取 $\\theta = \\frac{\\pi}{4})$ 。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":113,"id":"dda7e23a-302d-4b46-be41-891cbb35586e","page_size":[595,841],"block_position":"113-7"},{"bbox":[57,291,180,306],"type":"title","angle":0,"index":8,"level":2,"text":"## 13.1.3 梯度与方向导数","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":113,"id":"7b45425d-7cad-468a-b5ae-9934b6a67fa6","page_size":[595,841],"block_position":"113-8"},{"bbox":[73,318,146,331],"type":"title","angle":0,"index":9,"level":2,"text":"## 定义 13.3 (梯度)","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":113,"id":"cda494d4-a31f-4abe-ac98-20e7e43de4e2","page_size":[595,841],"block_position":"113-9"},{"bbox":[68,333,323,360],"type":"text","angle":0,"index":10,"merge_prev":false,"text":"对于 $f: R^{n} \\rightarrow R$ ，定义梯度 $\\nabla f := [\\frac{\\partial f}{\\partial x_{1}}, \\frac{\\partial f}{\\partial x_{2}}, \\cdots, \\frac{\\partial f}{\\partial x_{n}}]^{T}$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":113,"id":"f25f8da4-90cf-4e59-a544-3fd4b7b8afea","page_size":[595,841],"block_position":"113-10"},{"bbox":[73,380,166,393],"type":"title","angle":0,"index":11,"level":2,"text":"## 定义 13.4 (方向导数)","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":113,"id":"79bf5550-e822-4ffb-96f0-8dc6427df7e8","page_size":[595,841],"block_position":"113-11"},{"bbox":[67,396,477,423],"type":"text","angle":0,"index":12,"merge_prev":false,"text":"开集 $D \\subset \\mathbb{R}^n, f: D \\to \\mathbb{R}$ , $\\mathbf{u} \\in \\mathbb{R}^n$ 为单位向量, $\\mathbf{x}_0 \\in D$ , 定义 $\\frac{\\partial f}{\\partial \\mathbf{u}} := \\lim_{t \\to 0} \\frac{f(\\mathbf{x}_0 + t\\mathbf{u}) - f(\\mathbf{x}_0)}{t}$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":113,"id":"cab9573d-5953-43e6-aece-4ae457a49f1a","page_size":[595,841],"block_position":"113-12"},{"bbox":[52,434,538,481],"type":"text","angle":0,"index":13,"merge_prev":false,"text":"笔记 要计算方向导数，则可取方向为 $(\\cos\\alpha,\\cos\\beta)$ ，且 $\\cos^{2}\\alpha+\\cos^{2}\\beta=1$ 。原因在于 $\\frac{(x-x_{0},y-y_{0})}{|(x-x_{0},y-y_{0})|}=\\frac{1}{r}(x-x_{0},y-y_{0})=(\\cos\\alpha,\\cos\\beta)$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":113,"id":"1fa95b07-b4b8-4231-bd4f-c2536f5a925e","page_size":[595,841],"block_position":"113-13"},{"bbox":[35,492,468,505],"type":"text","angle":0,"index":14,"merge_prev":false,"text":"练习 13.4 常用反例 (1) 证明 $f(x,y)=\\sqrt[3]{x^{2}+y^{2}}$ 在原点连续，但不存在任意方向导数和偏导数 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":113,"id":"c619aa8a-8a3c-4b9b-9555-d4e5ccbcbb71","page_size":[595,841],"block_position":"113-14"},{"bbox":[51,506,495,530],"type":"text","angle":0,"index":15,"merge_prev":false,"text":"证明 (1) 考虑方向 $l = (\\cos \\alpha, \\cos \\beta)$ ，则 $f_{l}(0,0) = \\lim_{r\\to 0^{+}}\\frac{f(r\\cos\\alpha,r\\cos\\beta) - f(0,0)}{r} = \\lim_{r\\to 0^{+}}\\frac{1}{r^{1/3}} = \\infty$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":113,"id":"dde6aad2-2025-455c-bdd5-cadb2370f821","page_size":[595,841],"block_position":"113-15"},{"bbox":[73,545,255,558],"type":"title","angle":0,"index":16,"level":2,"text":"## 定理 13.1 (可微函数方向导数的梯度表达)","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":113,"id":"3b0fdb81-bd53-4d43-85d0-9ce53b2e4f9a","page_size":[595,841],"block_position":"113-16"},{"bbox":[69,562,421,576],"type":"text","angle":0,"index":17,"merge_prev":false,"text":"若 $f(x,y)$ 在 $P_0(x_0,y_0)$ 可微，则 $f(x,y)$ 在 $P_0$ 沿任一方向 $l$ 的方向导数存在，且 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":113,"id":"9d44b31a-a0a9-4a03-8bf0-4d6a44ae8e72","page_size":[595,841],"block_position":"113-17"},{"bbox":[196,582,396,595],"type":"interline_equation","angle":0,"index":18,"text":"$$\nf _ {l} (P _ {0}) = f _ {x} (P _ {0}) \\cos \\alpha + f _ {y} (P _ {0}) \\cos \\beta = \\nabla f \\cdot \\mathbf {n}\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":113,"id":"f2044f44-2ff7-466c-82b3-1096b89dd6a4","page_size":[595,841],"block_position":"113-18"},{"bbox":[68,602,259,613],"type":"text","angle":0,"index":20,"merge_prev":false,"text":"这里 $\\mathbf{n} = (\\cos \\alpha, \\cos \\beta)$ 是 $l$ 方向的单位向量 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":113,"id":"53e9a6a0-fb72-43dc-a6f5-dcd871d63104","page_size":[595,841],"block_position":"113-19"},{"bbox":[54,625,432,650],"type":"text","angle":0,"index":21,"merge_prev":false,"text":"证明 考虑方向导数 $f_{l}(P_{0}) = \\lim_{r\\to 0^{+}}\\frac{f(x_0 + r\\cos\\alpha, y_0 + r\\cos\\beta) - f(x_0, y_0)}{r}$ ，可化简为： ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":113,"id":"74a16ca3-628f-4194-8628-5072992f91af","page_size":[595,841],"block_position":"113-20"},{"bbox":[114,652,478,677],"type":"interline_equation","angle":0,"index":22,"text":"$$\n\\lim _ {r \\rightarrow 0 ^ {+}} \\frac {f _ {x} (x _ {0} , y _ {0}) r \\cos \\alpha + f _ {y} (x _ {0} , y _ {0}) r \\cos \\beta + o (r)}{r} = f _ {x} (x _ {0}, y _ {0}) \\cos \\alpha + f _ {y} (x _ {0}, y _ {0}) \\cos \\beta\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":113,"id":"db3a5791-c201-4351-accb-af6e52d037c1","page_size":[595,841],"block_position":"113-21"},{"bbox":[52,683,500,697],"type":"text","angle":0,"index":23,"merge_prev":false,"text":"笔记设 $\\theta$ 是 $\\nabla f$ 与 $l$ 的夹角，则 $f_{l}(\\mathbf{x}_{0}) = |\\nabla f||\\mathbf{n}|\\cos \\theta = |\\nabla f|\\cos \\theta$ 。可知梯度方向是增长最快的方向。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":113,"id":"dea6ef7b-096e-4971-8a37-6ac082ffc487","page_size":[595,841],"block_position":"113-22"},{"bbox":[35,715,442,728],"type":"text","angle":0,"index":24,"merge_prev":false,"text":"练习 13.5 梯度与方向导数计算 (1) 计算 u = xyz 在 $A(5,1,2)$ 朝向 $B(9,4,14)$ 的方向导数 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":113,"id":"4162721f-1b9a-49a1-a02a-1bad9cb559f2","page_size":[595,841],"block_position":"113-23"},{"bbox":[73,731,385,744],"type":"text","angle":0,"index":25,"merge_prev":false,"text":"(2) $u(x,y,z)=xy^{2}z^{3}$ ，求 $u(x,y,z)$ 在 $(1,1,1)$ 处函数值增长最快的方向 解(1)可微函数具体计算方向导数时一般用 $\\nabla f\\cdot \\mathbf{n}$ 计算。 $\\overrightarrow{AB} = (4,3,12)$ ，因此 $\\mathbf{n} = \\frac{1}{13} (4,3,12)$ ， $A$ 处梯度为(2,10,5)，因此 $\\nabla f\\cdot \\mathbf{n} = \\frac{1}{13} (8 + 30 + 60) = \\frac{98}{13}$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":113,"id":"ed6e11f4-ebe3-4472-9871-24bf520e779f","page_size":[595,841],"block_position":"113-24"},{"bbox":[289,802,305,812],"type":"page_number","angle":0,"index":1000018,"text":"106 ","color":{"line":"rgba(164, 164, 164, 1)","fill":"rgba(164, 164, 164, 1)"},"page_idx":113,"is_discarded":true,"id":"f064d89f-1ca6-45e3-965a-17bec8709914","page_size":[595,841],"block_position":"113-25"}],[{"bbox":[435,47,537,58],"type":"header","angle":0,"index":-99999,"text":"13.1 多变量导数与微分 ","color":{"line":"rgba(164, 164, 164, 1)","fill":"rgba(164, 164, 164, 1)"},"page_idx":114,"is_discarded":true,"id":"00b6ec3d-99aa-469b-9652-3ed9e3a46b71","page_size":[595,841],"block_position":"114-0"},{"bbox":[53,72,539,111],"type":"text","angle":0,"index":1,"merge_prev":false,"lines_deleted":true,"text":"","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":114,"id":"ed6e11f4-ebe3-4472-9871-24bf520e779f","page_size":[595,841],"block_position":"114-1"},{"bbox":[73,111,401,124],"type":"text","angle":0,"index":2,"merge_prev":false,"text":"(2) 即梯度方向， $u_{x}=y^{2}z^{3},u_{y}=2xyz^{3},u_{z}=3xy^{2}z^{2}$ ，因此梯度为 $(1,2,3)$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":114,"id":"6a41e514-1d60-4932-9719-9454921c9e5b","page_size":[595,841],"block_position":"114-2"},{"bbox":[36,142,539,171],"type":"text","angle":0,"index":3,"merge_prev":false,"text":"练习13.6方向导数的应用已知 $f(x,y)$ 可微，若 $l_{1},l_{2}$ 是 $\\mathbb{R}^2$ 上一组线性无关的向量，若 $f_{l_1}(x,y) = f_{l_2}(x,y)\\equiv 0$ ，证明 $f(x,y)$ 为常值函数。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":114,"id":"0bd66e7c-90ca-46b6-8e37-100ac887373c","page_size":[595,841],"block_position":"114-3"},{"bbox":[52,173,349,186],"type":"text","angle":0,"index":4,"merge_prev":false,"text":"证明 不妨设 $n_{l_1} = (\\cos \\alpha, \\sin \\alpha), n_{l_2} = (\\cos \\beta, \\sin \\beta)$ ，此时根据条件 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":114,"id":"d9248fd8-23b2-4c18-ba21-52236cd7697a","page_size":[595,841],"block_position":"114-4"},{"bbox":[225,190,358,227],"type":"interline_equation","angle":0,"index":5,"text":"$$\n\\left\\{ \\begin{array}{l} f _ {l _ {1}} = f _ {x} \\cos \\alpha + f _ {y} \\sin \\alpha \\equiv 0 \\\\ f _ {l _ {2}} = f _ {x} \\cos \\beta + f _ {y} \\sin \\beta \\equiv 0 \\end{array} \\right.\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":114,"id":"8c1be961-a92b-4866-9691-da802e6e8054","page_size":[595,841],"block_position":"114-5"},{"bbox":[54,231,305,243],"type":"text","angle":0,"index":6,"merge_prev":false,"text":"根据线性无关条件可解出 $f_{x} \\equiv f_{y} \\equiv 0$ ，因此为常值函数。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":114,"id":"6e00768b-0ff3-4bab-8c43-b62762fdcfc4","page_size":[595,841],"block_position":"114-6"},{"bbox":[57,262,300,276],"type":"title","angle":0,"index":7,"level":2,"text":"## 13.1.4 可微、偏导、方向导数、连续关系总结","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":114,"id":"65b453b2-22c0-4cc5-8779-7b2e07755722","page_size":[595,841],"block_position":"114-7"},{"bbox":[72,290,176,301],"type":"title","angle":0,"index":8,"level":2,"text":"## 定理 13.2 (可微的性质)","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":114,"id":"57b1c15c-129a-4385-8879-9f5e41ef8477","page_size":[595,841],"block_position":"114-8"},{"bbox":[69,306,158,318],"type":"text","angle":0,"index":9,"merge_prev":false,"text":"若 $f$ 在 $\\mathbf{x}_0$ 可微，则 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":114,"id":"ae2b6453-5f77-4c0f-933e-4eff07f4c568","page_size":[595,841],"block_position":"114-9"},{"bbox":[83,322,161,334],"type":"text","angle":0,"index":10,"merge_prev":false,"text":"- $f$ 在 $\\mathbf{x}_0$ 处连续 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":114,"id":"e7950e7d-5e4b-4c90-a71f-7c4d920cf615","page_size":[595,841],"block_position":"114-10"},{"bbox":[83,337,260,349],"type":"text","angle":0,"index":12,"merge_prev":false,"text":"- $f$ 在 $\\mathbf{x}_0$ 处存在偏导数和任意方向导数 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":114,"id":"86591d17-1798-45ba-8ac9-effe1838c8e2","page_size":[595,841],"block_position":"114-11"},{"bbox":[53,367,359,379],"type":"text","angle":0,"index":13,"merge_prev":false,"text":"证明 (1) 根据定义显然 $||\\mathbf{h}|| \\to 0$ 时， $f(\\mathbf{x} + \\mathbf{h}) - f(\\mathbf{x}) \\to 0$ ，因此连续 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":114,"id":"4189375e-194e-4681-b014-26b9b990fddd","page_size":[595,841],"block_position":"114-12"},{"bbox":[72,389,186,401],"type":"title","angle":0,"index":14,"level":2,"text":"## 定理 13.3 (可微充分条件)","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":114,"id":"7f3a2711-4668-401d-b842-f245d3586422","page_size":[595,841],"block_position":"114-13"},{"bbox":[67,406,362,419],"type":"text","angle":0,"index":15,"merge_prev":false,"text":"若 $f(x,y)$ 偏导数 $f_{x},f_{y}$ 在 $(x_0,y_0)$ 连续，则 $f(x,y)$ 在 $(x_0,y_0)$ 可微 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":114,"id":"00552751-456f-4835-8719-c531948668dc","page_size":[595,841],"block_position":"114-14"},{"bbox":[54,435,539,464],"type":"text","angle":0,"index":16,"merge_prev":false,"text":"证明 首先写为 $\\Delta z = [f(x_0 + \\Delta x, y_0 + \\Delta y) - f(x_0, y_0 + \\Delta y)] + [f(x_0, y_0 + \\Delta y) - f(x_0, y_0)]$ ，根据Lagrange中值定理得到 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":114,"id":"b06f5015-00d5-40ec-a366-d6c690343d23","page_size":[595,841],"block_position":"114-15"},{"bbox":[172,468,421,481],"type":"interline_equation","angle":0,"index":17,"text":"$$\n\\Delta z = f _ {x} (x _ {0} + \\theta_ {1} \\Delta x, y _ {0} + \\Delta y) \\Delta x + f _ {y} (x _ {0}, y _ {0} + \\theta_ {2} \\Delta y) \\Delta y\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":114,"id":"b37345c7-663b-4aea-bc99-896a7814d0a5","page_size":[595,841],"block_position":"114-16"},{"bbox":[54,491,324,503],"type":"text","angle":0,"index":18,"merge_prev":false,"text":"根据偏导连续， $\\exists \\alpha, \\beta$ 在 $(\\Delta x, \\Delta y) \\to (0, 0)$ 时 $\\alpha, \\beta \\to 0$ ，满足 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":114,"id":"06543f4b-5d88-4e55-a3c4-c581bef2a4c7","page_size":[595,841],"block_position":"114-17"},{"bbox":[198,507,386,543],"type":"interline_equation","angle":0,"index":19,"text":"$$\n\\left\\{ \\begin{array}{l} f _ {x} (x _ {0} + \\theta_ {1} \\Delta x, y _ {0} + \\Delta y) = f _ {x} (x _ {0}, y _ {0}) + \\alpha \\\\ f _ {y} (x _ {0}, y _ {0} + \\Delta y) = f _ {y} (x _ {0}, y _ {0}) + \\beta \\end{array} \\right.\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":114,"id":"3207eb54-eb39-414f-b9e3-51d49915a336","page_size":[595,841],"block_position":"114-18"},{"bbox":[54,548,338,561],"type":"text","angle":0,"index":20,"merge_prev":false,"text":"得到 $\\Delta z = f_{x}(x_{0}, y_{0})\\Delta x + f_{y}(x_{0}, y_{0})\\Delta y + \\alpha\\Delta x + \\beta\\Delta y$ ，即可微。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":114,"id":"e6b700e7-bd48-4bb2-b761-ef3b5011ffb9","page_size":[595,841],"block_position":"114-19"},{"bbox":[54,565,267,577],"type":"text","angle":0,"index":21,"merge_prev":false,"text":"笔记 若 $f_{x}, f_{y}$ 连续，则一般称 $f(x, y)$ 连续可微。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":114,"id":"914a7645-e5a3-41f2-b11d-0628be07f36d","page_size":[595,841],"block_position":"114-20"},{"bbox":[72,587,215,599],"type":"title","angle":0,"index":22,"level":2,"text":"## 定理 13.4 (偏导与方向导的性质)","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":114,"id":"5804d523-32f0-40df-aa83-9ba6065c47e8","page_size":[595,841],"block_position":"114-21"},{"bbox":[69,603,186,615],"type":"text","angle":0,"index":23,"merge_prev":false,"text":"偏导与方向导数相关性质： ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":114,"id":"67ccf938-0b55-4e0c-8f8e-7f35c042eb5a","page_size":[595,841],"block_position":"114-22"},{"bbox":[83,619,332,632],"type":"text","angle":0,"index":24,"merge_prev":false,"text":"- 若 $f$ 偏导数在 $(x_0, y_0)$ 邻域有界，则 $f$ 在 $(x_0, y_0)$ 连续 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":114,"id":"393d89b5-d871-4fee-8e11-744350a97fd7","page_size":[595,841],"block_position":"114-23"},{"bbox":[83,634,354,647],"type":"text","angle":0,"index":25,"merge_prev":false,"text":"- $f$ 的偏导数存在或任意方向导数存在并不能推出连续或可微 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":114,"id":"cd388e7a-eeaf-4c98-9646-f72629d49c81","page_size":[595,841],"block_position":"114-24"},{"bbox":[54,664,513,677],"type":"text","angle":0,"index":26,"merge_prev":false,"text":"证明 (1) 首先 $f(x, y) - f(x_0, y_0) = f(x, y) - f(x_0, y) + f(x_0, y) - f(x_0, y_0)$ ，根据 Lagrange 中值定理得到 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":114,"id":"093ed98d-b857-4564-abff-cd3afded9a0a","page_size":[595,841],"block_position":"114-25"},{"bbox":[124,684,467,697],"type":"interline_equation","angle":0,"index":27,"text":"$$\nf (x, y) - f \\left(x _ {0}, y\\right) = f _ {x} (\\xi , y) \\left(x - x _ {0}\\right), \\quad f \\left(x _ {0}, y\\right) - f \\left(x _ {0}, y _ {0}\\right) = f _ {y} \\left(x _ {0}, \\eta\\right) \\left(y - y _ {0}\\right)\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":114,"id":"b70d31b3-85ad-453c-8be7-c1c6de7db5bb","page_size":[595,841],"block_position":"114-26"},{"bbox":[54,703,329,716],"type":"text","angle":0,"index":28,"merge_prev":false,"text":"根据偏导有界条件，因此 $|f(x,y) - f(x_0,y_0)|\\to 0$ ，因此连续。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":114,"id":"bf123d62-944e-4aac-a5f5-389698f67ae5","page_size":[595,841],"block_position":"114-27"},{"bbox":[73,719,182,731],"type":"text","angle":0,"index":29,"merge_prev":false,"text":"(2) 均见下面的重要反例 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":114,"id":"27c0b3cd-3446-4d27-ad1b-30d61b766ea5","page_size":[595,841],"block_position":"114-28"},{"bbox":[289,802,305,812],"type":"page_number","angle":0,"index":1000010,"text":"107 ","color":{"line":"rgba(164, 164, 164, 1)","fill":"rgba(164, 164, 164, 1)"},"page_idx":114,"is_discarded":true,"id":"c8e5fbb6-25c7-4f4c-9b82-3d8732a981e3","page_size":[595,841],"block_position":"114-29"}],[{"bbox":[435,47,538,58],"type":"header","angle":0,"index":-99999,"text":"13.1 多变量导数与微分 ","color":{"line":"rgba(164, 164, 164, 1)","fill":"rgba(164, 164, 164, 1)"},"page_idx":115,"is_discarded":true,"id":"3bc38d77-ba28-4160-9db1-5add3dad6bb4","page_size":[595,841],"block_position":"115-0"},{"bbox":[35,71,539,100],"type":"text","angle":0,"index":1,"merge_prev":false,"text":"练习 13.7 方向导数、偏导数存在，但不可微，不连续例子证明下面的 $f(x,y)$ 在原点不连续，但是 $f_{x}(0,0)=f_{y}(0,0)=0$ ，任何方向 l 满足 $f_{l}(0,0)=0$ ， ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":115,"id":"dffe53ca-55dc-40d6-bdc5-8f9269b5cb69","page_size":[595,841],"block_position":"115-1"},{"bbox":[202,104,389,141],"type":"interline_equation","angle":0,"index":2,"text":"$$\nf (x, y) = \\left\\{ \\begin{array}{l l} 1, & 0 < y < x ^ {2}, - \\infty < x < \\infty \\\\ 0, & \\text {其他} \\end{array} \\right.\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":115,"id":"2cc44db7-1af2-4361-9860-c68dbff06d7b","page_size":[595,841],"block_position":"115-2"},{"bbox":[52,145,539,174],"type":"text","angle":0,"index":3,"merge_prev":false,"text":"证明 不连续的原因是可以选择 0 趋近，也可以选择 1 趋近。但方向导数都为 0 的原因是 $y = x^2$ 在原点处切线是水平的，无论原点处切线取什么，都会经过 0 区域。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":115,"id":"0404033b-5a62-43e7-ad0d-1b4c1d050c96","page_size":[595,841],"block_position":"115-3"},{"bbox":[56,208,204,222],"type":"title","angle":0,"index":4,"level":2,"text":"## 13.1.5 复合求导：链式法则","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":115,"id":"ddb5cf44-2592-4093-8bd7-85260aed4575","page_size":[595,841],"block_position":"115-4"},{"bbox":[73,235,166,248],"type":"title","angle":0,"index":5,"level":2,"text":"## 定理 13.5 (链式法则)","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":115,"id":"13aaee76-8e04-43e1-8f78-98f35e209aff","page_size":[595,841],"block_position":"115-5"},{"bbox":[67,251,524,280],"type":"text","angle":0,"index":6,"merge_prev":false,"text":"若 $x = \\varphi(x, t), y = \\psi(s, t)$ 在 $(s, t) \\in D$ 可微， $z = f(x, y)$ 在 $(x, y) = (\\varphi(s, t), \\varphi(s, t))$ 可微，则复合函数 $z = f(\\varphi(s, t), \\psi(s, t))$ 在 $(s, t)$ 可微，且 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":115,"id":"d633716d-649e-4398-a834-653f677b7aa2","page_size":[595,841],"block_position":"115-6"},{"bbox":[189,282,394,338],"type":"interline_equation","angle":0,"index":7,"text":"$$\n\\left\\{ \\begin{array}{l} \\frac {\\partial z}{\\partial s} \\bigg | _ {(s, t)} = \\frac {\\partial z}{\\partial x} \\bigg | _ {(x, y)} \\frac {\\partial x}{\\partial s} \\bigg | _ {(s, t)} + \\frac {\\partial z}{\\partial y} \\bigg | _ {(x, y)} \\frac {\\partial y}{\\partial s} \\bigg | _ {(s, t)} \\\\ \\frac {\\partial z}{\\partial t} \\bigg | _ {(s, t)} = \\frac {\\partial z}{\\partial x} \\bigg | _ {(x, y)} \\frac {\\partial x}{\\partial t} \\bigg | _ {(s, t)} + \\frac {\\partial z}{\\partial y} \\bigg | _ {(x, y)} \\frac {\\partial y}{\\partial t} \\bigg | _ {(s, t)} \\end{array} \\right.\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":115,"id":"e135517b-58d3-40fc-a356-7bfb487e51be","page_size":[595,841],"block_position":"115-7"},{"bbox":[35,364,396,388],"type":"text","angle":0,"index":8,"merge_prev":false,"text":"练习 13.8 链式法则练习 (1) 设 $z = uv + \\sin t$ ，其中 $u = e^{t}, v = \\cos t$ ，计算 $\\frac{dz}{dt}$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":115,"id":"93054691-a27a-4f0a-95a5-f60b03ada0e3","page_size":[595,841],"block_position":"115-8"},{"bbox":[73,386,158,397],"type":"text","angle":0,"index":10,"merge_prev":false,"text":"(2) 计算 $x^{x}$ 的导数 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":115,"id":"3a8fad13-4322-4ce7-a4d6-408607597368","page_size":[595,841],"block_position":"115-9"},{"bbox":[53,401,138,412],"type":"text","angle":0,"index":11,"merge_prev":false,"text":"解(1)展开后如下: ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":115,"id":"4f4be209-f4a2-48f9-98e6-be7a3fd40ecb","page_size":[595,841],"block_position":"115-10"},{"bbox":[178,414,415,455],"type":"interline_equation","angle":0,"index":12,"text":"$$\n\\begin{array}{r l} & {\\frac {\\mathrm{d} z}{\\mathrm{d} t} = \\frac {\\partial z}{\\partial u} \\frac {\\mathrm{d} u}{\\mathrm{d} t} + \\frac {\\partial z}{\\partial v} \\frac {\\mathrm{d} v}{\\mathrm{d} t} + \\frac {\\partial z}{\\partial t}} \\\\ & {\\qquad = v e ^ {t} + u (- \\sin t) + \\cos t = e ^ {t} (\\cos t - \\sin t) + \\cos t} \\end{array}\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":115,"id":"2e78aac2-dc1e-47a4-808c-e4e0d01423b5","page_size":[595,841],"block_position":"115-11"},{"bbox":[73,461,470,483],"type":"text","angle":0,"index":13,"merge_prev":false,"text":"(2) 可以两边取对数隐函数求导。或者令 $y = u^v$ ，则 $\\frac{\\mathrm{d}y}{\\mathrm{d}x} = u^v \\ln u + v(u^{v-1}) = x^x (1 + \\ln x)$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":115,"id":"641df91f-fd28-4c39-b26d-4bfd93153800","page_size":[595,841],"block_position":"115-12"},{"bbox":[35,489,303,515],"type":"text","angle":0,"index":14,"merge_prev":false,"text":"练习13.9高阶偏导(1)设 $z = f\\left(x,\\frac{x}{y}\\right)$ ，求 $\\frac{\\partial^2z}{\\partial x^2},\\frac{\\partial^2z}{\\partial x\\partial y}$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":115,"id":"e4e6cc8c-fb26-4705-b9b2-243167a9fe65","page_size":[595,841],"block_position":"115-13"},{"bbox":[53,515,329,527],"type":"text","angle":0,"index":15,"merge_prev":false,"text":"解 把每个位置当作一层参数即可，注意偏导的参数仍保持一致 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":115,"id":"3350c9b4-6f47-42b4-872c-31760efc1520","page_size":[595,841],"block_position":"115-14"},{"bbox":[73,527,237,550],"type":"text","angle":0,"index":16,"merge_prev":false,"text":"(1) $\\frac{\\partial z}{\\partial x} = f_{1}(x, \\frac{x}{y}) + f_{2}(x, \\frac{x}{y})\\frac{1}{y}$ ，因此 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":115,"id":"f0a8f718-8322-4a08-8767-7cacbcf2c708","page_size":[595,841],"block_position":"115-15"},{"bbox":[168,552,425,580],"type":"interline_equation","angle":0,"index":17,"text":"$$\n\\frac {\\partial^ {2} z}{\\partial x ^ {2}} = f _ {1 1} + f _ {1 2} \\frac {1}{y} + \\left[ f _ {2 1} + f _ {2 2} \\frac {1}{y} \\right] \\frac {1}{y} = f _ {1 1} + \\frac {2}{y} f _ {1 2} + \\frac {1}{y ^ {2}} f _ {2 2}\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":115,"id":"e93c51a4-90cb-44ac-a57d-5275430d9e21","page_size":[595,841],"block_position":"115-16"},{"bbox":[57,595,145,609],"type":"title","angle":0,"index":18,"level":2,"text":"## 13.1.6 混合偏导","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":115,"id":"af7c2008-c9ea-46fc-bc6e-9c62299187d3","page_size":[595,841],"block_position":"115-17"},{"bbox":[73,623,186,634],"type":"title","angle":0,"index":19,"level":2,"text":"## 定理 13.6 (混合偏导定理)","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":115,"id":"bee1baa9-0a24-4e92-ad1d-cff81022bc8b","page_size":[595,841],"block_position":"115-18"},{"bbox":[69,640,275,652],"type":"text","angle":0,"index":20,"merge_prev":false,"text":"若 $f_{xy}(x,y), f_{yx}(x,y)$ 均在 $x_0, y_0$ 邻域连续，则 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":115,"id":"a8311dd3-7432-48be-b822-1b139aadb5ad","page_size":[595,841],"block_position":"115-19"},{"bbox":[242,660,351,672],"type":"interline_equation","angle":0,"index":21,"text":"$$\nf _ {x y} (x _ {0}, y _ {0}) = f _ {y x} (x _ {0}, y _ {0})\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":115,"id":"62b6e094-ee97-4ee8-8ad1-696c53ed82af","page_size":[595,841],"block_position":"115-20"},{"bbox":[54,689,477,702],"type":"text","angle":0,"index":22,"merge_prev":false,"text":"证明 定义 $F(\\Delta x, \\Delta y) = f(x_0 + \\Delta x, y_0 + \\Delta y) - f(x_0 + \\Delta x, y_0) - f(x_0, y_0 + \\Delta y) + f(x_0, y_0)$ ，则 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":115,"id":"ed68f8d4-0d7b-4c29-a820-1072a28d81f4","page_size":[595,841],"block_position":"115-21"},{"bbox":[148,707,434,744],"type":"interline_equation","angle":0,"index":23,"text":"$$\nF (\\Delta x, \\Delta y) = \\left\\{ \\begin{array}{l} [ f _ {x} (x _ {0} + \\theta_ {1} \\Delta x, y _ {0} + \\Delta y) - f _ {x} (x _ {0} + \\theta_ {1} \\Delta x, y _ {0}) ] \\Delta x \\\\ [ f _ {y} (x _ {0} + \\Delta x, y _ {0} + \\theta_ {2} \\Delta y) - f _ {y} (x _ {0}, y _ {0} + \\theta_ {2} \\Delta y) ] \\Delta y \\end{array} \\right.\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":115,"id":"96891693-cb7f-4a1c-bf7a-1168e1df62ce","page_size":[595,841],"block_position":"115-22"},{"bbox":[289,803,305,812],"type":"page_number","angle":0,"index":1000008,"text":"108 ","color":{"line":"rgba(164, 164, 164, 1)","fill":"rgba(164, 164, 164, 1)"},"page_idx":115,"is_discarded":true,"id":"ec77122a-a0a2-48df-af2a-a2268cd88f2a","page_size":[595,841],"block_position":"115-23"}],[{"bbox":[405,47,539,58],"type":"header","angle":0,"index":-99999,"text":"13.2 多元中值定理与泰勒展开 ","color":{"line":"rgba(164, 164, 164, 1)","fill":"rgba(164, 164, 164, 1)"},"page_idx":116,"is_discarded":true,"id":"ef6a8659-56a3-4882-9e6d-1438654c7d1e","page_size":[595,841],"block_position":"116-0"},{"bbox":[54,72,174,84],"type":"text","angle":0,"index":1,"merge_prev":false,"text":"再根据 Lagrange 中值定理: ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":116,"id":"5763cb73-699d-44e9-a0cb-5e8d2dddb8b8","page_size":[595,841],"block_position":"116-1"},{"bbox":[181,88,401,125],"type":"interline_equation","angle":0,"index":3,"text":"$$\nF (\\Delta x, \\Delta y) = \\left\\{ \\begin{array}{l} f _ {x y} (x _ {0} + \\theta_ {1} \\Delta x, y _ {0} + \\theta_ {3} \\Delta y) \\Delta x \\Delta y \\\\ f _ {y x} (x _ {0} + \\theta_ {4} \\Delta x, y _ {0} + \\theta_ {2} \\Delta y) \\Delta x \\Delta y \\end{array} \\right.\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":116,"id":"3576c4ed-c04d-42ce-a73c-7ad535fb0ef8","page_size":[595,841],"block_position":"116-2"},{"bbox":[54,128,530,155],"type":"text","angle":0,"index":4,"merge_prev":false,"text":"因此 $\\frac{F(\\Delta x, \\Delta y)}{\\Delta x \\Delta y} = f_{xy}(x_0 + \\theta_1 \\Delta x, y_0 + \\theta_3 \\Delta y) = f_{yx}(x_0 + \\theta_4 \\Delta x, y_0 + \\theta_2 \\Delta y)$ ，令 $\\Delta x, \\Delta y \\to 0$ 即可得到结果。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":116,"id":"959e660c-b2ed-4b65-99c3-57e470669eb5","page_size":[595,841],"block_position":"116-3"},{"bbox":[35,166,377,179],"type":"text","angle":0,"index":5,"merge_prev":false,"text":"练习 13.10 混合偏导不相同的例子 对于下面的函数， $f_{xy}(0,0) \\neq f_{yx}(0,0)$ : ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":116,"id":"61565ffc-c055-4b71-9ca7-2d46d4454621","page_size":[595,841],"block_position":"116-4"},{"bbox":[211,181,380,225],"type":"interline_equation","angle":0,"index":6,"text":"$$\nf (x, y) = \\left\\{ \\begin{array}{l l} x y \\frac {x ^ {2} - y ^ {2}}{x ^ {2} + y ^ {2}} & (x, y) \\neq (0, 0) \\\\ 0, & (x, y) = (0, 0) \\end{array} \\right.\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":116,"id":"6f76f98f-eed4-47b1-99c1-a6f16cf00593","page_size":[595,841],"block_position":"116-5"},{"bbox":[52,229,274,243],"type":"text","angle":0,"index":7,"merge_prev":false,"text":"解 直接计算可以得到 $f_{xy}(0,0) = -1, f_{yx}(0,0) = 1$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":116,"id":"4968f9f3-435a-4154-9a52-257fc4cf08b2","page_size":[595,841],"block_position":"116-6"},{"bbox":[35,259,539,303],"type":"text","angle":0,"index":8,"merge_prev":false,"text":"练习13.11混合偏导相关练习(1)设 $f_{x}, f_{y}, f_{yx}$ 在 $(x_0, y_0)$ 邻域存在， $f_{yx}$ 在 $(x_0, y_0)$ 连续，证明 $f_{xy}(x_0, y_0)$ 也存在，且 $f_{xy}(x_0, y_0) = f_{yx}(x_0, y_0)$ (2)设 $f_{x}, f_{y}$ 在 $(x_0, y_0)$ 某邻域存在且在 $(x_0, y_0)$ 可微，证明： $f_{xy}(x_0, y_0) = f_{yx}(x_0, y_0)$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":116,"id":"8dacb181-2fe4-451e-a860-624de128dc14","page_size":[595,841],"block_position":"116-7"},{"bbox":[57,327,251,343],"type":"title","angle":0,"index":9,"level":2,"text":"## 13.2 多元中值定理与泰勒展开","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":116,"id":"a79fe450-7f7a-418c-a09f-6bfaac32f93d","page_size":[595,841],"block_position":"116-8"},{"bbox":[57,359,168,373],"type":"title","angle":0,"index":10,"level":2,"text":"## 13.2.1 多元中值定理","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":116,"id":"7d432114-1785-4409-ab64-d2ac3c6eb836","page_size":[595,841],"block_position":"116-9"},{"bbox":[73,386,155,399],"type":"title","angle":0,"index":11,"level":2,"text":"## 定义 13.5 (凸区域)","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":116,"id":"d67b2f7e-25fc-45a2-8281-7e928e960663","page_size":[595,841],"block_position":"116-10"},{"bbox":[69,404,346,417],"type":"text","angle":0,"index":12,"merge_prev":false,"text":"区域 $D \\subset \\mathbb{R}^2$ ，若 $\\forall P_1(x_1, y_1), P_2(x_2, y_2) \\in D$ 和 $\\forall \\lambda \\in [0, 1]$ 满足 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":116,"id":"a70ce2eb-f10e-4a7c-9673-85bbd80ad5a7","page_size":[595,841],"block_position":"116-11"},{"bbox":[211,424,382,436],"type":"interline_equation","angle":0,"index":13,"text":"$$\n\\left(x _ {1} + \\lambda \\left(x _ {2} - x _ {1}\\right), y _ {1} + \\lambda \\left(y _ {2} - y _ {1}\\right)\\right) \\in D\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":116,"id":"9c71d68d-3c71-4a9e-9333-8b019f5a0303","page_size":[595,841],"block_position":"116-12"},{"bbox":[69,444,280,455],"type":"text","angle":0,"index":14,"merge_prev":false,"text":"即任意两点的连线都含于 D，则称 D 为凸区域。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":116,"id":"f74d6443-d2f5-4102-8c74-43e2f284aad6","page_size":[595,841],"block_position":"116-13"},{"bbox":[193,498,270,558],"type":"image","angle":0,"index":15,"img_path":"/2512492e270f8e5160ea87da7c79597fffe8f0ac2bd4813aa99fd215e988f5ce.jpg","color":{"line":"rgba(89, 92, 220, 1)","fill":"rgba(89, 92, 220, 1)"},"img_ratio":0.2,"page_idx":116,"id":"3d8bc557-b0fc-40c3-9f0f-1203909e11ba","page_size":[595,841],"block_position":"116-14"},{"bbox":[217,561,245,572],"type":"image_caption","angle":0,"index":16,"text":"凸区域","id":"fdf8ab5e-b8d5-4cbe-843c-0ac8962cad9d","color":{"line":"rgba(13, 83, 222 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":116,"page_size":[595,841],"block_position":"116-15"},{"bbox":[288,483,390,556],"type":"image","angle":0,"index":17,"img_path":"/5ee26be4f33a285c0c1af4e019ea2e8a584421a961c78fd1c940bb689030f299.jpg","color":{"line":"rgba(89, 92, 220, 1)","fill":"rgba(89, 92, 220, 1)"},"img_ratio":0.2,"page_idx":116,"id":"6a38ecb0-9af7-4630-b396-0ea191f3efe7","page_size":[595,841],"block_position":"116-16"},{"bbox":[323,560,359,572],"type":"image_caption","angle":0,"index":18,"text":"非凸区域","id":"238a3aa4-a174-49bc-bd31-55e55d2697e4","color":{"line":"rgba(13, 83, 222 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":116,"page_size":[595,841],"block_position":"116-17"},{"bbox":[262,582,332,595],"type":"image_caption","angle":0,"index":19,"text":"图 13.1: 凸区域","id":"c6199b28-2c7e-49bf-b490-77de92940a1f","color":{"line":"rgba(13, 83, 222 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":116,"page_size":[595,841],"block_position":"116-18"},{"bbox":[73,618,186,630],"type":"title","angle":0,"index":20,"level":2,"text":"## 定理 13.7 (多元中值定理)","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":116,"id":"e23f932b-849a-4474-b2a4-b83566b88e97","page_size":[595,841],"block_position":"116-19"},{"bbox":[68,634,524,662],"type":"text","angle":0,"index":21,"merge_prev":false,"text":"二元函数 $f(x,y)$ 在凸开域 $D\\subset \\mathbb{R}^2$ 连续，在 $D$ 内点均可微，则对 $\\forall P(a,b),Q(a + h,b + k)\\in D,\\exists \\theta \\in (0,1)$ 满足 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":116,"id":"77829c2b-ef89-4e37-9954-0cad0c2fc2bb","page_size":[595,841],"block_position":"116-20"},{"bbox":[148,666,446,679],"type":"interline_equation","angle":0,"index":22,"text":"$$\nf (a + h, b + k) - f (a, b) = f _ {x} (a + \\theta h, b + \\theta k) h + f _ {y} (a + \\theta h, b + \\theta k) k\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":116,"id":"7d0b06f7-fc9c-4592-920e-69b4f5849fbf","page_size":[595,841],"block_position":"116-21"},{"bbox":[54,697,509,709],"type":"text","angle":0,"index":23,"merge_prev":false,"text":"证明 令 $\\Phi(t) = f(a + th, b + tk)$ ，根据 Lagrange 中值定理得到 $\\exists \\theta \\in (0,1)$ 使得 $\\Phi(1) - \\Phi(0) = \\Phi'( \\theta)$ ，而 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":116,"id":"8cb70f0f-d1de-4ee9-a656-248608fe0df5","page_size":[595,841],"block_position":"116-22"},{"bbox":[188,716,406,729],"type":"interline_equation","angle":0,"index":24,"text":"$$\n\\Phi^ {\\prime} (\\theta) = f _ {x} (a + \\theta h, b + \\theta k) h + f _ {y} (a + \\theta h, b + \\theta k) k\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":116,"id":"2be63035-c507-4dde-ab16-823c81c6249e","page_size":[595,841],"block_position":"116-23"},{"bbox":[55,735,326,748],"type":"text","angle":0,"index":25,"merge_prev":false,"text":"由于 $D$ 是凸区域，因此 $(a + \\theta h, b + \\theta k) \\in D$ ，从而结论成立。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":116,"id":"220d6475-c23c-4635-8b33-8c0162ee9164","page_size":[595,841],"block_position":"116-24"},{"bbox":[289,802,305,812],"type":"page_number","angle":0,"index":1000001,"text":"109 ","color":{"line":"rgba(164, 164, 164, 1)","fill":"rgba(164, 164, 164, 1)"},"page_idx":116,"is_discarded":true,"id":"d9bf47a9-10ed-411a-b4de-8ece9bc7378c","page_size":[595,841],"block_position":"116-25"}],[{"bbox":[465,47,539,58],"type":"header","angle":0,"index":-99999,"text":"13.3 隐函数定理 ","color":{"line":"rgba(164, 164, 164, 1)","fill":"rgba(164, 164, 164, 1)"},"page_idx":117,"is_discarded":true,"id":"b7e70cdd-eedb-44db-ba87-27de27b63714","page_size":[595,841],"block_position":"117-0"},{"bbox":[57,72,184,87],"type":"title","angle":0,"index":1,"level":2,"text":"## 13.2.2 多元 Taylor 展开","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":117,"id":"9b59e8a5-11e1-484f-8478-07281863a7e4","page_size":[595,841],"block_position":"117-1"},{"bbox":[73,100,199,113],"type":"title","angle":0,"index":2,"level":2,"text":"## 定理 13.8 (多元 Taylor 展开)","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":117,"id":"00c54b9f-f9e3-4e2b-a626-1baa272426ae","page_size":[595,841],"block_position":"117-2"},{"bbox":[69,117,272,130],"type":"text","angle":0,"index":3,"merge_prev":false,"text":"多元泰勒 $f(x + \\Delta x, y + \\Delta y)$ 展开的公式如下： ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":117,"id":"4e4061ca-7947-451a-8179-b71ae4168db3","page_size":[595,841],"block_position":"117-3"},{"bbox":[68,132,525,165],"type":"interline_equation","angle":0,"index":4,"text":"$$\nf (x, y) + \\sum_ {l = 1} ^ {n} \\frac {1}{l !} \\left(\\binom {l} {0} \\frac {\\partial^ {l} f}{\\partial y ^ {l}} (x, y) \\Delta y ^ {l} + \\binom {l} {1} \\frac {\\partial^ {l} f}{\\partial x \\partial y ^ {l - 1}} (x, y) \\Delta x \\Delta y ^ {l - 1} + \\dots + \\binom {l} {l} \\frac {\\partial^ {l} f}{\\partial x ^ {l}} (x, y) \\Delta x ^ {l}\\right) + R (\\Delta x, \\Delta y)\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":117,"id":"ccea025f-a2e0-442e-942e-22c932739173","page_size":[595,841],"block_position":"117-4"},{"bbox":[54,181,322,195],"type":"text","angle":0,"index":5,"merge_prev":false,"text":"证明 设 $\\Phi(t)=f(x_{0}+th,y_{0}+tk)$ ，用一元 Taylor 公式证明。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":117,"id":"fd6c8a55-5095-4c50-b2ba-6e370db6bdda","page_size":[595,841],"block_position":"117-5"},{"bbox":[57,216,165,233],"type":"title","angle":0,"index":6,"level":2,"text":"## 13.3 隐函数定理","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":117,"id":"67a7ef3e-0be8-4d96-88e6-0ba73013baa7","page_size":[595,841],"block_position":"117-6"},{"bbox":[56,250,240,264],"type":"title","angle":0,"index":7,"level":2,"text":"## 13.3.1 隐函数存在定理与可微定理","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":117,"id":"98428731-c97d-4598-9230-d8e6364f8b80","page_size":[595,841],"block_position":"117-7"},{"bbox":[53,274,539,317],"type":"text","angle":0,"index":8,"merge_prev":false,"text":"隐函数研究的问题是给定一个二维曲线表达式 $F(x,y)=0$ ，希望研究 $y=y(x)$ 的存在性，并计算 $y'(x)$ ，而计算 $y'(x)$ 其实不需要得到 $y(x)$ 的具体表达式，因此隐函数求导大幅度简化了求导过程，因此作为多元微分学的一部分。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":117,"id":"491bb223-e680-4bae-a06b-669e2bfcc621","page_size":[595,841],"block_position":"117-8"},{"bbox":[73,325,156,338],"type":"title","angle":0,"index":9,"level":2,"text":"## 定义 13.6 (隐函数)","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":117,"id":"f53f762e-1133-4b00-b725-dfccc54acd64","page_size":[595,841],"block_position":"117-9"},{"bbox":[68,342,524,372],"type":"text","angle":0,"index":10,"merge_prev":false,"text":"$E \\subset \\mathbb{R}^2$ ， $F: E \\to \\mathbb{R}$ ，对于方程 $F(x, y) = 0$ ，若存在集合 $I, J \\subset \\mathbb{R}$ 使得 $\\forall x \\in I$ 存在唯一 $y \\in \\mathbb{R}$ ，满足 $F(x, y) = 0$ ，则称 $F(x, y) = 0$ 定义了隐函数。