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            "content": "时间：September 22, 2022",
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            "content": "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",
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            "content": "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"
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            "content": "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"
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            "content": "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"
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            "content": "第7章 反常积分"
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            "content": "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"
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            "content": "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"
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            "content": "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"
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            "content": "1.1 三角函数"
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            "content": "\\[\n\\sin (\\alpha + \\beta) = \\sin \\alpha \\cos \\beta + \\cos \\alpha \\sin \\beta\n\\]"
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            "content": "\\[\n\\sin (\\alpha - \\beta) = \\sin \\alpha \\cos \\beta - \\cos \\alpha \\sin \\beta\n\\]"
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            "content": "\\[\n\\cos (\\alpha + \\beta) = \\cos \\alpha \\cos \\beta - \\sin \\alpha \\sin \\beta\n\\]"
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            "content": "\\[\n\\tan (\\alpha + \\beta) = \\frac {\\tan \\alpha + \\tan \\beta}{1 - \\tan \\alpha \\tan \\beta}\n\\]"
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            "content": "\\[\n\\tan (\\alpha - \\beta) = \\frac {\\tan \\alpha - \\tan \\beta}{1 + \\tan \\alpha \\tan \\beta}\n\\]"
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            "content": "\\[\n\\cot (\\alpha + \\beta) = \\frac {\\cot \\alpha \\cot \\beta - 1}{\\cot \\beta + \\cot \\alpha}\n\\]"
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            "content": "\\[\n\\cot (\\alpha - \\beta) = \\frac {\\cot \\alpha \\cot \\beta + 1}{\\cot \\beta - \\cot \\alpha}\n\\]"
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            "content": "\\[\n\\sin \\alpha = \\frac {2 \\tan {\\frac {\\alpha}{2}}}{1 + \\tan^ {2} {\\frac {\\alpha}{2}}}\n\\]"
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            "content": "\\[\n\\sin {\\frac {\\alpha}{2}} = \\pm \\sqrt {\\frac {1 - \\cos \\alpha}{2}}\n\\]"
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            "content": "\\[\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\\]"
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            "content": "\\[\n\\cos \\alpha + \\cos \\beta = 2 \\cos {\\frac {\\alpha + \\beta}{2}} \\cos {\\frac {\\alpha - \\beta}{2}}\n\\]"
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            "content": "\\[\n\\cos \\alpha - \\cos \\beta = - 2 \\sin {\\frac {\\alpha + \\beta}{2}} \\sin {\\frac {\\alpha - \\beta}{2}}\n\\]"
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            "content": "\\[\n\\sin \\alpha + \\sin \\beta = 2 \\sin {\\frac {\\alpha + \\beta}{2}} \\cos {\\frac {\\alpha - \\beta}{2}}\n\\]"
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            "content": "\\[\n\\sin \\alpha \\sin \\beta = - \\frac {1}{2} [ \\cos (\\alpha + \\beta) - \\cos (\\alpha - \\beta) ]\n\\]"
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            "content": "\\[\n\\sin \\alpha \\cos \\beta = \\frac {1}{2} [ \\sin (\\alpha + \\beta) + \\sin (\\alpha - \\beta) ]\n\\]"
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            "content": "\\[\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\\]"
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            "content": "\\[\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\\]"
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            "content": "\\[\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\\]"
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            "content": "4. \\(\\arctan \\frac{1}{2k^2} = \\arctan \\frac{k}{k + 1} - \\arctan \\frac{k - 1}{k}\\)",
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            "content": "\\[\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\\]"
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            "content": "1.3 不等式"
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            "content": "\\[\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\\]"
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            "content": "\\[\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\\]"
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            "content": "\\[\n\\frac {x}{1 + x} \\leq \\ln (1 + x) \\leq x\n\\]"
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            "content": "若 \\(\\sum_{i=1}^{n} t_i = 1\\), \\(f\\) 是 (下) 凸函数, 则",
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            "content": "\\[\n\\sum_ {i = 1} ^ {n} t _ {i} f (x _ {i}) \\geq f (\\sum_ {i = 1} ^ {n} t _ {i} x _ {i})\n\\]"
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            "content": "\\[\n1 + \\frac {1}{2} + \\dots + \\frac {1}{n} <   1 + \\ln \\frac {2}{1} + \\dots + \\ln \\frac {n}{n - 1} = 1 + \\ln n\n\\]"
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            "content": "\\(ax^{2} + by^{2} + cz^{2}\\leq r^{2}\\) 的体积为 \\(\\frac{4\\pi r^3}{3\\sqrt{abc}}\\)",
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            "content": "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",
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            "content": "\\[\nS = \\frac {S ^ {\\prime}}{\\sqrt {a b c}} = \\frac {4 \\pi r ^ {3}}{3 \\sqrt {a b c}}\n\\]"
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            "content": "第 2 章 数列极限"
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            "content": "2.1 数列极限的概念与性质"
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            "content": "定义 2.1 (数列极限)"
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            "content": "\\(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\\)",
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            "content": "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)\\) 。",
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            "content": "定理 2.2 (cauchy 收敛准则)"
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            "content": "数列 \\(a_{n}\\) 收敛当且仅当 \\(\\forall \\epsilon, \\exists N, \\forall m, n > N\\) 有 \\(|a_{n} - a_{m}| < \\epsilon\\)",
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            "content": "练习 2.1 几个经典极限 (1) 证明  \\( \\sin n \\)  发散",
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            "content": "(2) 证明 \\(a_{n} = 1 + \\frac{1}{2} + \\cdots + \\frac{1}{n}\\) 发散",
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            "content": "\\(a_{n}\\) 收敛到 \\(A\\) 当且仅当 \\(a_{2k}, a_{2k + 1}\\) 收敛到 \\(A\\)",
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            "content": "2.2 一些重要极限结论"
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            "content": "2.2 一些重要极限结论"
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            "content": "2.2.1 几个基本极限"
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            "content": "定理 2.5 (常用基本极限)"
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            "content": "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",
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            "content": "- 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\\)",
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            "content": "- 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\\)",
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            "content": "- 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\\)",
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            "content": "- 指数： \\(\\forall k, \\lim_{x\\to +\\infty}\\frac{x^k}{e^x} = 0\\)",
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            "content": "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}\\)",
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            "content": "定理 2.6 (重要极限)"
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            "content": "\\[\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\\]"
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            "content": "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\\)",