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":117,"id":"d1b60862-c836-47bf-8c8e-fee908e33b8f","page_size":[595,841],"block_position":"117-10"},{"bbox":[73,391,226,405],"type":"title","angle":0,"index":11,"level":2,"text":"## 定理 13.9 (隐函数存在唯一性定理)","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":117,"id":"7e6525b1-2007-4856-bbcb-f6af5c2f7652","page_size":[595,841],"block_position":"117-11"},{"bbox":[69,409,138,421],"type":"title","angle":0,"index":12,"level":2,"text":"## 若 $F(x,y)$ 满足","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":117,"id":"de596164-5810-4236-bbe4-6bbe067bc221","page_size":[595,841],"block_position":"117-12"},{"bbox":[80,424,337,437],"type":"text","angle":0,"index":13,"merge_prev":false,"text":"1. $F$ 在 $P_0(x_0, y_0)$ 为内点的某一区域 $D$ 存在 $x, y$ 连续偏导 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":117,"id":"d21750b5-8786-4055-9c2e-37901475e5fd","page_size":[595,841],"block_position":"117-13"},{"bbox":[69,440,239,467],"type":"text","angle":0,"index":14,"merge_prev":false,"text":"2. $F(x_0,y_0) = 0$ ，且 $F_{y}(x_{0},y_{0})\\neq 0$ 则： ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":117,"id":"c8db9817-b6b3-4868-b287-3ffccf898bec","page_size":[595,841],"block_position":"117-14"},{"bbox":[81,470,362,484],"type":"text","angle":0,"index":15,"merge_prev":false,"text":"1. $F(x,y) = 0$ 唯一定义了 $\\forall x\\in U(x_0,\\delta)$ 上的连续函数 $y = f(x)$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":117,"id":"fd2066e7-4176-4b78-820d-de5241f7eff4","page_size":[595,841],"block_position":"117-15"},{"bbox":[80,484,504,509],"type":"text","angle":0,"index":16,"merge_prev":false,"text":"2. $f(x)$ 在 $U(x_0,\\delta)$ 上连续可微，且 $f^{\\prime}(x) = -\\frac{F_x(x,y)}{F_y(x,y)}$ (这里 $F_{x}$ 求导时把 $y$ 当作常数而非函数！） ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":117,"id":"c2219af1-0d2b-4926-b49e-f3e2ab07a547","page_size":[595,841],"block_position":"117-16"},{"bbox":[80,508,471,534],"type":"text","angle":0,"index":17,"merge_prev":false,"text":"3. 特别地，对于 $F(x_{1},\\cdots,x_{n},y)=0,\\quad y=f(x_{1},\\cdots,x_{n})$ ，得到 $f_{x_{i}}(x_{1},\\cdots,x_{n})=-\\frac{F_{x_{i}}}{F_{y}}$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":117,"id":"6c2c739c-4d10-447d-9413-e51792717fbb","page_size":[595,841],"block_position":"117-17"},{"bbox":[54,545,434,570],"type":"text","angle":0,"index":18,"merge_prev":false,"text":"证明 (2) $F(x,y)=0$ 两侧对 x 求导得到 $F_{x}(x,y)+F_{y}(x,y)y'=0$ ，因此 $y'=-\\frac{F_{x}(x,y)}{F_{y}(x,y)}$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":117,"id":"0c58ef4d-f299-4db8-9abd-3f503668c7ac","page_size":[595,841],"block_position":"117-18"},{"bbox":[54,573,538,602],"type":"text","angle":0,"index":19,"merge_prev":false,"text":"笔记 如果只希望得到 $f(x)$ 连续，则条件可减弱为 $F(x, y)$ 在 $P_0(x_0, y_0)$ 某邻域连续， $F(x_0, y_0) = 0$ ，固定 $x$ 时 $F(x, y)$ 关于 $y$ 严格单调。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":117,"id":"038447bc-5d3f-49be-b995-c2a6a0e9acef","page_size":[595,841],"block_position":"117-19"},{"bbox":[52,602,539,637],"type":"text","angle":0,"index":20,"merge_prev":false,"text":"笔记 若要求高阶导，则可以对 $y' = -\\frac{F_{x}}{F_{y}}$ 两边求导，再将 $y'(x)$ 表达式带进去，或者实际计算时可以 $F(x, y) = 0$ 直接两边求两次导。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":117,"id":"4f0177e0-4c8e-4f3b-a58b-f7a665620b95","page_size":[595,841],"block_position":"117-20"},{"bbox":[35,650,384,674],"type":"text","angle":0,"index":21,"merge_prev":false,"text":"练习13.12隐函数求导 $(1)y = y(x)$ 由 $\\ln \\sqrt{x^2 + y^2} = \\arctan \\frac{y}{x}$ 确定，求 $\\frac{\\mathrm{dy}}{\\mathrm{dx}}$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":117,"id":"1f12143d-7878-4fab-ad16-b8ecbdc7ab3e","page_size":[595,841],"block_position":"117-21"},{"bbox":[73,671,371,685],"type":"text","angle":0,"index":23,"merge_prev":false,"text":"(2) $z = z(x,y)$ 由 $F(x - y,y - z) = 0$ 所确定的隐函数，求 $z_{x},z_{y},z_{xy}$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":117,"id":"610302fb-6da5-4130-a598-619d1809a4f1","page_size":[595,841],"block_position":"117-22"},{"bbox":[73,684,463,708],"type":"text","angle":0,"index":24,"merge_prev":false,"text":"(3) $F(x,y,z)=0$ 可以确定 $z=z(x,y),y=y(z,x),x=x(y,z)$ ，证明： $\\frac{\\partial x}{\\partial y}\\cdot\\frac{\\partial y}{\\partial z}\\cdot\\frac{\\partial z}{\\partial x}=-1$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":117,"id":"3f2f5ee2-daae-4da9-9038-d4b71201e57f","page_size":[595,841],"block_position":"117-23"},{"bbox":[53,708,401,719],"type":"text","angle":0,"index":25,"merge_prev":false,"text":"解(1)可以用隐函数求导公式，此时 $F$ 对 $x, y$ 求导时，将它们视为彼此无关的： ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":117,"id":"cb6aca67-8cff-43fa-b6c6-ef7b64200d67","page_size":[595,841],"block_position":"117-24"},{"bbox":[171,722,422,774],"type":"interline_equation","angle":0,"index":26,"text":"$$\n\\left\\{ \\begin{array}{l} F _ {x} = \\frac {x}{x ^ {2} + y ^ {2}} - \\frac {- \\frac {y}{x ^ {2}}}{1 + \\left(\\frac {y}{x}\\right) ^ {2}} = \\frac {x + y}{x ^ {2} + y ^ {2}} \\\\ F _ {y} = \\frac {y - x}{x ^ {2} + y ^ {2}} \\end{array} \\right. \\Rightarrow \\frac {\\mathrm{d} y}{\\mathrm{d} x} = \\frac {x + y}{x - y}\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":117,"id":"d51cf549-11f1-44d4-ae3f-937b10250e46","page_size":[595,841],"block_position":"117-25"},{"bbox":[289,802,305,812],"type":"page_number","angle":0,"index":1000021,"text":"110 ","color":{"line":"rgba(164, 164, 164, 1)","fill":"rgba(164, 164, 164, 1)"},"page_idx":117,"is_discarded":true,"id":"610143c2-cdf3-46a4-b275-c20d5e7730b3","page_size":[595,841],"block_position":"117-26"}],[{"bbox":[465,47,539,58],"type":"header","angle":0,"index":-99998,"text":"13.3 隐函数定理 ","color":{"line":"rgba(164, 164, 164, 1)","fill":"rgba(164, 164, 164, 1)"},"page_idx":118,"is_discarded":true,"id":"d5bba756-df97-417b-a5c9-f2524454036f","page_size":[595,841],"block_position":"118-0"},{"bbox":[54,72,264,84],"type":"text","angle":0,"index":2,"merge_prev":false,"text":"或者直接两侧求导，此时视 $y = y(x)$ 与 x 有关： ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":118,"id":"cdc6c55c-31fb-4875-8094-81333a097b48","page_size":[595,841],"block_position":"118-1"},{"bbox":[183,87,411,111],"type":"interline_equation","angle":0,"index":3,"text":"$$\n{\\frac {x}{x ^ {2} + y ^ {2}}} + {\\frac {y}{x ^ {2} + y ^ {2}}} \\cdot y ^ {\\prime} + {\\frac {y}{x ^ {2} + y ^ {2}}} + {\\frac {x}{x ^ {2} + y ^ {2}}} \\cdot y ^ {\\prime} = 0\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":118,"id":"c5edbf47-c5dd-4ceb-b198-997514336441","page_size":[595,841],"block_position":"118-2"},{"bbox":[73,113,540,126],"type":"text","angle":0,"index":4,"merge_prev":false,"text":"(2) 比较难用隐函数求导公式，只能两边求导。视 $z = z(x, y)$ ， $x, y$ 彼此无关，则两侧分别对 $x, y$ 求导得到： ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":118,"id":"32127a50-fd52-4b03-ae48-7020ea3f61f4","page_size":[595,841],"block_position":"118-3"},{"bbox":[204,128,379,174],"type":"interline_equation","angle":0,"index":5,"text":"$$\n\\left\\{ \\begin{array}{l} F _ {1} + F _ {2} (- z _ {x}) = 0 \\Rightarrow z _ {x} = \\frac {F _ {1}}{F _ {2}} \\\\ - F _ {1} + F _ {2} (1 - z _ {y}) = 0 \\Rightarrow z _ {y} = 1 - \\frac {F _ {1}}{F _ {2}} \\end{array} \\right.\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":118,"id":"f3c4c520-e38d-4b0e-afa8-bcc395daf78d","page_size":[595,841],"block_position":"118-4"},{"bbox":[54,176,406,190],"type":"text","angle":0,"index":6,"merge_prev":false,"text":"要求 $z_{xy}$ ，可以对 $F_{1}(x - y, y - z) - z_{x} F_{2}(x - y, y - z) = 0$ 两侧对 $y$ 求导，得到 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":118,"id":"8033dd3b-29e5-4ec3-9857-4a09d6fbb491","page_size":[595,841],"block_position":"118-5"},{"bbox":[73,191,517,218],"type":"interline_equation","angle":0,"index":7,"text":"$$\n\\left(- F _ {1 1} + F _ {1 2} \\left(1 - z _ {y}\\right)\\right) - \\left(F _ {2 1} (- 1) + F _ {2 2} \\left(1 - z _ {y}\\right)\\right) z _ {x} - F _ {2} z _ {x y} = 0 \\Rightarrow z _ {x y} = \\frac {2 F _ {1} F _ {2} F _ {1 2} - F _ {1 1} F _ {2} ^ {2} - F _ {2 2} F _ {1} ^ {2}}{F _ {2} ^ {3}}\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":118,"id":"dbc05b5a-9e1b-48c8-a363-c402eed65c31","page_size":[595,841],"block_position":"118-6"},{"bbox":[73,221,539,248],"type":"text","angle":0,"index":8,"merge_prev":false,"text":"(3) 注意这里的符号含义，不能简简单单乘起来变成 1。例如 $\\frac{\\partial x}{\\partial y}$ 的含义为 $\\frac{\\partial x(y, z)}{y}$ ，因此根据隐函数定理即 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":118,"id":"84bd917d-debd-4850-a296-e299190231cf","page_size":[595,841],"block_position":"118-7"},{"bbox":[249,248,351,275],"type":"interline_equation","angle":0,"index":9,"text":"$$\n- \\frac {F _ {y}}{F _ {x}} \\cdot - \\frac {F _ {z}}{F _ {y}} \\cdot - \\frac {F _ {x}}{F _ {z}} = - 1\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":118,"id":"8470de0c-66b0-4626-95da-c9908c7392bb","page_size":[595,841],"block_position":"118-8"},{"bbox":[57,291,253,306],"type":"title","angle":0,"index":10,"level":2,"text":"## 13.3.2 隐函数组存在定理与微分定理","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":118,"id":"e0b7cfce-e12a-481b-a7d2-598d6b325527","page_size":[595,841],"block_position":"118-9"},{"bbox":[73,319,190,331],"type":"title","angle":0,"index":11,"level":2,"text":"## 定理 13.10 (隐函数组定理)","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":118,"id":"726be7a4-a3bb-4fe9-80e3-534aaace9eb3","page_size":[595,841],"block_position":"118-10"},{"bbox":[69,336,217,348],"type":"text","angle":0,"index":12,"merge_prev":false,"text":"若 $F(x,y,u,v),G(x,y,u,v)$ 满足： ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":118,"id":"5e0a1957-c2d3-486f-badf-aedbc232110f","page_size":[595,841],"block_position":"118-11"},{"bbox":[83,351,312,363],"type":"text","angle":0,"index":13,"merge_prev":false,"text":"- 在 $P_0(x_0, y_0, u_0, v_0)$ 邻域中对所有变量有连续偏导 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":118,"id":"7ddc1f3a-af0f-42db-aac1-d066c8acc965","page_size":[595,841],"block_position":"118-12"},{"bbox":[83,363,278,390],"type":"interline_equation","angle":0,"index":14,"text":"$$\nF (P _ {0}) = G (P _ {0}) = 0, J = \\left. \\frac {\\partial (F , G)}{\\partial (u , v)} \\right| _ {P _ {0}} \\neq 0\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":118,"id":"518c3069-d43c-4fbf-808e-d9890cc77326","page_size":[595,841],"block_position":"118-13"},{"bbox":[69,389,112,400],"type":"text","angle":0,"index":15,"merge_prev":false,"text":"则方程组 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":118,"id":"4dc044bb-d2a9-4135-ae78-6f6da928968a","page_size":[595,841],"block_position":"118-14"},{"bbox":[252,400,333,437],"type":"interline_equation","angle":0,"index":16,"text":"$$\n\\left\\{ \\begin{array}{l} F (x, y, u, v) = 0 \\\\ G (x, y, u, v) = 0 \\end{array} \\right.\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":118,"id":"d721117f-8509-488a-873f-0ab73b518e6c","page_size":[595,841],"block_position":"118-15"},{"bbox":[67,445,465,470],"type":"text","angle":0,"index":17,"merge_prev":false,"text":"在 $P_0$ 邻域中唯一确定了两个二元连续隐函数 $u = f(x,y), v = g(x,y)$ ，设 $J = \\frac{\\partial(F,G)}{\\partial(u,v)}$ ，则 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":118,"id":"df3c6cff-b3b0-4924-884b-fe4bc0af75e4","page_size":[595,841],"block_position":"118-16"},{"bbox":[138,472,454,499],"type":"interline_equation","angle":0,"index":18,"text":"$$\nu _ {x} = - \\frac {1}{J} \\frac {\\partial (F , G)}{\\partial (x , v)}, u _ {y} = - \\frac {1}{J} \\frac {\\partial (F , G)}{\\partial (y , v)}, v _ {x} = - \\frac {1}{J} \\frac {\\partial (F , G)}{\\partial (u , x)}, v _ {y} = - \\frac {1}{J} \\frac {\\partial (F , G)}{\\partial (u , y)}\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":118,"id":"ad0c759f-9d02-4805-a0cb-a9a4e8dd8ebe","page_size":[595,841],"block_position":"118-17"},{"bbox":[68,501,433,529],"type":"text","angle":0,"index":19,"merge_prev":false,"text":"记忆口诀为求什么就换什么，例如求 $u_{x}$ 就把 $\\frac{\\partial(F,G)}{\\partial(u,v)}$ 中的 $u$ 换成 $x$ ，得到 $\\frac{\\partial(F,G)}{\\partial(x,v)}$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":118,"id":"a2a30e50-1fdd-4d16-9b11-0ab995e2ba4f","page_size":[595,841],"block_position":"118-18"},{"bbox":[53,544,208,556],"type":"text","angle":0,"index":20,"merge_prev":false,"text":"证明 求导结论使用了 Cramer 法则 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":118,"id":"56ad3413-96bf-49f5-823e-91ca981aed7f","page_size":[595,841],"block_position":"118-19"},{"bbox":[33,559,323,572],"type":"text","angle":0,"index":21,"merge_prev":false,"text":"笔记 一般而言有几个方程就有几个因变量，其余的为自变量。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":118,"id":"eb905473-85d7-4b0b-b818-ea6805ee8419","page_size":[595,841],"block_position":"118-20"},{"bbox":[35,570,499,605],"type":"text","angle":0,"index":22,"merge_prev":false,"text":"练习 13.13 容易分不清的题 (1) 错过：已知 $x = \\varphi(t), y = \\phi(x)$ ， $\\varphi'(t) = e^{t} + 1, \\frac{\\mathrm{d}^{3}y}{\\mathrm{d}x^{3}} = e^{t} + 2t$ ，求 $\\frac{d^{4}y}{dx^{4}}$ 解 (1) 直接求： ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":118,"id":"0c829950-3ad2-48cb-a12f-205ec50d6449","page_size":[595,841],"block_position":"118-21"},{"bbox":[236,604,358,637],"type":"interline_equation","angle":0,"index":23,"text":"$$\n{\\frac {\\mathrm{d} ^ {4} y}{\\mathrm{d} x ^ {4}}} = {\\frac {\\left({\\frac {\\mathrm{d} ^ {3} y}{\\mathrm{d} x ^ {3}}}\\right) / \\mathrm{d} t}{\\mathrm{d} x / \\mathrm{d} t}} = {\\frac {e ^ {t} + 2}{e ^ {t} + 1}}\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":118,"id":"7e58200c-e975-40c7-9179-29ce17df769a","page_size":[595,841],"block_position":"118-22"},{"bbox":[35,656,204,669],"type":"text","angle":0,"index":24,"merge_prev":false,"text":"练习 13.14 隐函数组求导 (1) 证明 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":118,"id":"63c20f74-491b-41cc-8010-cbac39432695","page_size":[595,841],"block_position":"118-23"},{"bbox":[238,667,345,706],"type":"interline_equation","angle":0,"index":25,"text":"$$\n\\left\\{ \\begin{array}{l} x ^ {2} + y ^ {2} + u ^ {2} - v ^ {2} = 2 \\\\ x + y + u + v = 0 \\end{array} \\right.\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":118,"id":"efea26d5-06a2-4683-8f61-ca1be9477a7b","page_size":[595,841],"block_position":"118-24"},{"bbox":[52,713,424,738],"type":"text","angle":0,"index":26,"merge_prev":false,"text":"在 $P_0(1,1, - 1, - 1)$ 的某邻域中能确定隐函数组 $u = u(x,y),v = v(x,y)$ ，并求 $\\frac{\\partial u}{\\partial x},\\frac{\\partial v}{\\partial y}$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":118,"id":"aaa2fbdc-76c5-4863-aa0f-f2d55510cb3d","page_size":[595,841],"block_position":"118-25"},{"bbox":[289,803,304,812],"type":"page_number","angle":0,"index":999999,"text":"111 ","color":{"line":"rgba(164, 164, 164, 1)","fill":"rgba(164, 164, 164, 1)"},"page_idx":118,"is_discarded":true,"id":"af2378a0-f604-4609-9b20-978a6332ad80","page_size":[595,841],"block_position":"118-26"}],[{"bbox":[425,47,539,58],"type":"header","angle":0,"index":-99999,"text":"13.4 多元无条件极值问题 ","color":{"line":"rgba(164, 164, 164, 1)","fill":"rgba(164, 164, 164, 1)"},"page_idx":119,"is_discarded":true,"id":"799be257-de7a-4654-97b3-738333af9d5a","page_size":[595,841],"block_position":"119-0"},{"bbox":[73,72,245,84],"type":"text","angle":0,"index":1,"merge_prev":false,"text":"(2) 设 $u = u(x, y), v = v(x, y)$ 由方程组 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":119,"id":"57ec0935-6705-43ee-b050-cc31928e5607","page_size":[595,841],"block_position":"119-1"},{"bbox":[248,88,334,126],"type":"interline_equation","angle":0,"index":2,"text":"$$\n\\left\\{ \\begin{array}{l} u = f (u x, v + y) \\\\ v = g (u - x, v ^ {2} y) \\end{array} \\right.\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":119,"id":"ac587020-bec0-4615-b537-307c6bdc8e0c","page_size":[595,841],"block_position":"119-2"},{"bbox":[53,129,164,142],"type":"text","angle":0,"index":3,"merge_prev":false,"text":"给出的隐函数，求 $u_{x}, v_{x}$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":119,"id":"c9029dc7-c065-4257-838e-4fe359dffdcf","page_size":[595,841],"block_position":"119-3"},{"bbox":[73,145,196,157],"type":"text","angle":0,"index":4,"merge_prev":false,"text":"(3) 已知如下方程组，求 $z_{x}$ : ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":119,"id":"96948fe4-1330-4c6f-bb6c-04e3285e3f69","page_size":[595,841],"block_position":"119-4"},{"bbox":[261,156,323,212],"type":"interline_equation","angle":0,"index":5,"text":"$$\n\\left\\{ \\begin{array}{l} x = u + v \\\\ y = u ^ {2} + v ^ {2} \\\\ z = u ^ {3} + v ^ {3} \\end{array} \\right.\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":119,"id":"343d6836-51d7-4edb-80f1-3481e48a7554","page_size":[595,841],"block_position":"119-5"},{"bbox":[71,221,334,234],"type":"text","angle":0,"index":6,"merge_prev":false,"text":"(4) 令 $u = f(x - ut, y - ut, z - ut), g(x, y, z) = 0$ ，求 $u_x, u_y$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":119,"id":"53cf93d1-eee6-49d3-a08b-c32e0ec3efdd","page_size":[595,841],"block_position":"119-6"},{"bbox":[54,235,539,263],"type":"text","angle":0,"index":7,"merge_prev":false,"text":"解(1) $F=x^{2}+y^{2}+u^{2}-v^{2}-2,G=x+y+u+v$ ，则显然F,G对每个分量都有连续偏导数， $F(P_{0})=G(P_{0})=0$ ，方便起见，直接求出所有偏导： ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":119,"id":"099a2b3f-a60b-4820-b62b-5a0c1eae1574","page_size":[595,841],"block_position":"119-7"},{"bbox":[105,265,490,299],"type":"interline_equation","angle":0,"index":8,"text":"$$\n\\left[ \\begin{array}{c c c c} F _ {x} & F _ {y} & F _ {u} & F _ {v} \\\\ G _ {x} & G _ {y} & G _ {u} & G _ {v} \\end{array} \\right] = \\left[ \\begin{array}{c c c c} 2 x & 2 y & 2 u & - 2 v \\\\ 1 & 1 & 1 & 1 \\end{array} \\right] \\Rightarrow \\frac {\\partial (F , G)}{\\partial (u , v)} = \\left| \\begin{array}{c c} 2 u & - 2 v \\\\ 1 & 1 \\end{array} \\right| = 2 v + 2 u\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":119,"id":"09f1daab-0eb7-49a9-8689-f334ed26daf8","page_size":[595,841],"block_position":"119-8"},{"bbox":[54,302,287,314],"type":"text","angle":0,"index":9,"merge_prev":false,"text":"因此 $J(P_0) \\neq 0$ ，隐函数组存在，根据隐函数组求导： ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":119,"id":"e0688239-9497-437a-a66b-3fe56a99069c","page_size":[595,841],"block_position":"119-9"},{"bbox":[148,317,445,343],"type":"interline_equation","angle":0,"index":10,"text":"$$\n\\frac {\\partial u}{\\partial x} = - \\frac {1}{J} \\frac {\\partial (F , G)}{\\partial (x , v)} = - \\frac {1}{2 u + 2 v} \\cdot (2 x + 2 v) = - \\frac {x + v}{u + v}, \\frac {\\partial v}{\\partial y} = - \\frac {y - u}{u + v}\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":119,"id":"7fc1c122-f7bd-4b66-b69a-25d583b377ed","page_size":[595,841],"block_position":"119-10"},{"bbox":[73,346,181,358],"type":"text","angle":0,"index":11,"merge_prev":false,"text":"(2) 直接两侧求导，得到 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":119,"id":"cb243cab-2c71-434f-bffd-da321901ca47","page_size":[595,841],"block_position":"119-11"},{"bbox":[160,362,423,399],"type":"interline_equation","angle":0,"index":12,"text":"$$\n\\left\\{ \\begin{array}{l} u _ {x} = f _ {1} \\cdot (u _ {x} x + u) + f _ {2} v _ {x} \\Rightarrow (1 - x f _ {1}) u _ {x} - f _ {2} v _ {x} = u f _ {1} \\\\ v _ {x} = g _ {1} \\cdot (u _ {x} - 1) + g _ {2} 2 v v _ {x} y \\Rightarrow g _ {1} u _ {x} + (2 v y g _ {2} - 1) v _ {x} = g _ {1} \\end{array} \\right.\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":119,"id":"9a314d16-4178-4652-b463-1162d9ac3e78","page_size":[595,841],"block_position":"119-12"},{"bbox":[54,402,137,414],"type":"text","angle":0,"index":13,"merge_prev":false,"text":"解线性方程组得到 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":119,"id":"589b60dd-b3ec-4083-b347-fb3cf9c717ed","page_size":[595,841],"block_position":"119-13"},{"bbox":[145,416,446,442],"type":"interline_equation","angle":0,"index":14,"text":"$$\nu _ {x} = \\frac {(2 v y g _ {2} - 1) u f _ {1} + f _ {2} g _ {1}}{(1 - x f _ {1}) (2 v y g _ {2} - 1) + f _ {2} g _ {1}}, v _ {x} = \\frac {(1 - x f _ {1}) g _ {1} - u f _ {1} g _ {1}}{(1 - x f _ {1}) (2 v y g _ {2} - 1) + f _ {2} g _ {1}}\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":119,"id":"7b466ea1-6727-48fc-9869-c6bd5058f270","page_size":[595,841],"block_position":"119-14"},{"bbox":[73,444,258,456],"type":"text","angle":0,"index":15,"merge_prev":false,"text":"(3) 令 $x, y$ 为自变量， $u, v, z$ 为因变量，则 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":119,"id":"2c5fd90f-426a-49c2-8e3f-71f2a12bf9f2","page_size":[595,841],"block_position":"119-15"},{"bbox":[260,459,332,495],"type":"interline_equation","angle":0,"index":16,"text":"$$\nz _ {x} = - \\frac {\\frac {\\partial (F , G , H)}{\\partial (x , u , v)}}{\\frac {\\partial (F , G , H)}{\\partial (z , u , v)}}\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":119,"id":"c1ed22a3-eb74-4d63-892f-63b07eb846c9","page_size":[595,841],"block_position":"119-16"},{"bbox":[57,513,222,530],"type":"title","angle":0,"index":17,"level":2,"text":"## 13.4 多元无条件极值问题","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":119,"id":"233a1540-cb8a-4140-b6a4-b5f4ebd86fd6","page_size":[595,841],"block_position":"119-17"},{"bbox":[73,548,186,560],"type":"title","angle":0,"index":18,"level":2,"text":"## 定义 13.7 (极值与极值点)","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":119,"id":"c35c158d-d3a8-4193-9590-08ed2c268bfd","page_size":[595,841],"block_position":"119-18"},{"bbox":[67,565,524,594],"type":"text","angle":0,"index":19,"merge_prev":false,"text":"开集 $D \\subset \\mathbb{R}^n$ ，函数 $f: D \\to \\mathbb{R}$ ， $x_0 \\in D$ ，若存在球 $B_r(x_0) \\subset D$ 使得 $\\forall x \\in B_r(x_0)$ 有 $f(x) \\geq f(x_0)$ ，则称 $x_0$ 为 $f$ 的一个极小值点。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":119,"id":"e7993bd2-45da-4773-bf1f-d7c5bf578b06","page_size":[595,841],"block_position":"119-19"},{"bbox":[73,613,186,627],"type":"title","angle":0,"index":20,"level":2,"text":"## 定义 13.8 (驻点、稳定点)","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":119,"id":"8bbc07f6-0e74-46c8-920f-d54c58158757","page_size":[595,841],"block_position":"119-20"},{"bbox":[68,630,418,645],"type":"text","angle":0,"index":21,"merge_prev":false,"text":"若 $P_0(x_0,y_0)$ 满足 $f_{x}(x_{0},y_{0}) = f_{y}(x_{0},y_{0}) = 0$ ，则称其为 $f(x,y)$ 的驻点/稳定点。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":119,"id":"13c3dfc1-36a9-44b8-be7c-daa55129bcfb","page_size":[595,841],"block_position":"119-21"},{"bbox":[289,803,305,812],"type":"page_number","angle":0,"index":1000021,"text":"112 ","color":{"line":"rgba(164, 164, 164, 1)","fill":"rgba(164, 164, 164, 1)"},"page_idx":119,"is_discarded":true,"id":"da86f3f0-40de-46ff-92a5-6efba71409b5","page_size":[595,841],"block_position":"119-22"}],[{"bbox":[426,47,539,58],"type":"header","angle":0,"index":-99999,"text":"13.4 多元无条件极值问题 ","color":{"line":"rgba(164, 164, 164, 1)","fill":"rgba(164, 164, 164, 1)"},"page_idx":120,"is_discarded":true,"id":"cb805904-7452-4de6-84d5-7f701ac5164a","page_size":[595,841],"block_position":"120-0"},{"bbox":[73,73,173,85],"type":"title","angle":0,"index":1,"level":2,"text":"## 定义 13.9 (Hesse 矩阵)","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":120,"id":"45225b09-3ec6-4a9d-9632-db5d4c2641c6","page_size":[595,841],"block_position":"120-1"},{"bbox":[69,90,214,102],"type":"text","angle":0,"index":2,"merge_prev":false,"text":"$f:\\mathbb{R}^n\\to \\mathbb{R}$ ，则其Hesse矩阵为： ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":120,"id":"bdcf26ff-c3a8-48ef-b91b-ac5e60e737d1","page_size":[595,841],"block_position":"120-2"},{"bbox":[199,104,392,200],"type":"interline_equation","angle":0,"index":3,"text":"$$\nH = \\left[ \\begin{array}{c c c c} \\frac {\\partial^ {2} f}{\\partial x _ {1} ^ {2}} & \\frac {\\partial^ {2} f}{\\partial x _ {1} \\partial x _ {2}} & \\dots & \\frac {\\partial^ {2} f}{\\partial x _ {1} \\partial x _ {n}} \\\\ \\frac {\\partial^ {2} f}{\\partial x _ {2} \\partial x _ {1}} & \\frac {\\partial^ {2} f}{\\partial x _ {2} ^ {2}} & \\dots & \\frac {\\partial^ {2} f}{\\partial x _ {2} \\partial x _ {n}} \\\\ \\vdots & \\vdots & \\ddots & \\vdots \\\\ \\frac {\\partial^ {2} f}{\\partial x _ {n} \\partial x _ {1}} & \\frac {\\partial^ {2} f}{\\partial x _ {n} \\partial x _ {2}} & \\dots & \\frac {\\partial^ {2} f}{\\partial x _ {n} ^ {2}} \\end{array} \\right]\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":120,"id":"fc5fb7c7-0a8f-4c76-b3dd-0c953b7b973e","page_size":[595,841],"block_position":"120-3"},{"bbox":[73,221,211,233],"type":"title","angle":0,"index":4,"level":2,"text":"## 定义 13.10 (正定、负定、不定)","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":120,"id":"17c8d2fc-3441-4623-a660-5f6f07da7e83","page_size":[595,841],"block_position":"120-4"},{"bbox":[69,238,341,250],"type":"text","angle":0,"index":5,"merge_prev":false,"text":"矩阵 $A$ 是对称矩阵，考虑 $Q(x) = \\mathbf{x}^T A\\mathbf{x}$ ，分为以下三种情况： ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":120,"id":"58e6ec52-48de-4113-a674-61dc983f6a18","page_size":[595,841],"block_position":"120-5"},{"bbox":[83,254,233,266],"type":"text","angle":0,"index":6,"merge_prev":false,"text":"- 正(负)定: 对 $\\forall \\mathbf{x}, Q(\\mathbf{x}) \\geq (\\leq)0$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":120,"id":"b6e7370f-ebcb-47d0-971a-c1b13ead0ac1","page_size":[595,841],"block_position":"120-6"},{"bbox":[83,269,260,282],"type":"text","angle":0,"index":7,"merge_prev":false,"text":"- 不定： $\\exists \\mathbf{p}, \\mathbf{q} \\in \\mathbb{R}^n$ ， $Q(\\mathbf{p}) \\leq 0 \\leq Q(\\mathbf{p})$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":120,"id":"21e46ac5-bf89-4720-9fbd-8fa1d8acadcf","page_size":[595,841],"block_position":"120-7"},{"bbox":[73,303,190,315],"type":"title","angle":0,"index":8,"level":2,"text":"## 定理 13.11 (正定的判别法)","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":120,"id":"344aa428-e60d-4db6-afb5-c20afe27648e","page_size":[595,841],"block_position":"120-8"},{"bbox":[68,320,361,333],"type":"text","angle":0,"index":9,"merge_prev":false,"text":"$A$ 为对称矩阵，则 $A$ 严格正定当且仅当其各阶顺序主子式均大于0 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":120,"id":"cb12f867-5290-4923-b44e-04d2bd239315","page_size":[595,841],"block_position":"120-9"},{"bbox":[57,364,157,379],"type":"title","angle":0,"index":10,"level":2,"text":"## 13.4.1 无条件极值","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":120,"id":"4c3447fd-bda2-481b-80da-0d5fe77b7ecf","page_size":[595,841],"block_position":"120-10"},{"bbox":[73,391,200,404],"type":"title","angle":0,"index":11,"level":2,"text":"## 定理 13.12 (极值点必要条件)","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":120,"id":"c4004339-d978-4e57-b888-f7a94a7f05ec","page_size":[595,841],"block_position":"120-11"},{"bbox":[68,408,447,422],"type":"text","angle":0,"index":12,"merge_prev":false,"text":"若 $f(x,y)$ 在 $P_0(x_0,y_0)$ 存在偏导数，且在 $P_0$ 取到极值，则 $f_{x}(x_{0},y_{0}) = f_{y}(x_{0},y_{0}) = 0$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":120,"id":"5fed0cb5-73bc-4ac0-96c1-05cd7b109234","page_size":[595,841],"block_position":"120-12"},{"bbox":[54,439,146,451],"type":"text","angle":0,"index":13,"merge_prev":false,"text":"证明 用反证法即可。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":120,"id":"17435aa5-7589-4adc-adea-6b1cc406c7f0","page_size":[595,841],"block_position":"120-13"},{"bbox":[73,460,200,472],"type":"title","angle":0,"index":14,"level":2,"text":"## 定理 13.13 (极值点充分条件)","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":120,"id":"77a32707-f1cb-4b30-8605-5cf672155c1a","page_size":[595,841],"block_position":"120-14"},{"bbox":[68,476,406,490],"type":"text","angle":0,"index":15,"merge_prev":false,"text":"$f:\\mathbb{R}^2\\to \\mathbb{R},(x_0,y_0)$ 为其驻点，且 $f$ 在 $(x_0,y_0)$ 某一邻域中有二阶连续导数， ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":120,"id":"a2609219-5930-4279-8bf9-fe87699d32a9","page_size":[595,841],"block_position":"120-15"},{"bbox":[83,493,408,505],"type":"text","angle":0,"index":16,"merge_prev":false,"text":"- 若Hesse矩阵 $H$ 是严格正(负)定矩阵，则 $\\mathbf{x}_0$ 是 $f$ 的严格极小(大)值点 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":120,"id":"e729e873-ef9b-4e25-b0be-35d601ffa213","page_size":[595,841],"block_position":"120-16"},{"bbox":[83,508,356,520],"type":"text","angle":0,"index":17,"merge_prev":false,"text":"- 若Hesse矩阵 $H$ 可逆但为不定方阵，则 $\\mathbf{x}_0$ 不是 $f$ 的极值点 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":120,"id":"ad76aa3c-fbbc-48bf-82c0-eb0a6ff37b17","page_size":[595,841],"block_position":"120-17"},{"bbox":[83,524,260,536],"type":"text","angle":0,"index":18,"merge_prev":false,"text":"- 若 Hesse 矩阵 $H$ 不可逆，则不能判别 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":120,"id":"db10c278-6417-4d9f-ae7e-d9d3b916c6b8","page_size":[595,841],"block_position":"120-18"},{"bbox":[54,553,157,565],"type":"text","angle":0,"index":19,"merge_prev":false,"text":"证明 根据 Taylor 公式: ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":120,"id":"ac81a056-af7c-4439-9c8c-f0c9887280b7","page_size":[595,841],"block_position":"120-19"},{"bbox":[97,568,495,600],"type":"interline_equation","angle":0,"index":20,"text":"$$\nf (x, y) = f (x _ {0}, y _ {0}) + \\frac {1}{2} \\left( \\begin{array}{c c} \\Delta x & \\Delta y \\end{array} \\right) \\left( \\begin{array}{c c} f _ {x x} (P _ {0}) & f _ {x y} (P _ {0}) \\\\ f _ {x y} (P _ {0}) & f _ {y y} (P _ {0}) \\end{array} \\right) \\binom{\\Delta x}{\\Delta y} + o ((\\Delta x) ^ {2} + (\\Delta y) ^ {2})\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":120,"id":"3da66897-515e-4a36-8706-b3f0159326d0","page_size":[595,841],"block_position":"120-20"},{"bbox":[54,604,539,632],"type":"text","angle":0,"index":21,"merge_prev":false,"text":"因此如果黑塞矩阵为正定，则 $f(x,y) > f(x_0,y_0)$ ，若负定则 $f(x,y) < f(x_0,y_0)$ ，若不定则不取极值，若黑塞矩阵行列式为0，则不能判别 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":120,"id":"09783588-e04e-485c-8367-c68d7401a153","page_size":[595,841],"block_position":"120-21"},{"bbox":[73,640,196,653],"type":"title","angle":0,"index":22,"level":2,"text":"## 推论 13.2 (二元极值判别法)","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":120,"id":"ab68f262-c59a-4929-a774-7d350dbd1fee","page_size":[595,841],"block_position":"120-22"},{"bbox":[69,657,405,670],"type":"text","angle":0,"index":23,"merge_prev":false,"text":"$(x_0, y_0)$ 是 $f: \\mathbb{R}^2 \\to \\mathbb{R}$ 的驻点， $f$ 在 $(x_0, y_0)$ 某个邻域中有连续二阶偏导，记 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":120,"id":"4d588806-58c1-4c59-9de6-521dbbbdd912","page_size":[595,841],"block_position":"120-23"},{"bbox":[185,672,408,699],"type":"interline_equation","angle":0,"index":24,"text":"$$\na = \\frac {\\partial^ {2} f}{\\partial x ^ {2}} (x _ {0}, y _ {0}), b = \\frac {\\partial^ {2} f}{\\partial x \\partial y} (x _ {0}, y _ {0}), c = \\frac {\\partial^ {2} f}{\\partial y ^ {2}} (x _ {0}, y _ {0})\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":120,"id":"98afcb10-dd2a-4a61-839b-6c3ffe468ccc","page_size":[595,841],"block_position":"120-24"},{"bbox":[84,702,224,714],"type":"text","angle":0,"index":25,"merge_prev":false,"text":"- $ac - b^{2} > 0$ 且 $a > 0$ : 极小值 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":120,"id":"3c389788-eb6c-45f2-8710-782efbbe3afc","page_size":[595,841],"block_position":"120-25"},{"bbox":[84,717,224,729],"type":"text","angle":0,"index":26,"merge_prev":false,"text":"- $ac - b^{2} > 0$ 且 $a < 0$ : 极大值 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":120,"id":"6ada93c7-23f1-48d8-ad0d-d832b2b89b65","page_size":[595,841],"block_position":"120-26"},{"bbox":[84,733,186,745],"type":"text","angle":0,"index":28,"merge_prev":false,"text":"- $ac - b^{2} < 0$ ：无极值 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":120,"id":"36c871df-bf53-4b84-8ed0-d56318f66034","page_size":[595,841],"block_position":"120-27"},{"bbox":[84,749,185,761],"type":"text","angle":0,"index":29,"merge_prev":false,"text":"- $ac - b^{2} = 0$ ：不确定 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":120,"id":"273d0033-4a4b-4e81-8b83-5449387f7448","page_size":[595,841],"block_position":"120-28"},{"bbox":[289,803,305,812],"type":"page_number","angle":0,"index":1000026,"text":"113 ","color":{"line":"rgba(164, 164, 164, 1)","fill":"rgba(164, 164, 164, 1)"},"page_idx":120,"is_discarded":true,"id":"d2b4b6c7-1ccb-4943-b49c-329a6359e0bb","page_size":[595,841],"block_position":"120-29"}],[{"bbox":[475,47,539,58],"type":"header","angle":0,"index":-99999,"text":"13.5 条件极值 ","color":{"line":"rgba(164, 164, 164, 1)","fill":"rgba(164, 164, 164, 1)"},"page_idx":121,"is_discarded":true,"id":"e2d613b3-714c-4ad4-803f-28354dd890d7","page_size":[595,841],"block_position":"121-0"},{"bbox":[52,71,539,100],"type":"text","angle":0,"index":1,"merge_prev":false,"text":"证明 极小值即顺序主子式大于0；极大值需要负定，而负定等价于所有元素取负号后为正定，因此注意 $ac - b^{2} > 0$ 。 $ac - b^{2} < 0$ 等价于不负定也不正定，而 $ac - b^{2} = 0$ 等价于不可逆。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":121,"id":"6420b437-00dd-4715-8f5c-33acb2f76480","page_size":[595,841],"block_position":"121-1"},{"bbox":[36,118,539,162],"type":"text","angle":0,"index":2,"merge_prev":false,"text":"练习13.15无穷极大、无极小证明 $f(x,y) = (1 + e^{y})\\cos x - ye^{y}$ 有无穷个极大值点，但无极小值点。证明(1)先计算驻点： $f_{x} = -(1 + e^{y})\\sin x = 0,f_{y} = e^{y}\\cos x - e^{y} - ye^{y} = 0$ ，需要 $\\sin x = 0,y = \\cos x - 1$ ，即 $x = n\\pi ,y = (-1)^{n} - 1$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":121,"id":"db074308-4a0f-4c0d-8e86-78aa0f31129a","page_size":[595,841],"block_position":"121-2"},{"bbox":[52,164,539,210],"type":"text","angle":0,"index":3,"merge_prev":false,"text":"(2) 判断极值类型: $f_{xx} = -(1 + e^{y})\\cos x, f_{xy} = -e^{y}\\sin x, f_{yy} = e^{y}(\\cos x - 2 - y)$ , 考虑 $n = 2m$ 时, $P_{m}(2m\\pi,0)$ , $f_{xx}(P_{m}) = -2, f_{xy} = 0, f_{yy} = -1, f_{xx}f_{yy} - f_{xy}^{2} = 2 > 0$ , 因此为极大值点。考虑 $n = 2m + 1, P_{m}((2m + 1)\\pi, -2), f_{xx}(P_{m}) = 1 + e^{-2} > 0, f_{xy} = 0, f_{yy} = -e^{-2}, f_{xx}f_{yy} - f_{xy}^{2} = -e^{-2}(1 + e^{-2}) < 0$ 不取极值。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":121,"id":"06cb2478-e465-4eae-808c-ffdde0ec55a0","page_size":[595,841],"block_position":"121-3"},{"bbox":[54,227,539,255],"type":"text","angle":0,"index":4,"merge_prev":false,"text":"当无法直接使用 Hesse 矩阵进行判断时，可以先求出驻点，再使用配方等方法判断驻点是否最小，下面的题目便有用到 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":121,"id":"ba57e821-5f9f-4df2-b6db-5011e5851750","page_size":[595,841],"block_position":"121-4"},{"bbox":[35,267,483,290],"type":"text","angle":0,"index":5,"merge_prev":false,"text":"练习13.16计算极值点与极值(1)求 $f(x,y) = \\frac{1}{2} x^2 +xy + \\frac{1}{2} y^2 -2x - 2y + 5$ 的全部极值点与极值 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":121,"id":"9ba49115-108f-4226-a274-07635cd02101","page_size":[595,841],"block_position":"121-5"},{"bbox":[51,288,539,349],"type":"text","angle":0,"index":6,"merge_prev":false,"text":"(2) 求 $f(x, y, z) = x^2 + y^2 + z^2 - xz - yz - 3x + y + 4z + 7$ 的极值点与极值解 (1) 先算驻点: $f_x = f_y = x + y - 2$ , 因此 $x + y = 2$ 的点为驻点。再判断极值性: $f_{xx} = 1, f_{xy} = 1, f_{yy} = 1$ , 因此 $f_{xx}f_{yy} - f_{xy}^2 = 0$ , 无法判断用 Hesse 矩阵判断。而 $f(x, y) = \\frac{1}{2}(x + y)^2 - 2(x + y) + 5 = \\frac{1}{2}[(x + y) - 2]^2 + 3$ , 因此 $x + y = 2$ 时不仅是极值点, 而且是最值点。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":121,"id":"3a878d77-5b1e-4b1c-8f06-8647265e2799","page_size":[595,841],"block_position":"121-6"},{"bbox":[54,353,539,381],"type":"text","angle":0,"index":7,"merge_prev":false,"text":"(2) 先算驻点: $f_x = 2x - z - 3, f_y = 2y - z + 1, f_z = 2z - x - y + 4$ , 解线性方程组得到 $x = 0, y = -2, z = -3$ 。先尝试用 Hesse 矩阵: $f_{xx} = 2, f_{xy} = 0, f_{xz} = -1, f_{yy} = 2, f_{yz} = -1, f_{zz} = 2$ , 即 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":121,"id":"242ad32a-8fd9-4829-b2bb-e899c2e971a5","page_size":[595,841],"block_position":"121-7"},{"bbox":[242,384,348,434],"type":"interline_equation","angle":0,"index":8,"text":"$$\nH = \\left[ \\begin{array}{r r r} 2 & 0 & - 1 \\\\ 0 & 2 & - 1 \\\\ - 1 & - 1 & 2 \\end{array} \\right]\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":121,"id":"44277317-de62-46f9-bded-5c49d73acc22","page_size":[595,841],"block_position":"121-8"},{"bbox":[54,437,229,449],"type":"text","angle":0,"index":9,"merge_prev":false,"text":"顺序主子式都大于0，因此为极小值点。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":121,"id":"7feb1f6b-db9e-4483-a8c6-c3edeb31fd45","page_size":[595,841],"block_position":"121-9"},{"bbox":[56,468,229,481],"type":"title","angle":0,"index":10,"level":2,"text":"## 13.4.2 最值问题：极值点 + 边界","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":121,"id":"64e81aae-f6d5-46be-86b1-5ef58bc72a4f","page_size":[595,841],"block_position":"121-10"},{"bbox":[73,491,390,503],"type":"text","angle":0,"index":11,"merge_prev":false,"text":"要求某区域内函数的最值，只需要考虑区域内部的极值点以及边界即可。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":121,"id":"99b10891-6963-48bb-9dcb-4eaf01e580fa","page_size":[595,841],"block_position":"121-11"},{"bbox":[57,523,193,536],"type":"title","angle":0,"index":12,"level":2,"text":"## 13.4.3 多元极值理论证明","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":121,"id":"e98913fb-be6c-42a5-9d76-56e8f7f117dc","page_size":[595,841],"block_position":"121-12"},{"bbox":[35,548,539,600],"type":"text","angle":0,"index":13,"merge_prev":false,"text":"练习 13.17 $f(x,y)$ 在 D 上连续且有连续偏导，若 $f_{x} + f_{y} = f, \\quad f\\bigg|_{\\partial D} = 0$ ，证明 $f(x,y)$ 在 D 上恒等于 0\n\n证明 反证：若 $f(x,y)$ 不恒为 0，则 $\\exists(x_{0},y_{0}) \\in D^{o}$ 满足 $f(x_{0},y_{0}) \\neq 0$ ，不妨设 $f(x_{0},y_{0}) > 0$ ，设 f 在 $(x_{0},y_{0})$ 处取正最大值，由于在区域内部，最大值一定是极值点，即 $f_{x}(x_{0},y_{0}) = f_{x}(x_{0},y_{0}) = 0$ ，则与 $f(x_{0},y_{0}) > 0$ 矛盾。