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            "content": "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\\)",
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            "content": "解(1)等价于 \\(e^{\\frac{\\ln(1 + x + x^2)}{x}} = e^{\\frac{2x + 1}{1 + x + x^2}} = e\\)",
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            "content": "2.2.2 Stirling 公式与 Wallis 公式"
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            "content": "定理 2.7 (Stirling 公式)"
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            "content": "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}\\)",
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            "content": "定理 2.8 (Wallis 公式)"
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            "content": "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}\\)",
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            "content": "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} \\)",
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            "content": "\\[\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\\]"
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            "content": "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}\\)",
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            "content": "2.2.3 Euler 常数"
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            "content": "定理 2.9 (Euler 常数)"
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            "content": "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。",
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            "content": "推论 2.1 (Euler 常数的应用)"
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            "content": "当 \\(n\\to \\infty\\) 时， \\(\\sum_{k = 1}^{n}\\frac{1}{k} = \\ln n + \\gamma = \\ln n + o(n)\\)",
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            "content": "2.3 夹逼准则"
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            "content": "\\(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\\)。",
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            "content": "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\\)",
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            "content": "(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。",
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            "content": "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。",
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            "content": "2.4.1 有限项幂次根号平均"
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            "content": "\\( 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)",
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            "content": "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。",
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            "content": "2.4 平均值定理"
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            "content": "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。",
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            "content": "(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。",
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            "content": "2.4.2 平均值定理：无穷项算术平均与几何平均"
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            "content": "9"
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            "content": "定理 2.12 (平均值定理)"
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            "content": "已知 \\(\\lim_{n\\to \\infty}a_n = a\\) （有限数或正负无穷），则",
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            "content": "1. 算术平均： \\( \\lim_{n\\to\\infty}\\frac{a_{1}+a_{2}+\\cdots+a_{n}}{n}=a \\)",
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            "content": "2. 几何平均：若  \\( a_{n} > 0 \\) ，则  \\( \\lim_{n \\to \\infty} \\sqrt[n]{a_{1}a_{2}\\cdots a_{n}} = a \\)",
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            "content": "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",
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            "content": "\\[\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\\]"
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            "content": "(2) 取对数则转换为 (1)",
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            "content": "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} \\) \n证明 分为奇偶数项进行讨论：",
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            "content": "\\[\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\\]"
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            "content": "另一方面",
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            "content": "\\[\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\\]"
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            "content": "因此得到结论成立。",
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            "content": "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：",
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            "content": "\\[\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\\]"
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            "content": "因此得到：",
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            "content": "\\[\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\\]"
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            "content": "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。",
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            "content": "2.5 使用定积分定义"
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            "content": "2.4.3 技巧：一招解决 n 次根号与除以 n"
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            "content": "定理 2.13 (n 次根号与除以 n)"
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            "content": "\\(a_{n}\\) 为数列，则",
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            "content": "若 \\(\\lim_{n\\to \\infty}(a_n - a_{n - 1}) = a\\) ，则 \\(\\lim_{n\\to \\infty}\\frac{a_n}{n} = a\\)",
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            "content": "若 \\(\\lim_{n\\to \\infty}\\frac{a_n}{a_{n - 1}} = a\\) ，则 \\(\\lim_{n\\to \\infty}\\sqrt[n]{a_n} = a\\)",
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            "content": "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",
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            "content": "(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}\\).",
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            "content": "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!} \\)",
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            "content": "解 (1) \\(\\frac{a_n}{a_{n-1}} = 1\\)，因此极限为 \\(1\\)。",
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            "content": "(2) \\( \\frac{a_{n}}{a_{n-1}}=\\frac{n}{n-1}\\rightarrow1 \\) ，极限为1",
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            "content": "(3) \\(\\frac{a_n}{a_{n-1}} = n \\to \\infty\\)，因此不收敛",
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            "content": "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} \\)",
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            "content": "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}\\)",
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            "content": "(2) \\( a_{n}=\\frac{n(n+1)\\cdots(2n-1)}{n^{n}} \\) ，因此",
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            "content": "\\[\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\\]"
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            "content": "因此极限为 \\(\\frac{4}{e}\\)",
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            "content": "2.5 使用定积分定义"
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            "content": "2.5.1 使用定积分计算极限"
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            "content": "定理 2.14 (用定积分定义计算极限)"
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            "content": "若 \\(f(x)\\) 在 \\([a,b]\\) 连续，则",
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            "content": "\\[\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\\]"
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            "content": "10"
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            "content": "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。",
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            "content": "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) \\)",
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            "content": "2.5 使用定积分定义"
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            "content": "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",
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            "content": "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)",
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            "content": "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\\)",
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            "content": "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) \\)",
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            "content": "(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)\\)",
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            "content": "(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)\\)",
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            "content": "\\[\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\\]"
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            "content": "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：  \n\\(\\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.\\)",
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            "content": "因此根据夹逼可知极限为 \\(\\frac{2}{\\pi}\\)",
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            "content": "(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\n\\(\\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.\\)",
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            "content": "2.6 单调有界定理"
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            "content": "(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。",
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            "content": "2.5.2 利用定积分定义构造等价无穷小"
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            "content": "定理 2.15 (利用积分构造等价无穷小)"
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            "content": "对于正整数 k， \\( 1^{k} + 2^{k} + \\cdots + n^{k} \\sim \\frac{1}{k+1} n^{k+1} \\)",
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            "content": "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。",
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            "content": "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})} \\)",
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            "content": "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",
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            "content": "\\[\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\\]"
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            "content": "定理 2.16 (已知极限计算求和极限)"
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            "content": "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\\)",
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            "content": "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：",
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            "content": "\\[\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\\]"
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            "content": "2.6 单调有界定理"
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            "content": "2.6.1 单调有界定理及其推广"
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            "content": "定理 2.17 (单调有界定理)"
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            "content": "若数列 \\(a_{n}\\) 单调且有界，则其必收敛。",
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            "content": "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} \\)\n\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}<b_{n}+1 \\)，于是 \\( a_{1}\\leq a_{2}\\leq\\cdots\\leq a_{n}<b_{n}+1\\leq b_{n-1}+1\\leq\\cdots\\leq b_{1}+1 \\)，根据单调有界定理可知均收敛，根据极限四则运算可知极限相等。",
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            "content": "(2) 重点: \\(a_{n} = \\sqrt{1 + \\sqrt{2 + \\cdots + \\sqrt{n}}}\\), 证明 \\(a_{n}\\) 收敛",
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            "content": "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}<a \\) ，则 \\( a_{n+1}<a \\) ，而根据 \\( a_{1}<a \\) 得到 \\( a_{n}<a \\) ，因此有界，根据单调有界定理可知收敛。",
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            "content": "(2) 显然 \\(a_{n}\\) 单增，下证其有界。由于 \\(n < 2^{2^{n}}\\) ，于是",
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            "content": "\\[\na _ {n} <   \\sqrt {2 ^ {2 ^ {1}} + \\sqrt {2 ^ {2 ^ {2}} + \\cdots + \\sqrt {2 ^ {2 ^ {n}}}}} <   2 \\sqrt {1 + \\sqrt {1 + \\cdots + \\sqrt {1}}}\n\\]"
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            "content": "根据(1)可知  \\( \\sqrt{1+\\sqrt{1+\\cdots+\\sqrt{1}}} \\)  有界，故根据单调有界定理收敛。",
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            "content": "笔记 不动点迭代一般有界性都和不动点本身去比。",
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            "content": "2.6.2 相减法"
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            "content": "练习 2.17Euler 常数 重点：证明  \\( a_{n}=1+\\frac{1}{2}+\\cdots+\\frac{1}{n}-\\ln n \\)  收敛\n\n证明 由于  \\( 1+\\frac{1}{2}+\\cdots+\\frac{1}{n}-\\ln 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\n\n\\( \\frac{1}{n+1}<\\ln(1+\\frac{1}{n}) \\)",
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            "content": "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。",
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            "content": "\\(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\\)",
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            "content": "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",
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            "content": "\\[\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\\]"
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            "content": "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}\\)",
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            "content": "2.6.3 相除法"
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            "content": "(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",
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            "content": "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\\)。",
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            "content": "(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}\\) 。",
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            "content": "2.7 递推问题"
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            "content": "2.6.4 不等式放缩"
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            "content": "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。",
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            "content": "(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",
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            "content": "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",
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            "content": "\\[\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\\]"
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            "content": "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。",
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            "content": "(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",
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            "content": "\\[\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\\]"
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            "content": "\\( 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",
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            "content": "\\[\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\\]"
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            "content": "因此 \\(ab = a_{1}b_{1}\\) ，即极限为 \\(a = b = \\sqrt{a_1b_1}\\)",
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            "content": "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",
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            "content": "(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}\\)",
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            "content": "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",
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            "content": "(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。",
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            "content": "笔记 前后项不等式先用同项进行放缩，可得出单调性结论",
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            "content": "2.7.1 极限压缩定理"
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            "content": "定理 2.18 (极限压缩定理)"
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            "content": "\\(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\\)。",