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":121,"id":"15d5d459-1c09-4038-ac04-12b27a784b73","page_size":[595,841],"block_position":"121-13"},{"bbox":[35,617,539,694],"type":"text","angle":0,"index":14,"merge_prev":false,"text":"练习13.18构造辅助函数设 $f(x,y)$ 在单位圆盘 $D:\\{x^{2} + y^{2}\\leq 1\\}$ 内具有连续一阶偏导，且对 $\\forall (x,y)\\in D$ ，有 $|f(x,y)|\\leq 1$ ，证明： $\\exists (x_0,y_0)\\in D^o$ 使得 $f_{x}^{2}(x_{0},y_{0}) + f_{y}^{2}(x_{0},y_{0}) < 16$ 证明构造 $g(x,y) = f(x,y) + 2(x^{2} + y^{2})$ ， $g(0,0) = f(0,0)\\leq 1$ ，而在边界上 $g(x,y) = f(x,y) + 2\\geq 1$ ，因此要么 $g(x,y)$ 在 $D$ 上恒等于1，要么在 $D^o$ 某点取极小值，总 $\\exists (x_0,y_0)$ 使得 $g_{x}(x_{0},y_{0}) = g_{y}(x_{0},y_{0}) = 0$ ，即 $f_{x}(x_{0},y_{0}) = -4x_{0},f_{y}(x_{0},y_{0}) = -4y_{0}$ ，进而 $f_{x}^{2}(x_{0},y_{0}) + f_{y}^{2}(x_{0},y_{0})\\leq 16(x_{0}^{2} + y_{0}^{2})\\leq 16$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":121,"id":"50526a93-19bd-4167-9315-d3c353083567","page_size":[595,841],"block_position":"121-14"},{"bbox":[289,803,305,812],"type":"page_number","angle":0,"index":1000014,"text":"114 ","color":{"line":"rgba(164, 164, 164, 1)","fill":"rgba(164, 164, 164, 1)"},"page_idx":121,"is_discarded":true,"id":"a2618177-712d-454e-8cf0-e0a922292190","page_size":[595,841],"block_position":"121-15"}],[{"bbox":[475,46,539,58],"type":"header","angle":0,"index":-99999,"text":"13.5 条件极值 ","color":{"line":"rgba(164, 164, 164, 1)","fill":"rgba(164, 164, 164, 1)"},"page_idx":122,"is_discarded":true,"id":"190469e9-d4ca-4c5f-8611-bd653a3d52dd","page_size":[595,841],"block_position":"122-0"},{"bbox":[57,74,151,90],"type":"title","angle":0,"index":1,"level":2,"text":"## 13.5 条件极值","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":122,"id":"9772fe64-d9aa-48f5-94dd-0cd797305971","page_size":[595,841],"block_position":"122-1"},{"bbox":[57,107,185,121],"type":"title","angle":0,"index":2,"level":2,"text":"## 13.5.1 Lagrange 乘数法","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":122,"id":"a302561a-7af9-4f9d-b00c-39278f21f8f2","page_size":[595,841],"block_position":"122-2"},{"bbox":[74,131,506,144],"type":"text","angle":0,"index":3,"merge_prev":false,"text":"下面考虑目标函数为 $y = f(x_{1}, x_{2}, \\cdots, x_{n})$ ，条件组 $\\varphi_{k}(x_{1}, x_{2}, \\cdots, x_{n}) = 0, k = 1, 2, \\cdots, m (m < n)$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":122,"id":"b209bbc4-729d-4121-b29a-c75d0cd4201d","page_size":[595,841],"block_position":"122-3"},{"bbox":[73,153,195,166],"type":"title","angle":0,"index":4,"level":2,"text":"## 定义 13.11 (Lagrange 函数)","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":122,"id":"417322fc-75bf-41d5-8f45-1b80b5b8485e","page_size":[595,841],"block_position":"122-4"},{"bbox":[69,171,285,183],"type":"text","angle":0,"index":5,"merge_prev":false,"text":"针对上述目标函数与条件组，定义 Lagrange 函数 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":122,"id":"f7568689-b6ee-4626-a220-375df0a2925a","page_size":[595,841],"block_position":"122-5"},{"bbox":[122,185,471,217],"type":"interline_equation","angle":0,"index":6,"text":"$$\nL (x _ {1}, x _ {2}, \\dots , x _ {n}, \\lambda_ {1}, \\lambda_ {2}, \\dots , \\lambda_ {m}) = f (x _ {1}, x _ {2}, \\dots , x _ {n}) + \\sum_ {k = 1} ^ {m} \\lambda_ {k} \\varphi_ {k} (x _ {1}, x _ {2}, \\dots , x _ {n})\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":122,"id":"27d5d428-09d5-4ab9-a59f-dee08847af43","page_size":[595,841],"block_position":"122-6"},{"bbox":[73,238,204,251],"type":"title","angle":0,"index":7,"level":2,"text":"## 定理 13.14 (Lagrange 乘数法)","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":122,"id":"8690332a-fa46-472d-b9d9-948f646fa727","page_size":[595,841],"block_position":"122-7"},{"bbox":[67,255,525,299],"type":"text","angle":0,"index":8,"merge_prev":false,"text":"若 $f, \\varphi_{i}$ 在 D 上有一阶连续偏导， $P_{0}(x_{1}^{(0)}, \\cdots, x_{n}^{(0)})$ 是 f 极值点，Jacobi 矩阵 $\\left(\\frac{\\partial \\varphi_{k}}{\\partial x_{i}}(P_{0})\\right)_{m \\times n}$ 秩为 m，则存在 $\\lambda_{1}^{(0)}, \\lambda_{2}^{(1)}, \\cdots, \\lambda_{m}^{(0)}$ ，使得 $(x_{1}^{(0)}, x_{2}^{(0)}, \\cdots, x_{n}^{(0)}, \\lambda_{1}^{(0)}, \\cdots, \\lambda_{m}^{(0)})$ 是 Lagrange 函数的稳定点： ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":122,"id":"e2d9acc4-21a2-4d1b-a12b-d63bfecb8720","page_size":[595,841],"block_position":"122-8"},{"bbox":[225,301,358,339],"type":"interline_equation","angle":0,"index":9,"text":"$$\n\\left\\{ \\begin{array}{l} L _ {x _ {1}} = L _ {x _ {2}} = \\dots = L _ {x _ {n}} = 0 \\\\ L _ {\\lambda_ {1}} = L _ {\\lambda_ {2}} = \\dots = L _ {\\lambda_ {m}} = 0 \\end{array} \\right.\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":122,"id":"c7132c1d-629c-4eb6-ac6f-54585ba37037","page_size":[595,841],"block_position":"122-9"},{"bbox":[35,371,539,399],"type":"text","angle":0,"index":10,"merge_prev":false,"text":"练习 13.19Lagrange 乘数法的直接使用 (1) 讨论 $f(x,y,z)=xyz$ 在条件 $x^{2}+y^{2}+z^{2}=1$ 和 $x+y+z=0$ 条件下的最值。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":122,"id":"22814eb2-a808-44e4-8cc4-b34304cadf09","page_size":[595,841],"block_position":"122-10"},{"bbox":[35,399,477,420],"type":"text","angle":0,"index":11,"merge_prev":false,"text":"练习 13.20Lagrange 乘数法证明不等式 (1) 用 Lagrange 乘数法证明 $\\sqrt[n]{x_1 x_2 \\cdots x_n} \\leq \\frac{x_1 + \\cdots + x_n}{n}$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":122,"id":"448b5517-00bf-4856-b450-80779fb742a3","page_size":[595,841],"block_position":"122-11"},{"bbox":[73,418,249,444],"type":"text","angle":0,"index":12,"merge_prev":false,"text":"(2)x,y>0，证明 $\\frac{x^{n}+y^{n}}{2}\\geq\\left(\\frac{x+y}{2}\\right)^{n}$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":122,"id":"1e566e76-5d18-4d66-baa0-e17976f7ff7f","page_size":[595,841],"block_position":"122-12"},{"bbox":[52,444,539,473],"type":"text","angle":0,"index":13,"merge_prev":false,"text":"证明 (1) 设 $x_{1} + \\cdots + x_{n} = a$ ，目标即求 $f = x_{1} \\cdots x_{n}$ 在 $x_{1} + \\cdots + x_{n} = a$ 条件下的最值。则构造 $L = x_{1}x_{2} \\cdots x_{n} + \\lambda(x_{1} + \\cdots + x_{n} - a)$ ，解方程组： ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":122,"id":"8be78231-bf74-43df-ace5-6fa237693903","page_size":[595,841],"block_position":"122-13"},{"bbox":[215,476,368,557],"type":"interline_equation","angle":0,"index":14,"text":"$$\n\\left\\{ \\begin{array}{l} L _ {x _ {1}} = \\frac {x _ {1} \\cdots x _ {n}}{x _ {1}} + \\lambda = 0 \\\\ \\vdots \\\\ L _ {x _ {n}} = \\frac {x _ {1} \\cdots x _ {n}}{x _ {n}} + \\lambda = 0 \\\\ L _ {\\lambda} = x _ {1} + x _ {2} + \\dots + x _ {n} - a = 0 \\end{array} \\right.\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":122,"id":"7620e6d0-c486-4503-b661-acb38f4b453d","page_size":[595,841],"block_position":"122-14"},{"bbox":[54,562,442,583],"type":"text","angle":0,"index":15,"merge_prev":false,"text":"解得 $x_{1}=x_{2}=\\cdots=x_{n}=\\frac{a}{n}$ 。f 在区域边界上取到最小值，在极值点取到最大值，因此 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":122,"id":"37da3f7d-fada-4ff6-a375-7debf2b48903","page_size":[595,841],"block_position":"122-15"},{"bbox":[210,585,383,611],"type":"interline_equation","angle":0,"index":16,"text":"$$\nx _ {1} \\dots x _ {n} \\leq \\left(\\frac {a}{n}\\right) ^ {n} = \\left(\\frac {x _ {1} + \\cdots + x _ {n}}{n}\\right) ^ {n}\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":122,"id":"22d3b82c-65e2-44d5-9871-be777f0742d3","page_size":[595,841],"block_position":"122-16"},{"bbox":[35,626,539,654],"type":"text","angle":0,"index":17,"merge_prev":false,"text":"练习 13.21Lagrange 乘数法的应用 (1) 抛物面 $x^{2} + y^{2} = z$ 被平面 $x + y + z = 1$ 截成一个椭圆，求椭圆到原点的最近和最远距离 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":122,"id":"3e8f7661-9e34-462f-8196-37307e292d77","page_size":[595,841],"block_position":"122-17"},{"bbox":[53,657,539,671],"type":"text","angle":0,"index":18,"merge_prev":false,"text":"解(1)目标函数 $f(x,y,z) = x^{2} + y^{2} + z^{2}$ ，条件为 $x^{2} + y^{2} - z = 0$ 与 $x + y + z - 1 = 0$ 。根据Lagrange乘数法，令 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":122,"id":"21032ba8-71b4-4226-9845-efdfae727cec","page_size":[595,841],"block_position":"122-18"},{"bbox":[151,677,441,691],"type":"interline_equation","angle":0,"index":19,"text":"$$\nL (x, y, z, \\lambda , \\mu) = x ^ {2} + y ^ {2} + z ^ {2} + \\lambda (x ^ {2} + y ^ {2} - z) + \\mu (x + y + z - 1)\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":122,"id":"a37f1d6b-1477-4ae0-bdac-fa05902b6d59","page_size":[595,841],"block_position":"122-19"},{"bbox":[289,803,305,812],"type":"page_number","angle":0,"index":1000019,"text":"115 ","color":{"line":"rgba(164, 164, 164, 1)","fill":"rgba(164, 164, 164, 1)"},"page_idx":122,"is_discarded":true,"id":"7b5c9237-7f71-4af0-941c-d2284595ceca","page_size":[595,841],"block_position":"122-20"}],[{"bbox":[476,47,539,58],"type":"header","angle":0,"index":-99999,"text":"13.5 条件极值 ","color":{"line":"rgba(164, 164, 164, 1)","fill":"rgba(164, 164, 164, 1)"},"page_idx":123,"is_discarded":true,"id":"b4ede745-074d-4b6c-9dec-942b7a1731c8","page_size":[595,841],"block_position":"123-0"},{"bbox":[54,72,248,84],"type":"text","angle":0,"index":1,"merge_prev":false,"text":"对 $L$ 求一阶偏导数，且令它们都为零，则有 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":123,"id":"61011b5a-5b1f-4056-9a86-20ee81297c0b","page_size":[595,841],"block_position":"123-1"},{"bbox":[236,89,348,179],"type":"interline_equation","angle":0,"index":2,"text":"$$\n\\left\\{ \\begin{array}{l} L _ {x} = 2 x + 2 x \\lambda + \\mu = 0 \\\\ L _ {y} = 2 y + 2 y \\lambda + \\mu = 0 \\\\ L _ {z} = 2 z - \\lambda + \\mu = 0 \\\\ L _ {\\lambda} = x ^ {2} + y ^ {2} - z = 0 \\\\ L _ {\\mu} = x + y + z - 1 = 0 \\end{array} \\right.\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":123,"id":"4c232baa-219a-4c67-a780-912617e33f92","page_size":[595,841],"block_position":"123-2"},{"bbox":[54,184,540,241],"type":"text","angle":0,"index":3,"merge_prev":false,"text":"首先前两个方程有对称性，因此相减得到 $2(x - y) + 2(x - y)\\lambda = (x - y)(1 + \\lambda) = 0$ ，要求 $x = y$ 或者 $\\lambda = -1$ 。若 $\\lambda = -1$ ，代入第一个式子得到 $\\mu = 0$ ，代入第三个式子得到 $z = -\\frac{1}{2}$ ，而第四个式子要求 $x^{2} + y^{2} = z$ ，因此无解。若 $x = y$ ，则 $z = 2x^{2}, z = -2x + 1$ ，解出 $x = y = \\frac{-1 \\pm \\sqrt{3}}{2}, z = 2 \\mp \\sqrt{3}$ ，而 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":123,"id":"b3fc4919-52e4-4d2c-8b04-f9aa4b55caa7","page_size":[595,841],"block_position":"123-3"},{"bbox":[198,245,396,275],"type":"interline_equation","angle":0,"index":4,"text":"$$\nf \\left(\\frac {- 1 \\pm \\sqrt {3}}{2}, \\frac {- 1 \\pm \\sqrt {3}}{2}, 2 \\mp \\sqrt {3}\\right) = 9 \\mp 5 \\sqrt {3}\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":123,"id":"4d7276de-c230-47df-8476-a21f816392c9","page_size":[595,841],"block_position":"123-4"},{"bbox":[54,280,224,292],"type":"text","angle":0,"index":5,"merge_prev":false,"text":"最大值为 $9 + 5\\sqrt{3}$ ，最小值为 $9 - 5\\sqrt{3}$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":123,"id":"795cdf35-6ccf-4bf4-8e8f-bb46b8719762","page_size":[595,841],"block_position":"123-5"},{"bbox":[289,803,305,812],"type":"page_number","angle":0,"index":1000005,"text":"116 ","color":{"line":"rgba(164, 164, 164, 1)","fill":"rgba(164, 164, 164, 1)"},"page_idx":123,"is_discarded":true,"id":"188e1908-85f9-471d-9022-445cdba26647","page_size":[595,841],"block_position":"123-6"}],[{"bbox":[231,74,373,94],"type":"title","angle":0,"index":0,"level":2,"text":"## 第 14 章 含参积分","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":124,"id":"2933b966-895e-439f-be1f-3620989fb727","page_size":[595,841],"block_position":"124-0"},{"bbox":[57,122,180,139],"type":"title","angle":0,"index":1,"level":2,"text":"## 14.1 含参正常积分","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":124,"id":"7695d6ce-cdb0-4d53-b8d8-59ec213b3629","page_size":[595,841],"block_position":"124-1"},{"bbox":[56,155,253,170],"type":"title","angle":0,"index":2,"level":2,"text":"## 14.1.1 含参正常积分基本概念与定理","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":124,"id":"abc83c8c-7b71-4cae-9fa6-33ab07b5140e","page_size":[595,841],"block_position":"124-2"},{"bbox":[71,182,186,195],"type":"title","angle":0,"index":3,"level":2,"text":"## 定义 14.1 (含参正常积分)","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":124,"id":"95fd4a79-303f-475f-b09c-14e94820ebc8","page_size":[595,841],"block_position":"124-3"},{"bbox":[68,199,523,213],"type":"text","angle":0,"index":4,"merge_prev":false,"text":"$f(x,t)$ 是定义于 $G = \\{\\alpha (t)\\leq x\\leq \\beta (t),a\\leq t\\leq b\\}$ 区域的二元函数，当对 $x$ 积分时，该积分为 $t$ 的函数： ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":124,"id":"9aa10ac4-0b9a-44b6-a9be-42ab4deb327b","page_size":[595,841],"block_position":"124-4"},{"bbox":[226,215,367,245],"type":"interline_equation","angle":0,"index":5,"text":"$$\nF (t) = \\int_ {\\alpha (t)} ^ {\\beta (t)} f (x, t) \\mathrm{d} x, t \\in [ a, b ]\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":124,"id":"cec9d40e-8dec-44e2-b4a3-1752cd69a20a","page_size":[595,841],"block_position":"124-5"},{"bbox":[73,265,246,279],"type":"title","angle":0,"index":6,"level":2,"text":"## 定理 14.1 (规则区域含参正常积分性质)","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":124,"id":"54d6c55c-a78f-4122-aedd-17d0423d6217","page_size":[595,841],"block_position":"124-6"},{"bbox":[68,282,221,296],"type":"text","angle":0,"index":7,"merge_prev":false,"text":"若 $f(x,t)$ 在 $[a,b]\\times [c,d]$ 连续，则 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":124,"id":"30c4f4ba-23c3-44d3-9cf7-cf58a216b922","page_size":[595,841],"block_position":"124-7"},{"bbox":[82,294,436,322],"type":"text","angle":0,"index":8,"merge_prev":false,"text":"- 连续性：含参积分连续，即对 $\\forall t_0 \\in [c, d]$ 有 $\\lim_{t \\to t_0} \\int_a^b f(x, t) \\mathrm{d}x = \\int_a^b f(x, t_0) \\mathrm{d}x$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":124,"id":"52520ab8-0567-4281-812b-d9a4939aed56","page_size":[595,841],"block_position":"124-8"},{"bbox":[83,320,326,348],"type":"text","angle":0,"index":9,"merge_prev":false,"text":"- 积分交换： $\\int_{c}^{d}\\mathrm{d}t\\int_{a}^{b}f(x,t)\\mathrm{d}x = \\int_{a}^{b}\\mathrm{d}x\\int_{c}^{d}f(x,t)\\mathrm{d}x$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":124,"id":"4a33e86b-d7b3-4428-8af3-38a7d5862388","page_size":[595,841],"block_position":"124-9"},{"bbox":[82,347,444,379],"type":"text","angle":0,"index":10,"merge_prev":false,"text":"- 可导：若 $f_{t}(x,t)$ 在 $[a,b]\\times [c,d]$ 也连续，则 $\\frac{\\mathrm{d}}{\\mathrm{d}t}\\left(\\int_{a}^{b}f(x,t)\\mathrm{d}x\\right) = \\int_{a}^{b}f_{t}(x,t)\\mathrm{d}x$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":124,"id":"528c41cd-9f67-4cd1-a4fa-b9fb683ce73e","page_size":[595,841],"block_position":"124-10"},{"bbox":[73,399,246,412],"type":"title","angle":0,"index":11,"level":2,"text":"## 定理 14.2 (不规则区域的连续与可微性)","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":124,"id":"3deb4836-b4e0-4e72-9578-3c324379f5aa","page_size":[595,841],"block_position":"124-11"},{"bbox":[69,416,363,430],"type":"text","angle":0,"index":12,"merge_prev":false,"text":"若 $f(x,t)$ 在 $[\\alpha (t),\\beta (t)]\\times [c,d]$ 连续， $\\alpha (t),\\beta (t)$ 在 $[c,d]$ 上连续，则 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":124,"id":"bb17b360-f02c-46d6-a2a3-148a9c1376a5","page_size":[595,841],"block_position":"124-12"},{"bbox":[82,428,320,458],"type":"text","angle":0,"index":13,"merge_prev":false,"text":"- 连续性： $\\varphi (t) = \\int_{\\alpha (t)}^{\\beta (t)}f(x,t)\\mathrm{d}x$ 在 $[c,d]$ 上连续，即 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":124,"id":"fe8f53c0-06ea-4915-8428-e51bfc2519ac","page_size":[595,841],"block_position":"124-13"},{"bbox":[218,460,399,489],"type":"interline_equation","angle":0,"index":14,"text":"$$\n\\forall t _ {0} \\in [ c, d ], \\lim _ {t \\rightarrow t _ {0}} \\varphi (t) = \\int_ {\\alpha (t _ {0})} ^ {\\beta (t _ {0})} f (x, t _ {0}) \\mathrm{d} x\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":124,"id":"b12c1718-d482-45d8-8a65-d3418606f4c2","page_size":[595,841],"block_position":"124-14"},{"bbox":[83,491,525,520],"type":"text","angle":0,"index":15,"merge_prev":false,"text":"- 可微性：若 $f_{t}(x,t)$ 在 $[a,b]\\times [c,d]$ 上连续， $\\alpha (t),\\beta (t)$ 在 $[c,d]$ 上可导，则(将被积分变量用上下限代替) ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":124,"id":"b88f1a60-5bea-4fcf-9e9f-3425827e3fa4","page_size":[595,841],"block_position":"124-15"},{"bbox":[183,518,434,548],"type":"interline_equation","angle":0,"index":16,"text":"$$\n\\varphi^ {\\prime} (t) = f (\\beta (t), t) \\beta^ {\\prime} (t) - f (\\alpha (t), t) \\alpha^ {\\prime} (t) + \\int_ {\\alpha (t)} ^ {\\beta (t)} f _ {t} (x, t) \\mathrm{d} x\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":124,"id":"04f523aa-00d1-4ba1-9f08-a1ceb3787d38","page_size":[595,841],"block_position":"124-16"},{"bbox":[35,574,508,600],"type":"text","angle":0,"index":17,"merge_prev":false,"text":"练习 14.1 含参积分求导 (1) 重点: 已知 $f(x) \\in C[a, b]$ , $F(y) = \\int_{a}^{b} f(x) |y - x| \\mathrm{d}x, y \\in [a, b]$ , 计算 $F''(y)$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":124,"id":"5eaabe83-8aa5-453b-88a5-d76950b8cca2","page_size":[595,841],"block_position":"124-17"},{"bbox":[52,600,455,629],"type":"text","angle":0,"index":18,"merge_prev":false,"text":"解(1)首先得改写为不规则区域积分： $F(y)=\\int_{a}^{y}f(x)(y-x)\\mathrm{d}x+\\int_{y}^{b}f(x)(x-y)\\mathrm{d}x$ ，因此 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":124,"id":"6fe3bab9-c05e-4cea-8c4d-9b83637b875c","page_size":[595,841],"block_position":"124-18"},{"bbox":[95,630,497,661],"type":"interline_equation","angle":0,"index":19,"text":"$$\nF ^ {\\prime} (y) = f (y) (y - y) \\cdot 1 + \\int_ {a} ^ {y} f (x) \\mathrm{d} x - f (y) (y - y) - \\int_ {y} ^ {b} f (x) \\mathrm{d} x = \\int_ {a} ^ {y} f (x) \\mathrm{d} x - \\int_ {y} ^ {b} f (x) \\mathrm{d} x\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":124,"id":"352374d1-eb16-4381-a28e-41691894b1ff","page_size":[595,841],"block_position":"124-19"},{"bbox":[54,662,226,676],"type":"text","angle":0,"index":20,"merge_prev":false,"text":"进而得到 $F''(y) = f(y) + f(y) = 2f(y)$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":124,"id":"371e1723-1c2a-486e-a353-eda83693b61a","page_size":[595,841],"block_position":"124-20"},{"bbox":[35,688,539,744],"type":"text","angle":0,"index":21,"merge_prev":false,"text":"练习14.2含参积分求导的应用(1)计算极限 $\\lim_{x\\to 0}\\frac{\\int_0^{x^2}\\sqrt{1 + t^2}\\mathrm{d}t}{e^x\\sin x - \\sin x}$ 解(1)方法一：使用含参积分求导，分母等价于 $(1 + x)x - x = x^2$ ，使用洛必达法则得到(注意分子是含参积分求导): ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":124,"id":"67e890df-96aa-4f6c-81df-44dbe98a0605","page_size":[595,841],"block_position":"124-21"},{"bbox":[217,743,376,768],"type":"interline_equation","angle":0,"index":22,"text":"$$\n\\lim _ {x \\to 0} \\frac {2 x \\sqrt {1 + x ^ {4}}}{2 x} = \\lim _ {x \\to 0} \\sqrt {1 + x ^ {4}} = 1\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":124,"id":"2d81d227-72c1-4fe6-954e-87803e018ac6","page_size":[595,841],"block_position":"124-22"}],[{"bbox":[405,47,539,58],"type":"header","angle":0,"index":-99999,"text":"14.2 含参反常积分与一致收敛 ","color":{"line":"rgba(164, 164, 164, 1)","fill":"rgba(164, 164, 164, 1)"},"page_idx":125,"is_discarded":true,"id":"adf5c388-5ca0-4de0-a12c-aacf81f67ce4","page_size":[595,841],"block_position":"125-0"},{"bbox":[74,69,466,94],"type":"text","angle":0,"index":1,"merge_prev":false,"text":"方法二：Taylor展开 $+$ 直接积分：分母等价于 $x^{2}$ ，分子用 $\\sqrt{1 + t^2} = 1 + \\frac{t^2}{2} + o(t^2)$ 得到： ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":125,"id":"0403ef28-e32b-4080-bfbf-8e33e0e17a07","page_size":[595,841],"block_position":"125-1"},{"bbox":[171,96,421,126],"type":"interline_equation","angle":0,"index":2,"text":"$$\nI = \\lim _ {x \\to 0} \\frac {\\int_ {0} ^ {x ^ {2}} 1 + \\frac {t ^ {2}}{2} + o (t ^ {2}) \\mathrm{d} t}{x ^ {2}} = \\lim _ {x \\to 0} \\frac {x ^ {2} + \\frac {x ^ {6}}{6} + o (x ^ {6})}{x ^ {2}} = 1\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":125,"id":"79c513da-9001-4ab3-be65-20354da46d8f","page_size":[595,841],"block_position":"125-2"},{"bbox":[56,141,229,156],"type":"title","angle":0,"index":3,"level":2,"text":"## 14.1.2 常用含参正常积分的计算","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":125,"id":"4fa6aa3d-a257-413e-b053-58f60a68dc1f","page_size":[595,841],"block_position":"125-3"},{"bbox":[73,169,117,180],"type":"title","angle":0,"index":4,"level":2,"text":"## 引理14.1","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":125,"id":"fd9b42ea-251c-4883-aea4-848025a862ad","page_size":[595,841],"block_position":"125-4"},{"bbox":[66,183,525,216],"type":"text","angle":0,"index":5,"merge_prev":false,"text":"$|a| < 1$ 时， $\\int \\frac{\\mathrm{d}x}{1 + a\\cos x} = \\frac{2}{\\sqrt{1 - a^2}}\\arctan \\left(\\sqrt{\\frac{1 - a}{1 + a}}\\tan \\frac{x}{2}\\right) + C,$ 特别地， $\\int_0^\\pi \\frac{\\mathrm{d}x}{1 + a\\cos x} = \\frac{\\pi}{\\sqrt{1 - x^2}}$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":125,"id":"7972b480-371c-444a-a147-1af53cc4f998","page_size":[595,841],"block_position":"125-5"},{"bbox":[52,232,307,245],"type":"text","angle":0,"index":6,"merge_prev":false,"text":"证明 使用万能公式，在前面三角函数不定积分中涉及过。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":125,"id":"a7a7a569-072a-4821-95e9-653aa80d19b0","page_size":[595,841],"block_position":"125-6"},{"bbox":[35,258,349,285],"type":"text","angle":0,"index":7,"merge_prev":false,"text":"练习14.3积分：涉及 $\\frac{x^b - x^a}{\\ln x}$ 型(1)计算 $\\int_0^1\\frac{x^b - x^a}{\\ln x}\\mathrm{d}x,0 < a < b$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":125,"id":"31ccfb3e-f58d-40fd-9c66-6837f833e16f","page_size":[595,841],"block_position":"125-7"},{"bbox":[73,284,226,311],"type":"text","angle":0,"index":8,"merge_prev":false,"text":"(2) 计算 $\\int_0^1 \\sin \\left(\\ln \\frac{1}{x}\\right) \\frac{x^b - x^a}{\\ln x} \\, \\mathrm{d}x$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":125,"id":"26eb2aff-e9b8-48c8-9381-52ba5c91dca6","page_size":[595,841],"block_position":"125-8"},{"bbox":[73,312,345,336],"type":"text","angle":0,"index":9,"merge_prev":false,"text":"(3) 计算 $F(a) = \\int_{0}^{\\frac{\\pi}{2}} \\ln \\frac{1 + a \\cos x}{1 - a \\cos x} \\cdot \\frac{1}{\\cos x} \\, \\mathrm{d}x$ ，这里 $a \\in (-1, 1)$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":125,"id":"ef3c5b71-d590-4a3d-bb49-b88fa9b4286a","page_size":[595,841],"block_position":"125-9"},{"bbox":[52,337,360,364],"type":"text","angle":0,"index":10,"merge_prev":false,"text":"解 (1) 由于 $\\int_{a}^{b} x^{y} \\mathrm{d}y = \\frac{x^{y}}{\\ln x}\\bigg|_{a}^{b} = \\frac{x^{b} - x^{a}}{\\ln x}$ ，因此根据交换积分顺序得到 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":125,"id":"1a71bd95-74cf-46e4-a783-2c6c2d40f13e","page_size":[595,841],"block_position":"125-10"},{"bbox":[161,367,431,396],"type":"interline_equation","angle":0,"index":11,"text":"$$\n\\int_ {0} ^ {1} \\frac {x ^ {b} - x ^ {a}}{\\ln x} \\mathrm{d} t = \\int_ {0} ^ {1} \\mathrm{d} x \\int_ {a} ^ {b} x ^ {y} \\mathrm{d} y = \\int_ {a} ^ {b} \\mathrm{d} y \\int_ {0} ^ {1} x ^ {y} \\mathrm{d} x = \\ln \\frac {1 + b}{1 + a}\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":125,"id":"582e485e-2d3f-4e9e-9bc7-ad5305138077","page_size":[595,841],"block_position":"125-11"},{"bbox":[73,397,514,425],"type":"text","angle":0,"index":12,"merge_prev":false,"text":"(2) 由于 $\\int_{a}^{b} x^{y} \\mathrm{d}y = \\frac{x^{b} - x^{a}}{\\ln x}$ ，而 $x = 0$ 为瑕点，考虑 $\\lim_{x \\to 0} \\sin \\left(\\ln \\frac{1}{x}\\right) x^{y} = 0$ ，因此可视为连续。从而 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":125,"id":"50b0c4cd-2e62-4b5e-a72a-01b9df16bc1e","page_size":[595,841],"block_position":"125-12"},{"bbox":[175,428,418,455],"type":"interline_equation","angle":0,"index":13,"text":"$$\n\\int_ {0} ^ {1} \\sin \\left(\\ln {\\frac {1}{x}}\\right) \\frac {x ^ {b} - x ^ {a}}{\\ln x} \\mathrm{d} x = \\int_ {a} ^ {b} \\mathrm{d} y \\int_ {0} ^ {1} \\sin \\left(\\ln {\\frac {1}{x}}\\right) x ^ {y} \\mathrm{d} x\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":125,"id":"ff5d59f3-8027-4f49-bcdb-3c7eccdaba81","page_size":[595,841],"block_position":"125-13"},{"bbox":[54,458,453,486],"type":"text","angle":0,"index":14,"merge_prev":false,"text":"令 $u = \\ln \\frac{1}{x}$ 即可转化为 $\\int_{a}^{b}\\mathrm{d}y\\int_{0}^{+\\infty}\\sin ue^{-(y + 1)u}\\mathrm{d}u$ ，结果为 $\\arctan (b + 1) - \\arctan (a + 1)$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":125,"id":"200ea29d-d967-4cfd-866b-d7a85d2b882d","page_size":[595,841],"block_position":"125-14"},{"bbox":[74,484,121,496],"type":"text","angle":0,"index":15,"merge_prev":false,"text":"(3) 注意到 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":125,"id":"5680a30e-e4c2-4155-a401-6e6649838d34","page_size":[595,841],"block_position":"125-15"},{"bbox":[130,497,461,523],"type":"interline_equation","angle":0,"index":16,"text":"$$\n\\ln \\frac {1 + a \\cos x}{1 - a \\cos x} \\cdot \\frac {1}{\\cos x} \\mathrm{d} x = \\frac {\\ln (1 + a \\cos x) - \\ln (1 - a \\cos x)}{\\cos x} = \\int_ {- a} ^ {a} \\frac {\\mathrm{d} y}{1 + y \\cos x}\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":125,"id":"93a04ed8-fd4b-4d1b-896e-9ae4f185aea9","page_size":[595,841],"block_position":"125-16"},{"bbox":[54,528,170,550],"type":"text","angle":0,"index":17,"merge_prev":false,"text":"而 $\\frac{y}{1 + y\\cos x}$ 连续，因此 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":125,"id":"8fc93dea-d214-4921-b4aa-755bfdfb42e3","page_size":[595,841],"block_position":"125-17"},{"bbox":[223,552,370,580],"type":"interline_equation","angle":0,"index":18,"text":"$$\nF (a) = \\int_ {- a} ^ {a} \\mathrm{d} y \\int_ {0} ^ {\\frac {\\pi}{2}} \\frac {1}{1 + y \\cos x} \\mathrm{d} x\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":125,"id":"6fbffeec-0879-4cc8-8cdb-007567cb08ac","page_size":[595,841],"block_position":"125-18"},{"bbox":[54,583,171,595],"type":"text","angle":0,"index":19,"merge_prev":false,"text":"最终积分结果为 $\\pi \\arcsin a$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":125,"id":"8d9ca91c-992e-4924-9abe-0f5fcc4a8600","page_size":[595,841],"block_position":"125-19"},{"bbox":[289,802,305,812],"type":"page_number","angle":0,"index":1000019,"text":"118 ","color":{"line":"rgba(164, 164, 164, 1)","fill":"rgba(164, 164, 164, 1)"},"page_idx":125,"is_discarded":true,"id":"d202388a-e92d-4ee0-a3a1-dc56545605a6","page_size":[595,841],"block_position":"125-20"}],[{"bbox":[405,47,539,58],"type":"header","angle":0,"index":-99999,"text":"14.2 含参反常积分与一致收敛 ","color":{"line":"rgba(164, 164, 164, 1)","fill":"rgba(164, 164, 164, 1)"},"page_idx":126,"is_discarded":true,"id":"1a75c1eb-7135-4e9d-9e79-97cbb8e9568d","page_size":[595,841],"block_position":"126-0"},{"bbox":[57,74,252,90],"type":"title","angle":0,"index":1,"level":2,"text":"## 14.2 含参反常积分与一致收敛","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":126,"id":"8bbeb716-ff01-46cb-99de-9093ca919b95","page_size":[595,841],"block_position":"126-1"},{"bbox":[57,107,145,121],"type":"title","angle":0,"index":2,"level":2,"text":"## 14.2.1 基本概念","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":126,"id":"394b1767-916d-4155-aa6a-632f67d885cb","page_size":[595,841],"block_position":"126-2"},{"bbox":[73,134,186,148],"type":"title","angle":0,"index":3,"level":2,"text":"## 定义 14.2 (含参反常积分)","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":126,"id":"b568628e-8ccd-4dd2-885a-d1a4065acb42","page_size":[595,841],"block_position":"126-3"},{"bbox":[67,149,532,192],"type":"text","angle":0,"index":4,"merge_prev":false,"text":"$f(x,t)$ 在 $[c, +\\infty) \\times I$ 上有定义，对 $\\forall t \\in I$ ，反常积分 $\\int_{c}^{+\\infty} f(x,t) \\mathrm{d}x$ 收敛，则记 $\\Phi(t) = \\int_{c}^{+\\infty} f(x,t) \\mathrm{d}x$ 称其为含参反常积分。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":126,"id":"5c9fd55e-346c-4f93-adf4-f0625d3e1a21","page_size":[595,841],"block_position":"126-4"},{"bbox":[72,209,226,222],"type":"title","angle":0,"index":5,"level":2,"text":"## 定义 14.3 (含参反常积分一致收敛)","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":126,"id":"8ce34609-baaa-4607-aca5-cdafe7a88d91","page_size":[595,841],"block_position":"126-5"},{"bbox":[68,225,489,257],"type":"text","angle":0,"index":6,"merge_prev":false,"text":"若对 $\\forall \\epsilon, \\exists N > c, \\forall M > N, \\forall t \\in I$ 有 $|\\int_{M}^{+\\infty} f(x,t) \\mathrm{d}x| < \\epsilon$ ，则称 $\\int_{c}^{+\\infty} f(x,t) \\mathrm{d}x$ 一致收敛于 $\\Phi(t)$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":126,"id":"7c3eaa17-ffd0-485d-8950-5bf7b3d6149e","page_size":[595,841],"block_position":"126-6"},{"bbox":[72,275,226,287],"type":"title","angle":0,"index":7,"level":2,"text":"## 定理 14.3 (一致收敛基本充要条件)","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":126,"id":"856428f5-f355-40c7-b939-67d89ca67385","page_size":[595,841],"block_position":"126-7"},{"bbox":[68,291,286,303],"type":"text","angle":0,"index":8,"merge_prev":false,"text":"含参反常积分一致收敛有以下两个基本充要条件： ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":126,"id":"b2ab444f-746a-4e3a-af60-9c6f3bf1bd6f","page_size":[595,841],"block_position":"126-8"},{"bbox":[82,303,430,330],"type":"text","angle":0,"index":9,"merge_prev":false,"text":"- Cauchy 收敛准则： $\\forall \\epsilon, \\exists M > c, \\forall A_1, A_2 > M, \\forall t \\in I$ 满足 $|\\int_{A_1}^{A_2} f(x, t) \\mathrm{d}x| < \\epsilon$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":126,"id":"44f5ef51-8652-4be5-a487-22e6d3b59152","page_size":[595,841],"block_position":"126-9"},{"bbox":[83,329,309,357],"type":"text","angle":0,"index":10,"merge_prev":false,"text":"- 上确界判别法： $\\lim_{A\\to +\\infty}\\sup_{t\\in I}|\\int_{A}^{+\\infty}f(x,t)\\mathrm{d}x| = 0$ 。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":126,"id":"8ff5cb58-2af3-4886-934e-8fe3689f8de5","page_size":[595,841],"block_position":"126-10"},{"bbox":[52,373,329,386],"type":"title","angle":0,"index":11,"level":2,"text":"## 笔记 证明不一致收敛一般用上确界判别法，具体见下面的例子","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":126,"id":"666d9a14-a2ab-4dab-8f93-808d4523d0ea","page_size":[595,841],"block_position":"126-11"},{"bbox":[35,398,539,475],"type":"text","angle":0,"index":12,"merge_prev":false,"text":"练习 14.4 重要例子：确界判别 证明 $\\int_{0}^{+\\infty}\\frac{\\sin xy}{y}dy$ 在 $[\\delta,+\\infty)$ 上一致收敛，但在 $(0,+\\infty)$ 上不一致收敛\n\n证明 (1) 一致收敛：做变量替换 u=xy 得到 $\\int_{A}^{+\\infty}\\frac{\\sin xy}{y}dy=\\int_{Ax}^{+\\infty}\\frac{\\sin u}{u}du$ ，根据反常积分 Dirichlet 判别法可知 $\\int_{0}^{+\\infty}\\frac{\\sin u}{u}du$ 收敛，因此 $\\forall\\epsilon,\\exists M,\\forall A'>M$ 时，有 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":126,"id":"162b1fef-46cb-4399-bb0b-5ff2881c3508","page_size":[595,841],"block_position":"126-12"},{"bbox":[253,477,340,505],"type":"interline_equation","angle":0,"index":13,"text":"$$\n\\left| \\int_ {A ^ {\\prime}} ^ {+ \\infty} \\frac {\\sin u}{u} \\mathrm{d} u \\right| < \\epsilon\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":126,"id":"ea85b64a-0a8e-451d-88a4-961416726021","page_size":[595,841],"block_position":"126-13"},{"bbox":[54,505,230,530],"type":"text","angle":0,"index":14,"merge_prev":false,"text":"对题中的 $\\delta$ ，取 $A > \\frac{M}{\\delta}$ ，当 $x \\geq \\delta$ 时有 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":126,"id":"ce16765d-6778-4bd6-9adb-6a53520eb108","page_size":[595,841],"block_position":"126-14"},{"bbox":[250,528,343,556],"type":"interline_equation","angle":0,"index":15,"text":"$$\n\\left| \\int_ {A} ^ {+ \\infty} \\frac {\\sin x y}{y} \\mathrm{d} y \\right| < \\epsilon\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":126,"id":"0ed12069-6b79-4136-8b51-62788c99b1df","page_size":[595,841],"block_position":"126-15"},{"bbox":[54,562,212,574],"type":"text","angle":0,"index":16,"merge_prev":false,"text":"因此根据确界判别法可知一致收敛。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":126,"id":"21afaee0-cfe2-4820-893c-b7cba59fe840","page_size":[595,841],"block_position":"126-16"},{"bbox":[73,577,171,590],"type":"text","angle":0,"index":17,"merge_prev":false,"text":"(2) 不一致收敛：因为 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":126,"id":"b3959222-cfaa-47cf-a27e-5c0befd5b3a7","page_size":[595,841],"block_position":"126-17"},{"bbox":[130,592,462,621],"type":"interline_equation","angle":0,"index":18,"text":"$$\n\\sup _ {x \\in (0, + \\infty)} \\left| \\int_ {A} ^ {+ \\infty} \\frac {\\sin x y}{y} \\mathrm{d} y \\right| = \\sup _ {x \\in (0, + \\infty)} \\left| \\int_ {A x} ^ {+ \\infty} \\frac {\\sin u}{u} \\mathrm{d} u \\right| \\geq \\left| \\int_ {0} ^ {+ \\infty} \\frac {\\sin u}{u} \\mathrm{d} u \\right| = \\frac {\\pi}{2}\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":126,"id":"d88381be-34cd-4dfa-8060-14feb5f52fde","page_size":[595,841],"block_position":"126-18"},{"bbox":[54,624,365,636],"type":"text","angle":0,"index":19,"merge_prev":false,"text":"当 $A \\to +\\infty$ 时，极限不趋于0，因此根据上确界准则可知不一致收敛。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":126,"id":"bfaad511-28a3-4ca6-a5b7-f781200ec6cf","page_size":[595,841],"block_position":"126-19"},{"bbox":[39,634,539,676],"type":"text","angle":0,"index":20,"merge_prev":false,"text":"笔记 本题在 $[\\delta, +\\infty)$ 上也可以用 Dirichlet 判别法，在 $(0, +\\infty)$ 无法用的原因在于 $\\int_0^N \\sin xy \\, \\mathrm{d}y = \\left. \\frac{\\cos xy}{x} \\right|_0^N$ 对 $x \\in [0, +\\infty)$ 不一致有界。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":126,"id":"4b9a6425-9f1f-4498-a917-4e6ef199270b","page_size":[595,841],"block_position":"126-20"},{"bbox":[35,685,566,752],"type":"text","angle":0,"index":21,"merge_prev":false,"text":"练习14.5不一致收敛例子(1)讨论 $\\varphi (x) = \\int_0^{+\\infty}xe^{-xy}\\mathrm{d}y$ 在 $(0, + \\infty)$ 上的一致收敛性解 $(1)\\varphi (x) = \\int_0^{+\\infty}e^{-xy}d(xy) = e^{-xy}|_{+\\infty}^0 = 1$ ，但是其不一致收敛，原因是 $\\sup_{x\\in (0, + \\infty)}|\\int_M^{+\\infty}xe^{-xy}\\mathrm{d}y| = \\sup_{x\\in (0, + \\infty)}e^{-Mx} = 1$ 不趋于0。(如果 $x\\in (a, + \\infty),a > 0$ 则一致收敛) ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":126,"id":"feedf75b-1384-4c8f-87a2-f622839800a4","page_size":[595,841],"block_position":"126-21"},{"bbox":[289,802,305,812],"type":"page_number","angle":0,"index":1000021,"text":"119 ","color":{"line":"rgba(164, 164, 164, 1)","fill":"rgba(164, 164, 164, 1)"},"page_idx":126,"is_discarded":true,"id":"b9ade978-391c-46f5-aa22-19e8d6d721c1","page_size":[595,841],"block_position":"126-22"}],[{"bbox":[405,47,539,58],"type":"header","angle":0,"index":-99999,"text":"14.2 含参反常积分与一致收敛 ","color":{"line":"rgba(164, 164, 164, 1)","fill":"rgba(164, 164, 164, 1)"},"page_idx":127,"is_discarded":true,"id":"554210cb-63fd-4892-8cdd-775dc8df55f9","page_size":[595,841],"block_position":"127-0"},{"bbox":[73,73,168,86],"type":"title","angle":0,"index":1,"level":2,"text":"## 定理 14.4 (M 判别法)","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":127,"id":"b7b4b51e-d931-487c-a3cf-09c0510f9a3b","page_size":[595,841],"block_position":"127-1"},{"bbox":[67,88,530,127],"type":"text","angle":0,"index":2,"merge_prev":false,"text":"若 $g(x)$ 满足 $|f(x,t)| \\leq g(x), x \\in [c, +\\infty), t \\in I$ ，且 $\\int_{c}^{+\\infty} g(x) \\mathrm{d}x$ 收敛，则 $\\int_{c}^{+\\infty} f(x,t) \\mathrm{d}x$ 在 $I$ 上一致收敛。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":127,"id":"08a70b84-c8e8-45bd-9e58-915af62e49de","page_size":[595,841],"block_position":"127-2"},{"bbox":[73,146,176,158],"type":"title","angle":0,"index":3,"level":2,"text":"## 定理 14.5 (比较判别法)","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":127,"id":"71ee3e8e-9ed1-404d-96b5-3a6c9cc96e0b","page_size":[595,841],"block_position":"127-3"},{"bbox":[66,161,531,202],"type":"text","angle":0,"index":4,"merge_prev":false,"text":"$f(x,t)$ 在 $[c, +\\infty) \\times I$ 上连续， $|f(x,t)| \\leq F(x,t)$ ，若 $\\int_{c}^{+\\infty} F(x,t) \\mathrm{d}x$ 在 $I$ 上一致收敛，则 $\\int_{c}^{+\\infty} f(x,t) \\mathrm{d}x$ 在 $I$ 上绝对一致收敛 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":127,"id":"8db55c32-fbdc-4c61-b56a-fc5e20b6a6c1","page_size":[595,841],"block_position":"127-4"},{"bbox":[52,210,536,240],"type":"text","angle":0,"index":5,"merge_prev":false,"text":"证明 根据 Cauchy 收敛准则可知 $\\forall\\epsilon,\\exists M,\\forall A_{1},A_{2}>M,\\forall t,|\\int_{A_{1}}^{A_{2}}F(x,t)\\mathrm{d}x|<\\epsilon$ ，因此推出 $\\int_{A_{1}}^{A_{2}}|f(x,t)|\\mathrm{d}x<\\epsilon$ 。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":127,"id":"7832373e-ad1b-4c5b-9f41-a20b5be38cbd","page_size":[595,841],"block_position":"127-5"},{"bbox":[35,243,363,269],"type":"text","angle":0,"index":6,"merge_prev":false,"text":"练习14.6M判别法(1)证明 $\\int_0^{+\\infty}\\frac{\\cos xy}{1 + x^2}\\mathrm{d}x$ 在 $(- \\infty , + \\infty)$ 上一致收敛 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":127,"id":"69cad435-c043-4484-8882-9681a9c9576e","page_size":[595,841],"block_position":"127-6"},{"bbox":[72,268,270,295],"type":"text","angle":0,"index":7,"merge_prev":false,"text":"(2) 证明 $\\int_{0}^{+\\infty} e^{-x^2 y} \\, \\mathrm{d}y$ 在 $x \\in [a, b]$ 一致收敛 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":127,"id":"ac0fcdfd-83da-41f9-93ec-74365aea9ec0","page_size":[595,841],"block_position":"127-7"},{"bbox":[52,293,479,320],"type":"text","angle":0,"index":8,"merge_prev":false,"text":"证明 (1) 对任意 $y$ 有 $\\left|\\frac{\\cos xy}{1 + x^2}\\right| \\leq \\frac{1}{1 + x^2}$ ，而 $\\int_0^{+\\infty} \\frac{\\mathrm{d}x}{1 + x^2}$ 收敛，因此由 $M$ 判别法可知一致收敛。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":127,"id":"cfb826fc-0956-4e39-96bd-2e58ccf32798","page_size":[595,841],"block_position":"127-8"},{"bbox":[72,322,341,347],"type":"text","angle":0,"index":9,"merge_prev":false,"text":"(2) 由于 $e^{-x^2y} \\leq e^{-a^2y}$ , 且根据比较原则 $\\int_0^{+\\infty} e^{-a^2y} \\mathrm{d}y$ 收敛 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":127,"id":"95f0b467-e8b3-4485-82e7-ee0720634eb7","page_size":[595,841],"block_position":"127-9"},{"bbox":[52,349,293,362],"type":"text","angle":0,"index":10,"merge_prev":false,"text":"笔记 即使含参反常积分是常数，也不代表其一致收敛。