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            "content": "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",
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            "content": "\\[\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\\]"
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            "content": "定理 2.19 (压缩映射定理)"
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            "content": "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。",
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            "content": "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。",
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            "content": "练习 2.24 两个经典循环式 证明收敛并求极限:",
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            "content": "\\[\n(1) x _ {1} = 2, x _ {2} = 2 + \\frac {1}{2}, \\dots , x _ {n + 1} = 2 + \\frac {1}{a _ {n}}\n\\]"
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            "content": "\\[\n(2) x _ {1} = 1, x _ {2} = \\sqrt {2}, \\dots , x _ {n + 1} = \\sqrt {2 \\sqrt {x _ {n}}}\n\\]"
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            "content": "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} \\)",
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            "content": "(2) 设 \\( f(x) = \\sqrt{2x} \\), \\( |f'(x)| = \\frac{\\sqrt{2}}{2\\sqrt{x}} \\leq \\frac{\\sqrt{2}}{2} < 1 \\)。",
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            "content": "2.8 Stolz 定理"
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            "content": "定理 2.20 (Stolz 定理)"
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            "content": "Stolz 定理分为以下两种，下面的 \\(A\\) 可以为无穷。",
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            "content": "- \\(\\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\\)",
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            "content": "- \\(\\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\\)",
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            "content": "练习 2.25 直接使用 用 Stolz 定理证明平均值定理",
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            "content": "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 \\)",
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            "content": "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。",
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            "content": "练习 2.26 几道经典题目 (1) ZJU2022.2:  \\( x_{0} > 0, x_{n} = \\arctan x_{n-1} \\) ，证明：",
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            "content": "(a) \\(\\lim_{n\\to \\infty}x_n = 0\\)",
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            "content": "(b) \\( \\sqrt[n]{nx_{n}} \\) 收敛，并求极限",
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            "content": "\\[\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\\]"
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            "content": "2.9 积分形式极限：分段法"
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            "content": "2.9 积分形式极限：分段法"
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            "content": "2.9.1 定积分形极限"
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            "content": "练习 2.27 基础拆分法 求下列极限:",
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            "content": "(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\\)",
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            "content": "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：",
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            "content": "\\[\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\\]"
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            "content": "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。",
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            "content": "(2)(3) 同理都是 0",
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            "content": "(4) 可以考虑  \\( \\int_{0}^{1} e^{x^{n}} - 1 dx \\) ，结果为 1.",
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            "content": "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)\\)",
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            "content": "2.9.2 反常积分形极限"
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            "content": "定理 2.21 (函数形式平均值定理)"
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            "content": "\\(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",
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            "content": "\\[\n\\lim _ {x \\to + \\infty} \\frac {1}{x} \\int_ {0} ^ {x} f (t) \\mathrm{d} t = A\n\\]"
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            "content": "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",
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            "content": "\\[\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\\]"
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            "content": "\\(\\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",
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            "content": "\\[\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\\]"
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            "content": "因此结论成立",
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            "content": "2.9 积分形式极限：分段法"
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            "content": "2.9.3 周期函数积分极限"
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            "content": "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\\)",
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            "content": "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",
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            "content": "\\[\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\\]"
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            "content": "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",
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            "content": "\\[\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\\]"
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            "content": "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：",
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            "content": "\\[\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\\]"
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            "content": "因此 \\(\\lim_{x\\to \\infty}\\frac{1}{x}\\int_{n_xT}^x f(x)\\mathrm{d}x = 0\\)",
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            "content": "\\[\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\\]"
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            "content": "第 3 章 函数极限与连续性"
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            "content": "定义 3.1 (函数极限)"
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            "content": "\\(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。",
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            "content": "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\\)",
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            "content": "\\(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.\\)",
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            "content": "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",
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            "content": "\\[\n\\delta = \\min \\left\\{\\left| x _ {i} - x _ {0} \\right|: i = 1, 2, \\dots , k \\right\\}\n\\]"
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            "content": "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\\)",
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            "content": "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。",
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            "content": "定理 3.1 (极限的唯一性)"
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            "content": "若函数的极限存在，则它是唯一的。",
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            "content": "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。",
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            "content": "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。",
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            "content": "\\(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))\\)",
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            "content": "\\(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))\\)",
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            "content": "若 \\(f(n)\\) 是 \\(\\Omega (g(n))\\) 且 \\(O(g(n))\\) 的，则记 \\(f(n) = \\Theta (g(n))\\)",