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":127,"id":"d091f03d-e68e-412a-804b-13035abde625","page_size":[595,841],"block_position":"127-10"},{"bbox":[57,380,222,395],"type":"title","angle":0,"index":11,"level":2,"text":"## 14.2.2 Dirichlet 与 Abel 判别法","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":127,"id":"c84187b6-9828-44b6-800a-cbf70ce90197","page_size":[595,841],"block_position":"127-11"},{"bbox":[73,408,231,420],"type":"title","angle":0,"index":12,"level":2,"text":"## 定理 14.6 (Dirichlet 与 Abel 判别法)","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":127,"id":"053ce7f7-1656-4327-b045-ba92ff5948ee","page_size":[595,841],"block_position":"127-12"},{"bbox":[68,425,315,452],"type":"text","angle":0,"index":13,"merge_prev":false,"text":"对于 $\\int_{c}^{+\\infty}f(x,t)g(x,t)\\mathrm{d}x$ ，满足下面的条件则一致收敛 - Dirichlet 判别法: (1) $\\forall N > c, \\int_{c}^{N} f(x, t) \\mathrm{d}x$ 关于 $t$ 一致有界 (2) 固定 $t \\in I$ , $g(x, t)$ 关于 $x$ 单调, 且 $x \\to \\infty$ 时关于 $t$ 一致收敛 (即与 $t$ 的选取无关) 到 0 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":127,"id":"0e988018-56a6-44d9-8914-ebda0b77dced","page_size":[595,841],"block_position":"127-13"},{"bbox":[83,451,524,488],"type":"text","angle":0,"index":14,"merge_prev":false,"lines_deleted":true,"text":"","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":127,"id":"0e988018-56a6-44d9-8914-ebda0b77dced","page_size":[595,841],"block_position":"127-14"},{"bbox":[83,491,529,527],"type":"text","angle":0,"index":15,"merge_prev":false,"text":"- Abel 判别法: (1) $\\int_{c}^{+\\infty} f(x,t) \\mathrm{d}x$ 在 $I$ 上一致收敛 (2) 固定 $t \\in I$ , $g(x,t)$ 关于 $x$ 单调, $g(x,t)$ 关于 $t \\in I$ 一致有界 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":127,"id":"b4da6668-ab07-4e84-9893-3a7441ff51d5","page_size":[595,841],"block_position":"127-15"},{"bbox":[35,536,47,552],"type":"image","angle":0,"index":16,"img_path":"/3f126d83b6ea9697d9f0e348d015e579832e02fcc703d48ac76fdd9deaed2701.jpg","color":{"line":"rgba(89, 92, 220, 1)","fill":"rgba(89, 92, 220, 1)"},"img_ratio":0.2,"page_idx":127,"id":"39498e6e-b603-458f-97f8-25231f34e556","page_size":[595,841],"block_position":"127-16"},{"bbox":[52,540,292,553],"type":"text","angle":0,"index":17,"merge_prev":false,"text":"笔记 注意这里单调都是关于 x 的，一致都是关于 t 的。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":127,"id":"df601fc3-3a6c-44a2-981a-4d0c8c9bc819","page_size":[595,841],"block_position":"127-17"},{"bbox":[35,564,371,592],"type":"text","angle":0,"index":18,"merge_prev":false,"text":"练习14.7Dirichlet判别法(1)证明 $\\int_1^{+\\infty}\\frac{\\sin x}{1 + xe^y}\\mathrm{d}x$ 在 $[0, + \\infty)$ 一致收敛 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":127,"id":"9d338896-74bc-45af-a282-523981fb4192","page_size":[595,841],"block_position":"127-18"},{"bbox":[73,592,308,617],"type":"text","angle":0,"index":19,"merge_prev":false,"text":"(2) 证明 $\\int_{1}^{+\\infty} \\frac{y \\sin xy}{1 + y^2} \\, \\mathrm{d}y$ 在 $(0, +\\infty)$ 上内闭一致收敛 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":127,"id":"033497a9-b9f9-45b4-b9b5-e0a62a26e48e","page_size":[595,841],"block_position":"127-19"},{"bbox":[53,617,539,659],"type":"text","angle":0,"index":20,"merge_prev":false,"text":"证明 (1) $\\left|\\int_{1}^{N}\\sin x\\mathrm{d}x\\right|$ 关于 $y$ 一致有界。 $\\frac{1}{1 + xe^y}$ ，一旦固定 $y$ ，肯定关于 $x$ 单调。由于 $0 < \\frac{1}{1 + xe^y} \\leq \\frac{1}{1 + x} \\to 0$ ，因此一致收敛到0。由Dirichlet判别法可知。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":127,"id":"bb0723b1-f6e9-4868-b15c-3fcaab1b48a2","page_size":[595,841],"block_position":"127-20"},{"bbox":[73,662,267,675],"type":"text","angle":0,"index":22,"merge_prev":false,"text":"(2) 对于 $[a, b] \\subset (0, +\\infty)$ ，对 $\\forall x \\in [a, b]$ ，有 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":127,"id":"7d47bfb2-f68f-4351-9373-22cda4107135","page_size":[595,841],"block_position":"127-21"},{"bbox":[221,677,371,709],"type":"interline_equation","angle":0,"index":23,"text":"$$\n\\left| \\int_ {1} ^ {N} \\sin x y \\mathrm{d} y \\right| = \\left| - \\frac {\\cos x y}{x} \\right| _ {1} ^ {N} \\leq \\frac {2}{a}\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":127,"id":"84ed30da-5821-4b30-88f3-f1971aee6015","page_size":[595,841],"block_position":"127-22"},{"bbox":[54,712,189,739],"type":"text","angle":0,"index":24,"merge_prev":false,"text":"而 $\\left|\\frac{y}{1 + y^2}\\right| \\leq \\frac{1}{2}$ 一致有界，且 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":127,"id":"275f714b-d381-4e86-89fe-54c100fbbf5f","page_size":[595,841],"block_position":"127-23"},{"bbox":[235,736,359,766],"type":"interline_equation","angle":0,"index":25,"text":"$$\n\\left(\\frac {y}{1 + y ^ {2}}\\right) ^ {\\prime} = \\frac {1 - y ^ {2}}{(1 + y ^ {2}) ^ {2}} \\leq 0\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":127,"id":"3cf6e5c1-9618-4716-be5c-b063f1315b2a","page_size":[595,841],"block_position":"127-24"},{"bbox":[289,802,305,812],"type":"page_number","angle":0,"index":1000020,"text":"120 ","color":{"line":"rgba(164, 164, 164, 1)","fill":"rgba(164, 164, 164, 1)"},"page_idx":127,"is_discarded":true,"id":"c3785510-0bbd-4197-badd-d1595652743d","page_size":[595,841],"block_position":"127-25"}],[{"bbox":[425,47,538,58],"type":"header","angle":0,"index":-99998,"text":"14.3 含参反常积分的性质 ","color":{"line":"rgba(164, 164, 164, 1)","fill":"rgba(164, 164, 164, 1)"},"page_idx":128,"is_discarded":true,"id":"5efa83c1-9d33-4348-9242-881b6ff37582","page_size":[595,841],"block_position":"128-0"},{"bbox":[54,68,382,93],"type":"text","angle":0,"index":2,"merge_prev":false,"text":"因此 $\\frac{y}{1 + y^2}$ 单调减， $y \\to +\\infty$ 时， $\\frac{y}{1 + y^2} \\to 0$ ，根据Dirichlet判别法可知 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":128,"id":"b91054ef-feb6-41ad-a893-ef01df04207a","page_size":[595,841],"block_position":"128-1"},{"bbox":[36,96,372,123],"type":"text","angle":0,"index":3,"merge_prev":false,"text":"练习14.8Abel判别法(1)证明 $\\int_0^{+\\infty}e^{-xy}\\frac{\\sin x}{x}\\mathrm{d}x$ 在 $[0, + \\infty)$ 上一致收敛 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":128,"id":"b1f4c967-e855-4b5d-b5c0-ba5e9d7f44c5","page_size":[595,841],"block_position":"128-2"},{"bbox":[73,121,298,148],"type":"text","angle":0,"index":4,"merge_prev":false,"text":"(2) $\\varphi(y) = \\int_{0}^{+\\infty}\\frac{\\sin x^2}{1 + xy}\\mathrm{d}x$ 在 $y\\in [0, + \\infty)$ 一致收敛 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":128,"id":"c5790f02-14de-4aed-ae59-3329e3573f3f","page_size":[595,841],"block_position":"128-3"},{"bbox":[53,149,524,184],"type":"text","angle":0,"index":5,"merge_prev":false,"text":"证明 (1) $\\int_{0}^{+\\infty}\\frac{\\sin x}{x}dx$ 关于 y 一致收敛。 $e^{-xy}$ 关于 x 单调，且 $0 0$ ，则含参积分 $\\int_0^{+\\infty}e^{-px}\\frac{\\sin bx - \\sin ax}{x}\\mathrm{d}x = \\arctan \\frac{b}{p} -\\arctan \\frac{a}{p}$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":128,"id":"9dc41ffc-8f32-42ef-9160-4fccde20d8ad","page_size":[595,841],"block_position":"128-18"},{"bbox":[54,677,264,705],"type":"text","angle":0,"index":20,"merge_prev":false,"text":"证明 首先 $\\frac{\\sin bx - \\sin ax}{x} = \\int_{a}^{b}\\cos xy\\mathrm{d}y$ ，因此 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":128,"id":"ada2b976-b3fd-4ed0-9027-12ef63ef97e2","page_size":[595,841],"block_position":"128-19"},{"bbox":[229,707,365,735],"type":"interline_equation","angle":0,"index":21,"text":"$$\nI = \\int_ {0} ^ {+ \\infty} \\mathrm{d} x \\int_ {a} ^ {b} e ^ {- p x} \\cos x y \\mathrm{d} y\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":128,"id":"5869e3da-84a3-405f-a92e-a9010b74445c","page_size":[595,841],"block_position":"128-20"},{"bbox":[54,736,539,766],"type":"text","angle":0,"index":22,"merge_prev":false,"text":"由于 $|e^{-px}\\cos xy|\\leq e^{-px}$ ，显然 $\\int_0^{+\\infty}e^{-px}\\mathrm{d}x$ 绝对收敛，因此 $\\int_0^{+\\infty}e^{-px}\\cos xy\\mathrm{d}x,y\\in [a,b]$ 一致收敛，交换积 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":128,"id":"94252eac-cf18-49c2-b84d-1bb1d9d5bee5","page_size":[595,841],"block_position":"128-21"},{"bbox":[289,802,304,812],"type":"page_number","angle":0,"index":999999,"text":"121 ","color":{"line":"rgba(164, 164, 164, 1)","fill":"rgba(164, 164, 164, 1)"},"page_idx":128,"is_discarded":true,"id":"40197c55-21da-4935-b12e-cc8714ac306b","page_size":[595,841],"block_position":"128-22"}],[{"bbox":[425,47,539,58],"type":"header","angle":0,"index":-99999,"text":"14.3 含参反常积分的性质 ","color":{"line":"rgba(164, 164, 164, 1)","fill":"rgba(164, 164, 164, 1)"},"page_idx":129,"is_discarded":true,"id":"9a7fa1f8-b6ba-4104-b143-d6234d13914e","page_size":[595,841],"block_position":"129-0"},{"bbox":[54,72,107,84],"type":"text","angle":0,"index":1,"merge_prev":false,"text":"分顺序得到 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":129,"id":"c84f8fde-1532-4f10-b85e-8dcaca3757f6","page_size":[595,841],"block_position":"129-1"},{"bbox":[229,85,365,114],"type":"interline_equation","angle":0,"index":2,"text":"$$\nI = \\int_ {a} ^ {b} \\mathrm{d} y \\int_ {0} ^ {+ \\infty} e ^ {- p x} \\cos x y \\mathrm{d} x\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":129,"id":"f596da63-419a-48d9-963a-46c4f632721a","page_size":[595,841],"block_position":"129-2"},{"bbox":[54,116,131,127],"type":"text","angle":0,"index":3,"merge_prev":false,"text":"根据表格法得到: ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":129,"id":"ef40ff58-a64b-4424-a2db-6bf3f2c6f29c","page_size":[595,841],"block_position":"129-3"},{"bbox":[105,130,487,158],"type":"interline_equation","angle":0,"index":4,"text":"$$\n\\int_ {0} ^ {+ \\infty} e ^ {- p x} \\cos x y \\mathrm{d} x = - \\frac {1}{p} e ^ {- p x} \\cos x y \\bigg | _ {0} ^ {+ \\infty} + \\frac {y}{p ^ {2}} e ^ {- p x} \\sin x y \\bigg | _ {0} ^ {+ \\infty} - \\int_ {0} ^ {+ \\infty} \\frac {y ^ {2}}{p ^ {2}} e ^ {- p x} \\cos x y \\mathrm{d} x\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":129,"id":"a71e5bbb-1871-40c9-85b6-a7cacf295b49","page_size":[595,841],"block_position":"129-4"},{"bbox":[54,161,219,188],"type":"text","angle":0,"index":5,"merge_prev":false,"text":"因此得到 $\\frac{y^2 + p^2}{p^2}\\int_0^{+\\infty}\\cos xy\\mathrm{d}y = \\frac{1}{p}$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":129,"id":"02bbda64-5fae-4e53-b008-7a5ce372b2e7","page_size":[595,841],"block_position":"129-5"},{"bbox":[205,190,387,218],"type":"interline_equation","angle":0,"index":6,"text":"$$\nI = \\int_ {a} ^ {b} \\frac {p}{p ^ {2} + y ^ {2}} \\mathrm{d} y = \\arctan {\\frac {b}{p}} - \\arctan {\\frac {a}{p}}\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":129,"id":"5ab1fea5-7e29-4347-ad3c-06ca1db9141b","page_size":[595,841],"block_position":"129-6"},{"bbox":[73,231,191,243],"type":"title","angle":0,"index":7,"level":2,"text":"## 定理 14.11 (Dirichlet 积分)","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":129,"id":"a74edb05-692d-4aa5-bceb-2f7bd816d24f","page_size":[595,841],"block_position":"129-7"},{"bbox":[68,248,348,275],"type":"text","angle":0,"index":8,"merge_prev":false,"text":"积分 $\\int_0^{+\\infty}\\frac{\\sin x}{x}\\mathrm{d}x = \\frac{\\pi}{2}$ ，进一步地， $\\int_0^{+\\infty}\\frac{\\sin ax}{x}\\mathrm{d}x = \\frac{\\pi}{2}\\mathrm{sgn}(a)$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":129,"id":"93e3b1e4-0dc4-4163-a6c5-bfa40ac12e27","page_size":[595,841],"block_position":"129-8"},{"bbox":[73,295,190,307],"type":"title","angle":0,"index":9,"level":2,"text":"## 定理 14.12 (正态积分公式)","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":129,"id":"c55914bc-257e-4b86-8683-2aae75fd2793","page_size":[595,841],"block_position":"129-9"},{"bbox":[69,312,158,339],"type":"interline_equation","angle":0,"index":10,"text":"$$\n\\int_ {0} ^ {+ \\infty} e ^ {- x ^ {2}} \\mathrm{d} x = \\frac {\\sqrt {\\pi}}{2}\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":129,"id":"00ba588f-f690-4b80-a68d-9eddc8659b3b","page_size":[595,841],"block_position":"129-10"},{"bbox":[35,364,264,394],"type":"text","angle":0,"index":11,"merge_prev":false,"text":"练习14.9相关练习(1) $\\int_0^{+\\infty}\\frac{e^{-a^2x^2} - e^{-b^2x^2}}{x^e}\\mathrm{d}x$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":129,"id":"714ef506-9ad8-4ac4-80fd-5d268423f917","page_size":[595,841],"block_position":"129-11"},{"bbox":[57,409,276,423],"type":"title","angle":0,"index":12,"level":2,"text":"## 14.3.3 含参反常积分与函数项级数的关系","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":129,"id":"d0f4ba3c-26a4-4358-a96d-cf8691213c4d","page_size":[595,841],"block_position":"129-12"},{"bbox":[72,437,291,449],"type":"title","angle":0,"index":13,"level":2,"text":"## 定理 14.13 (一致收敛级数表示法 (ZJU 考研考过))","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":129,"id":"48704c69-d3a0-4447-81e6-102d17b8110d","page_size":[595,841],"block_position":"129-13"},{"bbox":[66,451,532,513],"type":"text","angle":0,"index":14,"merge_prev":false,"text":"$\\int_{c}^{+\\infty}f(x,y)\\mathrm{d}y$ 在 $I$ 上一致收敛的充要条件为对任一趋于无穷的递增数列 $A_{n}$ (其中 $A_{1} = c$ ), $\\sum_{n=1}^{\\infty}\\int_{A_n}^{A_{n+1}}f(x,y)\\mathrm{d}y$ 作为函数项级数 $\\sum_{n=1}^{\\infty}u_n(x)$ 在 $I$ 上一致收敛 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":129,"id":"a7339909-280c-4b83-a1b6-4d53c36ceb33","page_size":[595,841],"block_position":"129-14"},{"bbox":[52,524,539,587],"type":"text","angle":0,"index":15,"merge_prev":false,"text":"证明 (1) 左推右：由于一致收敛，因此 $\\forall\\epsilon,\\exists M>c$ 使得 $A^{\\prime\\prime}>A^{\\prime}>M$ 时， $\\forall x$ 有 $\\left|\\int_{A^{\\prime}}^{A^{\\prime\\prime}}f(x,y)\\mathrm{d}y\\right|<\\epsilon$ ，由于 $A_{n}$ 趋于无穷， $\\exists N,\\forall m>n>N$ 有 $A_{m}>A_{n}>M$ ，对 $\\forall x\\in I$ 有 $|u_{n}(x)+\\cdots+u_{m}(x)|=|\\int_{A_{n}}^{A_{m+1}}f(x,y)\\mathrm{d}y|<\\epsilon$ ，即右侧级数一致收敛 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":129,"id":"a1c127f5-597c-41d9-9dbd-8d4f014508eb","page_size":[595,841],"block_position":"129-15"},{"bbox":[73,590,426,602],"type":"text","angle":0,"index":16,"merge_prev":false,"text":"(2) 右推左：反设含参积分不一致收敛， $\\exists \\epsilon_0, \\forall M > c, \\exists A'' > A' > M, x' \\in I$ 使得 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":129,"id":"26ea4fb3-ce78-42c0-8155-3f1afe53c52a","page_size":[595,841],"block_position":"129-16"},{"bbox":[248,605,346,637],"type":"interline_equation","angle":0,"index":17,"text":"$$\n\\left| \\int_ {A ^ {\\prime}} ^ {A ^ {\\prime \\prime}} f (x ^ {\\prime}, y) \\mathrm{d} y \\right| \\geq \\epsilon_ {0}\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":129,"id":"9f4516d9-468a-4602-922c-8af6c739ece5","page_size":[595,841],"block_position":"129-17"},{"bbox":[53,640,376,653],"type":"text","angle":0,"index":18,"merge_prev":false,"text":"先取 $M_{1} = \\max \\{1,c\\}$ ，则 $\\exists A_2,A_1$ 满足 $A_{2} > A_{1} > M_{1}$ ，以及 $x_{1}\\in I$ 使得 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":129,"id":"1c309954-9654-4e20-8792-403cb8d3e56d","page_size":[595,841],"block_position":"129-18"},{"bbox":[248,656,346,687],"type":"interline_equation","angle":0,"index":20,"text":"$$\n\\left| \\int_ {A _ {1}} ^ {A _ {2}} f (x _ {1}, y) \\mathrm{d} y \\right| \\geq \\epsilon_ {0}\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":129,"id":"88b7a156-8044-4900-9839-fe4f16a6f2e1","page_size":[595,841],"block_position":"129-19"},{"bbox":[54,690,348,703],"type":"text","angle":0,"index":21,"merge_prev":false,"text":"同理取 $M_{n} = \\max \\{n,A_{2n - 2}\\}$ ，有 $A_{2n} > A_{2n - 1} > M_n,x_n\\in I$ 使得 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":129,"id":"be36372e-4597-4a58-9ed5-4110c3b5b803","page_size":[595,841],"block_position":"129-20"},{"bbox":[243,706,351,738],"type":"interline_equation","angle":0,"index":22,"text":"$$\n\\left| \\int_ {A _ {2 n - 1}} ^ {A _ {2 n}} f (x _ {n}, y) \\mathrm{d} y \\right| \\geq \\epsilon_ {0}\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":129,"id":"119a0a97-dae0-41ba-9501-29e76f17e213","page_size":[595,841],"block_position":"129-21"},{"bbox":[54,741,539,771],"type":"text","angle":0,"index":23,"merge_prev":false,"text":"此时 $\\exists \\epsilon_0, \\forall N, \\exists n > N, \\exists x_n \\in I$ 使得 $|u_{2n}(x_n)| = |\\int_{A_{2n}}^{A_{2n + 1}} f(x_n, y) \\mathrm{d}y| \\geq \\epsilon_0$ ，这与函数项级数一致收敛条件矛盾 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":129,"id":"b1c6783b-71d1-4e69-a637-2989f6fba0fc","page_size":[595,841],"block_position":"129-22"},{"bbox":[289,802,305,812],"type":"page_number","angle":0,"index":1000018,"text":"122 ","color":{"line":"rgba(164, 164, 164, 1)","fill":"rgba(164, 164, 164, 1)"},"page_idx":129,"is_discarded":true,"id":"3e963a00-7f87-4571-9a46-83bd72763c0f","page_size":[595,841],"block_position":"129-23"}],[{"bbox":[426,47,539,58],"type":"header","angle":0,"index":-99999,"text":"14.3 含参反常积分的性质 ","color":{"line":"rgba(164, 164, 164, 1)","fill":"rgba(164, 164, 164, 1)"},"page_idx":130,"is_discarded":true,"id":"7c62709f-60dd-4c69-ac90-76b759f0c3b1","page_size":[595,841],"block_position":"130-0"},{"bbox":[54,72,217,84],"type":"text","angle":0,"index":1,"merge_prev":false,"text":"(因为一致收敛要求所有 x 都趋于 0)。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":130,"id":"b2d128b5-7efc-4cca-8683-a7aae4872ca5","page_size":[595,841],"block_position":"130-1"},{"bbox":[289,802,305,812],"type":"page_number","angle":0,"index":1000001,"text":"123 ","color":{"line":"rgba(164, 164, 164, 1)","fill":"rgba(164, 164, 164, 1)"},"page_idx":130,"is_discarded":true,"id":"596320fc-cc62-4ad2-ac12-1360ba348c06","page_size":[595,841],"block_position":"130-2"}],[{"bbox":[231,74,373,94],"type":"title","angle":0,"index":0,"level":2,"text":"## 第 15 章 多重积分","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":131,"id":"7ed10db3-a91f-442e-9000-8ab998e3a8d8","page_size":[595,841],"block_position":"131-0"},{"bbox":[57,122,151,138],"type":"title","angle":0,"index":1,"level":2,"text":"## 15.1 二重积分","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":131,"id":"2e69e6b2-0beb-4553-b456-0e544d1c6e55","page_size":[595,841],"block_position":"131-1"},{"bbox":[56,155,217,169],"type":"title","angle":0,"index":2,"level":2,"text":"## 15.1.1 二重积分的概念与计算","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":131,"id":"ad90294c-3af8-4d20-935c-f1a3a30676d9","page_size":[595,841],"block_position":"131-2"},{"bbox":[72,182,166,195],"type":"title","angle":0,"index":3,"level":2,"text":"## 定义 15.1 (二重积分)","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":131,"id":"ff8baacc-a3a7-4eeb-b08c-47ca6ab42e12","page_size":[595,841],"block_position":"131-3"},{"bbox":[68,199,379,213],"type":"text","angle":0,"index":4,"merge_prev":false,"text":"$D \\subset R^{2}$ 是有界集，对 D 做分割 $T = \\{\\sigma_{1}, \\cdots, \\sigma_{n}\\}$ ，则二重积分定义为 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":131,"id":"360d3e72-4253-403c-a3b4-7905249f9ff3","page_size":[595,841],"block_position":"131-4"},{"bbox":[203,216,389,247],"type":"interline_equation","angle":0,"index":5,"text":"$$\n\\iint_ {D} f (x, y) \\mathrm{d} x \\mathrm{d} y = \\lim _ {| | T | | \\rightarrow 0} \\sum_ {i = 1} ^ {n} f (\\xi_ {i}, \\eta_ {i}) s (\\sigma_ {i})\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":131,"id":"b3e60ff1-c46e-42a1-ae41-1591ee4656c1","page_size":[595,841],"block_position":"131-5"},{"bbox":[73,268,216,281],"type":"title","angle":0,"index":6,"level":2,"text":"## 定理 15.1 (二重积分与累次积分)","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":131,"id":"2c13a960-6cfd-4ed6-a503-cbf3a5a54c9b","page_size":[595,841],"block_position":"131-6"},{"bbox":[68,284,525,312],"type":"text","angle":0,"index":7,"merge_prev":false,"text":"对于 $D = [a,b]\\times [c,d]$ 或者 $D = [a,b]\\times [f_1(x),f_2(x)]$ ，方形边界二重积分可随意转换为累次积分，函数边界先积分函数边界 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":131,"id":"205b7e0f-aa8a-4dc7-8285-e8d9dd017562","page_size":[595,841],"block_position":"131-7"},{"bbox":[113,315,480,344],"type":"interline_equation","angle":0,"index":8,"text":"$$\n\\iint_ {D} f (x, y) \\mathrm{d} x \\mathrm{d} y = \\int_ {a} ^ {b} \\mathrm{d} x \\int_ {c} ^ {d} f (x, y) \\mathrm{d} y, \\quad \\iint_ {D} f (x, y) \\mathrm{d} x \\mathrm{d} y = \\int_ {a} ^ {b} \\mathrm{d} x \\int_ {f _ {1} (x)} ^ {f _ {2} (x)} f (x, y) \\mathrm{d} y\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":131,"id":"77ef6e36-e3ae-4cae-96ff-486717e72cd4","page_size":[595,841],"block_position":"131-8"},{"bbox":[45,360,537,389],"type":"text","angle":0,"index":9,"merge_prev":false,"text":"笔记 二重积分的直接计算要特别关注其区域的对称性。而累次积分的顺序一般看函数，那边容易积就先积 (也得靠经验觉得哪个好积)。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":131,"id":"66565eec-5a1b-4fcf-92e3-8ff87b8ff4d2","page_size":[595,841],"block_position":"131-9"},{"bbox":[35,401,348,428],"type":"text","angle":0,"index":10,"merge_prev":false,"text":"练习15.1对称性计算(1)计算 $\\iint_{D}x^{2}y\\mathrm{d}x\\mathrm{d}y$ ，其中 $D:|x| + |y|\\leq 1$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":131,"id":"2576aa5c-97a7-4238-916b-4c95ad23e097","page_size":[595,841],"block_position":"131-10"},{"bbox":[52,426,437,451],"type":"text","angle":0,"index":11,"merge_prev":false,"text":"解(1)由于 D 关于 x 轴对称，且 $f(x,y)=x^{2}y$ 关于 y 为奇函数，因此 $\\iint_{D}x^{2}y\\,dx\\,dy=0$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":131,"id":"be91ac20-d36e-42e9-b15b-cd6595cc0bf6","page_size":[595,841],"block_position":"131-11"},{"bbox":[35,456,467,483],"type":"text","angle":0,"index":12,"merge_prev":false,"text":"练习 15.2 转换为累次积分 (1) 计算 $I = \\iint_{D} e^{-y^{2}} dx dy$ ，其中 D 是 y = x 与 $y = x^{\\frac{1}{3}}$ 围成的区域 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":131,"id":"2cf2403b-7fab-44f0-b978-9406d5a1b35c","page_size":[595,841],"block_position":"131-12"},{"bbox":[72,481,371,505],"type":"text","angle":0,"index":13,"merge_prev":false,"text":"(2) 计算 $I = \\iint_{D} \\frac{\\sin x}{x} \\, \\mathrm{d}x \\, \\mathrm{d}y$ ，其中 $D$ 是 $y = x$ 与 $y = x^2$ 围成的区域 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":131,"id":"d99bb605-b098-4451-bf74-75c90cbec566","page_size":[595,841],"block_position":"131-13"},{"bbox":[72,504,530,529],"type":"text","angle":0,"index":14,"merge_prev":false,"text":"(3) 圆周边界：计算 $I = \\iint_{D} \\frac{\\mathrm{d}x \\, \\mathrm{d}y}{\\sqrt{2a - x}}$ ，其中 $D$ 为 $x = 0, y = 0$ 与 $(x - a)^2 + (y - a)^2 = a^2$ 所围成的区域 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":131,"id":"efa4f34e-dac2-4aed-b035-ae3e225b33b1","page_size":[595,841],"block_position":"131-14"},{"bbox":[53,528,318,552],"type":"text","angle":0,"index":15,"merge_prev":false,"text":"解(1)注意要先积 $x$ ，因为 $e^{-y^2}$ 先积 $x$ 比较方便，答案为 $\\frac{1}{e}$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":131,"id":"2d8253e4-134c-4148-9d34-9c7f3d8a05ca","page_size":[595,841],"block_position":"131-15"},{"bbox":[72,550,334,568],"type":"text","angle":0,"index":16,"merge_prev":false,"text":"(2) 注意要先积 $y$ ，因为显然 $\\frac{\\sin x}{x}$ 不好积，答案为 $1 - \\sin 1$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":131,"id":"b2fe49e0-3d7a-42d4-9aec-19d55a2b9d5b","page_size":[595,841],"block_position":"131-16"},{"bbox":[73,570,459,595],"type":"text","angle":0,"index":17,"merge_prev":false,"text":"(3) 这里 $\\int \\frac{1}{\\sqrt{2a - x}} \\mathrm{d}x$ 看成 $\\int (2a - x)^{-\\frac{1}{2}} \\mathrm{d}x$ 就是个幂函数，不要被吓到。先积 $y$ ，得到 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":131,"id":"d676ca58-719f-4aaf-bd3b-f339dbc69dc1","page_size":[595,841],"block_position":"131-17"},{"bbox":[127,599,463,628],"type":"interline_equation","angle":0,"index":18,"text":"$$\nI = \\int_ {0} ^ {a} \\mathrm{d} x \\int_ {0} ^ {a - \\sqrt {2 a x - x ^ {2}}} \\frac {1}{\\sqrt {2 a - x}} \\mathrm{d} y = \\int_ {0} ^ {a} \\frac {a - \\sqrt {2 a x - x ^ {2}}}{\\sqrt {2 a - x}} \\mathrm{d} x = (2 \\sqrt {2} - \\frac {8}{3}) a ^ {\\frac {3}{2}}\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":131,"id":"9da6a593-ca0a-4f28-a04a-a6cb6f6f23b4","page_size":[595,841],"block_position":"131-18"},{"bbox":[35,636,443,662],"type":"text","angle":0,"index":19,"merge_prev":false,"text":"练习 15.3 几个进阶题 (1) 计算 $I = \\iint_{D} |xy - \\frac{1}{4}| \\, dx \\, dy$ ，其中 $D : [0, 1] \\times [0, 1]$ (补区域技巧) ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":131,"id":"ebdffe1d-902a-4531-b7aa-9820c94492fe","page_size":[595,841],"block_position":"131-19"},{"bbox":[73,662,398,687],"type":"text","angle":0,"index":20,"merge_prev":false,"text":"(2) 计算 $I = \\iint_{D} \\sqrt{|y - x^2|} \\, \\mathrm{d}x \\, \\mathrm{d}y$ ，其中 $D: -1 \\leq x \\leq 1, 0 \\leq y \\leq 2$ (对称性) ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":131,"id":"f44ea61a-cc85-485a-b920-05588b16b42b","page_size":[595,841],"block_position":"131-20"},{"bbox":[52,687,539,756],"type":"text","angle":0,"index":21,"merge_prev":false,"text":"解(1)要画出 $xy = \\frac{1}{4}$ 线，分成上下两个部分，记上面的为 $D_{1}$ ，下面的是 $D_{2}$ ，但是 $D_{2}$ 过于不规则，可以为 $D_{2}$ 补上 $D_{1}$ 的部分，这样算起来就方便不少。即 $I = \\iint_{D_1}(xy - \\frac{1}{4})\\mathrm{d}x\\mathrm{d}y - \\iint_{D_2}(xy - \\frac{1}{4})\\mathrm{d}x\\mathrm{d}y = 2\\iint_{D_1}(xy - \\frac{1}{4})\\mathrm{d}x\\mathrm{d}y - \\iint_{D}(xy - \\frac{1}{4})\\mathrm{d}x\\mathrm{d}y$ ，最终答案为 $\\frac{3}{32} +\\frac{1}{8}\\ln 2$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":131,"id":"8dd91bb5-f3a3-4ae6-a5f8-85104921a222","page_size":[595,841],"block_position":"131-21"}],[{"bbox":[475,47,538,58],"type":"header","angle":0,"index":-99998,"text":"15.1 二重积分 ","color":{"line":"rgba(164, 164, 164, 1)","fill":"rgba(164, 164, 164, 1)"},"page_idx":132,"is_discarded":true,"id":"0ec0979e-43e7-4d5b-a25a-99cfa569c8e1","page_size":[595,841],"block_position":"132-0"},{"bbox":[53,71,539,100],"type":"text","angle":0,"index":2,"merge_prev":false,"text":"(2) 画出 $y = x^2$ ，根据关于 $x$ 为偶，只需要考虑 $x > 0$ ，关于 $y = x^2$ 分为两个区域，上面记 $D_1$ ，下面记 $D_2$ ， $D_1, D_2$ 都算规则，直接计算即可。要注意 $(x^2)^{\\frac{2}{3}} = |x|^3!$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":132,"id":"2167d4b9-727e-45fb-a411-02ec6cb3c6c9","page_size":[595,841],"block_position":"132-1"},{"bbox":[56,119,181,132],"type":"title","angle":0,"index":3,"level":2,"text":"## 15.1.2 二重积分可积性","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":132,"id":"f43e7a8e-f3e8-4c08-889d-5055baeba91e","page_size":[595,841],"block_position":"132-2"},{"bbox":[35,142,539,169],"type":"text","angle":0,"index":4,"merge_prev":false,"text":"练习15.4两道经典题 $(1)D = [0,1]\\times [0,1]$ ，对 $x\\in \\mathbb{R}$ ， $q_{x}$ 表示 $x$ 化成既约分数后的分母，证明下面 $f(x,y)$ 在 $D$ 上二重积分存在但两个累次积分不存在 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":132,"id":"e048c3f7-c115-4d32-95a3-0173c4586c07","page_size":[595,841],"block_position":"132-3"},{"bbox":[211,171,380,214],"type":"interline_equation","angle":0,"index":5,"text":"$$\nf (x, y) = \\left\\{ \\begin{array}{l l} \\frac {1}{q _ {x}} + \\frac {1}{q _ {y}}, & (x, y) \\text {为有理点} \\\\ 0, & (x, y) \\text {非有理点} \\end{array} \\right.\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":132,"id":"56c9af92-c973-4dc8-80e4-e59df13a16d9","page_size":[595,841],"block_position":"132-4"},{"bbox":[73,218,439,232],"type":"text","angle":0,"index":6,"merge_prev":false,"text":"(2) $D=[0,1]\\times[0,1]$ ，证明下面 $f(x,y)$ 在D上二重积分不存在但两个累次积分存在 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":132,"id":"6324a25e-d02b-4ebe-8dae-372de988ab0e","page_size":[595,841],"block_position":"132-5"},{"bbox":[206,235,384,271],"type":"interline_equation","angle":0,"index":7,"text":"$$\nf (x, y) = \\left\\{ \\begin{array}{l l} 1, & (x, y) \\text {为有理点且} q _ {x} = q _ {y} \\\\ 0, & \\text {其他情形} \\end{array} \\right.\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":132,"id":"cd4c2896-68f0-493b-9910-7f384b04256c","page_size":[595,841],"block_position":"132-6"},{"bbox":[52,275,540,349],"type":"text","angle":0,"index":8,"merge_prev":false,"text":"证明 (1) $\\forall \\epsilon$ ，仅有有限个点满足 $f(x, y) > \\epsilon$ ，因此存在分割 $T$ 使得 $S(T) - s(T) < \\epsilon$ ，二重积分存在且等于0。下证先 $y$ 后 $x$ 不存在： $y$ 取无理时， $f(x, y) \\equiv 0$ ， $\\int_{0}^{1} f(x, y) \\mathrm{d}x = 0$ 积分存在。若 $y$ 取有理，则在 $x$ 无理处 $f(x, y) = 0$ ， $x$ 有理处 $f(x, y) = \\frac{1}{q_x} + \\frac{1}{q_y}$ ，因此 $f(x, y)$ 在任意区间上振幅总大于 $\\frac{1}{q_y}$ ，因此积分不存在。同理可证先 $x$ 后 $y$ 不存在 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":132,"id":"82164626-4a94-4262-a6a9-6916b9c0a830","page_size":[595,841],"block_position":"132-7"},{"bbox":[52,353,539,381],"type":"text","angle":0,"index":9,"merge_prev":false,"text":"(2) $D$ 中任意部分均有振幅为1，因此二重积分不存在。若固定 $y$ ，若 $y$ 无理，则 $f(x,y) = 0$ ，显然关于 $x$ 可积，若 $y$ 无理，则 $f(x,y) > \\epsilon$ 对于 $x$ 而言只有有限处，因此也关于 $x$ 可积，所以先 $y$ 后 $x$ 可积。另一侧同理。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":132,"id":"20b048c8-1a2c-48d7-a357-b4be151ffc30","page_size":[595,841],"block_position":"132-8"},{"bbox":[57,399,181,413],"type":"title","angle":0,"index":10,"level":2,"text":"## 15.1.3 二重积分换元法","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":132,"id":"d0a0c454-dc26-4de3-a8f8-956be4f9b69d","page_size":[595,841],"block_position":"132-9"},{"bbox":[72,426,196,439],"type":"title","angle":0,"index":11,"level":2,"text":"## 定理 15.2 (二重积分换元法)","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":132,"id":"4ef45d64-8a1e-4795-b459-c51c077e8446","page_size":[595,841],"block_position":"132-10"},{"bbox":[67,443,524,471],"type":"text","angle":0,"index":12,"merge_prev":false,"text":"原二重积分坐标轴 $x, y$ ，定义于 $R$ 上，令 $u = u(x, y), v = v(x, y)$ 反解出 $x = x(u, v), y = y(u, v)$ ，变换后的区域为 $(u, v) \\in \\Omega$ ，则 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":132,"id":"fa6e553c-cb42-44dc-b2c8-aae7a4c03cd0","page_size":[595,841],"block_position":"132-11"},{"bbox":[176,474,418,502],"type":"interline_equation","angle":0,"index":13,"text":"$$\n\\iint_ {R} f (x, y) \\mathrm{d} x \\mathrm{d} y = \\iint_ {\\Omega} f (x (u, v), y (u, v)) \\left| \\frac {\\partial (x , y)}{\\partial (u , v)} \\right| \\mathrm{d} u \\mathrm{d} v\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":132,"id":"7e795b12-29f3-49f5-926e-2635923f6a7a","page_size":[595,841],"block_position":"132-12"},{"bbox":[69,503,168,529],"type":"text","angle":0,"index":14,"merge_prev":false,"text":"定义 $J(u,v)=\\frac{\\partial(x,y)}{\\partial(u,v)}$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":132,"id":"ce596d21-c903-4ead-983a-78a651327479","page_size":[595,841],"block_position":"132-13"},{"bbox":[49,541,540,597],"type":"text","angle":0,"index":15,"merge_prev":false,"text":"笔记一、注意Jacobi行列式外面的是绝对值！二、可以先计算 $I = \\frac{\\partial(u,v)}{\\partial(x,y)}$ ，计算 $|J|$ 时只需要取倒数加绝对值，即 $|J| = |\\frac{1}{I}|$ 。三、计算变换后的范围可以尝试将原本的各个边界转化到新变量，即代入原边界，计算新边界各个变量需要满足的关系。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":132,"id":"76a454fd-dc80-45d3-bf6e-7b911b1e0201","page_size":[595,841],"block_position":"132-14"},{"bbox":[35,610,539,635],"type":"text","angle":0,"index":16,"merge_prev":false,"text":"练习15.5一般换元计算题(1)计算 $I = \\iint_{D}e^{\\frac{x - y}{x + y}}dxdy$ ，其中 $D$ 是由 $x = 0,y = 0$ 与 $x + y = a(a > 0)$ 围成的区域 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":132,"id":"03bc46e0-e427-40bb-a8e9-e7886dca1c25","page_size":[595,841],"block_position":"132-15"},{"bbox":[72,634,464,661],"type":"text","angle":0,"index":17,"merge_prev":false,"text":"(2) 计算 $I = \\iint_{D} \\frac{(x + y) \\ln(1 + \\frac{y}{x})}{\\sqrt{1 - x - y}} \\, \\mathrm{d}x \\, \\mathrm{d}y$ ，其中 $D$ 是由 $x = 0, y = 0, x + y = 1$ 围成的区域 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":132,"id":"c59337a6-a30d-46d2-99c4-2deb7e2a45c5","page_size":[595,841],"block_position":"132-16"},{"bbox":[73,661,465,684],"type":"text","angle":0,"index":18,"merge_prev":false,"text":"(3) 计算 $I = \\iint_{D} \\frac{3x}{y^2 + xy^3} \\, \\mathrm{d}x \\, \\mathrm{d}y$ ，其中 $D$ 是由 $xy = 1, xy = 3, y^2 = x, y^2 = 3x$ 围成的区域 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":132,"id":"efd37fb3-2cb0-4b87-b39d-2dc1e456cbe6","page_size":[595,841],"block_position":"132-17"},{"bbox":[53,684,538,716],"type":"text","angle":0,"index":19,"merge_prev":false,"text":"证明 (1) 设 $x + y = u, x - y = v$ ，即 $x = \\frac{u + v}{2}, y = \\frac{u - v}{2}$ ，考虑边界转化， $x = 0, y = 0$ 等价于 $u = -v, u = v$ ，而 $y = -x + a$ 等价于： ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":132,"id":"730d68d3-97d0-412f-95ab-edbce2d14295","page_size":[595,841],"block_position":"132-18"},{"bbox":[257,720,337,730],"type":"interline_equation","angle":0,"index":20,"text":"$$\nu = a, v = 2 x + a\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":132,"id":"c17924b6-81d2-4f1f-9776-038416c6e945","page_size":[595,841],"block_position":"132-19"},{"bbox":[54,741,159,754],"type":"text","angle":0,"index":21,"merge_prev":false,"text":"此时 u = a，而 v 任意。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":132,"id":"640059b9-92cf-4e6c-a51f-ca89a9aa9407","page_size":[595,841],"block_position":"132-20"},{"bbox":[289,803,305,812],"type":"page_number","angle":0,"index":999999,"text":"125 ","color":{"line":"rgba(164, 164, 164, 1)","fill":"rgba(164, 164, 164, 1)"},"page_idx":132,"is_discarded":true,"id":"efcb6e42-0daf-4312-ac00-caa4c6e2f221","page_size":[595,841],"block_position":"132-21"}],[{"bbox":[476,47,538,58],"type":"header","angle":0,"index":-99999,"text":"15.1 二重积分 ","color":{"line":"rgba(164, 164, 164, 1)","fill":"rgba(164, 164, 164, 1)"},"page_idx":133,"is_discarded":true,"id":"8439055f-97d4-4f7b-846d-5c3df080999d","page_size":[595,841],"block_position":"133-0"},{"bbox":[54,69,539,136],"type":"text","angle":0,"index":1,"merge_prev":false,"text":"因此区域变为 $u + v = 0, u - v = 0, u = a$ 围成的区域，即 $0 \\leq u \\leq a, -u \\leq v \\leq u, J = \\left| \\begin{array}{cc} \\frac{1}{2} & \\frac{1}{2} \\\\ \\frac{1}{2} & -\\frac{1}{2} \\end{array} \\right| = -\\frac{1}{2}$ ，得到最终结果为 $\\frac{a^2}{4}(e - e^{-1})$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":133,"id":"d269e9a6-5940-48f3-b68f-511d3e1b105e","page_size":[595,841],"block_position":"133-1"},{"bbox":[72,135,530,158],"type":"text","angle":0,"index":2,"merge_prev":false,"text":"(2) 设 $x + y = u, x = v$ ，即 $x = v, y = u - v$ 即可，区域变为 $v = 0, u = v, u = 1$ 围成的区域，答案为 $\\frac{16}{15}$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":133,"id":"7d16761e-29e2-4e08-bf2a-1a1a9d74b745","page_size":[595,841],"block_position":"133-2"},{"bbox":[73,156,170,181],"type":"text","angle":0,"index":3,"merge_prev":false,"text":"(3) 做 $xy = u, \\frac{y^2}{x} = v$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":133,"id":"4e3cec76-0cd7-4a26-b1d8-46edadfee6d5","page_size":[595,841],"block_position":"133-3"},{"bbox":[57,196,157,211],"type":"title","angle":0,"index":4,"level":2,"text":"## 15.1.4 极坐标换元","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":133,"id":"e81cb5fb-c4a1-4e2b-a502-b5d520b6f22f","page_size":[595,841],"block_position":"133-4"},{"bbox":[73,223,216,236],"type":"title","angle":0,"index":5,"level":2,"text":"## 定理 15.3 (二重积分极坐标换元)","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":133,"id":"2f11bf67-ccc9-42f8-958c-be945b73521c","page_size":[595,841],"block_position":"133-5"},{"bbox":[67,242,531,278],"type":"text","angle":0,"index":6,"merge_prev":false,"text":"对于二重积分 $\\iint_{S} f(x, y) \\mathrm{d}x \\mathrm{~d}y$ ，做极坐标换元 $x = r \\cos \\theta, y = r \\sin \\theta$ ，此时 $J(r, \\theta) = r$ ；极坐标换元 $x = ar \\cos \\theta, y = br \\sin \\theta$ 时 $J(r, \\theta) = abr$ 。积分时先 $\\mathrm{d}r$ 后 $\\mathrm{d}\\theta$ ，具体如下图所示 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":133,"id":"e9114a33-c0f5-4011-a552-87cd2cbaa769","page_size":[595,841],"block_position":"133-6"},{"bbox":[104,300,489,457],"type":"image","angle":0,"index":7,"img_path":"/5b7b42b6a99544f36af749cad867071bd860f5a296b8bf3c0ba808703c397c2b.jpg","color":{"line":"rgba(89, 92, 220, 1)","fill":"rgba(89, 92, 220, 1)"},"img_ratio":0.6470588235294118,"page_idx":133,"id":"812e4e27-c374-4800-98ac-e16610084ab2","page_size":[595,841],"block_position":"133-7"},{"bbox":[231,460,362,472],"type":"image_caption","angle":0,"index":8,"text":"图 15.1: 二重积分极坐标换元","id":"f0bf1db0-0786-488d-bcc5-01815ff3bf84","color":{"line":"rgba(13, 83, 222 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":133,"page_size":[595,841],"block_position":"133-8"},{"bbox":[35,501,432,527],"type":"text","angle":0,"index":9,"merge_prev":false,"text":"练习15.6规则极坐标与广义极坐标(1)计算 $I = \\iint_{D}\\frac{\\mathrm{d}x\\mathrm{d}y}{\\sqrt{1 - x^2 - y^2}}$ ， $D$ 为 $x^{2} + y^{2}\\leq 1$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":133,"id":"bb6e351f-973c-4ef7-a497-f6d8b1f68ef6","page_size":[595,841],"block_position":"133-9"},{"bbox":[71,525,286,552],"type":"text","angle":0,"index":10,"merge_prev":false,"text":"(2)广义极坐标：计算 $\\frac{x^2}{a^2} +\\frac{y^2}{b^2} +\\frac{z^2}{c^2}\\leq 1$ 的体积 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":133,"id":"fd69034e-ec34-4a56-946a-986e25f12a6a","page_size":[595,841],"block_position":"133-10"},{"bbox":[72,560,380,587],"type":"text","angle":0,"index":11,"merge_prev":false,"text":"(2)是8倍第一卦限积分，该区域顶面为 $z = c\\sqrt{1 - \\frac{x^2}{a^2} - \\frac{y^2}{b^2}}$ ，底面为 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":133,"id":"dd5d0bfb-8254-41ad-bb9a-6fcf0c8852f8","page_size":[595,841],"block_position":"133-11"},{"bbox":[196,589,396,620],"type":"interline_equation","angle":0,"index":12,"text":"$$\nD = \\left\\{(x, y): 0 \\leq y \\leq b \\sqrt {1 - \\frac {x ^ {2}}{a ^ {2}}}, 0 \\leq x \\leq a \\right\\}\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":133,"id":"f29638a5-f1ae-4a52-94d3-bb83b4d10863","page_size":[595,841],"block_position":"133-12"},{"bbox":[54,624,417,652],"type":"text","angle":0,"index":13,"merge_prev":false,"text":"因此 $V = 8\\iint_{D}c\\sqrt{1 - \\frac{x^2}{a^2} - \\frac{y^2}{b^2}}\\mathrm{d}x\\mathrm{d}y$ ，利用广义极坐标变换得到 $z = c\\sqrt{1 - r^2}$ ，故 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":133,"id":"daa3e983-6e29-4d50-94d3-00201c7ec98c","page_size":[595,841],"block_position":"133-13"},{"bbox":[204,655,390,682],"type":"interline_equation","angle":0,"index":14,"text":"$$\nV = 8 \\int_ {0} ^ {\\frac {\\pi}{2}} \\mathrm{d} \\theta \\int_ {0} ^ {1} c \\sqrt {1 - r ^ {2}} a b r \\mathrm{d} r = \\frac {4 \\pi}{3} a b c\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":133,"id":"b1b59058-8577-4d84-a7d5-f50946b84e01","page_size":[595,841],"block_position":"133-14"},{"bbox":[35,690,534,738],"type":"text","angle":0,"index":15,"merge_prev":false,"text":"练习15.7几道非规则边界(1) $f$ 为连续函数，将 $I = \\iint_{D}f\\left(\\frac{y}{x}\\right)\\mathrm{d}x\\mathrm{d}y$ 转化为定积分，其中 $D:\\frac{1}{4}\\leq x^2 +y^2\\leq x$ (2)将 $I = \\iint_{D}f(x,y)\\mathrm{d}x\\mathrm{d}y$ 化为极坐标形式，其中 $D:[0,1]\\times [0,1]$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":133,"id":"97a0f3f6-f7c4-450e-a69f-fec72a023d99","page_size":[595,841],"block_position":"133-15"},{"bbox":[52,739,539,763],"type":"text","angle":0,"index":16,"merge_prev":false,"text":"解 (1) 考虑 $x = r\\cos \\theta, y = r\\sin \\theta$ ，则 $\\frac{1}{4} \\leq r^2 \\leq r\\cos \\theta$ ，即 $\\frac{1}{2} \\leq r \\leq \\cos \\theta$ ，因此 $-\\frac{\\pi}{3} \\leq \\theta \\leq \\frac{\\pi}{3}$ 。