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            "content": "3.3 函数的连续性"
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            "content": "3.2.2 Taylor 展开公式"
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            "content": "指数、对数:",
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            "content": "\\[\ne ^ {x} = \\sum_ {n = 0} ^ {\\infty} \\frac {x ^ {n}}{n !}\n\\]"
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            "content": "\\[\n\\ln (1 + x) = \\sum_ {n = 1} ^ {\\infty} (- 1) ^ {n - 1} \\frac {x ^ {n}}{n}\n\\]"
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            "content": "三角：",
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            "content": "\\[\n\\sin x = \\sum_ {n = 0} ^ {\\infty} (- 1) ^ {n} \\frac {x ^ {2 n + 1}}{(2 n + 1) !}\n\\]"
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            "content": "\\[\n\\cos x = \\sum_ {n = 0} ^ {\\infty} (- 1) ^ {n} \\frac {x ^ {2 n}}{(2 n) !}\n\\]"
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            "content": "\\[\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\\]"
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            "content": "\\[\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\\]"
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            "content": "分式:",
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            "content": "\\[\n{\\frac {1}{1 + x}} = \\sum_ {n = 0} ^ {\\infty} (- 1) ^ {n} x ^ {n}\n\\]"
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            "content": "\\[\n{\\frac {1}{1 - x}} = \\sum_ {n = 0} ^ {\\infty} x ^ {n}\n\\]"
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            "content": "其他：",
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            "content": "\\[\n(1 + x) ^ {\\alpha} = 1 + \\alpha x + \\frac {\\alpha (\\alpha - 1)}{2} x ^ {2} + o (x ^ {2})\n\\]"
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            "content": "\\[\n(1 + x) ^ {\\alpha} = \\sum_ {n = 0} ^ {\\infty} {\\binom {\\alpha} {n}} x ^ {n}\n\\]"
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            "content": "3.3 函数的连续性"
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            "content": "3.3.1 函数连续的定义"
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            "content": "定义 3.5 (单点连续)"
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            "content": "\\(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。",
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            "content": "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。",
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            "content": "若 \\(f(x_0^+) = f(x_0)\\) 则称 \\(f(x)\\) 右连续。左连续同理。",
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            "content": "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！",
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            "content": "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",
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            "content": "\\[\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\\]"
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            "content": "解(1)周期: 任意有理数(显然)",
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            "content": "(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。",
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            "content": "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。",
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            "content": "\\[\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\\]"
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            "content": "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。",
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            "content": "3.3 函数的连续性"
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            "content": "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)|]\\)",
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            "content": "(2) 绝对值: \\( |f(x)| = \\sqrt{|f(x)|^2} \\) 连续",
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            "content": "3.3.2 间断点的概念"
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            "content": "定义 3.7 (第一类间断点)"
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            "content": "跳跃点与可去间断点统称为 \\(f(x)\\) 的第一类间断点",
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            "content": "- 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，",
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            "content": "- 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^+) \\)",
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            "content": "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。",
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            "content": "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。",
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            "content": "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",
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            "content": "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",
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            "content": "\\[\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\\]"
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            "content": "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。",
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            "content": "\\[\n| g (x) - g (x _ {0}) | <   \\epsilon\n\\]"
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            "content": "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",
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            "content": "3.3 函数的连续性"
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            "content": "因此  \\( g(x) \\)  在  \\( x_{0} \\)  连续，由  \\( x_{0} \\)  任意性可知结论成立。",
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            "content": "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",
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            "content": "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：",
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            "content": "\\[\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\\]"
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            "content": "前者取  \\( h \\rightarrow 0^{+} \\) ，后者分别取  \\( h \\rightarrow 0^{+}, h \\rightarrow 0^{-} \\)  则",
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            "content": "\\[\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\\]"
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            "content": "综合得到 \\( f(x_0 - 0) = f(x_0 + 0) = f(x_0) \\)，因此在任意点连续。",
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            "content": "3.3.3 介值性定理"
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            "content": "定理 3.4 (零点存在定理)"
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            "content": "\\[\nf (a) f (b) <   0\n\\]"
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            "content": "\\[\n\\exists \\xi \\in (a, b)\n\\]"
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            "content": "\\[\nf (\\xi) = 0\n\\]"
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            "content": "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\\)",
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            "content": "(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。",
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            "content": "22"
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            "content": "定理 3.5 (介值性定理)"
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            "content": "定理 3.6 (连续函数平均值定理)"
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            "content": "\\(f(x)\\) 在 \\([a,b]\\) 连续， \\(x_{1},\\dots ,x_{n}\\in [a,b]\\) ，证明： \\(\\exists \\xi \\in [a,b]\\) ，使得",
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            "content": "\\[\nf (\\xi) = \\frac {1}{n} [ f (x _ {1}) + \\dots + f (x _ {n}) ]\n\\]"
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            "content": "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",
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            "content": "\\[\nf (x _ {1}) \\leq \\frac {1}{n} [ f (x _ {1}) + \\dots + f (x _ {n}) ] \\leq f (x _ {n})\n\\]"
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            "content": "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。",
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            "content": "3.5 一致连续"
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            "content": "(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。",
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            "content": "(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。",
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            "content": "(2) 和 (1) 同理，只是把 \\(\\exists \\delta\\) 换成 \\(\\exists N\\)",
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            "content": "(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",
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            "content": "3.5 一致连续"
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            "content": "3.5.1 一致连续的概念"
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            "content": "定义 3.9 (一致连续)"
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            "content": "\\(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",
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            "content": "定理 3.7 (常用充要条件)"
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            "content": "\\(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\\)",
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            "angle": 0,
            "content": "24"
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            "content": "证明 (1) 左推右: 显然",
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            "content": "(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。",
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            "content": "笔记 证明不一致连续绝大多数情况都用上述定理。",
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            "content": "(2) 取 \\(x_{n} = \\sqrt{2n\\pi}, y_{n} = \\sqrt{2n\\pi + \\frac{\\pi}{2}}\\)",
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            "content": "(3) 取 \\(x_{n} = \\frac{1}{2n\\pi}, y_{n} = \\frac{1}{2n\\pi + \\frac{\\pi}{2}}\\)",
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            "content": "(4) 取 \\(x_{n} = \\frac{1}{n}, y_{n} = \\frac{1}{n + 1}\\)",
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            "content": "(2) 证明 \\( f(x) = \\sqrt{x} \\) 在 \\([0, +\\infty)\\) 一致连续",
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            "content": "(3) 证明 \\( f(x) = \\sin x \\) 在 \\([0, +\\infty)\\) 一致连续",
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            "content": "\\[\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\\]"
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            "content": "3.5 一致连续"
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            "content": "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",
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            "content": "3.5.2 Cantor 定理及其推广"
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            "content": "定理 3.8 (Cantor 定理)"
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            "content": "<table><tr><td>定理 3.8 (Cantor 足理)</td></tr><tr><td>若 f(x) 在闭区间 [a,b] 连续,则其在 [a,b] 上一致连续。</td></tr></table>"
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            "content": "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。",
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            "content": "笔记 Cantor 定理的其他证明方式:"
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            "content": "- 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\\)",
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            "content": "- 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",
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            "content": "- 单调有界定理：和闭区间套一种构造方式",
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            "content": "定理 3.9 (开区间一致连续充要条件)"
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            "content": "<table><tr><td>f(x) 在 (a,b) 连续,则 f(x) 在 (a,b) 一致连续的充要条件是 <eq>\\lim_{x\\to a^{+}}f(x),\\lim_{x\\to b^{-}}f(x)</eq> 存在(有限)</td></tr></table>"
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            "content": "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。",
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            "content": "(2) 右推左：补充定义两个端点，根据 Cantor 定理可知。",
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            "content": "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。",
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            "content": "定理 3.10 (无穷区间一致连续)"
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            "content": "<table><tr><td>定理3.10(元为区间一致连续)</td></tr><tr><td>f(x)在[a,+∞)连续,limx→+∞f(x)=A(有限),则f(x)在[a,+∞)一致连续</td></tr></table>"
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            "content": "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。",
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            "content": "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。",
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            "content": "(2) 因为 \\(\\lim_{x\\to 0}\\sin \\frac{1}{x}\\) 不存在，故不一致连续。",
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            "content": "3.5 一致连续"
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            "content": "3.5.3 Lip 连续与一致连续"
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            "content": "定理 3.11 (Lip 连续与一致连续)"
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            "angle": 0,
            "content": "\\(f(x)\\) 在区间 \\(I\\) 上满足Lip条件，则 \\(f(x)\\) 在 \\(I\\) 上一致连续",
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            "content": "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。",
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            "content": "推论 3.1 (导数有界一定一致连续)",
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            "angle": 0,
            "content": "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",
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            "content": "推论 3.2 (导数在无穷处趋于无穷则不一致连续)"
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            "angle": 0,
            "content": "若 \\(\\lim_{x\\to +\\infty}f'(x) = \\infty\\) ，则 \\(f(x)\\) 在 \\([a, + \\infty)\\) 不一致连续",
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            "content": "证明 根据 \\(\\lim_{x\\to +\\infty}|f'(x)| = \\infty\\) 可知 \\(|x' - x''|\\) 控制不住导数的增长。",
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            "content": "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。",
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            "content": "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) \\)",
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            "content": "解(1)一致连续，因为导数有界",
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            "angle": 0,
            "content": "(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",
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            "content": "(3) 导数有界，一致连续",
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            "content": "(4) 导数在无穷处趋于无穷，则不一致连续。",
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            "content": "3.5.4 区间合并问题"
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            "content": "定理 3.12 (一致连续区间合并定理)"
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            "content": "3.6 一致连续的经典问题"
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            "content": "3.5.5 四则运算、复合的一致连续性"
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            "content": "定理 3.13 (四则运算的一致连续性)"
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            "content": "设 \\(f(x), g(x)\\) 在区间 \\(I\\) 上一致连续，则",
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            "content": "- 数乘、加减：\\(\\forall k \\in \\mathbb{R}\\)，\\(kf(x), f(x) \\pm g(x)\\) 在 \\(I\\) 一致连续",
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            "content": "- 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",
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            "content": "- 除法： \\( f(x) \\neq 0 \\)，\\( \\frac{1}{f(x)} \\)，\\( \\frac{g(x)}{f(x)} \\) 不一定一致连续",
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            "content": "证明 (1) 直接三角不等式放缩"
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            "content": "(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。",
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            "content": "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",
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            "content": "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",
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            "content": "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。",
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            "content": "笔记 上述练习本身也是一个重要的结论。",
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            "content": "定理 3.14 (复合的一致连续性)"