即 $I =$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":133,"id":"94d0e05b-91fe-4c89-a561-7ca605c026b6","page_size":[595,841],"block_position":"133-16"},{"bbox":[289,802,305,812],"type":"page_number","angle":0,"index":1000016,"text":"126 ","color":{"line":"rgba(164, 164, 164, 1)","fill":"rgba(164, 164, 164, 1)"},"page_idx":133,"is_discarded":true,"id":"14d77f58-7ebb-40a3-b6d0-198dc37d5d1f","page_size":[595,841],"block_position":"133-17"}],[{"bbox":[475,47,539,58],"type":"header","angle":0,"index":-99998,"text":"15.2 三重积分 ","color":{"line":"rgba(164, 164, 164, 1)","fill":"rgba(164, 164, 164, 1)"},"page_idx":134,"is_discarded":true,"id":"9ea30056-56fc-4938-9a21-369fab038e54","page_size":[595,841],"block_position":"134-0"},{"bbox":[53,68,309,98],"type":"interline_equation","angle":0,"index":2,"text":"$$\n\\int_ {- \\frac {\\pi}{3}} ^ {\\frac {\\pi}{3}} \\mathrm{d} \\theta \\int_ {\\frac {1}{2}} ^ {\\cos \\theta} f (\\tan \\theta) r \\mathrm{d} r = \\frac {1}{2} \\int_ {- \\frac {\\pi}{3}} ^ {\\frac {\\pi}{3}} f (\\tan \\theta) (\\cos^ {2} \\theta - \\frac {1}{4}) \\mathrm{d} \\theta\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":134,"id":"ff0edcbe-763c-4fa3-a080-2f82d418f875","page_size":[595,841],"block_position":"134-1"},{"bbox":[73,99,342,111],"type":"text","angle":0,"index":3,"merge_prev":false,"text":"(2)一般都是先 $r$ 后 $\\theta$ ，将 $D$ 分为左上，右下两个三角形，则 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":134,"id":"6f69cb45-5680-46b6-8a74-69d276dc0be2","page_size":[595,841],"block_position":"134-2"},{"bbox":[134,112,458,143],"type":"interline_equation","angle":0,"index":4,"text":"$$\nI = \\int_ {0} ^ {\\frac {\\pi}{4}} \\mathrm{d} \\theta \\int_ {0} ^ {\\frac {1}{\\cos \\theta}} f (r \\cos \\theta , r \\sin \\theta) r \\mathrm{d} r + \\int_ {\\frac {\\pi}{4}} ^ {\\frac {\\pi}{2}} \\mathrm{d} \\theta \\int_ {0} ^ {\\frac {1}{\\sin \\theta}} f (r \\cos \\theta , r \\sin \\theta) r \\mathrm{d} r\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":134,"id":"42f43e43-6788-4ce3-be41-9ec0372ffaca","page_size":[595,841],"block_position":"134-3"},{"bbox":[54,147,147,158],"type":"text","angle":0,"index":5,"merge_prev":false,"text":"先 $\\theta$ 后 $r$ 会比较麻烦 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":134,"id":"bfe71ed6-3099-4a6d-bd11-69d680578125","page_size":[595,841],"block_position":"134-4"},{"bbox":[54,177,117,190],"type":"title","angle":0,"index":6,"level":2,"text":"## 三类特殊的圆","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":134,"id":"cc4f4ba7-1a18-49b1-a982-8e0dc1c554b8","page_size":[595,841],"block_position":"134-5"},{"bbox":[73,198,205,211],"type":"title","angle":0,"index":7,"level":2,"text":"## 定理 15.4 (三类特殊圆极坐标)","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":134,"id":"2e4ba281-34ab-4115-a0f9-8c85e4610133","page_size":[595,841],"block_position":"134-6"},{"bbox":[68,215,368,227],"type":"text","angle":0,"index":8,"merge_prev":false,"text":"设 $a > 0$ ，极坐标 $x = r\\cos \\theta ,y = r\\sin \\theta$ ，以下特殊圆的极坐标公式： ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":134,"id":"95e0782e-c537-4589-87e6-ae625155a205","page_size":[595,841],"block_position":"134-7"},{"bbox":[83,228,389,246],"type":"text","angle":0,"index":9,"merge_prev":false,"text":"- $(x - a)^2 + y^2 = a^2$ ，即 $x^2 + y^2 = 2ax$ ，则 $r = 2a\\cos \\theta, \\theta \\in [-\\frac{\\pi}{2}, \\frac{\\pi}{2}]$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":134,"id":"e94e93b8-28d1-47bc-aa9c-5219518675ee","page_size":[595,841],"block_position":"134-8"},{"bbox":[83,245,376,259],"type":"text","angle":0,"index":10,"merge_prev":false,"text":"- $x^{2} + (y - a)^{2} = a^{2}$ ，即 $x^{2} + y^{2} = 2ay$ ，则 $r = 2a\\sin \\theta, \\theta \\in [0, \\pi]$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":134,"id":"d8414cb8-0d58-47b9-ab72-ab0784c6f9b3","page_size":[595,841],"block_position":"134-9"},{"bbox":[83,259,487,280],"type":"text","angle":0,"index":11,"merge_prev":false,"text":"- $(x - a)^2 + (y - a)^2 = 2a^2$ ，即 $x^2 + y^2 = 2a(x + y)$ ，则 $r = 2a(\\cos \\theta + \\sin \\theta), \\theta \\in [-\\frac{\\pi}{4}, \\frac{3\\pi}{4}]$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":134,"id":"36d33340-f69a-4c9c-b6d6-b419caeda11a","page_size":[595,841],"block_position":"134-10"},{"bbox":[152,302,442,427],"type":"image","angle":0,"index":12,"img_path":"/b54fa69c6f03c09c287c78d1c9f720c960f3f0469a6172b2465da531dea1784b.jpg","color":{"line":"rgba(89, 92, 220, 1)","fill":"rgba(89, 92, 220, 1)"},"img_ratio":0.48739495798319327,"page_idx":134,"id":"bcfa9570-b737-453f-a5fc-7fb5daaba8c7","page_size":[595,841],"block_position":"134-11"},{"bbox":[246,430,346,443],"type":"image_caption","angle":0,"index":13,"text":"图 15.2: 三个特殊的圆","id":"7f4de072-7fe3-4a0b-835a-e07d47eaedf1","color":{"line":"rgba(13, 83, 222 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":134,"page_size":[595,841],"block_position":"134-12"},{"bbox":[57,481,151,497],"type":"title","angle":0,"index":14,"level":2,"text":"## 15.2 三重积分","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":134,"id":"8c2d3e5d-c2ef-4be0-b6e6-1d1670fcbd49","page_size":[595,841],"block_position":"134-13"},{"bbox":[57,513,133,528],"type":"title","angle":0,"index":15,"level":2,"text":"## 15.2.1 投影法","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":134,"id":"255c0abc-b909-4e53-839f-c78c6a9ddd05","page_size":[595,841],"block_position":"134-14"},{"bbox":[72,537,377,550],"type":"text","angle":0,"index":16,"merge_prev":false,"text":"将三重积分转换为累次积分进行计算，常用的方法有投影法与截面法 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":134,"id":"8e5b73d5-ae39-496b-a061-a7a60d819be3","page_size":[595,841],"block_position":"134-15"},{"bbox":[73,558,157,571],"type":"title","angle":0,"index":17,"level":2,"text":"## 定理 15.5 (投影法)","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":134,"id":"abb895e2-beb2-4e3f-ace6-34eff6a9d49f","page_size":[595,841],"block_position":"134-16"},{"bbox":[68,575,525,602],"type":"text","angle":0,"index":18,"merge_prev":false,"text":"积分区域 $V \\subset R^{3}$ ，对应 x, y 坐标平面投影区域 D，且 $V = \\{(x, y, z) : (x, y) \\in D, z_{1}(x, y) \\leq z \\leq z_{2}(x, y)\\}$ ，则： ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":134,"id":"982dae01-4935-41bc-88af-bbba15c1beaf","page_size":[595,841],"block_position":"134-17"},{"bbox":[187,602,406,631],"type":"interline_equation","angle":0,"index":19,"text":"$$\n\\iiint_ {V} f (x, y, z) \\mathrm{d} V = \\iint_ {D} \\mathrm{d} x \\mathrm{d} y \\int_ {z _ {1} (x, y)} ^ {z _ {2} (x, y)} f (x, y, z) \\mathrm{d} z\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":134,"id":"284634b5-b233-489e-b85c-5e32210bfb50","page_size":[595,841],"block_position":"134-18"},{"bbox":[35,642,47,658],"type":"image","angle":0,"index":20,"img_path":"/ad5fd2105a5935951c414ac9ded62f37636325f2db703c2dd4021028281fbc93.jpg","color":{"line":"rgba(89, 92, 220, 1)","fill":"rgba(89, 92, 220, 1)"},"img_ratio":0.2,"page_idx":134,"id":"e3d4f012-3d6a-404b-89b7-a71c004ab620","page_size":[595,841],"block_position":"134-19"},{"bbox":[52,647,239,660],"type":"text","angle":0,"index":21,"merge_prev":false,"text":"笔记 一般而言投影法比截面法更常用一点 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":134,"id":"844c69b6-3b71-43ab-8d65-205b158aa128","page_size":[595,841],"block_position":"134-20"},{"bbox":[35,673,477,698],"type":"text","angle":0,"index":22,"merge_prev":false,"text":"练习 15.8 投影法 (1) 计算 $I = \\iiint_{V} (x + y + z) \\, \\mathrm{d}V$ ，其中 V 是 $x + y + z = 1$ 与坐标面围成的区域 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":134,"id":"b3181d37-7413-41b2-ab0b-969904af3f44","page_size":[595,841],"block_position":"134-21"},{"bbox":[72,698,524,735],"type":"text","angle":0,"index":23,"merge_prev":false,"text":"(2) 计算 $\\iiint_{V} (x^{2} + y^{2} + z) \\mathrm{d}V$ ，其中 $V$ 是 $\\left\\{ \\begin{array}{l} z = y \\\\ x = 0 \\end{array} \\right.$ 绕着 $z$ 轴旋转一圈而成的曲面与 $z = 1$ 围成的区域 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":134,"id":"1d7031d7-d269-4d7e-9aba-083f599c217a","page_size":[595,841],"block_position":"134-22"},{"bbox":[52,735,296,763],"type":"text","angle":0,"index":24,"merge_prev":false,"text":"解 (1) 用投影法 $I = \\iint_{D} \\mathrm{d}x \\, \\mathrm{d}y \\int_{0}^{1 - x - y} x + y + z \\, \\mathrm{d}z = \\frac{1}{8}$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":134,"id":"eb61924f-2363-4880-8f73-d1bc5d32b3ca","page_size":[595,841],"block_position":"134-23"},{"bbox":[289,802,305,812],"type":"page_number","angle":0,"index":999999,"text":"127 ","color":{"line":"rgba(164, 164, 164, 1)","fill":"rgba(164, 164, 164, 1)"},"page_idx":134,"is_discarded":true,"id":"236f1679-b77f-4cd8-8907-26b1888d5304","page_size":[595,841],"block_position":"134-24"}],[{"bbox":[476,47,538,58],"type":"header","angle":0,"index":-99999,"text":"15.2 三重积分 ","color":{"line":"rgba(164, 164, 164, 1)","fill":"rgba(164, 164, 164, 1)"},"page_idx":135,"is_discarded":true,"id":"a446580d-7c15-4ccf-a142-fddb0c6f982f","page_size":[595,841],"block_position":"135-0"},{"bbox":[73,71,433,84],"type":"text","angle":0,"index":1,"merge_prev":false,"text":"(2) 旋转面为 $z = \\sqrt{x^2 + y^2}$ (画个图就看得出来)，投影为 $x^2 + y^2 \\leq 1$ ，因此积分为 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":135,"id":"17502cec-f42e-4c10-ae79-283528abfcb6","page_size":[595,841],"block_position":"135-1"},{"bbox":[151,88,441,197],"type":"interline_equation","angle":0,"index":2,"text":"$$\n\\begin{array}{l} I = \\iint_ {D} \\mathrm{d} x \\mathrm{d} y \\int_ {\\sqrt {x ^ {2} + y ^ {2}}} ^ {1} (x ^ {2} + y ^ {2} + z) \\mathrm{d} z \\\\ = \\iint_ {D} (x ^ {2} + y ^ {2}) (1 - \\sqrt {x ^ {2} + y ^ {2}}) \\mathrm{d} x \\mathrm{d} y + \\frac {1}{2} \\iint_ {D} [ 1 - (x ^ {2} + y ^ {2}) ] \\mathrm{d} x \\mathrm{d} y \\\\ = \\int_ {0} ^ {2 \\pi} \\mathrm{d} \\theta \\int_ {0} ^ {1} r ^ {3} (1 - r) \\mathrm{d} r + \\frac {1}{2} \\int_ {0} ^ {2 \\pi} \\mathrm{d} \\theta \\int_ {0} ^ {1} (1 - r ^ {2}) r \\mathrm{d} r \\\\ = \\frac {\\pi}{1 0} + \\frac {\\pi}{4} = \\frac {7 \\pi}{2 0} \\end{array}\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":135,"id":"d4e1a390-aee8-4585-89d8-55a95c9200c5","page_size":[595,841],"block_position":"135-2"},{"bbox":[57,212,133,226],"type":"title","angle":0,"index":3,"level":2,"text":"## 15.2.2 截面法","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":135,"id":"8bf11b35-1340-46cd-8aca-672e444f7ade","page_size":[595,841],"block_position":"135-3"},{"bbox":[73,239,156,252],"type":"title","angle":0,"index":4,"level":2,"text":"## 定理 15.6 (截面法)","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":135,"id":"97c541d7-4f38-4ad6-ac52-56c1f87a1ac3","page_size":[595,841],"block_position":"135-4"},{"bbox":[68,256,343,269],"type":"text","angle":0,"index":5,"merge_prev":false,"text":"积分区域 $V \\subset \\mathbb{R}^3$ ， $V$ 被垂直于 $z$ 轴平面截取的平面为 $D_z$ ，则 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":135,"id":"76649980-d8fb-4c0e-b036-a8b56a2106ca","page_size":[595,841],"block_position":"135-5"},{"bbox":[196,272,398,301],"type":"interline_equation","angle":0,"index":6,"text":"$$\n\\iiint_ {V} f (x, y, z) \\mathrm{d} V = \\int_ {a} ^ {b} \\mathrm{d} z \\iint_ {D _ {z}} f (x, y, z) \\mathrm{d} x \\mathrm{d} y\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":135,"id":"e39708cc-cbee-48ad-afcd-f227af3faa8e","page_size":[595,841],"block_position":"135-6"},{"bbox":[57,332,180,346],"type":"title","angle":0,"index":7,"level":2,"text":"## 15.2.3 三重积分换元法","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":135,"id":"426237af-7a6d-4324-9eb3-edb2c64b9273","page_size":[595,841],"block_position":"135-7"},{"bbox":[73,359,186,372],"type":"title","angle":0,"index":8,"level":2,"text":"## 定理 15.7 (三重积分换元)","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":135,"id":"4f909998-bab5-43e7-8219-373913304749","page_size":[595,841],"block_position":"135-8"},{"bbox":[69,376,329,389],"type":"text","angle":0,"index":10,"merge_prev":false,"text":"若坐标替换 $x = x(u,v,w),y = y(u,v,w),z = z(u,v,w)$ ，则 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":135,"id":"32781977-0f88-4aa3-915b-2dc78ab10817","page_size":[595,841],"block_position":"135-9"},{"bbox":[166,391,427,418],"type":"interline_equation","angle":0,"index":11,"text":"$$\n\\iiint_ {V} f (x, y, z) \\mathrm{d} x \\mathrm{d} y \\mathrm{d} z = \\iiint_ {V ^ {\\prime}} f (x, y, z) \\left| \\frac {\\partial (x , y , z)}{\\partial (u , v , w)} \\right| \\mathrm{d} u \\mathrm{d} v \\mathrm{d} w\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":135,"id":"65f96f45-66b1-4fc4-aa72-df098e8fe5fe","page_size":[595,841],"block_position":"135-10"},{"bbox":[224,420,364,486],"type":"interline_equation","angle":0,"index":12,"text":"$$\n\\left| \\frac {\\partial (x , y , z)}{\\partial (u , v , w)} \\right| = \\left| \\begin{array}{l l l} \\frac {\\partial x}{\\partial u} & \\frac {\\partial x}{\\partial v} & \\frac {\\partial x}{\\partial w} \\\\ \\frac {\\partial y}{\\partial u} & \\frac {\\partial y}{\\partial v} & \\frac {\\partial y}{\\partial w} \\\\ \\frac {\\partial z}{\\partial u} & \\frac {\\partial z}{\\partial v} & \\frac {\\partial z}{\\partial w} \\end{array} \\right|\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":135,"id":"dba5b554-7c61-48ec-9678-a2ef2f00b203","page_size":[595,841],"block_position":"135-11"},{"bbox":[53,497,539,547],"type":"text","angle":0,"index":13,"merge_prev":false,"text":"笔记 一方面还是注意 Jacobi 行列式外面的是绝对值。另一方面可以先算 $\\frac{\\partial(u, v, w)}{\\partial(x, y, z)}$ ，再取倒数和绝对值计算 $\\left|\\frac{\\partial(x, y, z)}{\\partial(u, v, w)}\\right|$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":135,"id":"2e48ec5f-1e65-4351-a8da-aca1f454cc71","page_size":[595,841],"block_position":"135-12"},{"bbox":[57,562,240,576],"type":"title","angle":0,"index":14,"level":2,"text":"## 15.2.4 柱面坐标变换与球坐标变换","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":135,"id":"f996ab83-47a3-43a9-9fa0-4a93ea5e468a","page_size":[595,841],"block_position":"135-13"},{"bbox":[73,589,166,602],"type":"title","angle":0,"index":15,"level":2,"text":"## 定理 15.8 (柱面变换)","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":135,"id":"e6c1e43d-8209-4a87-9e9f-fb2cb2db749f","page_size":[595,841],"block_position":"135-14"},{"bbox":[68,607,381,618],"type":"text","angle":0,"index":16,"merge_prev":false,"text":"做 $x = r\\cos \\theta ,y = r\\sin \\theta ,z = z$ 变换时， $J(u,v,w) = r$ 。往往确定区域： ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":135,"id":"d5c12f70-2dd1-4e73-965e-f62e243da954","page_size":[595,841],"block_position":"135-15"},{"bbox":[68,625,403,671],"type":"text","angle":0,"index":17,"merge_prev":false,"text":"$V^{\\prime} = \\{a\\leq r\\leq b,c\\leq \\theta \\leq d,\\alpha (r,\\theta)\\leq z\\leq \\beta (r,\\theta)\\}$ 因此用投影法 $\\iint_{D}\\mathrm{d}r\\mathrm{d}\\theta \\int_{\\alpha (r,\\theta)}^{\\beta (r,\\theta)}z\\mathrm{d}z$ ，即一般积分顺序按照 $\\mathrm{dz},\\mathrm{dr},\\mathrm{d}\\theta$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":135,"id":"87138e15-645d-4876-8128-4cf41a7d2e86","page_size":[595,841],"block_position":"135-16"},{"bbox":[35,700,535,734],"type":"text","angle":0,"index":18,"merge_prev":false,"text":"练习 15.9 柱坐标变换 (1) 计算 $I = \\iiint_{V} (x^{2} + y^{2}) \\, \\mathrm{d}V$ ，其中 V 是由曲面 $2(x^{2} + y^{2}) = z$ 与 z = 4 为界面的区域。\n解 (1) 为抛物面与平面围成的区域，令 $x = r \\cos \\theta, y = r \\sin \\theta, z = z$ 得到区域为 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":135,"id":"8e8cd71c-de52-4294-a691-2b49969dfc82","page_size":[595,841],"block_position":"135-17"},{"bbox":[182,740,409,755],"type":"interline_equation","angle":0,"index":19,"text":"$$\nV ^ {\\prime} = \\left\\{(r, \\theta , z): 0 \\leq r \\leq 2, 0 \\leq \\theta \\leq 2 \\pi , 2 r ^ {2} \\leq z \\leq 4 \\right\\}\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":135,"id":"2482b7f9-b949-4f66-a3af-4e571fa5b68d","page_size":[595,841],"block_position":"135-18"},{"bbox":[289,802,305,812],"type":"page_number","angle":0,"index":1000008,"text":"128 ","color":{"line":"rgba(164, 164, 164, 1)","fill":"rgba(164, 164, 164, 1)"},"page_idx":135,"is_discarded":true,"id":"dc9a0300-61b6-42bd-b93a-d8f7f9de8872","page_size":[595,841],"block_position":"135-19"}],[{"bbox":[475,47,538,58],"type":"header","angle":0,"index":-99999,"text":"15.2 三重积分 ","color":{"line":"rgba(164, 164, 164, 1)","fill":"rgba(164, 164, 164, 1)"},"page_idx":136,"is_discarded":true,"id":"61fba8b6-ea4a-4521-be90-822571048a9e","page_size":[595,841],"block_position":"136-0"},{"bbox":[54,71,306,100],"type":"text","angle":0,"index":1,"merge_prev":false,"text":"对应积分 $\\iiint_{V'} r^2 \\cdot r \\, \\mathrm{d}V = \\int_0^{2\\pi} \\, \\mathrm{d}\\theta \\int_0^{\\sqrt{2}} \\, \\mathrm{d}r \\int_{2r^2}^4 r^3 \\, \\mathrm{d}z = \\frac{8\\pi}{3}$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":136,"id":"d7c0d90c-fc02-48b8-8d1c-bcfff3ebe95a","page_size":[595,841],"block_position":"136-1"},{"bbox":[73,114,176,126],"type":"title","angle":0,"index":2,"level":2,"text":"## 定理 15.9 (球坐标变换)","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":136,"id":"75032ad4-69aa-4b83-8573-428a078b3164","page_size":[595,841],"block_position":"136-2"},{"bbox":[68,131,495,145],"type":"text","angle":0,"index":3,"merge_prev":false,"text":"做 $x = r\\sin \\varphi \\cos \\theta ,y = r\\sin \\varphi \\sin \\theta ,z = r\\cos \\varphi$ 时 $J = r^2\\sin \\varphi$ 。积分时先按照 $\\mathrm{d}r,\\mathrm{d}\\varphi ,\\mathrm{d}\\theta$ 顺序积分 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":136,"id":"9e5ebc16-0263-4ac7-b25c-552db0069f2a","page_size":[595,841],"block_position":"136-3"},{"bbox":[103,166,489,302],"type":"image","angle":0,"index":4,"img_path":"/93315663037de58e7e2eda9f87b39ab445a1eaf012a8cff06679212e4a2f18a2.jpg","color":{"line":"rgba(89, 92, 220, 1)","fill":"rgba(89, 92, 220, 1)"},"img_ratio":0.6487394957983194,"page_idx":136,"id":"1e34cce4-65b0-439a-8ed2-1124c64015f6","page_size":[595,841],"block_position":"136-4"},{"bbox":[231,305,362,317],"type":"image_caption","angle":0,"index":5,"text":"图 15.3: 三重积分球坐标换元","id":"b94e301f-fd9a-467b-85d7-527e7e714623","color":{"line":"rgba(13, 83, 222 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":136,"page_size":[595,841],"block_position":"136-5"},{"bbox":[34,330,48,347],"type":"image","angle":0,"index":6,"img_path":"/fc68779cbf73ff0e464878fe775b9e52163a5b057fb1e986b9ed4d5fe07d16e1.jpg","color":{"line":"rgba(89, 92, 220, 1)","fill":"rgba(89, 92, 220, 1)"},"img_ratio":0.2,"page_idx":136,"id":"ee86c8c2-c931-4210-a15f-3e89f26005f5","page_size":[595,841],"block_position":"136-6"},{"bbox":[35,333,526,349],"type":"text","angle":0,"index":7,"merge_prev":false,"text":"笔记 使用球坐标变换的原则：(1) 被积函数为 $f(x^{2} + y^{2} + z^{2}), f(x^{2} + y^{2})(2)$ 积分区域为球、锥或者其一部分 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":136,"id":"e340d6c3-2569-43d5-8b41-4de7f5a90257","page_size":[595,841],"block_position":"136-7"},{"bbox":[73,370,201,383],"type":"title","angle":0,"index":8,"level":2,"text":"## 定理 15.10 (广义球坐标变换)","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":136,"id":"0cc10849-b155-4f93-9c32-6b249d8013b3","page_size":[595,841],"block_position":"136-8"},{"bbox":[68,387,461,400],"type":"text","angle":0,"index":9,"merge_prev":false,"text":"对于以 $(x_0, y_0, z_0)$ ，以 $a, b, c$ 为半轴长的椭球面所围成的区域 $V$ ，做如下广义球坐标变换： ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":136,"id":"df0b8624-1dda-4d39-9bc2-74b6f0cf0209","page_size":[595,841],"block_position":"136-9"},{"bbox":[163,405,429,460],"type":"interline_equation","angle":0,"index":10,"text":"$$\nT: \\left\\{ \\begin{array}{l} x = x _ {0} + a r \\sin \\varphi \\cos \\theta \\\\ y = y _ {0} + b r \\sin \\varphi \\sin \\theta \\\\ z = z _ {0} + c r \\cos \\varphi \\end{array} , \\quad r \\geq 0, \\varphi \\in [ 0, \\pi ], \\theta \\in [ 0, 2 \\pi ] \\right.\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":136,"id":"9fe89595-4647-42e0-b357-593fae8eb222","page_size":[595,841],"block_position":"136-10"},{"bbox":[68,464,92,475],"type":"text","angle":0,"index":11,"merge_prev":false,"text":"得到 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":136,"id":"a45a4657-9b6f-4a0d-b8e4-4979718f32fd","page_size":[595,841],"block_position":"136-11"},{"bbox":[232,472,362,499],"type":"interline_equation","angle":0,"index":12,"text":"$$\n| J | = \\frac {\\partial (x , y , z)}{\\partial (r , \\varphi , \\theta)} = a b c r ^ {2} \\sin \\varphi\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":136,"id":"51f41850-1877-4545-96ba-de93f9983de7","page_size":[595,841],"block_position":"136-12"},{"bbox":[35,523,518,562],"type":"text","angle":0,"index":13,"merge_prev":false,"text":"练习15.10广义球坐标变换(1)计算 $I = \\iiint_{V}z\\mathrm{d}x\\mathrm{d}y\\mathrm{d}z$ ，其中 $V$ 是 $\\frac{x^2}{a^2} +\\frac{y^2}{b^2} +\\frac{z^2}{c^2}\\leq 1$ 与 $z\\geq 0$ 围成的区域解(1)做球坐标变换，积分变为： ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":136,"id":"35da1362-e078-45fd-b317-508893ab803d","page_size":[595,841],"block_position":"136-13"},{"bbox":[113,565,480,592],"type":"interline_equation","angle":0,"index":14,"text":"$$\n\\iiint_ {V ^ {\\prime}} c r \\cos \\varphi \\cdot a b c r ^ {2} \\sin \\varphi \\mathrm{d} r \\mathrm{d} \\varphi \\mathrm{d} \\theta = a b c ^ {2} \\int_ {0} ^ {\\frac {\\pi}{2}} \\mathrm{d} \\varphi \\int_ {0} ^ {2 \\pi} \\mathrm{d} \\theta \\int_ {0} ^ {1} r ^ {3} \\sin \\varphi \\cos \\varphi \\mathrm{d} r = \\frac {\\pi}{4} a b c ^ {2}\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":136,"id":"ae753d36-5c46-4f5c-b86d-a3a096aa2a79","page_size":[595,841],"block_position":"136-14"},{"bbox":[289,802,305,812],"type":"page_number","angle":0,"index":1000014,"text":"129 ","color":{"line":"rgba(164, 164, 164, 1)","fill":"rgba(164, 164, 164, 1)"},"page_idx":136,"is_discarded":true,"id":"01a917a0-cb69-4384-a35f-0ace0b98f235","page_size":[595,841],"block_position":"136-15"}],[{"bbox":[231,74,373,94],"type":"title","angle":0,"index":0,"level":2,"text":"## 第 16 章 曲线积分","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":137,"id":"ceb7b70b-509a-4295-9a75-6160f874c861","page_size":[595,841],"block_position":"137-0"},{"bbox":[57,122,180,139],"type":"title","angle":0,"index":1,"level":2,"text":"## 16.1 曲线的参数化","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":137,"id":"25428ca7-dc17-4532-b28d-15032c783b61","page_size":[595,841],"block_position":"137-1"},{"bbox":[57,156,193,173],"type":"title","angle":0,"index":2,"level":2,"text":"## 16.2 第一型曲线积分","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":137,"id":"f0148538-c912-4e75-8aa8-0ef6097b75be","page_size":[595,841],"block_position":"137-2"},{"bbox":[56,189,240,203],"type":"title","angle":0,"index":3,"level":2,"text":"## 16.2.1 第一型曲线积分的基本概念","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":137,"id":"0a2754de-6423-4b30-a6ed-e78bb8bedc36","page_size":[595,841],"block_position":"137-3"},{"bbox":[73,216,196,229],"type":"title","angle":0,"index":4,"level":2,"text":"## 定义 16.1 (第一型曲线积分)","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":137,"id":"8b4f945e-df3a-4337-a035-a2e40f83ab14","page_size":[595,841],"block_position":"137-4"},{"bbox":[67,232,525,261],"type":"text","angle":0,"index":5,"merge_prev":false,"text":"$\\Gamma$ 是 $R^{3}$ 中的可求长曲线, $f:\\Gamma\\to R$ , 将 $\\Gamma$ 划分为分段 $s_{1},\\cdots,s_{n}$ , 每段的弧长为 $\\Delta s_{i},\\xi_{i}=(x_{i},y_{i},z_{i})\\in s_{i}$ , 定义 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":137,"id":"38d8bbbf-71f7-410f-92a3-a99c3d84bd2b","page_size":[595,841],"block_position":"137-5"},{"bbox":[203,259,389,290],"type":"interline_equation","angle":0,"index":6,"text":"$$\n\\int_ {\\Gamma} f (x, y, z) \\mathrm{d} s = \\lim _ {\\Delta s _ {i} \\rightarrow 0} \\sum_ {i = 1} ^ {n} f (x _ {i}, y _ {i}, z _ {i}) \\Delta s _ {i}\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":137,"id":"bdcfd6ad-6150-40b0-9746-ff6ed32d7f3a","page_size":[595,841],"block_position":"137-6"},{"bbox":[73,312,226,324],"type":"title","angle":0,"index":7,"level":2,"text":"## 定理 16.1 (第一型曲线积分的计算)","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":137,"id":"b4d8296a-4c5e-4d0f-ac5b-a7c8687a3912","page_size":[595,841],"block_position":"137-7"},{"bbox":[67,328,463,342],"type":"text","angle":0,"index":8,"merge_prev":false,"text":"$\\Gamma$ 是 $\\mathbb{R}^3$ 中的光滑曲线，且曲线可表示为 $\\mathbf{r} = x(t)\\mathbf{i} + y(t)\\mathbf{j} + z(t)\\mathbf{k}, t \\in [a, b]$ ，则曲线积分为 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":137,"id":"ac73d050-e7e6-4887-9d20-eaae7475854f","page_size":[595,841],"block_position":"137-8"},{"bbox":[155,345,437,374],"type":"interline_equation","angle":0,"index":9,"text":"$$\n\\int_ {\\Gamma} f (x, y, z) \\mathrm{d} s = \\int_ {a} ^ {b} f (x (t), y (t), z (t)) \\sqrt {x ^ {\\prime 2} (t) + y ^ {\\prime 2} (t) + z ^ {\\prime 2} (t)} \\mathrm{d} t\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":137,"id":"613abc2e-cbeb-4674-b4b8-108908b03981","page_size":[595,841],"block_position":"137-9"},{"bbox":[53,388,539,470],"type":"text","angle":0,"index":10,"merge_prev":false,"text":"证明 设 $\\Gamma$ 上的分点对应 $r_{0}, r_{1}, \\cdots, r_{n}$ ，对应 $a = t_{0} < t_{1} < \\cdots < t_{n-1} < t_{n} = b$ 。首先根据弧长公式得到 $\\Delta s_{i} = \\int_{t_{i-1}}^{t_{i}} \\sqrt{x'^{2}(t) + y'^{2}(t) + z'^{2}(t)} \\mathrm{d}t$ ，根据积分中值定理得到 $\\Delta s_{i} = \\sqrt{x'^{2}(\\zeta_{i}) + y'^{2}(\\zeta_{i}) + z'^{2}(\\zeta_{i})} \\Delta t_{i}, \\zeta_{i} \\in [t_{i-1}, t_{i}]$ ，代入第一型曲线积分的定义得到 $\\int_{\\Gamma} f(x, y, z) \\, ds = \\lim_{||T|| \\to 0} \\sum_{i=1}^{n} f(x_{i}, y_{i}, z_{i}) \\sqrt{x'^{2}(\\zeta_{i}) + y'^{2}(\\zeta_{i}) + z'^{2}(\\zeta_{i})} \\Delta t_{i}$ ，当 $\\Delta t_{i} \\to 0$ 时即 Riemann 积分，因此得到结论。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":137,"id":"f8f4b2aa-1b4f-4b5d-a1fa-9514ed5cae95","page_size":[595,841],"block_position":"137-10"},{"bbox":[73,478,246,491],"type":"title","angle":0,"index":11,"level":2,"text":"## 推论 16.1 (二维第一型曲线积分的计算)","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":137,"id":"9a62e3af-dead-42d4-bc31-995b6d7f72de","page_size":[595,841],"block_position":"137-11"},{"bbox":[68,496,371,509],"type":"text","angle":0,"index":12,"merge_prev":false,"text":"对于 $\\Gamma \\subset \\mathbb{R}^2$ ，若平面曲线 $\\Gamma$ 可表示为 $y = \\varphi (x)$ ， $\\varphi$ 在 $[a,b]$ 连续，则 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":137,"id":"787a0e36-7349-4bf9-bff0-de07b986f382","page_size":[595,841],"block_position":"137-12"},{"bbox":[199,513,393,540],"type":"interline_equation","angle":0,"index":13,"text":"$$\n\\int_ {\\Gamma} f (x, y) \\mathrm{d} s = \\int_ {a} ^ {b} f (x, \\varphi (x)) \\sqrt {1 + \\varphi^ {\\prime 2} (x)} \\mathrm{d} x\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":137,"id":"d5306bdb-a6ae-49ae-9a5a-4b975fc1f017","page_size":[595,841],"block_position":"137-13"},{"bbox":[52,555,538,597],"type":"text","angle":0,"index":14,"merge_prev":false,"text":"笔记 第一型曲线积分没有方向一说，例如 $A, B$ 曲线上参数 $t \\in [a, b]$ ，那么积分时 $AB$ 和 $BA$ 的积分都是 $\\int_{a}^{b} f(x, y) \\mathrm{d}t$ ，不能写为 $\\int_{b}^{a} f(x, y) \\mathrm{d}t$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":137,"id":"3e3d2969-a733-4dc7-bdf4-366b7039fd6c","page_size":[595,841],"block_position":"137-14"},{"bbox":[35,600,386,627],"type":"text","angle":0,"index":15,"merge_prev":false,"text":"练习16.1极坐标格式 $(1)L:x^{2} + y^{2} = 1$ ，求 $\\int_{L}(x^{2} + y^{2})\\mathrm{d}s,\\int_{L}(x^{2} + 2y^{2})\\mathrm{d}s$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":137,"id":"ad3789d0-f557-4e58-bc14-d7842a0a31e4","page_size":[595,841],"block_position":"137-15"},{"bbox":[72,625,381,650],"type":"text","angle":0,"index":16,"merge_prev":false,"text":"(2) $\\int_{L} \\sqrt{2y^2 + z^2} \\, \\mathrm{d}s$ ，其中 $L$ 是 $x^2 + y^2 + z^2 = a^2$ 与 $x = y$ 相交的圆周 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":137,"id":"2817940d-163b-45e3-95b8-beef998ccbb6","page_size":[595,841],"block_position":"137-16"},{"bbox":[73,650,462,673],"type":"text","angle":0,"index":17,"merge_prev":false,"text":"(3) $\\int_{L} (x^{2} + y^{2} + z^{2}) \\, \\mathrm{d}s$ ，这里 $L$ 是螺旋线 $x = a \\cos t, y = a \\sin t, z = bt, t \\in [0, 2\\pi]$ 的一段 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":137,"id":"4edff6b9-1689-4e8f-a8f0-afc264f97c4e","page_size":[595,841],"block_position":"137-17"},{"bbox":[73,671,377,698],"type":"text","angle":0,"index":18,"merge_prev":false,"text":"(4) 积分较难算： $\\int_{L} xy \\, \\mathrm{d}s$ ，其中 $L$ 为 $\\frac{x^2}{a^2} + \\frac{y^2}{b^2} = 1$ 在第一象限的部分 解 (1) $\\int_{L} (x^{2} + y^{2}) \\, \\mathrm{d}s = \\int_{L} 1 \\, \\mathrm{d}s = 2\\pi$ 。记 $x = \\cos \\theta, y = \\sin \\theta$ ，此时 $(x'(\\theta))^{2} + (y'(\\theta))^{2} = 1$ ，因此 $\\int_{L} (x^{2} + 2y^{2}) \\, \\mathrm{d}s = \\int_{0}^{2\\pi} (1 + \\sin^{2}\\theta) \\, \\mathrm{d}\\theta = 2\\pi + 4 \\cdot \\frac{1}{2} \\cdot \\frac{\\pi}{2} = 3\\pi$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":137,"id":"f1574d8c-8062-45eb-9ac8-7dbd7df6e550","page_size":[595,841],"block_position":"137-18"},{"bbox":[52,697,539,748],"type":"text","angle":0,"index":19,"merge_prev":false,"lines_deleted":true,"text":"","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":137,"id":"f1574d8c-8062-45eb-9ac8-7dbd7df6e550","page_size":[595,841],"block_position":"137-19"},{"bbox":[73,747,352,771],"type":"text","angle":0,"index":20,"merge_prev":false,"text":"(2) 显然 $2y^{2} + z^{2} = a^{2}$ ，因此 $\\int_{L} \\sqrt{2y^2 + z^2} \\, \\mathrm{d}s = a \\cdot 2a\\pi = 2\\pi a^2$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":137,"id":"a3065406-cec6-4079-85d7-18d004a7dec2","page_size":[595,841],"block_position":"137-20"}],[{"bbox":[445,47,538,58],"type":"header","angle":0,"index":-99998,"text":"16.2 第一型曲线积分 ","color":{"line":"rgba(164, 164, 164, 1)","fill":"rgba(164, 164, 164, 1)"},"page_idx":138,"is_discarded":true,"id":"ae09e0c7-6445-48ca-8662-1c9e6478fb86","page_size":[595,841],"block_position":"138-0"},{"bbox":[54,71,539,116],"type":"text","angle":0,"index":2,"merge_prev":false,"text":"(3) $x^{2}+y^{2}+z^{2}=a^{2}+b^{2}t^{2},\\quad(x^{\\prime})^{2}+(y^{\\prime})^{2}+(z^{\\prime})^{2}=a^{2}+b^{2}$ ，因此 $I=\\int_{0}^{2\\pi}(a^{2}+b^{2}t^{2})\\sqrt{a^{2}+b^{2}}dt=$ $\\sqrt{a^{2}+b^{2}}(2\\pi a^{2}+\\frac{8}{3}\\pi^{3}b^{2})$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":138,"id":"2cdfe812-4b91-4838-8683-9def467a27c0","page_size":[595,841],"block_position":"138-1"},{"bbox":[73,116,492,144],"type":"text","angle":0,"index":3,"merge_prev":false,"text":"(4) 令 $x = a \\cos t, y = b \\sin t$ ，则 $\\int_{L} xy \\, \\mathrm{d}s = ab \\int_{0}^{\\frac{\\pi}{2}} \\sin \\theta \\cos \\theta \\sqrt{a^2 \\sin^2 \\theta + b^2 \\cos^2 \\theta} \\, \\mathrm{d}\\theta$ ，可化简为： ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":138,"id":"ddadb7bc-0cca-4235-bffb-7ccf96ead176","page_size":[595,841],"block_position":"138-2"},{"bbox":[135,147,458,174],"type":"interline_equation","angle":0,"index":4,"text":"$$\n{\\frac {a b}{2 (a ^ {2} - b ^ {2})}} \\int_ {0} ^ {\\frac {\\pi}{2}} \\sqrt {(a ^ {2} - b ^ {2}) \\sin^ {2} \\theta + b ^ {2}} \\mathrm{d} [ (a ^ {2} - b ^ {2}) \\sin^ {2} \\theta ] = {\\frac {a b (a ^ {2} + a b + b ^ {2})}{3 (a + b)}}\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":138,"id":"d09362bd-6255-4b20-8a9a-bb05a3dcb381","page_size":[595,841],"block_position":"138-3"},{"bbox":[35,183,404,208],"type":"text","angle":0,"index":5,"merge_prev":false,"text":"练习16.2折线形式(1)计算 $\\int_{L}y\\mathrm{d}s$ ，这里 $L$ 为 $y^{2} = 4x$ 从(0,0)到(1,2)的一段 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":138,"id":"413d4f63-00d6-4d26-a469-17eee274c7e3","page_size":[595,841],"block_position":"138-4"},{"bbox":[71,208,402,232],"type":"text","angle":0,"index":6,"merge_prev":false,"text":"(2) 计算 $\\int_{L} (x + y) \\, \\mathrm{d}s$ ，这里 $L$ 是以 $O(0,0), A(1,0), B(0,1)$ 为顶点的三角形 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":138,"id":"9d66a9a7-500b-4c19-95e7-56a8063b4541","page_size":[595,841],"block_position":"138-5"},{"bbox":[52,232,374,259],"type":"text","angle":0,"index":7,"merge_prev":false,"text":"解 $(1)x = \\frac{1}{4} y^2, y \\in [0,2]$ ，因此 $\\int_{L} y \\, \\mathrm{d}s = \\int_{0}^{2} y\\sqrt{1 + \\frac{1}{4} y^2} \\, \\mathrm{d}y = \\frac{4}{3}(2\\sqrt{2} - 1)$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":138,"id":"91bd0a6d-8dda-4c0e-91ad-3e5ea8a0cd13","page_size":[595,841],"block_position":"138-6"},{"bbox":[52,259,539,311],"type":"text","angle":0,"index":8,"merge_prev":false,"text":"(2) 分三段进行考虑。 $\\int_{OA} (x + y) \\mathrm{d}s = \\int_{0}^{1} (x) \\mathrm{d}x = \\frac{1}{2}$ , $\\int_{OB} (x + y) \\mathrm{d}s = \\frac{1}{2}$ 。 $AB$ 上 $y = 1 - x, x \\in [0,1]$ , 因此 $\\int_{AB} (x + y) \\mathrm{d}s = \\int_{0}^{1} 1\\sqrt{1 + 1} \\mathrm{d}x = \\sqrt{2}$ , 结果为 $1 + \\sqrt{2}$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":138,"id":"826445a4-2c6a-4c55-963c-ea507e9dcc48","page_size":[595,841],"block_position":"138-7"},{"bbox":[56,326,157,340],"type":"title","angle":0,"index":9,"level":2,"text":"## 16.2.2 使用对称性","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":138,"id":"0b9c26d1-19e2-423a-ac48-1c891087289a","page_size":[595,841],"block_position":"138-8"},{"bbox":[35,348,405,375],"type":"text","angle":0,"index":10,"merge_prev":false,"text":"练习 16.3 镜面对称性 (1) 计算 $\\int_{L} xy ds$ ，其中 L 为上半椭圆 $\\frac{x^{2}}{a^{2}} + \\frac{y^{2}}{b^{2}} = 1 (y \\geq 0)$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":138,"id":"a3fe2d90-1df3-42be-ac71-34801b891d7e","page_size":[595,841],"block_position":"138-9"},{"bbox":[73,376,369,396],"type":"text","angle":0,"index":11,"merge_prev":false,"text":"(2)重要：计算 $\\int_{L}|y|\\mathrm{d}s$ ，其中 $L$ 为双组线 $(x^{2} + y^{2})^{2} = a^{2}(x^{2} - y^{2})$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":138,"id":"d63ddc66-a024-4bbf-9eac-9800fe3cd101","page_size":[595,841],"block_position":"138-10"},{"bbox":[73,399,299,423],"type":"text","angle":0,"index":12,"merge_prev":false,"text":"(3) 计算 $\\oint_{L} |x|^{\\frac{1}{3}} \\mathrm{d}s$ ，这里 $L$ 为星形线 $x^{\\frac{2}{3}} + y^{\\frac{2}{3}} = a^{\\frac{2}{3}}$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":138,"id":"6faec91d-2afa-459d-ab63-e62e765e5bd0","page_size":[595,841],"block_position":"138-11"},{"bbox":[52,422,358,446],"type":"text","angle":0,"index":13,"merge_prev":false,"text":"解(1)由于 $L$ 关于 $y$ 轴对称，而 $xy$ 关于 $y$ 轴对称，因此 $\\int_{L} xy \\, \\mathrm{d}s = 0$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":138,"id":"6eac76f2-5676-453b-82a1-73d6324185db","page_size":[595,841],"block_position":"138-12"},{"bbox":[53,446,539,487],"type":"text","angle":0,"index":14,"merge_prev":false,"text":"(2) 显然 $L$ 关于 $x, y$ 轴均对称，因此将 $L$ 分为四段，设 $L_{1}$ 是第一象限的部分，因此 $\\int_{L} |y| \\mathrm{~d}s = 4 \\int_{L_{1}} y \\mathrm{~d}s$ 考虑 $x = r \\cos \\theta, y = r \\sin \\theta$ ，则 $L$ 极坐标方程为 $r^{2} = a^{2} \\cos 2\\theta$ ， $L_{1}$ 方程为 $r = a \\sqrt{\\cos 2\\theta}$ ，且 $\\theta \\in [0, \\frac{\\pi}{4}]$ ，即： ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":138,"id":"5060899d-5af1-4771-9b46-5e9640310f16","page_size":[595,841],"block_position":"138-13"},{"bbox":[98,493,494,524],"type":"interline_equation","angle":0,"index":15,"text":"$$\n4 \\int_ {L _ {1}} y \\mathrm{d} s = 4 \\int_ {0} ^ {\\frac {\\pi}{4}} a \\sqrt {\\cos 2 \\theta} \\sin \\theta \\sqrt {a ^ {2} \\cos 2 \\theta + a ^ {2} \\frac {\\sin^ {2} 2 \\theta}{\\cos 2 \\theta}} \\mathrm{d} \\theta = 4 a ^ {2} \\int_ {0} ^ {\\frac {\\pi}{4}} \\sin \\theta \\mathrm{d} \\theta = (4 - 2 \\sqrt {2}) a ^ {2}\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":138,"id":"0e53532c-c621-44cc-8820-22d54de7e60b","page_size":[595,841],"block_position":"138-14"},{"bbox":[52,528,537,565],"type":"text","angle":0,"index":16,"merge_prev":false,"text":"(3) 显然关于 $x, y$ 轴对称，因此 $\\int_{L} |x|^{\\frac{1}{3}} \\mathrm{d}s = 4 \\int_{L_1} x^{\\frac{1}{3}} \\mathrm{d}s$ 。设 $x = a \\cos^3 \\theta, y = a \\sin^3 \\theta, \\theta \\in [0, \\frac{\\pi}{2}]$ ，此时 $x'^2 y'^2 = 9a^2 \\sin^2 \\cos^2 \\theta$ ，结果为： ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":138,"id":"0c8cfcdc-dfdb-468a-b074-af567f961dee","page_size":[595,841],"block_position":"138-15"},{"bbox":[192,568,401,596],"type":"interline_equation","angle":0,"index":17,"text":"$$\nI = 4 \\int_ {L _ {1}} x ^ {\\frac {1}{3}} \\mathrm{d} s = 1 2 a ^ {\\frac {4}{3}} \\int_ {0} ^ {\\frac {\\pi}{2}} \\cos^ {2} \\theta \\sin \\theta \\mathrm{d} \\theta = 4 a ^ {\\frac {4}{3}}\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":138,"id":"fa3471c3-c012-4c93-9c59-73ce25fa41e4","page_size":[595,841],"block_position":"138-16"},{"bbox":[35,604,514,629],"type":"text","angle":0,"index":18,"merge_prev":false,"text":"练习 16.4 轮换对称性 (1) L 是 $x + y + z = 0, x^{2} + y^{2} + z^{2} = 1$ 的交线，计算 $\\int_{L} x ds, \\int_{L} x^{2} ds, \\int_{L} y dz ds$ 的值 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":138,"id":"f07ae412-6fa1-4800-bc38-223bb25c4bb0","page_size":[595,841],"block_position":"138-17"},{"bbox":[72,629,368,652],"type":"text","angle":0,"index":19,"merge_prev":false,"text":"(2)L是 $x + y + z = 0,x^{2} + y^{2} + z^{2} = 1$ 的交线，计算 $\\int_{L}x^{2} + yz\\mathrm{d}s$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":138,"id":"cf8c0459-83ec-4c28-9760-9c39681457b9","page_size":[595,841],"block_position":"138-18"},{"bbox":[52,652,413,665],"type":"text","angle":0,"index":20,"merge_prev":false,"text":"解(1)考虑轮换对称性，查看交线，发现 $x, y, z$ 轴都等价，因此变量可以随意轮换 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":138,"id":"e844339e-4988-4600-8b64-4b84f29c9c4e","page_size":[595,841],"block_position":"138-19"},{"bbox":[186,667,397,714],"type":"interline_equation","angle":0,"index":21,"text":"$$\n\\left\\{ \\begin{array}{l} \\int_ {L} x \\mathrm{d} s = \\frac {1}{3} \\int_ {L} x + y + z \\mathrm{d} s = 0 \\\\ \\int_ {L} x ^ {2} \\mathrm{d} s = \\frac {1}{3} \\int_ {L} x ^ {2} + y ^ {2} + z ^ {2} \\mathrm{d} s = \\frac {1}{3} \\cdot 2 \\pi = \\frac {2 \\pi}{3} \\end{array} \\right.\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":138,"id":"21e7db50-29a8-4f4a-9e41-5a2f5ec4d87b","page_size":[595,841],"block_position":"138-20"},{"bbox":[54,717,539,731],"type":"text","angle":0,"index":22,"merge_prev":false,"text":"第三个：由于 $(x + y + z) = 0$ ，因此 $(x + y + z)^2 = 0$ ，得到 $(x + y + z)^2 -(x^2 +y^2 +z^2) = 2(xy + yz + xz) = -1,$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":138,"id":"58ee5fb0-1b13-4f61-b232-3154846ef79e","page_size":[595,841],"block_position":"138-21"},{"bbox":[289,803,304,812],"type":"page_number","angle":0,"index":999999,"text":"131 ","color":{"line":"rgba(164, 164, 164, 1)","fill":"rgba(164, 164, 164, 1)"},"page_idx":138,"is_discarded":true,"id":"38899026-db9d-49b7-81be-98ac5f6cbacd","page_size":[595,841],"block_position":"138-22"}],[{"bbox":[445,46,538,58],"type":"header","angle":0,"index":-99999,"text":"16.3 第二型曲线积分 ","color":{"line":"rgba(164, 164, 164, 1)","fill":"rgba(164, 164, 164, 1)"},"page_idx":139,"is_discarded":true,"id":"8e4f30f5-67fc-46c9-bfab-f91507c700a5","page_size":[595,841],"block_position":"139-0"},{"bbox":[54,68,189,93],"type":"text","angle":0,"index":1,"merge_prev":false,"text":"即 $xy + yz + xz = -\\frac{1}{2}$ ，因此： ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":139,"id":"ce601c88-f1a8-439e-a05d-9f300737f444","page_size":[595,841],"block_position":"139-1"},{"bbox":[181,94,412,121],"type":"interline_equation","angle":0,"index":2,"text":"$$\n\\int_ {L} y z \\mathrm{d} s = \\frac {1}{3} \\int_ {L} x y + y z + x z \\mathrm{d} s = \\frac {1}{3} \\cdot - \\frac {1}{6} \\cdot 2 \\pi = - \\frac {\\pi}{3}\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":139,"id":"8b6adbc3-17b0-4dc0-b89e-1f8646cef3fb","page_size":[595,841],"block_position":"139-2"},{"bbox":[73,121,171,134],"type":"text","angle":0,"index":3,"merge_prev":false,"text":"(2) 和第一题本质一样 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":139,"id":"c3b48e90-4b1b-4a31-9634-039da229c705","page_size":[595,841],"block_position":"139-3"},{"bbox":[57,153,193,167],"type":"title","angle":0,"index":4,"level":2,"text":"## 16.2.3 曲线积分中值问题","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":139,"id":"9a2ec0d5-a4e0-4b8c-84c8-53460e40d26d","page_size":[595,841],"block_position":"139-4"},{"bbox":[72,180,206,193],"type":"title","angle":0,"index":5,"level":2,"text":"## 定理 16.2 (曲线积分中值定理)","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":139,"id":"1544c2c1-8d11-459c-919a-aadfcdeaad7c","page_size":[595,841],"block_position":"139-5"},{"bbox":[67,196,524,225],"type":"text","angle":0,"index":6,"merge_prev":false,"text":"$f(x,y)$ 是光滑曲线 $L:x = x(t),y = y(t),t\\in [a,b]$ 上的连续函数，设 $\\Delta L$ 是曲线 $L$ 的弧长，则 $\\exists (x_0,y_0)\\in L,$ 使得 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":139,"id":"63dc8917-e5ff-49c5-ba2f-8d9ed298984b","page_size":[595,841],"block_position":"139-6"},{"bbox":[235,222,358,249],"type":"interline_equation","angle":0,"index":7,"text":"$$\n\\int_ {L} f (x, y) \\mathrm{d} s = f (x _ {0}, y _ {0}) \\Delta L\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":139,"id":"2adcfd7a-7808-4c15-a214-c3e93cf3dff6","page_size":[595,841],"block_position":"139-7"},{"bbox":[53,259,539,314],"type":"text","angle":0,"index":8,"merge_prev":false,"text":"证明 根据计算公式得到 $\\int_{L} f(x, y) \\mathrm{d}s = \\int_{a}^{b} f(x(t), y(t)) \\sqrt{(x'(t))^2 + (y'(t))^2} \\mathrm{d}t$ ，根据积分第一中值定理得到等于 $f(x(t_0), y(t_0)) \\int_{a}^{b} \\sqrt{(x'(t))^2 + (y'(t))^2} \\mathrm{d}t = f(x_0, y_0) \\Delta L$ ，这里 $(x_0, y_0) = (x(t_0), y(t_0))$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":139,"id":"52a8beaa-c915-4c0e-bd08-4e0b74f57152","page_size":[595,841],"block_position":"139-8"},{"bbox":[35,322,539,352],"type":"text","angle":0,"index":9,"merge_prev":false,"text":"练习 16.5 积分第一中值定理的曲线推广 $f(x,y,z)$ , $g(x,y,z)$ 在可求长连续曲线 L: $x = x(t)$ , $y = y(t)$ , $z = z(t)$ , $t \\in [a,b]$ (曲线不一定光滑！)，上连续，且 $g(x,y,z)$ 非负，证明： $\\exists(\\xi,\\eta,\\zeta) \\in L$ 使得 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":139,"id":"3cb56e5c-cb29-4bb7-adbd-32892cab2a56","page_size":[595,841],"block_position":"139-9"},{"bbox":[187,354,406,380],"type":"interline_equation","angle":0,"index":10,"text":"$$\n\\int_ {L} f (x, y, z) g (x, y, z) \\mathrm{d} s = f (\\xi , \\eta , \\zeta) \\int_ {L} g (x, y, z) \\mathrm{d} s\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":139,"id":"baace474-e83d-4131-9f87-e9adb9335271","page_size":[595,841],"block_position":"139-10"},{"bbox":[53,381,539,411],"type":"text","angle":0,"index":11,"merge_prev":false,"text":"证明 由于不一定光滑，因此不能用含参形式计算。由于 $f(x,y,z)$ 连续，设在 $L$ 上 $m \\leq f(x(t), y(t), z(t)) \\leq M$ ，由于 $g$ 非负，因此 $mg(x,y,z) \\leq f(x,y,z)g(x,y,z) \\leq Mg(x,y,z)$ ，两侧积分得到： ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":139,"id":"dfd0e299-1abe-435e-9707-76f565045e49","page_size":[595,841],"block_position":"139-11"},{"bbox":[158,413,434,439],"type":"interline_equation","angle":0,"index":12,"text":"$$\nm \\int_ {L} g (x, y, z) \\mathrm{d} s \\leq \\int_ {L} f (x, y, z) g (x, y, z) \\mathrm{d} s \\leq M \\int_ {L} g (x, y, z) \\mathrm{d} s\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":139,"id":"903d1020-21c2-429e-b7a6-dc4e4b8eeb33","page_size":[595,841],"block_position":"139-12"},{"bbox":[53,441,567,481],"type":"text","angle":0,"index":13,"merge_prev":false,"text":"若 $\\int_{L} g(x, y, z) \\mathrm{d}s = 0$ ，则全部为零， $(\\xi, \\eta, \\zeta)$ 任选。若不为 0，则同除以 $\\int_{L} g(x, y, z) \\mathrm{d}s$ ，得到 $m \\leq \\frac{\\int_{L} f(x, y, z) g(x, y, z) \\mathrm{d}s}{\\int_{L} f(x, y, z) \\mathrm{d}s} \\leq M$ ，根据连续性介值性定理可证。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":139,"id":"d0514a25-6fbf-4294-8997-357ddef44d67","page_size":[595,841],"block_position":"139-13"},{"bbox":[56,502,193,519],"type":"title","angle":0,"index":15,"level":2,"text":"## 16.3 第二型曲线积分","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":139,"id":"42145403-2408-498b-8725-6f49b8ae5c6c","page_size":[595,841],"block_position":"139-14"},{"bbox":[57,535,205,550],"type":"title","angle":0,"index":16,"level":2,"text":"## 16.3.1 基本概念与参数计算","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":139,"id":"86f4e6fb-860a-4497-8460-9b576f8e4798","page_size":[595,841],"block_position":"139-15"},{"bbox":[73,563,196,575],"type":"title","angle":0,"index":17,"level":2,"text":"## 定义 16.2 (第二型曲线积分)","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":139,"id":"2ad5e512-5698-495a-a6d3-c24336183530","page_size":[595,841],"block_position":"139-16"},{"bbox":[67,579,524,608],"type":"text","angle":0,"index":18,"merge_prev":false,"text":"$\\Gamma$ 是 $\\mathbb{R}^3$ 中的有向曲线，其端点为 $A, B \\in \\mathbb{R}^3$ ， $F(x, y, z) = P(x, y, z)\\mathbf{i} + Q(x, y, z)\\mathbf{j} + R(x, y, z)\\mathbf{k}$ ，将 $\\Gamma$ 划分为 $A = \\mathbf{r}_0 \\to \\mathbf{r}_1 \\to \\mathbf{r}_{n-1} \\to \\mathbf{r}_n = B$ ，第二型曲线积分定义为： ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":139,"id":"bed9d7e3-ce19-4f89-8649-dd281fd674ff","page_size":[595,841],"block_position":"139-17"},{"bbox":[117,612,475,637],"type":"interline_equation","angle":0,"index":19,"text":"$$\n\\int_ {\\Gamma} P \\mathrm{d} x + Q \\mathrm{d} y + R \\mathrm{d} z = \\lim _ {| | T | | \\to 0} P (\\xi_ {i}, \\eta_ {i}, \\zeta_ {i}) \\Delta x _ {i} + Q (\\xi_ {i}, \\eta_ {i}, \\zeta_ {i}) \\Delta y _ {i} + R (\\xi_ {i}, \\eta_ {i}, \\zeta_ {i}) \\Delta z _ {i}\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":139,"id":"d185a4b5-688e-4434-b8b2-6c1c861b6e8a","page_size":[595,841],"block_position":"139-18"},{"bbox":[73,657,226,671],"type":"title","angle":0,"index":20,"level":2,"text":"## 定理 16.3 (第二型曲线积分的计算)","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":139,"id":"0b96c7e5-37e9-4390-9c19-32ec2f64f386","page_size":[595,841],"block_position":"139-19"},{"bbox":[68,674,324,687],"type":"text","angle":0,"index":21,"merge_prev":false,"text":"若曲线 $\\Gamma$ 由 $\\mathbf{r}(t) = x(t)\\mathbf{i} + y(t)\\mathbf{j} + z(t)\\mathbf{k}, t \\in [a, b]$ 给出，则 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":139,"id":"97fa7633-8725-4165-a21f-5167da260e4d","page_size":[595,841],"block_position":"139-20"},{"bbox":[174,692,418,719],"type":"interline_equation","angle":0,"index":22,"text":"$$\n\\int_ {\\Gamma} P \\mathrm{d} x + Q \\mathrm{d} y + R \\mathrm{d} z = \\int_ {a} ^ {b} [ P x ^ {\\prime} (t) + Q y ^ {\\prime} (t) + R z ^ {\\prime} (t) ] \\mathrm{d} t\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":139,"id":"684740d4-4a4e-4204-9da4-feaf995d6a29","page_size":[595,841],"block_position":"139-21"},{"bbox":[35,743,539,771],"type":"text","angle":0,"index":23,"merge_prev":false,"text":"练习16.6折线积分(1)计算 $\\int_{L}x\\mathrm{d}y - y\\mathrm{d}x$ ，其中 $L$ 为三角形 $O(0,0),A(1,0),B(1,2)$ 为顶点的三角形逆时针曲线 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":139,"id":"ad95df17-0791-450c-9f49-67077e6cf679","page_size":[595,841],"block_position":"139-22"},{"bbox":[289,802,305,812],"type":"page_number","angle":0,"index":1000013,"text":"132 ","color":{"line":"rgba(164, 164, 164, 1)","fill":"rgba(164, 164, 164, 1)"},"page_idx":139,"is_discarded":true,"id":"06d728cb-f980-46fe-bd2c-a81c76e8e3b1","page_size":[595,841],"block_position":"139-23"}],[{"bbox":[445,47,538,58],"type":"header","angle":0,"index":-99999,"text":"16.3 第二型曲线积分 ","color":{"line":"rgba(164, 164, 164, 1)","fill":"rgba(164, 164, 164, 1)"},"page_idx":140,"is_discarded":true,"id":"e76e0299-ac3f-4152-8855-06adc7ee1304","page_size":[595,841],"block_position":"140-0"},{"bbox":[53,71,539,146],"type":"text","angle":0,"index":1,"merge_prev":false,"text":"(2) 计算 $\\int_{L} x dx + y dy + z dz$ ，其中 L 是 $(1, 1, 1)$ 到 $(2, 3, 4)$ 的直线段\n\n解 (1) 分三段计算： $\\int_{OA} x \\, dx - y \\, dx = 0$ 。 $\\int_{AB} x \\, dy - y \\, dx$ 以 y 为参数，即 $\\int_{0}^{2} 1 \\, dy - y \\cdot 0 = 2$ 。 $\\int_{BO} x \\, dy - y \\, dx$ ，以 x 为参数，y = 2x，即 $\\int_{1}^{0} x \\cdot 2 - 2x \\, dx = 0$ ，最终结果为 2。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":140,"id":"145c4fe5-f4d8-4c20-959f-5d852ba1bf5b","page_size":[595,841],"block_position":"140-1"},{"bbox":[72,146,474,173],"type":"text","angle":0,"index":2,"merge_prev":false,"text":"(2) 以 $x$ 为参数, $y = 2x - 1, z = 3x - 2$ , 因此结果为 $\\int_{1}^{2} x + 2(2x - 1) + 3(3x - 2) \\mathrm{d}x = 13$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":140,"id":"aac3cc92-999f-4b07-99e6-583d37fd9093","page_size":[595,841],"block_position":"140-2"},{"bbox":[35,175,472,202],"type":"text","angle":0,"index":3,"merge_prev":false,"text":"练习 16.7 椭圆、圆周积分 (1) 计算 $\\int_{L}(x^{2}+2xy)\\mathrm{d}y$ ，其中 L 为 $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1(y\\geq0)$ 方向逆时针 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":140,"id":"18f7dedf-de74-4549-b5bb-496830e50d7b","page_size":[595,841],"block_position":"140-3"},{"bbox":[72,200,385,227],"type":"text","angle":0,"index":4,"merge_prev":false,"text":"(2) 计算 $I = \\oint_{L} \\frac{-x \\, \\mathrm{d}x + y \\, \\mathrm{d}y}{x^2 + y^2}$ ，其中 $L$ 为圆周 $x^2 + y^2 = 1$ ，方向逆时针 解 (1) 设 $x = a\\cos \\theta, y = b\\sin \\theta, \\theta \\in [0, \\pi]$ 。则 $\\int_{L} (x^2 + 2xy)\\mathrm{d}y = \\int_{0}^{\\pi} a^2\\cos^3\\theta + 2ab^2\\sin \\theta\\cos^2\\theta\\mathrm{d}\\theta = \\frac{4}{3}ab^2$ (2) 设 $x = \\cos \\theta, y = \\sin \\theta, \\theta \\in [0, 2\\pi]$ ，则 $I = \\int_{0}^{2\\pi} (\\cos \\theta\\sin \\theta + \\sin \\theta\\cos \\theta)\\mathrm{d}\\theta = 0$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":140,"id":"c2930bfe-f966-45ff-bce2-bab7505b7e3f","page_size":[595,841],"block_position":"140-4"},{"bbox":[54,223,512,277],"type":"text","angle":0,"index":5,"merge_prev":false,"lines_deleted":true,"text":"","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":140,"id":"c2930bfe-f966-45ff-bce2-bab7505b7e3f","page_size":[595,841],"block_position":"140-5"},{"bbox":[35,284,539,317],"type":"text","angle":0,"index":6,"merge_prev":false,"text":"练习 16.8 交线积分: 化为单变量 (1) 计算 $I = \\oint_{L} (y - z) \\, \\mathrm{d}x + (z - x) \\, \\mathrm{d}y + (x - y) \\, \\mathrm{d}z$ ，其中 L 为 $x^{2} + y^{2} = 1, x + z = 1$ ，从 z 轴正无穷看为顺时针 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":140,"id":"8121dbe7-7c53-4e2c-bc54-1c81faee9e9e","page_size":[595,841],"block_position":"140-6"},{"bbox":[71,318,493,339],"type":"text","angle":0,"index":7,"merge_prev":false,"text":"(2) 重要：计算 $\\oint_{L} xyz \\, \\mathrm{d}z$ ，其中 $L$ 为 $x^{2} + y^{2} + z^{2} = 1$ 与 $y = z$ 相交的圆，方向为 1,2,7,8 卦限。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":140,"id":"1c436930-c0b2-4ac4-abe0-26f6fdbee2fc","page_size":[595,841],"block_position":"140-7"},{"bbox":[53,339,388,353],"type":"text","angle":0,"index":8,"merge_prev":false,"text":"解 (1) 设 $x = \\cos \\theta, y = \\sin \\theta, z = 1 - \\cos \\theta$ ，则 $\\theta \\in [2\\pi, 0]$ (不是 $[0, 2\\pi]!$ 。)。则 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":140,"id":"f0ed0d79-3ef8-481a-af1b-46d01c28a7a5","page_size":[595,841],"block_position":"140-8"},{"bbox":[112,356,480,383],"type":"interline_equation","angle":0,"index":9,"text":"$$\nI = \\int_ {2 \\pi} ^ {0} [ (\\sin \\theta + \\cos \\theta - 1) (- \\sin \\theta) + (1 - 2 \\cos \\theta) \\cos \\theta + (\\cos \\theta - \\sin \\theta) \\sin \\theta ] \\mathrm{d} \\theta = 4 \\pi\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":140,"id":"373ecda7-f49e-45d6-998f-bbf5d0a4c100","page_size":[595,841],"block_position":"140-9"},{"bbox":[52,386,538,421],"type":"text","angle":0,"index":10,"merge_prev":false,"text":"(2) 由于 $x^{2} + 2y^{2} = 1$ ，设 $x = \\cos \\theta, y = z = \\frac{1}{\\sqrt{2}} \\sin \\theta$ ，由于从第一卦限出发， $x, y, z$ 均为正，而 $\\theta$ 为小正数时，位于第一卦限，因此 $\\theta \\in [0, 2\\pi]$ 。因此： ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":140,"id":"bcd99ba0-a70a-44a7-8df8-a06569faa1d7","page_size":[595,841],"block_position":"140-10"},{"bbox":[205,423,387,451],"type":"interline_equation","angle":0,"index":11,"text":"$$\nI = \\int_ {0} ^ {2 \\pi} \\frac {1}{2} \\cos \\theta \\sin^ {2} \\theta \\frac {1}{\\sqrt {2}} \\cos \\theta \\mathrm{d} \\theta = \\frac {\\sqrt {2} \\pi}{1 6}\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":140,"id":"ab82a5db-467b-44b7-8be8-74ea63428544","page_size":[595,841],"block_position":"140-11"},{"bbox":[35,456,473,470],"type":"text","angle":0,"index":12,"merge_prev":false,"text":"笔记 第二题虽然是球面的一部分，但是曲线积分都要化成一个变量的，因此不能用球坐标变换！ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":140,"id":"0d798ff2-e82a-4cae-bf0e-ea6f180d22b8","page_size":[595,841],"block_position":"140-12"},{"bbox":[35,483,398,509],"type":"text","angle":0,"index":13,"merge_prev":false,"text":"练习16.9二次曲面交线计算 $\\int_{L}y\\mathrm{d}x + z\\mathrm{d}y + x\\mathrm{d}z$ ，其中 $L$ 为椭球与柱面交线： ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":140,"id":"e2adc207-3d53-40d6-a144-4b5ed9dced22","page_size":[595,841],"block_position":"140-13"},{"bbox":[205,511,388,537],"type":"interline_equation","angle":0,"index":14,"text":"$$\n\\frac {x ^ {2}}{a ^ {2}} + \\frac {y ^ {2}}{b ^ {2}} + \\frac {z ^ {2}}{c ^ {2}} = 1, \\frac {x}{a} + \\frac {z}{c} = 1, x, y, z \\geq 0\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":140,"id":"454c0bc3-6889-47fa-a529-674113d35f2c","page_size":[595,841],"block_position":"140-14"},{"bbox":[53,537,167,550],"type":"text","angle":0,"index":15,"merge_prev":false,"text":"方向从 $(a,0,0)$ 到 $(0,0,c)$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":140,"id":"7e5b4686-4bb0-4244-8290-728949dfad62","page_size":[595,841],"block_position":"140-15"},{"bbox":[52,552,539,628],"type":"text","angle":0,"index":16,"merge_prev":false,"text":"解 (1) 直接用极坐标极度不方便，先令 $x = a\\overline{x}, y = b\\overline{y}, z = c\\overline{z}$ ，得到 $\\overline{x}^2 + \\overline{y}^2 + \\overline{z}^2 = 1, \\overline{x} + \\overline{z} = 1$ 。将前者配凑为 $(\\overline{x} + \\overline{z})^2 + (\\overline{x} - \\overline{z})^2 + 2\\overline{y}^2 = 2$ ，因此 $(\\overline{x} - \\overline{z})^2 + 2\\overline{y}^2 = 1$ ，因此设 $\\overline{x} - \\overline{z} = \\cos \\theta, \\sqrt{2}\\overline{y} = \\sin \\theta$ ，得出 $\\overline{x} = \\frac{1}{2}(1 + \\cos \\theta), \\overline{y} = \\frac{1}{\\sqrt{2}}\\sin \\theta, \\overline{z} = \\frac{1}{2}(1 - \\cos \\theta)$ ，根据方向以及端点可判断 $\\theta \\in [0, \\pi]$ （具体含义难以判断）。因此代入计算可知结果为 $\\frac{ac}{2} - \\frac{\\pi b}{4\\sqrt{2}}(a + c)$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":140,"id":"9516890b-5bec-45fe-8a3c-fa8a0ccae174","page_size":[595,841],"block_position":"140-16"},{"bbox":[57,645,157,659],"type":"title","angle":0,"index":18,"level":2,"text":"## 16.3.2 使用对称性","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":140,"id":"08a1368b-e274-4cea-9478-8911d7585693","page_size":[595,841],"block_position":"140-17"},{"bbox":[72,668,337,681],"type":"text","angle":0,"index":19,"merge_prev":false,"text":"和第一型曲线积分一样，下面考虑镜像对称以及轮换对称性 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":140,"id":"1106dec9-47d8-42d7-b3c2-656a078c72d7","page_size":[595,841],"block_position":"140-18"},{"bbox":[69,682,399,702],"type":"text","angle":0,"index":20,"merge_prev":false,"text":"- 镜像对称：如 $\\int_{S} f(x, y, z) \\mathrm{d}x$ ，要同时考虑 $f(x, y, z)$ 和 $\\mathrm{d}x$ 在 $S$ 上的符号 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":140,"id":"8c8f3564-5906-496b-a435-3c03597cd0f8","page_size":[595,841],"block_position":"140-19"},{"bbox":[68,703,539,731],"type":"text","angle":0,"index":21,"merge_prev":false,"text":"- 轮换对称：要保证积分区域和积分表达式均轮换对称。轮换有两种使用方法：(1) 保持积分区域不变，积分表达式轮换 (2) 保持积分表达式不变，积分区域轮换 (具体见轮换对称练习) ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":140,"id":"1b644cb0-c7ee-4a10-bf27-5b2e53a9c46b","page_size":[595,841],"block_position":"140-20"},{"bbox":[289,802,305,812],"type":"page_number","angle":0,"index":1000016,"text":"133 ","color":{"line":"rgba(164, 164, 164, 1)","fill":"rgba(164, 164, 164, 1)"},"page_idx":140,"is_discarded":true,"id":"a1e58968-6ca6-4861-b30a-6115365a2898","page_size":[595,841],"block_position":"140-21"}],[{"bbox":[445,47,537,58],"type":"header","angle":0,"index":-99999,"text":"16.3 第二型曲线积分 ","color":{"line":"rgba(164, 164, 164, 1)","fill":"rgba(164, 164, 164, 1)"},"page_idx":141,"is_discarded":true,"id":"bff5b438-805e-4a0d-af2e-7a5092794f77","page_size":[595,841],"block_position":"141-0"},{"bbox":[73,74,170,86],"type":"title","angle":0,"index":1,"level":2,"text":"## 定理 16.4 (dx 的符号)","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":141,"id":"678d9df6-95a8-4578-91b1-9627071097ba","page_size":[595,841],"block_position":"141-1"},{"bbox":[68,89,526,119],"type":"text","angle":0,"index":2,"merge_prev":false,"text":"若沿着给定方向， $x$ 由大变小，则 $\\mathrm{dx}$ 为负，同理若变大则 $\\mathrm{dx}$ 为正，以此判断对称性。如果不容易直接判断，则画出投影曲线，再在平面曲线上判断方向。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":141,"id":"1b1bf05c-7851-4dee-924d-3b6e3075e37c","page_size":[595,841],"block_position":"141-2"},{"bbox":[35,143,489,169],"type":"text","angle":0,"index":3,"merge_prev":false,"text":"练习16.10镜像对称性(1)证明 $I = \\oint_{L}\\frac{-x\\mathrm{d}x + y\\mathrm{d}y}{x^{2} + y^{2}} = 0$ ，其中 $L$ 为圆周 $x^{2} + y^{2} = 1$ ，方向逆时针。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":141,"id":"61991dba-1ce5-4054-9ad7-24779dd3895e","page_size":[595,841],"block_position":"141-3"},{"bbox":[52,170,538,238],"type":"text","angle":0,"index":4,"merge_prev":false,"text":"(2) 计算 $I = \\int_{L} (-2xe^{-x^2} \\sin y) \\, \\mathrm{d}x + (e^{-x^2} \\cos y + x^4) \\, \\mathrm{d}y$ ，其中 $L$ 为 (1,0) 到 (-1,0) 的半圆 $y = \\sqrt{1 - x^2}$ 解 (1) $I = -\\int_{L} x \\, \\mathrm{d}x + \\int_{L} y \\, \\mathrm{d}y$ ，前者分为 $x$ 轴上下两部分，上下 $x$ 符号大小均相等，由于上方 $x$ 从大变小，下方从小变大，因此 $\\mathrm{d}x$ 符号相反，得到 $-\\int_{L} x \\, \\mathrm{d}x = 0$ 。后者同理根据 $x$ 轴上下两部分得到结果为 0。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":141,"id":"959eaf01-8ab1-4201-906f-a9dc4e83b1ea","page_size":[595,841],"block_position":"141-4"},{"bbox":[53,240,539,269],"type":"text","angle":0,"index":5,"merge_prev":false,"text":"(2) 关于 $y$ 轴对称。第一项 $\\mathrm{d}x$ 同号， $-2xe^{-x^2}\\sin y$ 异号，因此第一个积分为0。第二个 $\\mathrm{d}y$ 异号， $e^{-x^2}\\cos y + x^4$ 同号等值，因此第二个积分也为0。综上整个积分为0。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":141,"id":"17be24c4-62c9-4504-b243-afd3110adfe9","page_size":[595,841],"block_position":"141-5"},{"bbox":[35,285,540,436],"type":"text","angle":0,"index":6,"merge_prev":false,"text":"练习16.11轮换对称性(1)计算 $\\int_{L}x\\mathrm{d}x + y\\mathrm{d}y + z\\mathrm{d}z$ ，其中 $L$ 为 $x^{2} + y^{2} + z^{2} = 1$ 与 $x + y + z = 0$ 相交圆周， $L$ 从第一卦限来看是逆时针的。(2)计算 $\\int_{L}(y^{2} - z^{2})\\mathrm{d}x + (z^{2} - x^{2})\\mathrm{d}y + (x^{2} - y^{2})\\mathrm{d}z$ ，其中 $L$ 为 $x^{2} + y^{2} + z^{2} = 1$ 在第一卦限的边界曲线，方向为 $xy$ 平面， $yz$ 平面， $zx$ 平面解(1)由于 $x,y,z$ 可以随便互换， $I$ 可以写出 $3\\int_{L}x\\mathrm{d}x$ 或 $y,z$ 对应积分，要选择一个好算的。考虑 $z = -(x + y)$ ，则 $x^{2} + y^{2} + (x + y)^{2} = 1$ ，得到 $(x + \\frac{1}{2} y)^{2} + \\frac{3}{4} y^{2} = \\frac{1}{2}$ 。因此令 $x + \\frac{1}{2} y = \\frac{1}{\\sqrt{2}}\\cos \\theta, \\frac{\\sqrt{3}}{2} y = \\frac{1}{\\sqrt{2}}\\sin \\theta$ ，因此得到参数方程为： $x = \\frac{1}{\\sqrt{2}}\\cos \\theta - \\frac{1}{\\sqrt{6}}\\sin \\theta, y = \\frac{2}{\\sqrt{6}}\\sin \\theta, z = -\\frac{1}{\\sqrt{2}}\\cos \\theta - \\frac{1}{\\sqrt{6}}\\sin \\theta, \\theta \\in [0,2\\pi]$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":141,"id":"1acd1ebb-d9b2-4fdf-ac1b-44e1cede28a4","page_size":[595,841],"block_position":"141-6"},{"bbox":[54,440,253,466],"type":"text","angle":0,"index":7,"merge_prev":false,"text":"可以算 $\\int_{L} y \\, \\mathrm{d}y$ 比较简单，因此最终结果为 0。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":141,"id":"a3855b79-bf55-4aa6-a270-ddc45bc3da3d","page_size":[595,841],"block_position":"141-7"},{"bbox":[52,465,540,534],"type":"text","angle":0,"index":8,"merge_prev":false,"text":"(2) 设三个部分曲线分别为 $L_{1}, L_{2}, L_{3}$ , 由于 $x, y, z$ 可以随便互换, 若 $(x, y, z) \\to (z, x, y)$ , 则积分表达式不变, 而 $L_{1}$ 部分在该变换下变为 $L_{2}$ , 因此 $L_{1}, L_{2}, L_{3}$ 积分值相等。因此 $I = 3 \\int_{L_{1}} (y^{2} - z^{2}) \\mathrm{d}x + (z^{2} - x^{2}) \\mathrm{d}y + (x^{2} - y^{2}) \\mathrm{d}z =$ , 对于 $L_{1}$ 而言, 虽然积分表达式轮换对称, 但是积分区域不轮换对称, 因此只能硬算了。取 $x = \\cos \\theta, y = \\sin \\theta, z = 0, \\theta \\in [0, \\frac{\\pi}{2}]$ , 得到结果为 $-4$ 。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":141,"id":"b927b6cf-4916-44dd-b08d-8bc131e6d31c","page_size":[595,841],"block_position":"141-8"},{"bbox":[35,543,539,663],"type":"text","angle":0,"index":9,"merge_prev":false,"text":"练习 16.12 双曲面交线 + 对称性 重点：计算 $I=\\int_{L}y^{2}dx+z^{2}dy+x^{2}dz$ ，其中 L 是维维安尼曲线 $x^{2}+y^{2}+z^{2}=a^{2}, x^{2}+y^{2}=ax (z\\geq0, a>0)$ ，从 x 轴 $+\\infty$ 看去，L 方向为逆时针\n\n解 注意 $x^{2}+y^{2}=ax$ 即 $(x-\\frac{a}{2})^{2}+y^{2}=\\frac{a^{2}}{4}$ 是圆柱面，且过原点。考虑关于 xz 平面对称，积分表达式显然不变，dx 相反，dy 相同（需要把投影的圆画出来看看），dz 相反，因此只需要计算 $I=\\int_{L}z^{2}dy=2\\int_{L_{1}}z^{2}dy$ ，这里 $L_{1}$ 是 L 在 $y\\geq0$ 的部分，设 $x=a\\sin\\varphi\\cos\\theta, y=a\\sin\\varphi\\sin\\theta, z=a\\cos\\varphi, \\theta\\in[0, \\frac{\\pi}{2}] (\\theta$ 的范围具体见三重积分球坐标变换），代入柱面公式得到 $\\sin\\varphi=\\cos\\theta$ ，即 $\\theta=\\frac{\\pi}{2}-\\varphi$ ，综上 $L_{1}$ 的参数方程为： ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":141,"id":"315bc5ee-0086-496d-bed7-ce4035c5c5fe","page_size":[595,841],"block_position":"141-9"},{"bbox":[183,667,409,687],"type":"interline_equation","angle":0,"index":11,"text":"$$\nx = a \\sin^ {2} \\varphi , y = a \\sin \\varphi \\cos \\varphi , z = a \\cos \\varphi , \\varphi \\in [ \\frac {\\pi}{2}, 0 ]\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":141,"id":"3215e39a-bb25-4b78-944e-b8688b4b5117","page_size":[595,841],"block_position":"141-10"},{"bbox":[55,689,155,710],"type":"text","angle":0,"index":12,"merge_prev":false,"text":"得到最终结果为 $-\\frac{\\pi}{4} a^3$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":141,"id":"0437794a-8219-4d41-8cf3-ee092e106a2f","page_size":[595,841],"block_position":"141-11"},{"bbox":[289,803,305,811],"type":"page_number","angle":0,"index":1000009,"text":"134 ","color":{"line":"rgba(164, 164, 164, 1)","fill":"rgba(164, 164, 164, 1)"},"page_idx":141,"is_discarded":true,"id":"2665d9c1-250a-40de-bc6d-9b2226f6d332","page_size":[595,841],"block_position":"141-12"}],[{"bbox":[468,47,539,58],"type":"header","angle":0,"index":-99999,"text":"16.4 Green 公式 ","color":{"line":"rgba(164, 164, 164, 1)","fill":"rgba(164, 164, 164, 1)"},"page_idx":142,"is_discarded":true,"id":"e699c242-7e95-46f6-abe4-12cd73913b43","page_size":[595,841],"block_position":"142-0"},{"bbox":[57,72,205,86],"type":"title","angle":0,"index":1,"level":2,"text":"## 16.3.3 两类曲线积分的关系","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":142,"id":"4c545544-c589-45ee-bb81-bbb5f151d59a","page_size":[595,841],"block_position":"142-1"},{"bbox":[73,100,196,112],"type":"title","angle":0,"index":2,"level":2,"text":"## 定理 16.5 (光滑曲线切向量)","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":142,"id":"2b21f833-0ff7-49f0-b22d-97034789215d","page_size":[595,841],"block_position":"142-2"},{"bbox":[68,116,524,147],"type":"text","angle":0,"index":3,"merge_prev":false,"text":"考虑光滑曲线 $\\mathbf{r}(s) = (x(s), y(s)), s \\in [0, l]$ ，满足 $(x'(s))^2 + (y'(x))^2 = 1$ 。则其单位切向量为 $\\mathbf{t} = (x'(s), y'(s)) = (\\cos(\\mathbf{t}, x), \\cos(\\mathbf{t}, y))$ ，即 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":142,"id":"792d8b66-8c02-448e-b440-49542c0da616","page_size":[595,841],"block_position":"142-3"},{"bbox":[217,153,376,166],"type":"interline_equation","angle":0,"index":4,"text":"$$\n\\mathrm{d} x = \\cos (\\mathbf {t}, x) \\mathrm{d} s, \\quad \\mathrm{d} y = \\cos (\\mathbf {t}, y) \\mathrm{d} s\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":142,"id":"cbd0ab58-007c-40b7-90c1-864f92f68c82","page_size":[595,841],"block_position":"142-4"},{"bbox":[72,186,216,199],"type":"title","angle":0,"index":5,"level":2,"text":"## 定理 16.6 (两类曲线积分的关系)","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":142,"id":"2ad77abf-03a1-4033-aec7-b88521b296b1","page_size":[595,841],"block_position":"142-5"},{"bbox":[68,203,226,215],"type":"text","angle":0,"index":6,"merge_prev":false,"text":"根据光滑曲线切向量的结论，可知： ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":142,"id":"22a10489-51c3-45b6-b3c0-5182de78b6c9","page_size":[595,841],"block_position":"142-6"},{"bbox":[166,216,427,243],"type":"interline_equation","angle":0,"index":7,"text":"$$\n\\int_ {L} P \\mathrm{d} x + Q \\mathrm{d} y = \\int_ {L} \\left[ P (x, y) \\cos (\\mathbf {t}, x) + Q (x, y) \\cos (\\mathbf {t}, y) \\right] \\mathrm{d} s\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":142,"id":"babc9681-faa0-439b-b826-5652a8e15aef","page_size":[595,841],"block_position":"142-7"},{"bbox":[57,280,164,296],"type":"title","angle":0,"index":8,"level":2,"text":"## 16.4 Green 公式","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":142,"id":"8c97e548-f0b9-4dd1-8218-e65d1b9cf170","page_size":[595,841],"block_position":"142-8"},{"bbox":[57,312,226,326],"type":"title","angle":0,"index":9,"level":2,"text":"## 16.4.1 Green 公式及其基本应用","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":142,"id":"33517858-0e88-4849-9aab-7233efdb2343","page_size":[595,841],"block_position":"142-9"},{"bbox":[73,339,176,352],"type":"title","angle":0,"index":10,"level":2,"text":"## 定理 16.7 (Green 公式)","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":142,"id":"35489c21-2b48-44ef-adac-0696a0c47194","page_size":[595,841],"block_position":"142-10"},{"bbox":[67,356,524,391],"type":"text","angle":0,"index":11,"merge_prev":false,"text":"$\\Omega \\subset \\mathbb{R}^2$ 是有限条分段光滑曲线连接成的闭曲线， $P(x,y), Q(x,y)$ 在 $\\Omega$ 上连续，且有连续偏导 $\\frac{\\partial Q}{\\partial x}, \\frac{\\partial P}{\\partial y}$ ，沿着 $\\partial \\Omega$ 的方向， $\\Omega$ 一直在左侧，则 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":142,"id":"8413f76d-7eb2-4cb1-bb12-41f8f38725cc","page_size":[595,841],"block_position":"142-11"},{"bbox":[209,393,384,420],"type":"interline_equation","angle":0,"index":12,"text":"$$\n\\int_ {\\partial \\Omega} P \\mathrm{d} x + Q \\mathrm{d} y = \\iint_ {\\Omega} (\\frac {\\partial Q}{\\partial x} - \\frac {\\partial P}{\\partial y}) \\mathrm{d} x \\mathrm{d} y\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":142,"id":"2b63965a-44e4-47d9-800a-ea7e8549c206","page_size":[595,841],"block_position":"142-12"},{"bbox":[52,430,539,465],"type":"text","angle":0,"index":13,"merge_prev":false,"text":"笔记 使用 Green 公式的基本条件是曲线光滑和偏导存在，一般在 P, Q 本身复杂，但 $\\frac{\\partial Q}{\\partial x} - \\frac{\\partial P}{\\partial y}$ 简单时使用。记忆时可记忆为： ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":142,"id":"dd39b7e2-d208-47b8-a5c2-e6d1f38a899f","page_size":[595,841],"block_position":"142-13"},{"bbox":[267,462,325,502],"type":"interline_equation","angle":0,"index":14,"text":"$$\n\\left| \\begin{array}{c c} \\frac {\\partial}{\\partial x} & \\frac {\\partial}{\\partial y} \\\\ P & Q \\end{array} \\right|\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":142,"id":"e2a14896-6f4e-400a-a4d9-2075a3cdc153","page_size":[595,841],"block_position":"142-14"},{"bbox":[35,517,491,530],"type":"text","angle":0,"index":15,"merge_prev":false,"text":"练习 16.13Green 公式基本应用 (1)a, b 为正数，L 为从 (2a, 0) 到 (0, 0) 的半圆 $y = \\sqrt{2ax - x^{2}}$ ，计算 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":142,"id":"46715e12-df9b-4664-975d-c43f6df947ad","page_size":[595,841],"block_position":"142-15"},{"bbox":[188,534,404,558],"type":"interline_equation","angle":0,"index":16,"text":"$$\nI = \\int_ {L} [ e ^ {x} \\sin y - b (x + y) ] \\mathrm{d} x + (e ^ {x} \\cos y - a x) \\mathrm{d} y\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":142,"id":"602ebefd-b964-4fff-b4fe-d0fad1cbb023","page_size":[595,841],"block_position":"142-16"},{"bbox":[72,560,272,573],"type":"text","angle":0,"index":17,"merge_prev":false,"text":"(2)L 为 $\\sin x$ 从 $(0,0)$ 到 $(\\pi,0)$ 的一段，计算： ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":142,"id":"2c6ac3fc-4889-42f4-a53c-196619a58b24","page_size":[595,841],"block_position":"142-17"},{"bbox":[179,576,414,601],"type":"interline_equation","angle":0,"index":18,"text":"$$\nI = \\int_ {L} \\sqrt {x ^ {2} + y ^ {2}} \\mathrm{d} x + y \\left[ x y + \\ln \\left(x + \\sqrt {x ^ {2} + y ^ {2}}\\right) \\right] \\mathrm{d} y\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":142,"id":"1c543d2b-7219-4d69-b0a5-141e13b9b6f1","page_size":[595,841],"block_position":"142-18"},{"bbox":[73,603,307,617],"type":"text","angle":0,"index":19,"merge_prev":false,"text":"(3) $L$ 为曲线 $y = \\sin x$ 从 $(0,0)$ 到 $(\\pi ,0)$ 的一段，计算 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":142,"id":"427a07bb-59be-42c4-9708-fad8fd30c443","page_size":[595,841],"block_position":"142-19"},{"bbox":[208,619,384,645],"type":"interline_equation","angle":0,"index":20,"text":"$$\nI = \\int_ {L} e ^ {x} \\left[ (1 - \\cos y) \\mathrm{d} x - (y - \\sin y) \\mathrm{d} y \\right]\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":142,"id":"21a2c0de-2c33-4fc1-8f93-f66a8c0bdd5a","page_size":[595,841],"block_position":"142-20"},{"bbox":[53,645,539,705],"type":"text","angle":0,"index":21,"merge_prev":false,"text":"解(1)显然积分本身很麻烦，考虑 $P = e^{x}\\sin y - b(x + y),Q = e^{x}\\cos y - ax$ ，则 $Q_{x} - P_{y} = b - a$ 很方便，因此考虑用Green公式把圆周补上，记 $L_{1}$ 为 $(0,0)\\rightarrow (2a,0)$ 直线。则 $I = \\int_{L + L_1} - \\int_{L_1} = \\iint_Db - a\\mathrm{d}x\\mathrm{d}y - \\int_{L_1} - bx\\mathrm{d}x =$ $\\frac{1}{2}\\pi a^2 (b - a) + 2a^2 b$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":142,"id":"2251f460-241e-48c8-a6d3-6563280b22f0","page_size":[595,841],"block_position":"142-21"},{"bbox":[72,704,539,738],"type":"text","angle":0,"index":23,"merge_prev":false,"text":"(2) 考虑 $Q_x - P_y = y^2 + \\frac{y + \\frac{xy}{\\sqrt{x^2 + y^2}}}{x + \\sqrt{x^2 + y^2}} - \\frac{y}{\\sqrt{x^2 + y^2}} = y^2$ ，记 $L_1$ 是 $(\\pi, 0)$ 到 $(0, 0)$ 的直线段，则 $\\int_L P \\, \\mathrm{d}x + Q \\, \\mathrm{d}y =$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":142,"id":"9c79363d-d6eb-494f-b446-fe383ce2aff9","page_size":[595,841],"block_position":"142-22"},{"bbox":[289,802,305,812],"type":"page_number","angle":0,"index":1000021,"text":"135 ","color":{"line":"rgba(164, 164, 164, 1)","fill":"rgba(164, 164, 164, 1)"},"page_idx":142,"is_discarded":true,"id":"79d44be1-aa8e-474a-b8ed-d8ceef781079","page_size":[595,841],"block_position":"142-23"}],[{"bbox":[468,47,539,58],"type":"header","angle":0,"index":-99998,"text":"16.4 Green 公式 ","color":{"line":"rgba(164, 164, 164, 1)","fill":"rgba(164, 164, 164, 1)"},"page_idx":143,"is_discarded":true,"id":"bbbce4be-4a59-4fdd-8a0a-19a58c1ff2b5","page_size":[595,841],"block_position":"143-0"},{"bbox":[526,515,536,525],"type":"text","angle":0,"index":0,"text":"# ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":143,"id":"da336295-b514-4dcf-af81-29e22fb4ae3b","page_size":[595,841],"block_position":"143-1"},{"bbox":[54,70,164,96],"type":"text","angle":0,"index":2,"merge_prev":false,"text":"$\\int_{\\pi}^{0}x\\mathrm{d}x = -\\frac{1}{2}\\pi^2$ ，因此： ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":143,"id":"241527ae-880a-4506-8030-006bc432d7b6","page_size":[595,841],"block_position":"143-2"},{"bbox":[116,100,477,127],"type":"interline_equation","angle":0,"index":3,"text":"$$\nI = \\int_ {L + L _ {1}} - \\int_ {L _ {1}} = - \\iint_ {D} y ^ {2} \\mathrm{d} x \\mathrm{d} y + \\frac {1}{2} \\pi^ {2} = - \\int_ {0} ^ {\\pi} \\mathrm{d} x \\int_ {0} ^ {\\sin x} y ^ {2} \\mathrm{d} y + \\frac {1}{2} \\pi^ {2} = \\frac {1}{2} \\pi^ {2} - \\frac {4}{9}\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":143,"id":"d1a30226-e53f-44af-ab39-37803f2a195e","page_size":[595,841],"block_position":"143-3"},{"bbox":[53,131,539,182],"type":"text","angle":0,"index":4,"merge_prev":false,"text":"(3) 首先 $Q_x - P_y = -ye^x$ ，记 $L_1$ 为 $(\\pi, 0) \\to (0, 0)$ 的直线，最终结果为 $\\frac{1}{5} (e^\\pi - 1)$ 。中间 $\\frac{1}{2} \\int_0^\\pi e^x \\sin^2 x \\, \\mathrm{d}x = \\frac{1}{4} \\int_0^\\pi e^x (1 - \\cos 2x) \\, \\mathrm{d}x = \\frac{1}{4} e^x \\bigg|_0^\\pi - \\frac{1}{4} \\frac{e^x}{4 + 1} (\\cos 2x + 2\\sin 2x) \\bigg|_0^\\pi$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":143,"id":"fc74258e-ef09-40ef-a99e-26616a9928f7","page_size":[595,841],"block_position":"143-4"},{"bbox":[35,186,539,238],"type":"text","angle":0,"index":5,"merge_prev":false,"text":"练习 16.14 一道经典题 (1) 计算 $I = \\oint_{L} \\frac{x \\, dy - y \\, dx}{x^{2} + y^{2}}$ ，其中 L 为任一不包含原点区域的边界 (包含原点情况在后面讨论)。\n解(1) 结果为 0 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":143,"id":"a671afd0-331d-4d3a-a202-d0a9b9faec6e","page_size":[595,841],"block_position":"143-5"},{"bbox":[72,261,236,274],"type":"title","angle":0,"index":6,"level":2,"text":"## 定理 16.8 (区域面积转换为曲线积分)","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":143,"id":"5bd78373-a4d6-4e4a-bc47-2e37291610c8","page_size":[595,841],"block_position":"143-6"},{"bbox":[69,277,169,289],"type":"text","angle":0,"index":7,"merge_prev":false,"text":"平面区域 $D$ 的面积为： ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":143,"id":"eeb7dd64-f87c-4a26-9e00-08d388e100df","page_size":[595,841],"block_position":"143-7"},{"bbox":[226,286,367,313],"type":"interline_equation","angle":0,"index":8,"text":"$$\nS _ {D} = \\iint_ {D} \\mathrm{d} S = \\frac {1}{2} \\oint_ {L} x \\mathrm{d} y - y \\mathrm{d} x\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":143,"id":"e126f893-8ff9-460e-b9e7-6d2084d51344","page_size":[595,841],"block_position":"143-8"},{"bbox":[35,344,395,378],"type":"text","angle":0,"index":9,"merge_prev":false,"text":"练习 16.15 曲面面积应用 计算抛物线 $(x+y)^{2}=ax(a>0)$ 与 x 轴围成的面积解结果： $\\frac{1}{6}a^{2}$ ，见华东师大课本 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":143,"id":"2dce890e-032f-4bd1-8b4c-29d33e1b893f","page_size":[595,841],"block_position":"143-9"},{"bbox":[57,395,156,409],"type":"title","angle":0,"index":10,"level":2,"text":"## 16.4.2 内部有瑕点","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":143,"id":"11adc8b5-2e60-4ed5-b855-04bd17829607","page_size":[595,841],"block_position":"143-10"},{"bbox":[73,423,198,434],"type":"title","angle":0,"index":11,"level":2,"text":"## 定理 16.9 (瑕点 Green 公式)","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":143,"id":"ef47867b-07a2-4c56-af89-83e5bd3eaf30","page_size":[595,841],"block_position":"143-11"},{"bbox":[68,439,524,478],"type":"text","angle":0,"index":12,"merge_prev":false,"text":"若 $P(x,y), Q(x,y)$ 除瑕点外满足 $Q_x - P_y = 0$ ，边界 $L$ 围成的区域 $D$ 包含瑕点 $x_0$ ，则可作包含 $x_0$ 的小区域 $D_1$ ，其边界 $L_1$ ，方向逆时针，根据 $\\int_L - \\int_{L_1} = \\iint_{D - D_1} = 0$ 得出： ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":143,"id":"4f7bf088-8711-4b40-bdab-af053255d6be","page_size":[595,841],"block_position":"143-12"},{"bbox":[223,480,370,506],"type":"interline_equation","angle":0,"index":13,"text":"$$\n\\int_ {L} P \\mathrm{d} x + Q \\mathrm{d} y = \\int_ {L _ {1}} P \\mathrm{d} x + Q \\mathrm{d} y\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":143,"id":"63aa6c71-5c81-4296-8176-597466189720","page_size":[595,841],"block_position":"143-13"},{"bbox":[68,508,292,521],"type":"text","angle":0,"index":14,"merge_prev":false,"text":"保证 $L_{1}$ 曲线表达式的特殊性即可快速计算出结果。