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            "content": "\\(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",
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            "content": "证明 根据定义套即可。外层取  \\( \\epsilon, \\delta \\) ，内层取  \\( \\delta, \\delta' \\) 。",
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            "content": "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。",
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            "content": "3.6 一致连续的经典问题"
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            "content": "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\\)",
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            "content": "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",
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            "content": "\\[\n| f (x) | \\leq M, x \\in [ - \\delta , \\delta ]\n\\]"
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            "content": "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",
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            "content": "27"
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            "content": "\\[\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\\]"
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            "content": "第 4 章 一元微分学"
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            "content": "4.1 导数与微分"
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            "content": "4.1.1 导数概念与可导性"
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            "content": "<table><tr><td>定义4.1(导数、左右导数)</td></tr><tr><td>导数定义为<eq>\\lim_{x\\to x_0}\\frac{f(x)-f(x_0)}{x-x_0}</eq>,左右导数定义为<eq>\\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}</eq></td></tr></table>"
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            "content": "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。",
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            "content": "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) \\)",
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            "content": "\\[\nf (x) = \\left\\{ \\begin{array}{l l} g (x) \\sin {\\frac {1}{x}}, & x \\neq 0 \\\\ 0, & x = 0 \\end{array} \\right.\n\\]"
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            "content": "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 \\)",
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            "content": "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\\)",
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            "content": "(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",
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            "content": "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：",
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            "content": "\\[\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\\]"
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            "content": "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}<A+\\frac{\\epsilon}{2} \\) ，因此  \\( f'(0) \\)  存在且等于 A。",
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            "content": "练习 4.3Dirichlet 函数的可微性  \\( D(x) \\)  为 Dirichlet 函数，判断",
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            "content": "(1) \\( f(x)=x^{2}D(x) \\)",
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            "content": "(2) \\( f(x)=xD(x) \\) 的可导性:",
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            "content": "\\[\nD (x) = \\left\\{ \\begin{array}{l l} 0, & x \\text {为无理数} \\\\ 1, & x \\text {为有理数} \\end{array} \\right.\n\\]"
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            "content": "解(1)先考察连续性：根据有理、无理数的稠密性，只有 \\(x = 0\\) 处连续。考察可导性时根据定义不管沿着无理数或有理数逼近，极限都为0，因此可导",
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            "content": "(2) 在 \\(x = 0\\) 连续，但不可导，因为沿着无理数逼近和沿着有理数逼近极限不一样。",
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            "content": "笔记不存在仅在一点可导且二阶可导的函数，原因在于二阶可导要求导数在一个小区间内存在。",
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            "content": "练习4.4Riemann函数的不可微性重点：证明Riemann函数 \\( R(x) \\) 在[0,1]处处不可微。",
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            "content": "\\[\nR (x) = \\left\\{ \\begin{array}{l l} \\frac {1}{q}, & x = \\frac {p}{q} (\\text {既约真分数}) \\\\ 0, & x \\text {为无理数或} 0, 1 \\end{array} \\right.\n\\]"
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            "content": "证明 (1) 有理点: 甚至不连续, 因此一定不可微 (具体见函数连续一节)",
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            "content": "4.1 导数与微分"
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            "content": "(2) 无理数：设 \\(x_0 = 0.a_1a_2\\cdots\\) 为无理数，若无理序列逼近，\\(\\frac{f(x_n) - f(x_0)}{x_n - x_0} = 0\\) 极限存在。有理序列逼近，取 \\(y_n = 0.a_1a_2\\cdots a_{n-1}a_n\\)，\\(R(y_n) = \\frac{1}{q} \\geq \\frac{1}{10^n}\\)，\\(|y_n - x_0| \\leq \\frac{1}{10^n}\\)，从而 \\(|\\frac{R(y_n) - R(x_0)}{y_n - x_0}| \\geq 1\\)，这说明 \\(\\lim_{n \\to \\infty} \\frac{R(y_n)}{y_n - x_0} \\neq 0\\)",
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            "content": "笔记 Riemann 函数性质总结:",
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            "content": "- 在 \\((0,1)\\) 每个点极限为 \\(0\\) (见函数极限)",
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            "bbox": [
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            "content": "- 在无理点连续，有理点间断",
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            "bbox": [
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            "angle": 0,
            "content": "- 在 \\([0,1]\\) 黎曼可积 (因为几乎处处连续)",
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            "content": "- 在 \\([0,1]\\) 处处不可微 (见上面例子)",
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            "content": "练习4.5利用导数概念 \\(f(x)\\) 对 \\(\\forall x_{1},x_{2}\\in \\mathbb{R}\\) 有 \\(f(x_{1} + x_{2}) = f(x_{1})f(x_{2})\\) ，若 \\(f^{\\prime}(0) = 1\\) ，证明： \\(\\forall x\\in \\mathbb{R}\\) 有\\(f^{\\prime}(x) = f(x)\\)",
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            "content": "证明 \\( f(x) = f(x) \\cdot f(0) \\)，因此 \\( f(x)[1 - f(0)] = 0 \\)，要么 \\( f(0) = 1 \\)，要么 \\( f(x) \\equiv 0 \\)，后者显然和 \\( f'(0) = 1 \\) 矛盾，因此 \\( f(0) = 1 \\)。因此",
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            "content": "\\[\nf ^ {\\prime} (x) = \\lim _ {\\Delta x \\to 0} \\frac {f (x) [ f (\\Delta x) - 1 ]}{\\Delta x} = f (x) \\lim _ {\\Delta x \\to 0} \\frac {f (\\Delta x) - f (0)}{\\Delta x} = f (x) \\cdot f ^ {\\prime} (0) = f (x)\n\\]"
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            "content": "4.1.2 微分的概念"
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            "content": "定义 4.2 (可微)"
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            "content": "\\(f(x)\\) 在 \\(x_0\\) 邻域 \\(U(x_0)\\) 中有定义，且增量可写为 \\(f(x_0 + \\Delta x) - f(x_0) = A\\Delta x + o(\\Delta x)\\)，此时称 \\(f(x)\\) 在 \\(x_0\\) 可微，记",
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            "content": "\\[\n\\mathrm{d} x := \\Delta x, \\quad \\mathrm{d} y := A \\Delta x = f ^ {\\prime} (x _ {0}) \\mathrm{d} x\n\\]"
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            "content": "练习 4.6 可导性与可微性 证明： \\( f(x) \\)  在  \\( x_{0} \\)  可导当且仅当  \\( f(x) \\)  在  \\( x_{0} \\)  邻域可写为  \\( f(x) = f(x_{0}) + \\varphi(x)(x - x_{0}) \\) ，其中  \\( \\varphi(x) \\)  在  \\( x_{0} \\)  处连续",
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            "content": "证明 (1) 右推左：根据  \\( \\lim_{x\\to x_{0}}\\frac{f(x)-f(x_{0})}{x-x_{0}}=\\lim_{x\\to x_{0}}\\varphi(x)=\\varphi(x_{0}) \\)  即可",
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            "content": "(2) 左推右：构造  \\( \\varphi(x)=\\left\\{\\begin{aligned}\\frac{f(x)-f(x_{0})}{x-x_{0}},&x\\neq x_{0}\\\\ f(x_{0})\\end{aligned}\\right. \\)",
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            "content": "4.1.3 常用的导数公式"
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            "content": "三角",
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            "content": "\\[\n(\\tan x) ^ {\\prime} = \\sec^ {2} x\n\\]"
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            "content": "\\[\n(\\sec x) ^ {\\prime} = \\sec x \\tan x\n\\]"
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            "content": "\\[\n(\\cot x) ^ {\\prime} = - \\csc^ {2} x\n\\]"
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            "content": "\\[\n(\\csc x) ^ {\\prime} = - \\csc x \\cot x\n\\]"
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            "content": "\\[\n(\\sin x) ^ {(n)} = \\sin \\left(x + \\frac {n \\pi}{2}\\right)\n\\]"
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            "content": "\\[\n(\\cos x) ^ {(n)} = \\cos \\left(x + \\frac {n \\pi}{2}\\right)\n\\]"