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":143,"id":"7b08a98d-f181-4dff-b78c-9b7da0ff9e5b","page_size":[595,841],"block_position":"143-14"},{"bbox":[35,533,539,644],"type":"text","angle":0,"index":15,"merge_prev":false,"text":"练习 16.16 最经典题目 重点：L 为简单封闭光滑曲线，且不经过原点，a, b > 0，计算 $I = \\oint_{L} \\frac{x \\, dy - y \\, dx}{ax^{2} + by^{2}}$ 。\n\n解 显然 P, Q 在不包含原点的区域上有连续偏导数， $P_{y} = \\frac{-(ax^{2} + by^{2}) + y \\cdot 2by}{(ax^{2} + by^{2})^{2}}$ ， $Q_{x} = \\frac{by^{2} - ax^{2}}{(ax^{2} + by^{2})^{2}}$ ，若 L 围成的区域不包含原点，则根据 $Q_{x} - P_{y} = 0$ 得到积分为 0。若 D 包含原点，考虑挖去原点的小区域，其边界 $L_{1}$ 为 $ax^{2} + by^{2} = r^{2}$ ，方向逆时针。则 $\\int_{L} P \\, dx + Q \\, dy = \\int_{L_{1}} P \\, dx + Q \\, dy = \\frac{1}{r^{2}} \\int_{L_{1}} x \\, dy - y \\, dx$ ，再次使用 Green 公式得到结果为 $\\frac{1}{r^{2}} \\iint_{D_{1}} 2 \\, dx \\, dy = \\frac{2\\pi}{\\sqrt{ab}}$ 。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":143,"id":"d5b5ff8f-f3dd-4be2-be24-485e1b8c94b3","page_size":[595,841],"block_position":"143-15"},{"bbox":[73,652,174,665],"type":"title","angle":0,"index":16,"level":2,"text":"## 推论 16.2 (常用函数 1)","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":143,"id":"8adcb927-eeff-4fa5-b33e-cb7cc40b0af6","page_size":[595,841],"block_position":"143-16"},{"bbox":[70,669,143,682],"type":"text","angle":0,"index":17,"merge_prev":false,"text":"若 $P_0(x_0,y_0)$ 则： ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":143,"id":"96dc0009-aeb9-4da7-90d7-110950e55852","page_size":[595,841],"block_position":"143-17"},{"bbox":[157,683,433,728],"type":"interline_equation","angle":0,"index":18,"text":"$$\n\\oint_ {L} \\frac {(x - x _ {0}) \\mathrm{d} y - (y - y _ {0}) \\mathrm{d} x}{a (x - x _ {0}) ^ {2} + b (y - y _ {0}) ^ {2}} = \\left\\{ \\begin{array}{l l} \\frac {2 \\pi}{\\sqrt {a b}}, & L \\text {围成的区域包含} P _ {0} \\\\ 0, & L \\text {围成的区域不包含} P _ {0} \\end{array} \\right.\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":143,"id":"d18fe46a-4396-4b15-aa95-227a5b64b548","page_size":[595,841],"block_position":"143-18"},{"bbox":[289,802,305,812],"type":"page_number","angle":0,"index":1000018,"text":"136 ","color":{"line":"rgba(164, 164, 164, 1)","fill":"rgba(164, 164, 164, 1)"},"page_idx":143,"is_discarded":true,"id":"03822835-0d81-4801-afed-09b53eb23d21","page_size":[595,841],"block_position":"143-19"}],[{"bbox":[468,47,539,58],"type":"header","angle":0,"index":-99999,"text":"16.4 Green 公式 ","color":{"line":"rgba(164, 164, 164, 1)","fill":"rgba(164, 164, 164, 1)"},"page_idx":144,"is_discarded":true,"id":"0f2124ac-e786-4be5-8098-5b31f02e19f9","page_size":[595,841],"block_position":"144-0"},{"bbox":[36,69,539,148],"type":"text","angle":0,"index":1,"merge_prev":false,"text":"练习16.17推广题1计算 $I = \\oint_{L}\\frac{(x - y)\\mathrm{d}x + (x + 4y)\\mathrm{d}y}{x^2 + 4y^2}$ ，其中 $L$ 为 $x^{2} + y^{2} = 1$ ，方向逆时针解 $P_{y} = Q_{x} = \\frac{-x^{2} - 8xy + 4y^{2}}{(x^{2} + 4y^{2})^{2}}$ ，记 $L_{1}$ 为 $x^{2} + 4y^{2} = r^{2}$ ，则 $I = \\int_{L_1} P\\mathrm{d}x + Q\\mathrm{d}y = \\frac{1}{r^2}\\int_{L_1}(x - y)\\mathrm{d}x + (x + 4y)\\mathrm{d}y,$ 再用Green公式得到 $I = \\frac{1}{r^2}\\iint_{D_1}2\\mathrm{d}x\\mathrm{d}y = \\frac{2}{r^2}\\cdot \\frac{\\pi r^2}{2} = \\pi$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":144,"id":"54f48046-fb4c-4e0d-9561-1ec933f27cf0","page_size":[595,841],"block_position":"144-1"},{"bbox":[74,156,173,169],"type":"title","angle":0,"index":2,"level":2,"text":"## 推论 16.3 (常用函数 2)","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":144,"id":"d1fd6893-bd4c-4e58-99c3-5872b0d65918","page_size":[595,841],"block_position":"144-2"},{"bbox":[68,174,232,198],"type":"text","angle":0,"index":3,"merge_prev":false,"text":"$c_{1},c_{2}>0$ 时， $a_{2}=-b_{1},\\frac{a_{1}}{c_{1}}=\\frac{b_{2}}{c_{2}}$ 有： ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":144,"id":"247ef48f-ea6d-4cfb-a5c9-09e59bf45ad0","page_size":[595,841],"block_position":"144-3"},{"bbox":[139,199,451,243],"type":"interline_equation","angle":0,"index":4,"text":"$$\n\\oint_ {L} \\frac {\\left(a _ {1} x + a _ {2} y\\right) \\mathrm{d} x + \\left(b _ {1} x + b _ {2} y\\right) \\mathrm{d} y}{c _ {1} x ^ {2} + c _ {2} y ^ {2}} = \\left\\{ \\begin{array}{l l} \\frac {2 b _ {1} \\pi}{\\sqrt {c _ {1} c _ {2}}}, & L \\text {围成区域包含原点} \\\\ 0, & L \\text {围成区域不包含原点} \\end{array} \\right.\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":144,"id":"96045315-d10f-4120-bba6-cdc2bab9bb3e","page_size":[595,841],"block_position":"144-4"},{"bbox":[74,280,186,294],"type":"title","angle":0,"index":5,"level":2,"text":"## 引理 16.1 (推广椭圆面积)","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":144,"id":"2b3e54ed-810a-470a-932f-b04b724f4c57","page_size":[595,841],"block_position":"144-5"},{"bbox":[68,296,487,323],"type":"text","angle":0,"index":6,"merge_prev":false,"text":"设 $a, c > 0, ac - b^2 > 0$ ，考虑 $ax^2 + 2bxy + cy^2 = 1$ ，则该曲线为椭圆，围成的面积为 $\\frac{\\pi}{\\sqrt{ac - b^2}}$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":144,"id":"006f8ade-f522-425f-931e-d120eeaf0b75","page_size":[595,841],"block_position":"144-6"},{"bbox":[37,329,537,380],"type":"text","angle":0,"index":7,"merge_prev":false,"text":"证明写为 $a(x + \\frac{b}{a} y)^2 +(c - \\frac{b^2}{a})y^2 = 1$ ，等价于 $\\frac{x^2}{1 / a} +\\frac{y^2}{a / (ac - b^2)} = 1$ ，根据椭圆面积公式 $(\\pi \\cdot ab)$ 得到 $\\frac{\\pi}{\\sqrt{ac - b^2}}$ 练习16.18推广题2重点：计算 $I = \\oint_{L}\\frac{x\\mathrm{d}y - y\\mathrm{d}x}{x^{2} + xy + y^{2}}$ ， $L$ 是圆周 $x^{2} + y^{2} = 1$ ，逆时针方向。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":144,"id":"1c2a8b55-a57c-4714-8b9a-dd935db5aa84","page_size":[595,841],"block_position":"144-7"},{"bbox":[50,379,539,433],"type":"text","angle":0,"index":8,"merge_prev":false,"text":"解分母即 $(x + \\frac{1}{2} y)^2 +\\frac{3}{4} y^2 = 0$ ，若分母为0，则 $x = y = 0$ 。 $P_{y} = Q_{x} = \\frac{y^{2} - x^{2}}{(x^{2} + xy + y^{2})^{2}}$ ，取 $L_{1}$ 为 $x^{2} + xy + y^{2} = r^{2}$ 该图形为椭圆，面积为 $\\frac{2\\pi r^2}{\\sqrt{3}}$ ，使用Green公式得到结果为 $\\frac{4\\pi}{\\sqrt{3}}$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":144,"id":"c4ecd219-68f4-4514-be63-b86c5a68c840","page_size":[595,841],"block_position":"144-8"},{"bbox":[73,440,173,453],"type":"title","angle":0,"index":9,"level":2,"text":"## 推论 16.4 (常用函数 3)","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":144,"id":"d994ac3c-0945-41cf-94b5-87c697c5d160","page_size":[595,841],"block_position":"144-9"},{"bbox":[69,457,236,470],"type":"text","angle":0,"index":11,"merge_prev":false,"text":"若 $a, b, c$ 满足 $a, c > 0, ac - b^2 > 0$ ，则 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":144,"id":"899991e8-b591-48ff-8e21-3bb92670b5f1","page_size":[595,841],"block_position":"144-10"},{"bbox":[223,472,370,498],"type":"interline_equation","angle":0,"index":12,"text":"$$\n\\oint_ {L} \\frac {x \\mathrm{d} y - y \\mathrm{d} x}{a x ^ {2} + 2 b x y + c y ^ {2}} = \\frac {2 \\pi}{\\sqrt {a c - b ^ {2}}}\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":144,"id":"aef2089d-8252-413e-989a-243cc1c4d4f9","page_size":[595,841],"block_position":"144-11"},{"bbox":[35,530,332,543],"type":"text","angle":0,"index":13,"merge_prev":false,"text":"练习 16.19 其他形式 L 为不过原点的简单光滑封闭曲线，计算： ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":144,"id":"af02069f-a28f-483c-8ea3-7aa6ad0987bd","page_size":[595,841],"block_position":"144-12"},{"bbox":[166,544,427,570],"type":"interline_equation","angle":0,"index":14,"text":"$$\nI = \\oint_ {L} {\\frac {e ^ {y}}{x ^ {2} + y ^ {2}}} \\left[ (x \\sin x + y \\cos x) \\mathrm{d} x + (y \\sin x - x \\cos x) \\mathrm{d} y \\right]\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":144,"id":"c3712250-ad48-4dfd-accd-0e940f3c2f7e","page_size":[595,841],"block_position":"144-13"},{"bbox":[57,585,216,600],"type":"title","angle":0,"index":15,"level":2,"text":"## 16.4.3 曲线积分与路径无关性","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":144,"id":"38aa6184-c397-4f8e-b9eb-e2dd9df9aae4","page_size":[595,841],"block_position":"144-14"},{"bbox":[73,613,180,625],"type":"title","angle":0,"index":16,"level":2,"text":"## 定理 16.10 (路径无关性)","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":144,"id":"80cd5101-6499-4426-beff-d71a770aefae","page_size":[595,841],"block_position":"144-15"},{"bbox":[68,631,524,666],"type":"text","angle":0,"index":17,"merge_prev":false,"text":"D 为单连通区域, 若 $P(x,y)$ , $Q(x,y)$ 在 D 连续且有一阶连续偏导, L 为 D 内任意按段光滑曲线则 $\\int_{L} P \\, dx + Q \\, dy$ 与路径无关当且仅当: ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":144,"id":"65216271-b1f9-496b-b3a9-546d30d7953e","page_size":[595,841],"block_position":"144-16"},{"bbox":[83,669,187,682],"type":"text","angle":0,"index":18,"merge_prev":false,"text":"- $D$ 内处处有 $P_{y} = Q_{x}$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":144,"id":"121d0f85-5d32-490c-b53e-5ab8107d556a","page_size":[595,841],"block_position":"144-17"},{"bbox":[83,681,331,704],"type":"text","angle":0,"index":19,"merge_prev":false,"text":"- $D$ 内任意按段光滑闭合曲线 $L$ , 有 $\\oint_{L} P \\mathrm{~d} x + Q \\mathrm{~d} y = 0$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":144,"id":"8abdfb16-dbb9-40e4-89c4-0a9413ea789a","page_size":[595,841],"block_position":"144-18"},{"bbox":[83,704,299,717],"type":"text","angle":0,"index":20,"merge_prev":false,"text":"- $D$ 内存在原函数 $u(x,y)$ 使得 $\\mathrm{du} = P\\mathrm{dx} + Q\\mathrm{dy}$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":144,"id":"63ef22b9-e89f-45aa-8617-3447c308b237","page_size":[595,841],"block_position":"144-19"},{"bbox":[289,802,305,812],"type":"page_number","angle":0,"index":1000009,"text":"137 ","color":{"line":"rgba(164, 164, 164, 1)","fill":"rgba(164, 164, 164, 1)"},"page_idx":144,"is_discarded":true,"id":"9643acab-4024-4815-9709-faecaf87bdd0","page_size":[595,841],"block_position":"144-20"}],[{"bbox":[232,74,372,94],"type":"title","angle":0,"index":0,"level":2,"text":"## 第 17 章 曲面积分","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":145,"id":"9e14effa-074f-45fd-8917-80ea1e6bd578","page_size":[595,841],"block_position":"145-0"},{"bbox":[57,122,166,139],"type":"title","angle":0,"index":1,"level":2,"text":"## 17.1 曲面的面积","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":145,"id":"d3f55b14-f66c-4e66-8b3e-c9b7b00d7b5e","page_size":[595,841],"block_position":"145-1"},{"bbox":[73,157,186,169],"type":"title","angle":0,"index":2,"level":2,"text":"## 定理 17.1 (曲面面积公式)","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":145,"id":"2bb3e539-effc-4146-b6f9-31385871dde6","page_size":[595,841],"block_position":"145-2"},{"bbox":[68,173,525,200],"type":"text","angle":0,"index":3,"merge_prev":false,"text":"$D$ 是可求面积平面有界区域， $z = f(x,y)$ 在 $D$ 上有连续一阶偏导，则曲面 $z = f(x,y),(x,y)\\in D$ 的面积为： ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":145,"id":"22564156-2db3-4c99-a150-a73d2677b45e","page_size":[595,841],"block_position":"145-3"},{"bbox":[230,200,363,225],"type":"interline_equation","angle":0,"index":4,"text":"$$\n\\Delta S = \\iint_ {D} \\sqrt {1 + f _ {x} ^ {2} + f _ {y} ^ {2}} \\mathrm{d} x \\mathrm{d} y\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":145,"id":"c1ad13af-17c3-4fae-8e7f-dba0c9b28250","page_size":[595,841],"block_position":"145-4"},{"bbox":[73,245,205,258],"type":"title","angle":0,"index":5,"level":2,"text":"## 定理 17.2 (参数方程曲面面积)","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":145,"id":"a4f8f6a1-8ce3-465a-b0c9-88c3fc923571","page_size":[595,841],"block_position":"145-5"},{"bbox":[67,262,524,291],"type":"text","angle":0,"index":6,"merge_prev":false,"text":"$D$ 为可求面积平面有界区域， $x = x(u,v),y = y(u,v),z = z(u,v)$ 在 $D$ 上有连续一阶偏导，则曲面 $\\mathbf{r}(u,v) = (x(u,v),y(u,v),z(u,v))$ 的面积公式为： ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":145,"id":"158ca8e7-5fe4-4d70-870b-f34ad2fe903d","page_size":[595,841],"block_position":"145-6"},{"bbox":[170,295,411,350],"type":"interline_equation","angle":0,"index":7,"text":"$$\n\\Delta S = \\iint_ {D} \\sqrt {E G - F ^ {2}} \\mathrm{d} u \\mathrm{d} v, \\left\\{ \\begin{array}{l} E = x _ {u} ^ {2} + y _ {u} ^ {2} + z _ {u} ^ {2} \\\\ F = x _ {u} x _ {v} + y _ {u} y _ {v} + z _ {u} z _ {v} \\\\ G = x _ {v} ^ {2} + y _ {v} ^ {2} + z _ {v} ^ {2} \\end{array} \\right.\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":145,"id":"3776257e-cd5f-49d4-96bd-283ff328606e","page_size":[595,841],"block_position":"145-7"},{"bbox":[36,383,539,427],"type":"text","angle":0,"index":8,"merge_prev":false,"text":"练习17.1曲面面积计算(1)证明 $x^{2} + y^{2} + z^{2} = R^{2}$ 的表面积为 $4\\pi R^2$ 解(1)考虑上半平面 $z = \\sqrt{R^2 - x^2 - y^2}$ ， $D$ 为 $x^{2} + y^{2} \\leq R^{2}$ ，则 $z_x = -\\frac{x}{z}, z_y = -\\frac{y}{z}$ ，因此 $\\sqrt{1 + z_x^2 + z_y^2} = \\frac{R}{z}$ 。得到球面面积为 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":145,"id":"4439c3ae-4031-4db5-8ade-5bb6909f09de","page_size":[595,841],"block_position":"145-8"},{"bbox":[129,428,463,459],"type":"interline_equation","angle":0,"index":9,"text":"$$\nS = 2 \\iint_ {D} \\frac {R}{z} \\mathrm{d} x \\mathrm{d} y = 2 \\int_ {0} ^ {R} \\mathrm{d} r \\int_ {0} ^ {2 \\pi} \\frac {R}{\\sqrt {R ^ {2} - r ^ {2}}} r \\mathrm{d} \\theta = 4 \\pi R \\sqrt {R ^ {2} - r ^ {2}} \\bigg | _ {R} ^ {0} = 4 \\pi R ^ {2}\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":145,"id":"b8884c7c-a3e3-4cca-bac1-3d31bf4c78de","page_size":[595,841],"block_position":"145-9"},{"bbox":[33,462,475,483],"type":"text","angle":0,"index":10,"merge_prev":false,"text":"笔记 对于 $x^{2} + y^{2} + z^{2} = a^{2}$ ，有 $\\sqrt{1 + z_x^2 + z_y^2} = \\frac{a}{z}$ ，这个结论在计算球面曲面积分时经常用到！ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":145,"id":"64c8abaa-9c27-4a0d-a63a-73fa286e720b","page_size":[595,841],"block_position":"145-10"},{"bbox":[57,501,193,518],"type":"title","angle":0,"index":11,"level":2,"text":"## 17.2 第一型曲面积分","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":145,"id":"9c87d7cf-3adc-4a7d-8502-5b8fb9b2619d","page_size":[595,841],"block_position":"145-11"},{"bbox":[57,534,204,548],"type":"title","angle":0,"index":12,"level":2,"text":"## 17.2.1 基本概念与计算公式","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":145,"id":"9ae46453-2af7-462e-8e80-deb476d4c6bf","page_size":[595,841],"block_position":"145-12"},{"bbox":[73,560,196,573],"type":"title","angle":0,"index":13,"level":2,"text":"## 定义 17.1 (第一型曲面积分)","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":145,"id":"8d202188-c3df-4fcb-b3aa-6d48e4d53eda","page_size":[595,841],"block_position":"145-13"},{"bbox":[68,577,451,591],"type":"text","angle":0,"index":14,"merge_prev":false,"text":"$S$ 为光滑曲面片，对 $S$ 做分割 $S = \\{\\Delta S_1, \\dots, \\Delta S_n\\}$ ，记 $\\| \\Delta \\| = \\max \\{s(\\Delta S_i)\\}$ ，则定义 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":145,"id":"13bd09af-1062-4efd-be3d-120f2525bf52","page_size":[595,841],"block_position":"145-14"},{"bbox":[199,595,394,624],"type":"interline_equation","angle":0,"index":15,"text":"$$\n\\iint_ {S} f (x, y, z) \\mathrm{d} S = \\lim _ {| | \\Delta | | \\rightarrow 0} \\sum_ {i = 1} ^ {n} f (x _ {i}, y _ {i}, z _ {i}) \\Delta S _ {i}\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":145,"id":"38b6b5fa-bfe8-40ac-a2c4-a5b1fa66b1fc","page_size":[595,841],"block_position":"145-15"},{"bbox":[73,645,155,659],"type":"title","angle":0,"index":16,"level":2,"text":"## 定理 17.3 (投影法)","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":145,"id":"95b2f673-1252-4829-86d4-4f895d0063f4","page_size":[595,841],"block_position":"145-16"},{"bbox":[69,662,251,676],"type":"text","angle":0,"index":17,"merge_prev":false,"text":"若 $S$ 可表示为 $z = z(x,y),(x,y)\\in D$ ，则 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":145,"id":"7e5ce225-c945-44de-9fab-bcbc598b177f","page_size":[595,841],"block_position":"145-17"},{"bbox":[158,681,435,710],"type":"interline_equation","angle":0,"index":18,"text":"$$\n\\iint_ {S} f (x, y, z) \\mathrm{d} S = \\iint_ {D} f (x, y, z (x, y)) \\sqrt {1 + (\\frac {\\partial z}{\\partial x}) ^ {2} + (\\frac {\\partial z}{\\partial y}) ^ {2}} \\mathrm{d} x \\mathrm{d} y\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":145,"id":"179917d8-ec71-49ff-bc95-2cc4cce62156","page_size":[595,841],"block_position":"145-18"}],[{"bbox":[445,46,538,58],"type":"header","angle":0,"index":-99998,"text":"17.2 第一型曲面积分 ","color":{"line":"rgba(164, 164, 164, 1)","fill":"rgba(164, 164, 164, 1)"},"page_idx":146,"is_discarded":true,"id":"68df4cb6-b8d0-4d37-82a2-43ada3cb8fc8","page_size":[595,841],"block_position":"146-0"},{"bbox":[73,73,195,85],"type":"title","angle":0,"index":2,"level":2,"text":"## 定理 17.4 (球坐标变换结论)","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":146,"id":"0cccb83b-10cd-4db2-9337-9538df35b76b","page_size":[595,841],"block_position":"146-1"},{"bbox":[68,90,336,105],"type":"text","angle":0,"index":3,"merge_prev":false,"text":"对于球面 $x^{2} + y^{2} + z^{2} = a^{2}$ ，令 $z = \\sqrt{a^2 - x^2 - y^2}$ ，则得到 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":146,"id":"25d8df71-dd02-44d6-b988-6f53c346eff8","page_size":[595,841],"block_position":"146-2"},{"bbox":[255,108,337,130],"type":"interline_equation","angle":0,"index":4,"text":"$$\n\\sqrt {1 + z _ {x} ^ {2} + z _ {y} ^ {2}} = \\frac {a}{z}\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":146,"id":"d31dcbca-d21b-4695-9186-436ae8b9f20e","page_size":[595,841],"block_position":"146-3"},{"bbox":[54,145,152,158],"type":"text","angle":0,"index":5,"merge_prev":false,"text":"证明 直接计算可验证 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":146,"id":"b33a4fc1-1e6a-4d22-93f4-142e220f2e87","page_size":[595,841],"block_position":"146-4"},{"bbox":[52,161,383,173],"type":"text","angle":0,"index":6,"merge_prev":false,"text":"笔记 球面可以使用投影法或极坐标，本质上两者都可以用，而且相差不大。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":146,"id":"97d0658a-d2d5-42a3-8c2d-28aea3edaeaf","page_size":[595,841],"block_position":"146-5"},{"bbox":[35,185,468,211],"type":"text","angle":0,"index":7,"merge_prev":false,"text":"练习17.2投影法(1)计算 $\\iint_{S} \\frac{\\mathrm{d} S}{z}$ ，其中 $S$ 是球面 $x^{2} + y^{2} + z^{2} = a^{2}$ 被平面 $z = h$ 所截的顶部。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":146,"id":"74c09d14-93ec-4537-bc2e-3c48c5d81617","page_size":[595,841],"block_position":"146-6"},{"bbox":[73,212,439,234],"type":"text","angle":0,"index":8,"merge_prev":false,"text":"(2) 计算 $A = \\int_{S} (x + y + z) \\, \\mathrm{d}S$ ，这里 $S$ 表示 $x^{2} + y^{2} + z^{2} = a^{2} (z \\geq 0)$ ，即球上表面 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":146,"id":"c0a73c65-f1ca-47d1-9832-4da37a38e1ef","page_size":[595,841],"block_position":"146-7"},{"bbox":[52,233,511,259],"type":"text","angle":0,"index":9,"merge_prev":false,"text":"解(1)做投影法 $z = \\sqrt{a^2 - x^2 - y^2}$ ，积分变为 $\\iint_{D} \\frac{a}{z^2} \\, \\mathrm{d}x \\, \\mathrm{d}y = \\iint_{D} \\frac{a}{a^2 - x^2 - y^2} \\, \\mathrm{d}x \\, \\mathrm{d}y$ ，做极坐标变换得到 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":146,"id":"21e023ea-7c2b-4431-b282-a03173c0c83a","page_size":[595,841],"block_position":"146-8"},{"bbox":[208,262,385,290],"type":"interline_equation","angle":0,"index":10,"text":"$$\n\\int_ {0} ^ {2 \\pi} \\mathrm{d} \\theta \\int_ {0} ^ {\\sqrt {a ^ {2} - h ^ {2}}} \\frac {a}{a ^ {2} - r ^ {2}} r \\mathrm{d} r = 2 \\pi a \\ln \\frac {a}{h}\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":146,"id":"873717c6-49fa-49b4-a247-0523fbe3e20f","page_size":[595,841],"block_position":"146-9"},{"bbox":[73,293,459,318],"type":"text","angle":0,"index":11,"merge_prev":false,"text":"(2) 根据对称性只剩一项，即 $A = \\iint_{S} z \\, \\mathrm{d}S$ ，因此根据 $z = \\sqrt{a^2 - x^2 - y^2}$ 代入计算得到： ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":146,"id":"a1773c01-1c92-4828-acc4-2131402b573c","page_size":[595,841],"block_position":"146-10"},{"bbox":[179,321,412,348],"type":"interline_equation","angle":0,"index":12,"text":"$$\nA = \\iint_ {x ^ {2} + y ^ {2} \\leq a ^ {2}} z \\cdot \\frac {a}{z} \\mathrm{d} x \\mathrm{d} y = \\iint_ {x ^ {2} + y ^ {2} \\leq a ^ {2}} a \\mathrm{d} x \\mathrm{d} y = a ^ {3} \\pi\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":146,"id":"a3bf3688-e93d-4737-b5be-d60415ce0cd5","page_size":[595,841],"block_position":"146-11"},{"bbox":[73,364,176,377],"type":"title","angle":0,"index":13,"level":2,"text":"## 定理 17.5 (参数方程法)","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":146,"id":"99e65c6c-69af-4339-9033-95fbd6736c3d","page_size":[595,841],"block_position":"146-12"},{"bbox":[69,381,336,395],"type":"text","angle":0,"index":14,"merge_prev":false,"text":"若 $S$ 由参数方程 $x(u,v),y(u,v),z(u,v)$ 给出， $(u,v)\\in D$ ，则 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":146,"id":"b2a337ae-1332-4ed6-850e-1b9760941077","page_size":[595,841],"block_position":"146-13"},{"bbox":[74,398,506,454],"type":"interline_equation","angle":0,"index":15,"text":"$$\n\\iint_ {S} f (x, y, z) \\mathrm{d} S = \\iint_ {D} f [ x (u, v), y (u, v), z (u, v) ] \\sqrt {E G - F ^ {2}} \\mathrm{d} u \\mathrm{d} v, \\text {其中:} \\left\\{ \\begin{array}{l l} E = x _ {u} ^ {2} + y _ {u} ^ {2} + z _ {u} ^ {2} \\\\ F = x _ {u} x _ {v} + y _ {u} y _ {v} + z _ {u} z _ {v} \\\\ G = x _ {v} ^ {2} + y _ {v} ^ {2} + z _ {v} ^ {2} \\end{array} \\right.\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":146,"id":"5b2d7d48-be7a-4b26-9054-123eaa5b2c09","page_size":[595,841],"block_position":"146-14"},{"bbox":[73,476,205,488],"type":"title","angle":0,"index":16,"level":2,"text":"## 定理 17.6 (球坐标极坐标变换)","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":146,"id":"f9f765bf-f1ae-4016-a239-c4e4f6cb6b69","page_size":[595,841],"block_position":"146-15"},{"bbox":[68,493,424,507],"type":"text","angle":0,"index":17,"merge_prev":false,"text":"对于球面 $x^{2} + y^{2} + z^{2} = a^{2}$ ，令 $x = a\\sin \\varphi \\cos \\theta ,y = a\\sin \\varphi \\sin \\theta ,z = a\\cos \\varphi$ ，则 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":146,"id":"14380f53-d75b-4d51-b512-19a1dfa9e01e","page_size":[595,841],"block_position":"146-16"},{"bbox":[202,510,393,565],"type":"interline_equation","angle":0,"index":18,"text":"$$\n\\left\\{ \\begin{array}{l l} E = a ^ {2} \\sin^ {2} \\varphi \\\\ F = 0 & \\Rightarrow \\sqrt {E G - F ^ {2}} = a ^ {2} \\sin \\varphi \\\\ G = a ^ {2} \\end{array} \\right.\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":146,"id":"e391d03b-1485-4573-ab51-fca6745b6b15","page_size":[595,841],"block_position":"146-17"},{"bbox":[35,593,537,692],"type":"text","angle":0,"index":19,"merge_prev":false,"text":"练习17.3参数方程法(1)重点：计算 $I = \\iint_{S}\\frac{\\mathrm{d}S}{\\sqrt{x^2 + y^2 + (z + a)^2}}$ ，其中 $S$ 为 $x^{2} + y^{2} + z^{2} = a^{2}(z\\geq 0)$ 解(1)采用球坐标变换，积分变为 $I = \\iint_{S'}\\frac{a^2\\sin\\varphi}{\\sqrt{a^2\\sin^2\\varphi + a^2(\\cos\\varphi + 1)^2}}\\mathrm{d}\\theta \\mathrm{d}\\varphi$ ，区域为 $\\varphi \\in [0,\\frac{\\pi}{2}],\\theta \\in [0,2\\pi ]$ ，进一步化简得到： $I = \\iint_{S'}\\frac{-a\\sin\\varphi}{\\sqrt{2}\\sqrt{1 + \\cos\\varphi}}\\mathrm{d}\\theta \\mathrm{d}\\varphi = -\\sqrt{2}\\pi a\\int_0^{\\frac{\\pi}{2}}\\frac{\\mathrm{d}(\\cos\\varphi)}{\\sqrt{1 + \\cos\\varphi}} = 4\\pi a - 2\\sqrt{2}\\pi a$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":146,"id":"179214d3-7c13-4d62-bbd3-c30b3751b25a","page_size":[595,841],"block_position":"146-18"},{"bbox":[57,708,157,721],"type":"text","angle":0,"index":20,"merge_prev":false,"text":"17.2.2 使用对称性 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":146,"id":"21a712e0-5887-4d86-b71e-bd07e0c957d8","page_size":[595,841],"block_position":"146-19"},{"bbox":[35,729,428,756],"type":"text","angle":0,"index":21,"merge_prev":false,"text":"练习17.4积分曲面对称(1)计算 $\\iint_{S} (x + \\sin yz) \\mathrm{d}S$ ，其中 $S$ 为球面 $x^{2} + y^{2} + z^{2} = a^{2}$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":146,"id":"f05e7b93-cabe-48ba-85b1-61307359898f","page_size":[595,841],"block_position":"146-20"},{"bbox":[289,802,305,812],"type":"page_number","angle":0,"index":999999,"text":"139 ","color":{"line":"rgba(164, 164, 164, 1)","fill":"rgba(164, 164, 164, 1)"},"page_idx":146,"is_discarded":true,"id":"b2a7607a-ee9f-4052-8251-093a26be0547","page_size":[595,841],"block_position":"146-21"}],[{"bbox":[445,47,537,58],"type":"header","angle":0,"index":-99999,"text":"17.3 第二型曲面积分 ","color":{"line":"rgba(164, 164, 164, 1)","fill":"rgba(164, 164, 164, 1)"},"page_idx":147,"is_discarded":true,"id":"3042c422-911f-4979-8a74-c85eafe97daf","page_size":[595,841],"block_position":"147-0"},{"bbox":[73,69,298,95],"type":"text","angle":0,"index":1,"merge_prev":false,"text":"(2) $\\iint_{S} (x + y + z)^{3} \\mathrm{d} S$ ，其中 $S$ 为 $|x| + |y| + |z| = 1$ 解(1)根据 $S$ 关于 $yz$ 平面对称得到 $\\iint_{S}x\\mathrm{d}s = 0$ ，根据 $S$ 关于 $xz$ 平面对称得到 $\\iint_{S}\\sin yz\\mathrm{d}s = 0$ (2)根据关于原点中心对称性得到为0。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":147,"id":"072c2862-547a-4658-bfe7-03657698dfef","page_size":[595,841],"block_position":"147-1"},{"bbox":[53,93,473,126],"type":"text","angle":0,"index":2,"merge_prev":false,"lines_deleted":true,"text":"","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":147,"id":"072c2862-547a-4658-bfe7-03657698dfef","page_size":[595,841],"block_position":"147-2"},{"bbox":[73,117,243,129],"type":"text","angle":0,"index":3,"merge_prev":false,"text":"(2) 根据关于原点中心对称性得到为 0。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":147,"id":"38468331-bf10-4816-be4f-7e1e44cece34","page_size":[595,841],"block_position":"147-3"},{"bbox":[35,141,539,203],"type":"text","angle":0,"index":4,"merge_prev":false,"text":"练习17.5轮换对称性(1)计算 $I = \\iint_{S}x^{2}\\mathrm{d}S$ 与 $J = \\iint xy\\mathrm{d}S$ ，其中 $S$ 为球面 $x^{2} + y^{2} + z^{2} = a^{2}$ 解(1)由于 $S$ 有轮换对称性，得到 $I = \\frac{1}{3}\\iint_{S}x^{2} + y^{2} + z^{2}\\mathrm{d}S = \\frac{4}{3}\\pi a^4$ 。 $J$ 没必要考虑轮换对称性，由于 $S$ 关于 $yz$ 平面对称，因此 $J = 0$ 。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":147,"id":"b2b9e802-f6d8-462b-8c1c-907347267917","page_size":[595,841],"block_position":"147-4"},{"bbox":[57,224,193,241],"type":"title","angle":0,"index":5,"level":2,"text":"## 17.3 第二型曲面积分","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":147,"id":"7d4524fe-b087-4d5a-b1ae-c25f67301419","page_size":[595,841],"block_position":"147-5"},{"bbox":[56,257,264,271],"type":"title","angle":0,"index":6,"level":2,"text":"## 17.3.1 基本概念、投影法、参数方程法","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":147,"id":"b6a12dc2-c4ed-47d4-85e8-4dd8f1587760","page_size":[595,841],"block_position":"147-6"},{"bbox":[73,284,196,297],"type":"title","angle":0,"index":7,"level":2,"text":"## 定义 17.2 (第二型曲面积分)","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":147,"id":"2060217d-0b51-4c5d-b485-55886128dacb","page_size":[595,841],"block_position":"147-7"},{"bbox":[67,301,524,329],"type":"text","angle":0,"index":8,"merge_prev":false,"text":"对于向量场 $F(x,y,z) = (P(x,y,z),Q(x,y,z),R(x,y,z))$ ，其通过曲面 $S$ 的通量定义为第二型曲面积分(向外为正): ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":147,"id":"e971f648-248a-4a19-9545-de5bf8f59ad3","page_size":[595,841],"block_position":"147-8"},{"bbox":[181,327,412,354],"type":"interline_equation","angle":0,"index":9,"text":"$$\n\\iint_ {S} P (x, y, z) \\mathrm{d} y \\mathrm{d} z + Q (x, y, z) \\mathrm{d} z \\mathrm{d} x + R (x, y, z) \\mathrm{d} x \\mathrm{d} y\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":147,"id":"b26110b6-bb67-415e-9200-18f61df3f900","page_size":[595,841],"block_position":"147-9"},{"bbox":[73,372,156,385],"type":"title","angle":0,"index":10,"level":2,"text":"## 定理 17.7 (投影法)","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":147,"id":"2de07d83-279e-40a9-810f-a8d43eb45e97","page_size":[595,841],"block_position":"147-10"},{"bbox":[69,389,348,402],"type":"text","angle":0,"index":11,"merge_prev":false,"text":"若曲面 $S$ 的方程为 $z = z(x,y)$ ，在 $xy$ 平面投影区域为 $D_{xy}$ ，则 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":147,"id":"856bf81a-89e6-4074-af3a-7bc1bbbb38e0","page_size":[595,841],"block_position":"147-11"},{"bbox":[188,406,405,431],"type":"interline_equation","angle":0,"index":12,"text":"$$\n\\iint_ {S} R (x, y, z) \\mathrm{d} x \\mathrm{d} y = \\pm \\iint_ {D _ {x y}} R (x, y, z (x, y)) \\mathrm{d} x \\mathrm{d} y\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":147,"id":"e2a37d01-9583-4c77-9da1-9b1a7584da01","page_size":[595,841],"block_position":"147-12"},{"bbox":[69,434,445,447],"type":"text","angle":0,"index":13,"merge_prev":false,"text":"正负取决于 $S$ 的法向量与 $z$ 轴正向的夹角，若锐角则取正，钝角取负，直角积分为0。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":147,"id":"0669216b-eb33-41da-a330-093a0005d346","page_size":[595,841],"block_position":"147-13"},{"bbox":[52,456,539,494],"type":"text","angle":0,"index":14,"merge_prev":false,"text":"笔记 若曲面 $S$ 为 $z = a$ 的一部分，则 $\\iint_{S} P \\, \\mathrm{d}y \\, \\mathrm{d}z + Q \\, \\mathrm{d}z \\, \\mathrm{d}x + R \\, \\mathrm{d}x \\, \\mathrm{d}y = \\iint_{S} R(x, y, z) \\, \\mathrm{d}x \\, \\mathrm{d}y = \\pm \\iint_{D} R(x, y, a) \\, \\mathrm{d}x \\, \\mathrm{d}y$ 。若计算 $\\mathrm{d}y \\, \\mathrm{d}z, \\mathrm{d}z \\, \\mathrm{d}x$ ，则需要考虑在 $yz, zx$ 平面的投影 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":147,"id":"0d25d8cb-6d32-4a7c-802d-9104b591c04a","page_size":[595,841],"block_position":"147-14"},{"bbox":[35,508,539,543],"type":"text","angle":0,"index":15,"merge_prev":false,"text":"练习17.6投影法练习(1)计算 $\\iint_{S}xyz\\mathrm{d}x\\mathrm{d}y$ ，其中 $S$ 是球面 $x^{2} + y^{2} + z^{2} = 1$ 在 $x\\geq 0,y\\geq 0$ 部分，且取球面外侧(2)考虑 $S$ 为 $x = y = z = 0,x = y = z = a$ 六个平面所围成的立方体表面，外侧为正，计算： ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":147,"id":"b91b1ebd-5acd-45ee-bc0b-d23d097b5a11","page_size":[595,841],"block_position":"147-15"},{"bbox":[196,545,396,570],"type":"interline_equation","angle":0,"index":17,"text":"$$\n\\iint_ {S} (x y - y z) \\mathrm{d} y \\mathrm{d} z + x ^ {2} \\mathrm{d} z \\mathrm{d} x + (y ^ {2} + x z) \\mathrm{d} x \\mathrm{d} y\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":147,"id":"6470b275-33b3-4c8a-a906-d3ae2a6caa15","page_size":[595,841],"block_position":"147-16"},{"bbox":[52,571,442,600],"type":"text","angle":0,"index":18,"merge_prev":false,"text":"解(1)积分等于 $2\\iint_{D_{xy}}xy\\sqrt{1 - x^2 - y^2}\\mathrm{d}x\\mathrm{d}y = 2\\int_0^{\\frac{\\pi}{2}}\\mathrm{d}\\theta \\int_0^1 r^3\\cos \\theta \\sin \\theta \\sqrt{1 - r^2}\\mathrm{d}r = \\frac{2}{15}$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":147,"id":"73c1bffe-1f69-4a5f-9304-62458f5da759","page_size":[595,841],"block_position":"147-17"},{"bbox":[53,601,539,647],"type":"text","angle":0,"index":19,"merge_prev":false,"text":"(2) 可以用 Gauss 公式，也可以一个一个考虑。例如考虑 $z = a$ 平面 $S_{1}$ ，即 $\\iint_{S_1}(y^2 + xz)\\mathrm{d}x\\mathrm{d}y = \\iint_D(y^2 + ax)\\mathrm{d}x\\mathrm{d}y = \\frac{5}{6} a^4$ ，其余同理。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":147,"id":"67798677-50db-4897-8ec7-75950d1362c3","page_size":[595,841],"block_position":"147-18"},{"bbox":[73,665,176,677],"type":"title","angle":0,"index":20,"level":2,"text":"## 定理 17.8 (参数方程法)","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":147,"id":"1be938a8-7984-4e7e-aa82-4fdaede92a61","page_size":[595,841],"block_position":"147-19"},{"bbox":[68,682,525,733],"type":"text","angle":0,"index":21,"merge_prev":false,"text":"若 $S$ 可表示为参数方程 $x = x(u,v),y = y(u,v),z = z(u,v),(u,v) \\in D$ ，记 $A = \\left|\\frac{\\partial(y,z)}{\\partial(u,v)}\\right|,B = \\left|\\frac{\\partial(z,x)}{\\partial(u,v)}\\right|,C = \\left|\\frac{\\partial(x,y)}{\\partial(u,v)}\\right|$ ，则 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":147,"id":"4cae3d48-6f1c-4805-8fca-40e2d30c02db","page_size":[595,841],"block_position":"147-20"},{"bbox":[160,735,433,761],"type":"interline_equation","angle":0,"index":22,"text":"$$\n\\iint_ {S} P \\mathrm{d} y \\mathrm{d} z + Q \\mathrm{d} z \\mathrm{d} x + R \\mathrm{d} x \\mathrm{d} y = \\pm \\iint_ {D} (P A + Q B + R C) \\mathrm{d} u \\mathrm{d} v\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":147,"id":"cb6976ad-47d8-4f35-8ae8-a76aa99e8d25","page_size":[595,841],"block_position":"147-21"},{"bbox":[289,802,305,812],"type":"page_number","angle":0,"index":1000015,"text":"140 ","color":{"line":"rgba(164, 164, 164, 1)","fill":"rgba(164, 164, 164, 1)"},"page_idx":147,"is_discarded":true,"id":"000b1e08-6168-4654-b3f4-155fe4d39996","page_size":[595,841],"block_position":"147-22"}],[{"bbox":[445,46,538,58],"type":"header","angle":0,"index":-99999,"text":"17.3 第二型曲面积分 ","color":{"line":"rgba(164, 164, 164, 1)","fill":"rgba(164, 164, 164, 1)"},"page_idx":148,"is_discarded":true,"id":"4675db80-18f8-4dbd-82cf-e782fd2aca54","page_size":[595,841],"block_position":"148-0"},{"bbox":[68,80,322,94],"type":"text","angle":0,"index":1,"merge_prev":false,"text":"若向量 $(A, B, C)$ 与 $S$ 法向量方向一致则取正，否则取负。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":148,"id":"c4a243f7-a0c9-41b9-96ea-c9cc9e57d46d","page_size":[595,841],"block_position":"148-1"},{"bbox":[35,117,420,157],"type":"text","angle":0,"index":2,"merge_prev":false,"text":"练习17.7参数方程法(1) $S$ 为 $\\frac{x^2}{a^2} +\\frac{y^2}{b^2} +\\frac{z^2}{c^2} = 1$ 上半部并取外侧，计算 $\\iint_{S}x^{3}\\mathrm{d}y\\mathrm{d}z$ 解 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":148,"id":"dad6f3a8-4ca0-4205-a20b-264da389018c","page_size":[595,841],"block_position":"148-2"},{"bbox":[56,175,157,190],"type":"title","angle":0,"index":3,"level":2,"text":"## 17.3.2 利用对称性","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":148,"id":"a3d3b774-1e6b-4e71-82e3-17b5d74947c8","page_size":[595,841],"block_position":"148-3"},{"bbox":[54,200,539,236],"type":"text","angle":0,"index":4,"merge_prev":false,"text":"以 $\\iint_{S} R(x, y, z) \\mathrm{d}x \\mathrm{~d}y$ 为例， $\\mathrm{d}x \\mathrm{~d}y$ 的正负取决于对称点处 $S$ 外法向量与 $z$ 轴正向的夹角，结合 $R(x, y, z)$ 关于 $xy$ 平面或者原点的对称性可分解 $S$ 上的曲面积分， $\\mathrm{d}y \\mathrm{~d}z$ 与 $\\mathrm{d}z \\mathrm{~d}x$ 同理。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":148,"id":"b971c4de-76d0-4621-9ec0-c0415283ba69","page_size":[595,841],"block_position":"148-4"},{"bbox":[35,253,437,266],"type":"text","angle":0,"index":5,"merge_prev":false,"text":"练习 17.8 方向判断训练 设 $S: |x| + |y| + |z| = 1$ ，方向朝外，判断下面哪些积分值为 0: ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":148,"id":"ff4cceb3-ba83-4815-b658-26cdc50881bb","page_size":[595,841],"block_position":"148-5"},{"bbox":[105,269,487,295],"type":"text","angle":0,"index":6,"merge_prev":false,"text":"(1) $\\iint_{S} z \\, \\mathrm{d}x \\, \\mathrm{d}y$ (2) $\\iint_{S} x^{2} \\sin y \\, \\mathrm{d}x \\, \\mathrm{d}y$ (3) $\\iint_{S} y^{2} \\sin z \\, \\mathrm{d}x \\, \\mathrm{d}y$ (4) $\\iint_{S} \\sin x \\sin^{2} y \\sin^{3} z \\, \\mathrm{d}x \\, \\mathrm{d}y$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":148,"id":"eecd7112-6fd6-4454-9236-ea58de5ca2c8","page_size":[595,841],"block_position":"148-6"},{"bbox":[53,296,461,309],"type":"text","angle":0,"index":7,"merge_prev":false,"text":"解(1)关于 $xy$ 平面对称， $z$ 相反，由于法向量与 $z$ 轴正向夹角相反， $\\mathrm{d}x\\mathrm{d}y$ 也为负，因此非零 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":148,"id":"eab9301c-a85f-4a15-a39e-b440102cf71e","page_size":[595,841],"block_position":"148-7"},{"bbox":[73,312,342,324],"type":"text","angle":0,"index":8,"merge_prev":false,"text":"(2) 关于 $xz$ 平面对称， $\\sin y$ 相反， $\\mathrm{d}x\\mathrm{d}y$ 为正，因此积分为0 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":148,"id":"80055346-af04-4ad7-a9c0-7e47b6a3ab73","page_size":[595,841],"block_position":"148-8"},{"bbox":[73,327,181,339],"type":"text","angle":0,"index":9,"merge_prev":false,"text":"(3) 无论怎么考虑都非零 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":148,"id":"4b5d3b19-972c-4d79-bae6-92658acfa758","page_size":[595,841],"block_position":"148-9"},{"bbox":[73,343,302,356],"type":"text","angle":0,"index":10,"merge_prev":false,"text":"(4) 考虑 yz， $\\sin x$ 相反，dxdy 为正，因此积分为 0。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":148,"id":"fa3eb1e2-c677-41b3-bd30-d3efb748a98d","page_size":[595,841],"block_position":"148-10"},{"bbox":[35,368,539,433],"type":"text","angle":0,"index":11,"merge_prev":false,"text":"练习 17.9 基本计算 (1) 计算 $I_{1} = \\iint_{S} z \\, dx \\, dy, I_{2} = \\iint_{S} z^{2} \\, dx \\, dy$ ，其中 S 为球面 $x^{2} + y^{2} + z^{2} = a^{2}$ ，方向取外侧解 (1) $I_{2}$ 考虑关于 xy 平面对称， $z^{2}$ 相等，dx dy 相反，因此积分为 0。 $I_{1}$ 可转换为两倍的上半球面， $I_{1} = 2 \\iint_{S_{1}} z \\, dx \\, dy$ ，由于 $z = \\sqrt{a^{2} - x^{2} - y^{2}}$ ， $D : x^{2} + y^{2} \\leq a^{2}$ ，因此 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":148,"id":"b97ccb5c-6d06-4b38-a314-fe6e0367c133","page_size":[595,841],"block_position":"148-11"},{"bbox":[150,433,442,461],"type":"interline_equation","angle":0,"index":12,"text":"$$\nI _ {1} = \\iint_ {D} \\sqrt {a ^ {2} - x ^ {2} - y ^ {2}} \\mathrm{d} x \\mathrm{d} y = 2 \\int_ {0} ^ {a} \\mathrm{d} r \\int_ {0} ^ {2 \\pi} \\sqrt {a ^ {2} - r ^ {2}} r \\mathrm{d} \\theta = \\frac {4 \\pi a ^ {3}}{3}\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":148,"id":"89a8b68c-0386-4026-ad6d-6e19f5c60e2b","page_size":[595,841],"block_position":"148-12"},{"bbox":[56,477,205,491],"type":"title","angle":0,"index":13,"level":2,"text":"## 17.3.3 两类曲面积分的联系","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":148,"id":"06da6f1e-93d3-4b6e-a93b-c87a6631d61b","page_size":[595,841],"block_position":"148-13"},{"bbox":[73,505,216,518],"type":"title","angle":0,"index":14,"level":2,"text":"## 定理 17.9 (两类曲面积分的联系)","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":148,"id":"c5faf780-f4a9-4e0d-939b-8dcb5c657ded","page_size":[595,841],"block_position":"148-14"},{"bbox":[67,521,525,550],"type":"text","angle":0,"index":15,"merge_prev":false,"text":"$S$ 为光滑曲面，正侧单位法向量 $\\mathbf{n} = (\\cos \\alpha, \\cos \\beta, \\cos \\gamma)$ ，则 $\\cos \\alpha \\mathrm{d}S = \\mathrm{d}y \\mathrm{d}z, \\cos \\beta \\mathrm{d}S = \\mathrm{d}z \\mathrm{d}x, \\cos \\gamma \\mathrm{d}S = \\mathrm{d}x \\mathrm{d}y$ ，即 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":148,"id":"f57d0e61-9ffb-404e-b3b4-f366a702f33a","page_size":[595,841],"block_position":"148-15"},{"bbox":[147,547,446,573],"type":"interline_equation","angle":0,"index":16,"text":"$$\n\\iint_ {S} P \\mathrm{d} y \\mathrm{d} z + Q \\mathrm{d} z \\mathrm{d} x + R \\mathrm{d} x \\mathrm{d} y = \\iint_ {S} (P \\cos \\alpha + Q \\cos \\beta + R \\cos \\gamma) \\mathrm{d} S\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":148,"id":"57841831-091f-4c9b-93ac-bd996c9df45b","page_size":[595,841],"block_position":"148-16"},{"bbox":[52,588,499,602],"type":"text","angle":0,"index":17,"merge_prev":false,"text":"笔记对一般曲面 $F(x,y,z) = 0$ ，其梯度 $(F_{x},F_{y},F_{z})$ 就是曲面法向量，因此单位法向量的计算方式为： ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":148,"id":"997c59f2-6296-469d-b593-3f070ff71f94","page_size":[595,841],"block_position":"148-17"},{"bbox":[218,605,374,637],"type":"interline_equation","angle":0,"index":18,"text":"$$\n\\mathbf {n} = \\pm \\frac {1}{\\sqrt {F _ {x} ^ {2} + F _ {y} ^ {2} + F _ {z} ^ {2}}} (F _ {x}, F _ {y}, F _ {z})\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":148,"id":"e8d83b5f-2cb2-42c4-a15b-021a3518fb66","page_size":[595,841],"block_position":"148-18"},{"bbox":[35,642,540,701],"type":"text","angle":0,"index":19,"merge_prev":false,"text":"练习17.10重新计算第二型曲面积分(1)重新计算 $\\iint_{S}xyz\\mathrm{d}x\\mathrm{d}y$ ，其中 $S$ 是球面 $x^{2} + y^{2} + z^{2} = 1$ 在 $x\\geq 0,y\\geq 0$ 部分，方向朝外解(1)法向量 $\\mathbf{n} = \\frac{1}{\\sqrt{3}} (1,1,1)$ ，因此可以转换为第一型曲面积分： ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":148,"id":"26114549-9c60-43ea-90d7-061cf1bdbe71","page_size":[595,841],"block_position":"148-19"},{"bbox":[188,703,405,729],"type":"interline_equation","angle":0,"index":21,"text":"$$\n\\iint_ {S} x y z \\cdot x \\cdot \\frac {a}{z} \\mathrm{d} S = 2 \\iint_ {D} r ^ {4} \\cos^ {2} \\theta \\sin \\theta \\mathrm{d} r \\mathrm{d} \\theta = \\frac {2}{1 5}\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":148,"id":"2e063fdd-8265-4fca-820d-7d609f0d25b2","page_size":[595,841],"block_position":"148-20"},{"bbox":[54,731,232,744],"type":"text","angle":0,"index":22,"merge_prev":false,"text":"这里乘上2的原因是关于 $z$ 正负有两块。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":148,"id":"5db83282-ffe4-4c33-b40f-19d90230c5ad","page_size":[595,841],"block_position":"148-21"},{"bbox":[289,803,304,812],"type":"page_number","angle":0,"index":1000019,"text":"141 ","color":{"line":"rgba(164, 164, 164, 1)","fill":"rgba(164, 164, 164, 1)"},"page_idx":148,"is_discarded":true,"id":"0a20bcff-aa8b-47be-9b3e-3e44411727c1","page_size":[595,841],"block_position":"148-22"}],[{"bbox":[468,47,539,58],"type":"header","angle":0,"index":-99998,"text":"17.4 Gauss 公式 ","color":{"line":"rgba(164, 164, 164, 1)","fill":"rgba(164, 164, 164, 1)"},"page_idx":149,"is_discarded":true,"id":"3f7c0107-e6a0-45c9-b26e-85696ca72c95","page_size":[595,841],"block_position":"149-0"},{"bbox":[35,71,539,99],"type":"text","angle":0,"index":2,"merge_prev":false,"text":"练习 17.11 用曲面积分联系计算第二型曲面积分 S 为 $x + y + z = 1$ 在 $x^{2} + y^{2} + z^{2} = 4x$ 内部的部分，取上侧为正，计算： ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":149,"id":"1c76926b-80e8-4d20-9282-b1821f26c9c8","page_size":[595,841],"block_position":"149-1"},{"bbox":[223,96,370,123],"type":"interline_equation","angle":0,"index":3,"text":"$$\nI = \\iint_ {S} y \\mathrm{d} y \\mathrm{d} z + z \\mathrm{d} z \\mathrm{d} x + x \\mathrm{d} x \\mathrm{d} y\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":149,"id":"88a87cbd-1c1b-4ae7-bec3-55e7c85dbc54","page_size":[595,841],"block_position":"149-2"},{"bbox":[52,129,540,179],"type":"text","angle":0,"index":4,"merge_prev":false,"text":"解 (1) 已知 $F(x, y, z) = x + y + z - 1$ ，则梯度为 $\\nabla F = (1, 1, 1)$ ，单位法向量 $\\mathbf{n} = \\frac{1}{\\sqrt{3}}(1, 1, 1)$ ，因此 $I = \\iint_{S} \\frac{1}{\\sqrt{3}}(x + y + z) dS = \\frac{1}{\\sqrt{3}} S$ ，计算截面圆面积即可，结果为 $\\frac{11}{3\\sqrt{3}}\\pi$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":149,"id":"e3f6eb25-b4bf-4e23-951a-de4164b6aa79","page_size":[595,841],"block_position":"149-3"},{"bbox":[56,196,164,212],"type":"title","angle":0,"index":5,"level":2,"text":"## 17.4 Gauss 公式","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":149,"id":"835b4e20-2561-4adf-b352-cfb3d2b105c5","page_size":[595,841],"block_position":"149-4"},{"bbox":[56,229,214,243],"type":"title","angle":0,"index":6,"level":2,"text":"## 17.4.1 Gauss 公式与基本使用","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":149,"id":"c4e3626c-476c-4368-91c6-964292959062","page_size":[595,841],"block_position":"149-5"},{"bbox":[72,256,181,269],"type":"title","angle":0,"index":7,"level":2,"text":"## 定理 17.10 (Gauss 公式)","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":149,"id":"6ce874a0-b27d-4542-907a-827c4f411385","page_size":[595,841],"block_position":"149-6"},{"bbox":[67,273,525,301],"type":"text","angle":0,"index":8,"merge_prev":false,"text":"$V \\subset \\mathbb{R}^3$ 是有界闭区域， $V$ 边界是分片光滑的曲面 $S$ ， $P(x, y, z), Q(x, y, z), R(x, y, z)$ 在 $V$ 内有连续偏导，则 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":149,"id":"caa03465-c0ff-4f98-9b6f-f1c85da0f467","page_size":[595,841],"block_position":"149-7"},{"bbox":[154,298,439,326],"type":"interline_equation","angle":0,"index":9,"text":"$$\n\\oiint_ {S _ {\\text {外}}} P \\mathrm{d} y \\mathrm{d} z + Q \\mathrm{d} z \\mathrm{d} x + R \\mathrm{d} x \\mathrm{d} y = \\iiint_ {V} (\\frac {\\partial P}{\\partial x} + \\frac {\\partial Q}{\\partial y} + \\frac {\\partial R}{\\partial z}) \\mathrm{d} x \\mathrm{d} y \\mathrm{d} z\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":149,"id":"3a979c58-d547-45e1-b276-a534662f075d","page_size":[595,841],"block_position":"149-8"},{"bbox":[51,343,359,356],"type":"text","angle":0,"index":10,"merge_prev":false,"text":"笔记 使用 Gauss 公式需要保证 V 内任意位置，P, Q, R 都有连续偏导。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":149,"id":"cb990750-8814-4321-9adf-4102683c5648","page_size":[595,841],"block_position":"149-9"},{"bbox":[35,368,539,442],"type":"text","angle":0,"index":11,"merge_prev":false,"text":"练习 17.12 补面计算 (1) 东南大学 2022: 计算 $I = \\iint_{S} \\frac{(z + 1)\\mathrm{d}y\\mathrm{d}z + (y + 1)\\mathrm{d}z\\mathrm{d}x + (x + 1)\\mathrm{d}x\\mathrm{d}y}{x^{2} + y^{2} + z^{2}}$ ，其中 S 为 $z = \\sqrt{9 - x^{2} - y^{2}}$ 曲面，取下侧\n\n解 (1) 补面 $S_{1} = \\{z = 0, x^{2} + y^{2} \\leq 9\\}$ ，方向朝上，则 $I = \\frac{1}{9}I_{1} - \\frac{1}{9}I_{2} = \\frac{1}{9}\\iint_{S + S_{1}} - \\frac{1}{9}\\iint_{S_{1}}$ ，根据 Gauss 公式计算 $I_{1}$ : ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":149,"id":"c390c90c-ac64-46b2-bbc9-1e13a1d4e46d","page_size":[595,841],"block_position":"149-10"},{"bbox":[210,443,382,468],"type":"interline_equation","angle":0,"index":12,"text":"$$\nI _ {1} = - \\iiint_ {V} (0 + 1 + 0) \\mathrm{d} x \\mathrm{d} y \\mathrm{d} z = - 1 8 \\pi\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":149,"id":"637aefd1-4dad-4e83-a880-905425f4999d","page_size":[595,841],"block_position":"149-11"},{"bbox":[53,470,349,483],"type":"text","angle":0,"index":13,"merge_prev":false,"text":"计算 $I_{2}$ ，由于在平面上，曲面积分直接变为二重积分，且 dz=0，： ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":149,"id":"21386999-405d-4bae-a3ad-985b2c1df299","page_size":[595,841],"block_position":"149-12"},{"bbox":[167,486,426,513],"type":"interline_equation","angle":0,"index":14,"text":"$$\nI _ {2} = \\iint_ {x ^ {2} + y ^ {2} \\leq 9} (x + 1) \\mathrm{d} x \\mathrm{d} y = \\int_ {0} ^ {2 \\pi} \\mathrm{d} \\theta \\int_ {0} ^ {3} r ^ {2} \\cos + r \\mathrm{d} r = 9 \\pi\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":149,"id":"bbda2b5f-c933-4511-b8e5-4183905da994","page_size":[595,841],"block_position":"149-13"},{"bbox":[54,516,223,538],"type":"text","angle":0,"index":15,"merge_prev":false,"text":"因此结果为 $I = \\frac{1}{9} (-18\\pi - 9\\pi) = -3\\pi$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":149,"id":"668c0afd-aaa0-46c6-afbe-6af6ba64a6a3","page_size":[595,841],"block_position":"149-14"},{"bbox":[35,536,539,572],"type":"text","angle":0,"index":16,"merge_prev":false,"text":"练习 17.13 菱形区域：变量替换！(1) 重点：S 为封闭曲面 $|x - y + z| + |y - z + x| + |z - x + y| = 1$ ，方向朝外，计算 $I = \\iint_{S} (x + y - z) \\, \\mathrm{d}y \\, \\mathrm{d}z + [2y + \\sin(z + x)] \\, \\mathrm{d}z \\, \\mathrm{d}x + (3z + e^{x + y}) \\, \\mathrm{d}x \\, \\mathrm{d}y$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":149,"id":"3a45cb36-828b-4454-ab32-f79f2696219c","page_size":[595,841],"block_position":"149-15"},{"bbox":[53,575,539,608],"type":"text","angle":0,"index":17,"merge_prev":false,"text":"(2)(东北师大) 求 $I = \\iint_{S} (x + \\cos(y - z)) \\, \\mathrm{d}y \\, \\mathrm{d}z + (2y + e^{x + z}) \\, \\mathrm{d}z \\, \\mathrm{d}x + (3z + \\sin(x - y) \\, \\mathrm{d}x \\, \\mathrm{d}y)$ ，这里 $S$ 为封闭曲面，方向取外侧 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":149,"id":"411ecc3d-ebf4-44f9-9091-a9462fcc5375","page_size":[595,841],"block_position":"149-16"},{"bbox":[204,611,389,624],"type":"interline_equation","angle":0,"index":18,"text":"$$\n\\vert x - y + z \\vert + \\vert y - z + x \\vert + \\vert z - x + y \\vert = 1\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":149,"id":"3d610162-991e-4120-ae4d-d757c408b650","page_size":[595,841],"block_position":"149-17"},{"bbox":[53,631,539,690],"type":"text","angle":0,"index":19,"merge_prev":false,"text":"解(1)使用 Gauss 公式即 $I=\\iiint_{V}(1+2+3)\\mathrm{d}x\\mathrm{d}y\\mathrm{d}z$ ，考虑坐标变换 $u=x-y+z, v=y-z+x, w=z-x+y$ ，即 $|u|+|v|+|w|\\leq1$ ，而 $\\frac{\\partial(u,v,w)}{\\partial(x,y,z)}=4$ ，因此 $J=\\frac{\\partial(x,y,z)}{\\partial(u,v,w)}=\\frac{1}{4}$ ，因此 $I=\\iint_{V'}\\frac{3}{2}\\mathrm{d}u\\mathrm{d}v\\mathrm{d}w=2$ ，这里 $V'=\\frac{4}{3}$ 是个锥体。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":149,"id":"6e63add8-dd99-4c38-a432-9246149aa6e1","page_size":[595,841],"block_position":"149-18"},{"bbox":[72,687,301,714],"type":"text","angle":0,"index":20,"merge_prev":false,"text":"(2) 使用 Gauss 公式即 $I = \\iiint_{V} 6 \\mathrm{~d} V$ ，使用坐标变换 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":149,"id":"12e5de0a-22ae-45bf-949e-777a5ca34297","page_size":[595,841],"block_position":"149-19"},{"bbox":[200,715,393,726],"type":"interline_equation","angle":0,"index":21,"text":"$$\nu = x + z - y, v = y + x - z, w = y + z - x\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":149,"id":"4da73ffd-1308-4bc3-a631-aac819634488","page_size":[595,841],"block_position":"149-20"},{"bbox":[54,728,539,754],"type":"text","angle":0,"index":22,"merge_prev":false,"text":"得到 $\\frac{\\partial(u,v,w)}{\\partial(x,y,z)} = 4$ ，即 $\\frac{\\partial(x,y,z)}{\\partial(u,v,w)} = \\frac{1}{4}$ ，积分区域为 $|u| + |v| + |w|\\leq 1$ ，分为8个小锥体，每个锥体体积 $\\frac{1}{6}$ ，因 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":149,"id":"62c24a48-ffa4-4d7d-a41c-b2a30ef62b4c","page_size":[595,841],"block_position":"149-21"},{"bbox":[289,802,305,811],"type":"page_number","angle":0,"index":999999,"text":"142 ","color":{"line":"rgba(164, 164, 164, 1)","fill":"rgba(164, 164, 164, 1)"},"page_idx":149,"is_discarded":true,"id":"5d08b946-3237-4d3f-b7b8-eaa79f00f08b","page_size":[595,841],"block_position":"149-22"}],[{"bbox":[467,47,539,58],"type":"header","angle":0,"index":-99999,"text":"17.5 Stokes 公式 ","color":{"line":"rgba(164, 164, 164, 1)","fill":"rgba(164, 164, 164, 1)"},"page_idx":150,"is_discarded":true,"id":"0f93eb06-91a3-4fb7-bde4-7de41793edb3","page_size":[595,841],"block_position":"150-0"},{"bbox":[54,72,122,84],"type":"text","angle":0,"index":1,"merge_prev":false,"text":"此最终结果为: ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":150,"id":"1aa4a343-6753-4452-a489-d559b15fd2e8","page_size":[595,841],"block_position":"150-1"},{"bbox":[253,81,340,105],"type":"interline_equation","angle":0,"index":2,"text":"$$\nI = \\frac {3}{2} \\times 8 \\times \\frac {1}{6} = 2\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":150,"id":"ac80805f-ed42-4aec-91a7-95db35717435","page_size":[595,841],"block_position":"150-2"},{"bbox":[56,120,157,134],"type":"title","angle":0,"index":3,"level":2,"text":"## 17.4.2 内部有瑕点","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":150,"id":"901879b8-3d89-43d2-87a4-0a8b495e4f6c","page_size":[595,841],"block_position":"150-3"},{"bbox":[35,143,318,156],"type":"text","angle":0,"index":4,"merge_prev":false,"text":"练习 17.14 经典题目 S 为不经过原点的光滑封闭曲面，计算 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":150,"id":"5a9e50d8-c937-470b-b5e3-d779bad11034","page_size":[595,841],"block_position":"150-4"},{"bbox":[105,158,487,185],"type":"interline_equation","angle":0,"index":5,"text":"$$\nI = \\iint_ {S} {\\frac {x}{(a x ^ {2} + b y ^ {2} + c z ^ {2}) ^ {\\frac {3}{2}}}} \\mathrm{d} y \\mathrm{d} z + {\\frac {y}{(a x ^ {2} + b y ^ {2} + c z ^ {2}) ^ {\\frac {3}{2}}}} \\mathrm{d} z \\mathrm{d} x + {\\frac {z}{(a x ^ {2} + b y ^ {2} + c z ^ {2}) ^ {\\frac {3}{2}}}} \\mathrm{d} x \\mathrm{d} y\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":150,"id":"170f1f32-54b0-421d-97d2-1539b1b3b9b1","page_size":[595,841],"block_position":"150-5"},{"bbox":[52,186,294,200],"type":"text","angle":0,"index":6,"merge_prev":false,"text":"解记 $r^2 = ax^2 + by^2 + cz^2$ ，则两侧对 $x$ 求导可以得到 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":150,"id":"660e641d-dfea-4724-afc7-5afa910f3f2d","page_size":[595,841],"block_position":"150-6"},{"bbox":[276,203,317,225],"type":"interline_equation","angle":0,"index":7,"text":"$$\nr _ {x} = \\frac {a x}{r}\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":150,"id":"486295db-48ca-448d-bf45-2024471822ac","page_size":[595,841],"block_position":"150-7"},{"bbox":[53,225,371,248],"type":"text","angle":0,"index":8,"merge_prev":false,"text":"同理可以得到 $r_y, r_z$ 。同时 $P = \\frac{x}{r^3}, Q = \\frac{y}{r^3}, R = \\frac{z}{r^3}$ ，因此可以计算得到 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":150,"id":"2c4fcf31-bcb0-470f-93e6-0cc008c4f3b1","page_size":[595,841],"block_position":"150-8"},{"bbox":[255,249,339,263],"type":"interline_equation","angle":0,"index":9,"text":"$$\nP _ {x} + Q _ {y} + R _ {z} = 0\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":150,"id":"0cdf6bf9-6451-4dd5-92de-f94604020bc5","page_size":[595,841],"block_position":"150-9"},{"bbox":[53,268,539,316],"type":"text","angle":0,"index":10,"merge_prev":false,"text":"如果 S 围成的区域不包含原点，则积分为 0。如果包含原点，则挖掉 $V_{1}: ax^{2} + by^{2} + cz^{2} \\leq \\epsilon^{2}$ ， $S_{1}$ 面方向定义为 $V - V_{1}$ 朝向 $V_{1}$ ，则 $\\iint_{S + S_{1}} P \\, dy \\, dz + Q \\, dz \\, dx + R \\, dx \\, dy = \\iiint_{V - V_{1}} 0 \\, dx \\, dy \\, dz = 0$ ，因此 $I = -\\iint_{S_{1}} P \\, dy \\, dz + Q \\, dz \\, dx + R \\, dx \\, dy$ ，用 Gauss 公式得到 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":150,"id":"83451f56-ea51-4c3a-b243-0dafc1fb3e33","page_size":[595,841],"block_position":"150-10"},{"bbox":[197,317,396,344],"type":"interline_equation","angle":0,"index":11,"text":"$$\nI = \\iiint_ {V _ {1}} \\frac {3}{\\epsilon^ {3}} \\mathrm{d} x \\mathrm{d} y \\mathrm{d} z = \\frac {3}{\\epsilon^ {3}} \\cdot \\frac {4}{3} \\cdot \\frac {\\pi \\epsilon^ {3}}{\\sqrt {a b c}} = \\frac {4 \\pi}{\\sqrt {a b c}}\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":150,"id":"e39d8652-1e70-4a1f-b937-683b3d8fa1a2","page_size":[595,841],"block_position":"150-11"},{"bbox":[33,349,426,386],"type":"text","angle":0,"index":12,"merge_prev":false,"text":"笔记最常见的形式为： $\\iint_{S}\\frac{x\\mathrm{d}y\\mathrm{d}z + y\\mathrm{d}z\\mathrm{d}x + z\\mathrm{d}x\\mathrm{d}y}{(x^2 + y^2 + z^2)^{\\frac{3}{2}}} = \\left\\{ \\begin{array}{ll}4\\pi , & S\\text{围成区域包含原点}\\\\ 0, & S\\text{围成区域不包含原点} \\end{array} \\right.$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":150,"id":"cdcbbe7d-9b26-48c4-80e5-750d230dfe06","page_size":[595,841],"block_position":"150-12"},{"bbox":[35,392,473,407],"type":"text","angle":0,"index":13,"merge_prev":false,"text":"练习 17.15 推广形式 S 为上半椭球面 $2021x^{2} + 2022y^{2} + 2023z^{2} = 1 (z \\geq 0)$ ，方向取向上，计算 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":150,"id":"780e4703-beca-4e2f-afd5-3b6560669e0e","page_size":[595,841],"block_position":"150-13"},{"bbox":[222,408,370,435],"type":"interline_equation","angle":0,"index":14,"text":"$$\nI = \\iint_ {S} \\frac {x \\mathrm{d} y \\mathrm{d} z + y \\mathrm{d} z \\mathrm{d} x + z \\mathrm{d} x \\mathrm{d} y}{\\left(x ^ {2} + y ^ {2} + z ^ {2}\\right) ^ {\\frac {3}{2}}}\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":150,"id":"eac32a9b-ee8c-4db3-8dbe-38e2099e7551","page_size":[595,841],"block_position":"150-14"},{"bbox":[52,439,539,492],"type":"text","angle":0,"index":15,"merge_prev":false,"text":"解 记 $r = \\sqrt{x^2 + y^2 + z^2}$ ，显然除原点外 $P_x + Q_y + R_z = 0$ ，记 $V: 2021x^2 + 2022y^2 + 2023z^2 \\leq 1 (z \\geq 0)$ ， $V_1: x^2 + y^2 + z^2 \\leq \\epsilon^2 (z \\geq 0)$ ，则 $V - V_1$ 的表面分为三个部分 $S + S_1 + S_2$ ，其中 $S_1$ 是小椭球体表面， $S_2$ 为 $z = 0$ 内两个直底面， $S_1, S_2$ 方向朝下。根据 Gauss 公式得到 $\\iint_{S + S_1 + S_2} P \\mathrm{d}y \\mathrm{d}z + Q \\mathrm{d}z \\mathrm{d}x + R \\mathrm{d}x \\mathrm{d}y = 0$ ，即 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":150,"id":"b5dadcf3-303b-4d33-a564-f8bf3778042a","page_size":[595,841],"block_position":"150-15"},{"bbox":[193,494,398,522],"type":"interline_equation","angle":0,"index":16,"text":"$$\n\\iint_ {S} P \\mathrm{d} y \\mathrm{d} z + Q \\mathrm{d} z \\mathrm{d} x + R \\mathrm{d} x \\mathrm{d} y = - \\iint_ {S _ {1}} - \\iint_ {S _ {2}}\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":150,"id":"8010832b-53d9-4c2a-ae73-ec453ad3039c","page_size":[595,841],"block_position":"150-16"},{"bbox":[52,523,539,576],"type":"text","angle":0,"index":17,"merge_prev":false,"text":"$\\iint_{S_2}$ 由于 $z = 0$ 得到积分为0， $\\iint_{S_1}$ 得到 $r = \\epsilon$ ，即 $P = \\frac{x}{\\epsilon^3},Q = \\frac{y}{\\epsilon^3},R = \\frac{z}{\\epsilon^3}$ ，再补上 $V_{1}$ 下面的小平面 $S_{3}$ ，用Gauss公式得到 $I = \\frac{1}{\\epsilon^3}\\iint_{-S_1 + S_3}x\\mathrm{d}y\\mathrm{d}z + y\\mathrm{d}z\\mathrm{d}x + z\\mathrm{d}x\\mathrm{d}y$ ，最终结果为 $2\\pi$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":150,"id":"72f2c787-8185-48ec-a705-09a6b8062a05","page_size":[595,841],"block_position":"150-17"},{"bbox":[35,577,528,590],"type":"text","angle":0,"index":18,"merge_prev":false,"text":"笔记 最后一步也可以用两类曲面积分的关系进行计算，当第二型不方便算时，要善于转化为第一型进行计算 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":150,"id":"bf528d3f-3140-4e05-8100-8b4e3467bdeb","page_size":[595,841],"block_position":"150-18"},{"bbox":[56,613,166,629],"type":"title","angle":0,"index":19,"level":2,"text":"## 17.5 Stokes 公式","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":150,"id":"cb423560-8f7d-43cc-85e8-6eedfd0826dc","page_size":[595,841],"block_position":"150-19"},{"bbox":[56,646,216,660],"type":"title","angle":0,"index":20,"level":2,"text":"## 17.5.1 Stokes 公式与基本使用","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":150,"id":"dadd9ea3-d496-4ed3-956c-35a497e8b677","page_size":[595,841],"block_position":"150-20"},{"bbox":[72,671,181,684],"type":"title","angle":0,"index":22,"level":2,"text":"## 定理 17.11 (Stokes 公式)","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":150,"id":"75c6d8b0-49c6-48c7-8690-01e458466a35","page_size":[595,841],"block_position":"150-21"},{"bbox":[67,689,453,702],"type":"text","angle":0,"index":23,"merge_prev":false,"text":"设曲面 $S$ 的边界 $L$ 是按段光滑的连续曲线，若 $P, Q, R$ 在 $S$ 上连续，且有一阶偏导，则 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":150,"id":"373f6b25-edbb-49ee-ad14-86e13513308e","page_size":[595,841],"block_position":"150-22"},{"bbox":[109,704,483,760],"type":"interline_equation","angle":0,"index":24,"text":"$$\n\\oint_ {L} P \\mathrm{d} x + Q \\mathrm{d} y + R \\mathrm{d} z = \\iint_ {S} \\left| \\begin{array}{c c c} \\mathrm{d} y \\mathrm{d} z & \\mathrm{d} z \\mathrm{d} x & \\mathrm{d} x \\mathrm{d} y \\\\ \\frac {\\partial}{\\partial x} & \\frac {\\partial}{\\partial y} & \\frac {\\partial}{\\partial z} \\\\ P & Q & R \\end{array} \\right| = \\iint_ {S} \\left| \\begin{array}{c c c} \\cos \\alpha & \\cos \\beta & \\cos \\gamma \\\\ \\frac {\\partial}{\\partial x} & \\frac {\\partial}{\\partial y} & \\frac {\\partial}{\\partial z} \\\\ P & Q & R \\end{array} \\right| \\mathrm{d} S\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":150,"id":"21635449-54b3-44fe-91c7-fe261c6ca414","page_size":[595,841],"block_position":"150-23"},{"bbox":[289,803,305,812],"type":"page_number","angle":0,"index":1000020,"text":"143 ","color":{"line":"rgba(164, 164, 164, 1)","fill":"rgba(164, 164, 164, 1)"},"page_idx":150,"is_discarded":true,"id":"cb5c8843-0ff3-4313-b65d-ee45b04bf13e","page_size":[595,841],"block_position":"150-24"}],[{"bbox":[467,47,539,58],"type":"header","angle":0,"index":-99999,"text":"17.5 Stokes 公式 ","color":{"line":"rgba(164, 164, 164, 1)","fill":"rgba(164, 164, 164, 1)"},"page_idx":151,"is_discarded":true,"id":"362a0cca-75a9-44ba-a9ac-faf70c5b9b30","page_size":[595,841],"block_position":"151-0"},{"bbox":[67,81,528,110],"type":"text","angle":0,"index":1,"merge_prev":false,"text":"即将空间第二型曲线积分转换为第二或第一型曲面积分，这里 $\\mathbf{n} = (\\cos \\alpha, \\cos \\beta, \\cos \\gamma)$ 为 $S$ 正侧的外单位法向量。注意求法向的曲面 $S$ 和最终积分的 $S$ 要是同一个曲面。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":151,"id":"4109642b-e211-4061-aa07-2ae678bb0af4","page_size":[595,841],"block_position":"151-1"},{"bbox":[52,125,539,154],"type":"text","angle":0,"index":2,"merge_prev":false,"text":"笔记Stokes公式一般用在三维第一型曲线积分，且积分曲线是两个曲面的交线，往往转换为第一型曲面积分更好算：外法向量能用平面就用平面(为常数)，圆面则为圆心到 $(x,y,z)$ 的单位向量。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":151,"id":"b8366391-9676-44e1-bad8-2aec8fab40a7","page_size":[595,841],"block_position":"151-2"},{"bbox":[51,156,502,169],"type":"text","angle":0,"index":3,"merge_prev":false,"text":"笔记 确定单位法向量的方向：若积分曲线方向是正向(区域在左)，则单位法向朝曲面朝外，否则朝内。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":151,"id":"2025c1dd-f46d-4ac5-bf0b-f1d1ebe5d2d6","page_size":[595,841],"block_position":"151-3"},{"bbox":[35,188,387,200],"type":"text","angle":0,"index":4,"merge_prev":false,"text":"练习 17.16 一次项 Stokes 公式 利用 Stokes 公式计算下面的第一型曲线积分: ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":151,"id":"d44ca6c8-61d6-4b42-88b4-8238b03c4e0a","page_size":[595,841],"block_position":"151-4"},{"bbox":[52,202,539,234],"type":"text","angle":0,"index":5,"merge_prev":false,"text":"(1) 平面与球面：计算 $I = \\oint_{L} y \\, \\mathrm{d}x + z \\, \\mathrm{d}y + x \\, \\mathrm{d}z$ ，其中 $L$ 为 $x^{2} + y^{2} + z^{2} = a^{2}$ 与 $x + y + z = 0$ 的交线，从 $z$ 轴 $+\\infty$ 看为逆时针。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":151,"id":"75674d40-f59b-49f3-8b70-3f21313ddb38","page_size":[595,841],"block_position":"151-5"},{"bbox":[53,234,539,268],"type":"text","angle":0,"index":6,"merge_prev":false,"text":"(2) 球面与柱面：计算 $\\oint_{L} (x - 1) \\mathrm{d}y - (y + 1) \\mathrm{d}x + z \\mathrm{d}z$ ，其中 $L$ 为 $x^{2} + y^{2} + z^{2} = 1$ 上半球与柱面 $x^{2} + y^{2} = x$ 的交线，从 $z$ 轴正向来看为逆时针方向。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":151,"id":"4c475f40-1648-4f44-a098-0c667f4afe9d","page_size":[595,841],"block_position":"151-6"},{"bbox":[52,267,540,290],"type":"text","angle":0,"index":7,"merge_prev":false,"text":"解(1)记 $S$ 为 $x + y + z = 0$ 位于球面 $x^{2} + y^{2} + z^{2} = a^{2}$ 内的部分，取上侧 $\\mathbf{n} = \\frac{1}{\\sqrt{3}} (1,1,1)$ ，则根据Stokes公式： ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":151,"id":"8c3f72ee-a1ed-4581-bba0-56af1908f81d","page_size":[595,841],"block_position":"151-7"},{"bbox":[153,292,439,354],"type":"interline_equation","angle":0,"index":8,"text":"$$\nI = \\iint_ {S} \\left| \\begin{array}{c c c} \\frac {1}{\\sqrt {3}} & \\frac {1}{\\sqrt {3}} & \\frac {1}{\\sqrt {3}} \\\\ \\frac {\\partial}{\\partial x} & \\frac {\\partial}{\\partial y} & \\frac {\\partial}{\\partial z} \\\\ y & z & x \\end{array} \\right| = \\frac {1}{\\sqrt {3}} \\iint_ {S} (- 1 - 1 - 1) \\mathrm{d} S = - \\sqrt {3} \\pi a ^ {2}\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":151,"id":"8170c8cf-bf41-434c-81ca-868b99b357fa","page_size":[595,841],"block_position":"151-8"},{"bbox":[73,359,262,370],"type":"text","angle":0,"index":9,"merge_prev":false,"text":"(2) $\\mathbf{n} = (x, y, z)$ ，因此根据Stokes公式得到 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":151,"id":"afb1781e-2145-4df1-af7d-14481a812cc1","page_size":[595,841],"block_position":"151-9"},{"bbox":[156,375,436,428],"type":"interline_equation","angle":0,"index":10,"text":"$$\nI = \\iint_ {S} \\left| \\begin{array}{c c c} x & y & z \\\\ \\frac {\\partial}{\\partial x} & \\frac {\\partial}{\\partial y} & \\frac {\\partial}{\\partial z} \\\\ - y - 1 & x - 1 & z \\end{array} \\right| \\mathrm{d} S = \\iint_ {S} 2 z \\mathrm{d} S = \\iint_ {D} 2 \\mathrm{d} S = \\frac {\\pi}{2}\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":151,"id":"7f5adaeb-bad5-4238-a2a4-46cf0328d888","page_size":[595,841],"block_position":"151-10"},{"bbox":[35,439,539,467],"type":"text","angle":0,"index":11,"merge_prev":false,"text":"练习 17.17 二次项 Stokes 公式 经典：L 是 $x^{2} + y^{2} + z^{2} = 4x$ 与 $x^{2} + y^{2} = 2x$ 的交线在 $z \\geq 0$ 的部分，方向为原点进入第一卦限，计算： ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":151,"id":"862f33bc-3a27-4dc5-99b0-16d32a3c22f7","page_size":[595,841],"block_position":"151-11"},{"bbox":[190,465,401,491],"type":"interline_equation","angle":0,"index":12,"text":"$$\nI = \\oint_ {L} (y ^ {2} + z ^ {2}) \\mathrm{d} x + (x ^ {2} + z ^ {2}) \\mathrm{d} y + (x ^ {2} + y ^ {2}) \\mathrm{d} z\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":151,"id":"4df83b87-7361-4343-b653-5b74edf6230e","page_size":[595,841],"block_position":"151-12"},{"bbox":[52,497,537,530],"type":"text","angle":0,"index":13,"merge_prev":false,"text":"解 曲面 S 是 $(x-2)^{2}+y^{2}+z^{2}=4$ 的一部分，注意原点进第一卦限时区域在右侧，因此方向朝内。单位法向量为 $\\mathbf{n}=-\\frac{1}{2}(x-2,y,z)$ ，因此根据 Stokes 公式： ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":151,"id":"f880695b-cf56-41e6-b2a5-f8a3864ccb97","page_size":[595,841],"block_position":"151-13"},{"bbox":[146,534,447,586],"type":"interline_equation","angle":0,"index":14,"text":"$$\n\\begin{array}{l} I = \\oint_ {L} (y ^ {2} + z ^ {2}) \\mathrm{d} x + (x ^ {2} + z ^ {2}) \\mathrm{d} y + (x ^ {2} + y ^ {2}) \\mathrm{d} z \\\\ = - \\iint_ {S} [ (x - 2) (y - z) + y (z - x) + z (x - y) ] \\mathrm{d} S = 2 \\iint_ {S} (y - z) \\mathrm{d} S \\end{array}\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":151,"id":"19154f6c-3244-4e54-9a1d-7cb0e3821c18","page_size":[595,841],"block_position":"151-14"},{"bbox":[54,588,279,613],"type":"text","angle":0,"index":15,"merge_prev":false,"text":"由于 S 关于 xz 平面对称，因此 $\\iint_{S} y \\, dS = 0$ ，此时 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":151,"id":"8cc43797-b7d4-48ab-9071-6b54b8cfe5d3","page_size":[595,841],"block_position":"151-15"},{"bbox":[261,614,332,640],"type":"interline_equation","angle":0,"index":16,"text":"$$\nI = - 2 \\iint_ {S} z \\mathrm{d} S\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":151,"id":"46b103d2-482e-44ea-add5-b74a0b90f83a","page_size":[595,841],"block_position":"151-16"},{"bbox":[54,644,525,665],"type":"text","angle":0,"index":18,"merge_prev":false,"text":"而 $S$ 方程为 $z = \\sqrt{4 - (x - 2)^2 - y^2}$ ，投影为 $D = \\{(x,y):(x - 1)^2 +y^2\\leq 1\\}$ ，且 $\\sqrt{1 + z_x^2 + z_y^2} = \\frac{2}{z}$ ，得到 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":151,"id":"30ea012b-8c15-4721-a07d-064edf3473df","page_size":[595,841],"block_position":"151-17"},{"bbox":[201,666,392,693],"type":"interline_equation","angle":0,"index":19,"text":"$$\nI = - 2 \\iint_ {S} z \\mathrm{d} S = - 2 \\iint_ {D} z \\cdot \\frac {2}{z} \\mathrm{d} x \\mathrm{d} y = - 4 \\pi\n$$","color":{"line":"rgba(230, 122, 171, 1)","fill":"rgba(230, 122, 171, 1)"},"page_idx":151,"id":"4f8cde04-894e-4370-ab11-b5aee521f737","page_size":[595,841],"block_position":"151-18"},{"bbox":[289,803,305,812],"type":"page_number","angle":0,"index":1000016,"text":"144 ","color":{"line":"rgba(164, 164, 164, 1)","fill":"rgba(164, 164, 164, 1)"},"page_idx":151,"is_discarded":true,"id":"560d3c12-dfe3-4a2a-9209-3cf93a742a91","page_size":[595,841],"block_position":"151-19"}],[{"bbox":[467,47,539,58],"type":"header","angle":0,"index":-99999,"text":"17.5 Stokes 公式 ","color":{"line":"rgba(164, 164, 164, 1)","fill":"rgba(164, 164, 164, 1)"},"page_idx":152,"is_discarded":true,"id":"a4b373ff-03a3-493e-ace0-f3de425cf09c","page_size":[595,841],"block_position":"152-0"},{"bbox":[57,72,276,86],"type":"title","angle":0,"index":1,"level":2,"text":"## 17.5.2 空间第二型曲线积分与路径无关性","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":152,"id":"4e7479bd-3e71-4db4-92ba-3d9bb6da5f8e","page_size":[595,841],"block_position":"152-1"},{"bbox":[73,100,281,113],"type":"title","angle":0,"index":2,"level":2,"text":"## 定理 17.12 (空间第二型曲线积分与路径无关性)","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":152,"id":"99ac595a-e109-45cd-93d8-c98b03239ee1","page_size":[595,841],"block_position":"152-2"},{"bbox":[68,118,524,153],"type":"text","angle":0,"index":3,"merge_prev":false,"text":"$V \\subseteq \\mathbb{R}^3$ 是单连通区域， $P, Q, R$ 在 $V$ 上连续，且有一阶偏导数， $L$ 为 $V$ 内任意光滑曲线，则 $\\int_{L} P \\mathrm{d}x + Q \\mathrm{d}y + R \\mathrm{d}z$ 与路径无关与下面条件等价： ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":152,"id":"d3619965-f967-4be5-bf92-5a54905a7c17","page_size":[595,841],"block_position":"152-3"},{"bbox":[83,153,354,174],"type":"text","angle":0,"index":4,"merge_prev":false,"text":"- 任意 $V$ 内光滑封闭曲线 $L$ , 满足 $\\oint_{L} P \\mathrm{~d} x + Q \\mathrm{~d} y + R \\mathrm{~d} z = 0$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":152,"id":"07ebfc0b-411c-4230-9dfb-eb4b18448636","page_size":[595,841],"block_position":"152-4"},{"bbox":[83,174,288,187],"type":"text","angle":0,"index":5,"merge_prev":false,"text":"- $V$ 内处处成立: $P_{y} = Q_{x}, Q_{z} = R_{y}, R_{x} = P_{z}$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":152,"id":"e94c8874-03fd-4368-ae6b-d11024ba8996","page_size":[595,841],"block_position":"152-5"},{"bbox":[289,803,305,811],"type":"page_number","angle":0,"index":1000005,"text":"145 ","color":{"line":"rgba(164, 164, 164, 1)","fill":"rgba(164, 164, 164, 1)"},"page_idx":152,"is_discarded":true,"id":"7725109d-4d06-4fea-acb7-e0d14294daab","page_size":[595,841],"block_position":"152-6"}],[{"bbox":[206,74,398,94],"type":"title","angle":0,"index":0,"level":1,"text":"# 第 18 章 实数完备性定理","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":153,"id":"171db42b-b225-4779-a745-5983ccfb3827","page_size":[595,841],"block_position":"153-0"},{"bbox":[73,122,146,135],"type":"title","angle":0,"index":1,"level":2,"text":"## 定义 18.1 (聚点)","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":153,"id":"903f82a2-9f9f-4737-b597-721bb953895d","page_size":[595,841],"block_position":"153-1"},{"bbox":[69,138,435,159],"type":"text","angle":0,"index":2,"merge_prev":false,"text":"若存在各项互异的收敛数列 $\\{x_{n}\\} \\subseteq S$ ，使得 $\\lim_{n\\to \\infty}x_n = \\xi$ ，则 $\\xi$ 称为 $S$ 的一个聚点。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":153,"id":"bf8db7de-2c76-44fa-a360-4432ce08cba4","page_size":[595,841],"block_position":"153-2"},{"bbox":[73,178,186,190],"type":"title","angle":0,"index":3,"level":2,"text":"## 定理 18.1 (实数七大定理)","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":153,"id":"5c413561-3c09-421c-a589-623445560ff9","page_size":[595,841],"block_position":"153-3"},{"bbox":[70,195,136,206],"type":"title","angle":0,"index":4,"level":2,"text":"## 七大定理如下：","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":153,"id":"64d9e73b-bee7-4509-aa13-00390663b6e9","page_size":[595,841],"block_position":"153-4"},{"bbox":[83,210,358,222],"type":"text","angle":0,"index":5,"merge_prev":false,"text":"- 确界定理： $S$ 是非空实数集，若 $S$ 有上界，则 $S$ 必有上确界 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":153,"id":"754c5459-df42-4a12-8f07-74ef845d7e57","page_size":[595,841],"block_position":"153-5"},{"bbox":[84,226,157,238],"type":"text","angle":0,"index":6,"merge_prev":false,"text":"- 单调有界定理 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":153,"id":"c23cae34-acd2-498e-a7c3-f59487bf5127","page_size":[595,841],"block_position":"153-6"},{"bbox":[83,242,169,253],"type":"text","angle":0,"index":7,"merge_prev":false,"text":"- Cauchy 收敛准则 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":153,"id":"329304a7-a706-46fb-aa32-afa137e86389","page_size":[595,841],"block_position":"153-7"},{"bbox":[84,257,157,268],"type":"text","angle":0,"index":8,"merge_prev":false,"text":"- 闭区间套定理 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":153,"id":"939dbfb5-eb73-472e-84a0-6c2b5951541a","page_size":[595,841],"block_position":"153-8"},{"bbox":[83,272,524,300],"type":"text","angle":0,"index":9,"merge_prev":false,"text":"- 有限开覆盖定理： $S$ 是实数子集， $H$ 为开区间族，若 $\\forall x \\in S$ 都含于 $H$ 至少一个开区间中，则称 $H$ 是 $S$ 开覆盖。若 $H$ 是 $[a, b]$ 的开覆盖，则可从 $H$ 中选出有限个区间覆盖 $[a, b]$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":153,"id":"2db6fff7-5aec-4d0a-a6c9-46e0c1144821","page_size":[595,841],"block_position":"153-9"},{"bbox":[83,302,339,315],"type":"text","angle":0,"index":10,"merge_prev":false,"text":"- 聚点定理：实轴上任一有界无限点集 $S$ 至少有一个聚点 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":153,"id":"13c1ee2a-cdd6-4d36-8ab7-fddc556093d6","page_size":[595,841],"block_position":"153-10"},{"bbox":[83,318,257,331],"type":"text","angle":0,"index":11,"merge_prev":false,"text":"致密性定理：有界数列必有收敛子列 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":153,"id":"dfedbbe8-5bc1-48d4-a36c-01f37d9d3353","page_size":[595,841],"block_position":"153-11"},{"bbox":[54,363,539,391],"type":"text","angle":0,"index":12,"merge_prev":false,"text":"下面采取确界原理-单调有界定理-闭区间套定理-有限覆盖定理-聚点定理/致密性定理-Cauchy收敛准则的形式循环证明其等价性： ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":153,"id":"3ba1751d-1c3a-4319-9931-ad37732f896b","page_size":[595,841],"block_position":"153-12"},{"bbox":[66,394,538,422],"type":"text","angle":0,"index":13,"merge_prev":false,"text":"1. 确界证明单调有界定理：设 $a_{n}$ 单增有上界，根据确界原理有上确界，取 $a = \\sup \\{a_{n}\\}$ ，只需证明 $a$ 是 $a_{n}$ 极限。根据上确界定义有 $\\exists N, a - \\epsilon < a_{N}$ ，再由单增和极限定义显然。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":153,"id":"db1173c4-538a-4a7a-b895-6e988823af40","page_size":[595,841],"block_position":"153-13"},{"bbox":[66,424,539,452],"type":"text","angle":0,"index":14,"merge_prev":false,"text":"2. 单调有界证明闭区间套： $a_{n}, b_{n}$ 都单调有极限，所以夹住的点存在。唯一性用 $|\\xi - \\xi'| \\leq \\lim_{n \\to \\infty} (b_{n} - a_{n}) = 0$ 证明 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":153,"id":"8814bc80-e2d9-4189-8504-6fc313828ac5","page_size":[595,841],"block_position":"153-14"},{"bbox":[66,454,539,484],"type":"text","angle":0,"index":15,"merge_prev":false,"text":"3. 闭区间套证明有限覆盖：反设不成立，则 $[a, \\frac{a + b}{2}], [\\frac{a + b}{2}, b]$ 中至少有一个没有有限覆盖，不停分下去，根据闭区间套有 $\\xi \\in [a_n, b_n]$ 。当 $n$ 足够大时，一定存在开覆盖集中的区间包含它，因此与假设矛盾。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":153,"id":"d8ebf2b3-a8c6-4fcc-81f9-38733eeeb30b","page_size":[595,841],"block_position":"153-15"},{"bbox":[66,486,539,534],"type":"text","angle":0,"index":16,"merge_prev":false,"text":"4. 有限覆盖证明致密性定理：设 $x_{n}$ 有界，若无穷项相等，则显然。若相等的只有有限项，可知 $\\exists x_0 \\in [a, b]$ 使得 $(x_0 - \\delta, x_0 + \\delta)$ 中有无限项，否则根据有限开覆盖，总共 $x_{n}$ 只有有限项，矛盾。因此在 $(x_0 - \\frac{1}{k}, x_0 + \\frac{1}{k})$ 中取 $x_{k}$ 即有 $\\lim_{k \\to \\infty} x_{k} = x_{0}$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":153,"id":"31648074-1f86-4e36-bdfd-336791eca2c8","page_size":[595,841],"block_position":"153-16"},{"bbox":[66,535,539,565],"type":"text","angle":0,"index":17,"merge_prev":false,"text":"5. 聚点定理证明 Cauchy 收敛准则：设 $x_{n}$ 收敛， $\\lim_{n \\to \\infty} x_{n} = a$ ， $\\exists N, \\forall n, m > N$ 有 $|x_{n} - a| < \\epsilon, |x_{m} - a| < \\epsilon$ ，因此 $|a_{n} - a_{m}| \\leq |a_{n} - a| + |a - a_{m}| = 2\\epsilon$ ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":153,"id":"181a45fc-6f1f-4fcf-b126-4eee2fd68b42","page_size":[595,841],"block_position":"153-17"},{"bbox":[66,567,539,595],"type":"text","angle":0,"index":18,"merge_prev":false,"text":"6. Cauchy 收敛准则证明确界原理: b 为上界, $a \\in S$ , 不断二分, 构造 $a_{n}, b_{n}$ , 最后 $|a_{n+p} - a_{n}| \\leq |b_{n} - a_{n}| < \\epsilon$ , 因此有极限 a, 满足 $\\exists N, a_{N} + \\epsilon > a$ 。 ","color":{"line":"rgba(13, 83, 222, 1)","fill":"rgba(13, 83, 222, 1)"},"page_idx":153,"id":"c3a91258-9a3d-4dcc-b25c-440d719da6bf","page_size":[595,841],"block_position":"153-18"}]],"mergeConnections":[{"id":"f9fba292-8dc0-417d-a2e8-54ae0ce54070","blocks":["15-19","16-1"],"type":"merge"},{"id":"71f48c5d-664a-460a-869a-2f89e68c31e2","blocks":["31-23","32-1"],"type":"merge"},{"id":"0a32a123-57b9-4990-a402-3685e17fe28c","blocks":["41-17","42-1"],"type":"merge"},{"id":"c8488733-2c81-42aa-953c-562a2f28ba9f","blocks":["56-16","56-17"],"type":"merge"},{"id":"89ba9e4f-4c7e-4395-95c3-88e85c3a8bef","blocks":["56-21","57-1"],"type":"merge"},{"id":"c3452e7b-96d5-45d7-b887-53b95111c19c","blocks":["63-13","64-1"],"type":"merge"},{"id":"4077de99-c60e-4387-9f21-c22b220629b4","blocks":["76-14","77-1"],"type":"merge"},{"id":"7fc6098b-6ff5-4c7a-8adf-ebc349c53ada","blocks":["86-9","86-10"],"type":"merge"},{"id":"ed6e11f4-ebe3-4472-9871-24bf520e779f","blocks":["113-24","114-1"],"type":"merge"},{"id":"0e988018-56a6-44d9-8914-ebda0b77dced","blocks":["127-13","127-14"],"type":"merge"},{"id":"f1574d8c-8062-45eb-9ac8-7dbd7df6e550","blocks":["137-18","137-19"],"type":"merge"},{"id":"c2930bfe-f966-45ff-bce2-bab7505b7e3f","blocks":["140-4","140-5"],"type":"merge"},{"id":"072c2862-547a-4658-bfe7-03657698dfef","blocks":["147-1","147-2"],"type":"merge"}]}