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            "content": "\\[\n(\\arcsin x) ^ {\\prime} = \\frac {1}{\\sqrt {1 - x ^ {2}}}\n\\]"
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            "content": "\\[\n(\\arccos x) ^ {\\prime} = - \\frac {1}{\\sqrt {1 - x ^ {2}}}\n\\]"
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            "content": "\\[\n(\\arctan x) ^ {\\prime} = \\frac {1}{1 + x ^ {2}}\n\\]"
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            "content": "\\[\n(\\operatorname{arccot} x) ^ {\\prime} = - \\frac {1}{1 + x ^ {2}}\n\\]"
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            "content": "4.1 导数与微分"
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            "content": "4.1.4 隐函数、反函数求导"
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            "content": "定理 4.1 (隐函数求导)"
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            "content": "等式两侧同时对同一个自变量求导。",
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            "content": "定理 4.2 (反函数求导)"
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            "content": "\\( y = y(x) \\) 的反函数为 \\( x = x(y(x)) \\)，则",
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            "content": "\\[\ny ^ {\\prime} (x) = \\frac {1}{x ^ {\\prime} (y)}\n\\]"
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            "content": "证明 若 \\(x = g(y)\\)，对 \\(x\\) 求导得到 \\(1 = \\frac{\\mathrm{d}g}{\\mathrm{d}y} \\cdot \\frac{\\mathrm{d}y}{\\mathrm{d}x}\\)，因此得到 \\(\\frac{\\mathrm{d}y}{\\mathrm{d}x} = \\frac{1}{g'(y)}\\)",
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            "content": "笔记一般都不是直接用结论，而是取 \\(x = g(y)\\) 后两侧对 \\(x\\) 求导进行计算的。",
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            "content": "练习4.7具体反函数求导练习求(1) \\(\\arcsin x(2)\\arccos x(3)\\arctan x\\) 的导数解 \\((1)y = \\arcsin x\\) ，因此 \\(x = \\sin y\\) ，两侧求导 \\(1 = \\cos y\\cdot y^{\\prime}\\) ，因此 \\(y^\\prime = \\frac{1}{\\cos y} = \\frac{1}{\\sqrt{1 - x^2}}\\) ，这里 \\(\\cos y\\) 是通过画直角三角形，三条边分别为 \\(x,\\sqrt{1 - x^2},1\\) 得到的。",
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            "content": "练习4.8反函数高阶求导 \\( y = f(x) \\) 有三阶导数， \\( y' = f'(x) \\neq 0 \\)，若 \\( f(x) \\) 有反函数 \\( x = f^{-1}(y) \\)，证明：",
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            "content": "\\[\n(f ^ {- 1}) ^ {\\prime} (y) = \\frac {1}{y ^ {\\prime}}, \\quad (f ^ {- 1}) ^ {\\prime \\prime} (y) = - \\frac {y ^ {\\prime \\prime}}{(y ^ {\\prime}) ^ {3}}, \\quad (f ^ {- 1}) ^ {\\prime \\prime \\prime} (y) = \\frac {3 (y ^ {\\prime \\prime}) ^ {2} - y ^ {\\prime} y ^ {\\prime \\prime \\prime}}{(y ^ {\\prime}) ^ {5}}\n\\]"
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            "content": "证明 建议采用隐函数求导。根据 \\( y = f(x) \\)，两侧对 \\( y \\) 求导得到 \\( 1 = f'(x)x' \\)，因此得到 \\( x' = \\frac{1}{f'(x)} = \\frac{1}{y'} \\)。\\( 0 = f''(x)(x')^2 + f'(x)x'' \\)，因此推出 \\( x'' = -\\frac{y''}{(y')^3} \\)",
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            "content": "笔记 这种题目一定要搞清楚导数的一撇代表的意思。例如 \\( y' = \\frac{\\mathrm{d}y}{\\mathrm{d}x} \\)，而 \\( x' = \\frac{\\mathrm{d}x}{\\mathrm{d}y} \\)，两者的求导对象不同。",
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            "content": "定理 4.3 (指数求导)"
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            "content": "对于 \\(y = u(x)^{v(x)}\\) ，取对数得到 \\(\\ln y = v(x)\\ln u(x)\\) ，两侧求导可以得到",
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            "content": "\\[\n\\frac {y ^ {\\prime}}{y} = v ^ {\\prime} (x) \\ln u (x) + v (x) \\frac {u ^ {\\prime} (x)}{u (x)} \\Rightarrow y ^ {\\prime} = u (x) ^ {v (x)} [ v (x) \\ln u (x) + v (x) \\frac {u ^ {\\prime} (x)}{u (x)} ]\n\\]"
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            "content": "练习 4.9 指数求导练习 求导：(1) \\( x^{\\frac{1}{x}} \\) (2) \\( x^{x}(3)x^{x^{x}} \\)",
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            "content": "\\[\n\\begin{array}{l l l} (\\sin x) ^ {(n)} = \\sin (x + \\frac {n \\pi}{2}) & (\\cos x) ^ {(n)} = \\cos (x + \\frac {n \\pi}{2}) \\\\ [ (1 + x) ^ {\\alpha ] ^ {(n)}} = \\alpha (\\alpha - 1) \\dots (\\alpha - n + 1) (1 + x) ^ {\\alpha - n} \\\\ (\\ln x) ^ {(n)} = \\frac {(- 1) ^ {n - 1} (n - 1) !}{x ^ {n}} & [ \\ln (1 + x) ] ^ {(n)} = \\frac {(- 1) ^ {n - 1} (n - 1) !}{(1 + x) ^ {n}} & [ \\ln (1 - x) ] ^ {(n)} = \\frac {(n - 1) !}{(1 - x) ^ {n}} \\\\ (\\frac {1}{x}) ^ {(n)} = \\frac {(- 1) ^ {n} n !}{x ^ {n + 1}} & (\\frac {1}{1 + x}) ^ {(n)} = \\frac {(- 1) ^ {n} n !}{(1 + x) ^ {n + 1}} & (\\frac {1}{1 - x}) ^ {(n)} = \\frac {n !}{(1 - x) ^ {n + 1}} \\end{array}\n\\]"
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            "content": "练习 4.11 三角函数与高阶导数 求 n 阶导数:",
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            "content": "\\[\n\\begin{array}{l} {(1) y = \\sin^ {6} x + \\cos^ {6} x} \\\\ {(2) y = \\sin a x \\cos b x} \\end{array}\n\\]"
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            "content": "解(1)这种形式在高阶导数、不定积分等中很常见！先用立方和公式：",
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            "content": "\\[\n\\begin{array}{c} \\sin^ {6} x + \\cos^ {6} x = \\sin^ {4} x - \\sin^ {2} x \\cos^ {2} x + \\cos^ {4} x = (\\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}{4} \\left(\\frac {1 - \\cos 4 x}{2}\\right) = \\frac {5}{8} + \\frac {3}{8} \\cos 4 x \\end{array}\n\\]"
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            "content": "因此 \\(y^{(n)} = \\frac{3\\cdot 4^n}{8}\\cos (4x + \\frac{n\\pi}{2})\\)",
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            "content": "(2) 积化和差即可",
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            "content": "练习 4.12 反三角函数高阶导数 (1) 重点:  \\( f(x) = \\arctan x \\) ，求  \\( f^{(n)}(0) \\)",
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            "content": "(2) 重点: \\(g(x) = \\arcsin x\\), 求 \\(g^{(n)}(0)\\)",
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            "content": "证明 (1) \\( f'(x)=\\frac{1}{1+x^{2}} \\) ，根据 Taylor 展开可知",
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            "content": "\\[\nf ^ {\\prime} (x) = \\frac {1}{1 + x ^ {2}} = 1 - x ^ {2} + x ^ {4} - \\dots + (- 1) ^ {m} x ^ {2 m} + \\dots\n\\]"
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            "content": "因此  \\( f(x)=x-\\frac{x^{3}}{3}+\\frac{x^{5}}{5}-\\cdots+\\frac{(-1)^{m}}{2m+1}x^{2m+1}+\\cdots \\) ，根据系数可知  \\( f^{(2m)}(x)=0, f^{(2m+1)}=(-1)^{m}(2m)! \\) \n\n(2) \\( g'(x)=\\frac{1}{\\sqrt{1-x^{2}}} \\) ，根据 Taylor 展开得到  \\( ((1-x)^{\\alpha} \\)  的展开式):",
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            "content": "\\[\ng ^ {\\prime} (x) = \\left(1 - x ^ {2}\\right) ^ {- \\frac {1}{2}} = 1 + \\sum_ {m = 1} ^ {\\infty} \\frac {(- \\frac {1}{2}) (- \\frac {3}{2}) \\cdots (- \\frac {1}{2} - m + 1)}{m !} (- x ^ {2}) ^ {m} = 1 + \\sum_ {m = 1} ^ {\\infty} \\frac {(2 m - 1) ! !}{(2 m) ! !} x ^ {2 m}\n\\]"
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            "content": "得到 \\(g(x) = x + \\sum_{m=1}^{\\infty} \\frac{(2m-1)!!}{(2m)!!(2m+1)} x^{2m+1}\\)，因此得到 \\(g^{(2m)}(0) = 0, g^{(2m+1)} = \\frac{(2m-1)!!}{(2m)!!}(2m+1)!\\)",
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            "content": "\\(f(x), g(x)\\) 为 \\(n\\) 阶可导函数，则 \\(f(x)g(x)\\) 也 \\(n\\) 阶可导，且 \\([f(x)g(x)]^{(n)} = \\sum_{k=0}^{n}\\binom{n}{k}f^{(n-k)}(x)g^{(k)}(x)\\)",
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            "content": "练习4.13Leibniz公式练习求 \\(n\\) 阶导数： \\((1)y = \\frac{2x}{1 - x^2}\\) (2) \\(y = x^n\\ln x\\) (3) \\(y = e^{ax}\\sin x\\)",
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            "content": "练习 4.14 使用导数构造 求  \\( (1)I_{1}=1+2x+3x^{2}+\\cdots+nx^{n-1} \\)",
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            "content": "(2) \\( I_{2}=1^{2}+2^{2}x+3^{2}x^{2}+\\cdots+n^{2}x^{n-1} \\)",
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            "content": "(3) 推论： \\( 1^{2} + 2^{2} + \\cdots + n^{2} = \\frac{n(n+1)(2n+1)}{6} \\)",
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            "content": "4.2 微分中值定理"
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            "content": "解(1)由于 \\(x + x^{2} + \\dots +x^{n} = \\frac{x - x^{n + 1}}{1 - x}\\) ，两边求导 \\(I_{1} = \\frac{n x^{n + 1} - (n + 1)x^{n} + 1}{(x - 1)^{2}}\\)",
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            "content": "(2) 根据 (1) 可知  \\( 1 + 2x + 3x^{2} + \\cdots + nx^{n-1} = \\frac{nx^{n+1} - (n+1)x^{n} + 1}{(x-1)^{2}} \\) ，同时乘 x 再求导即可得到结果\n\n\\( \\frac{n^{2}x^{n+2} - (2n^{2} + 2n - 1)x^{n+1} + (n+1)^{2}x^{n} - x - 1}{(x - 1)^{3}} \\)",
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            "content": "(3) 为 (2) 的推论",
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            "content": "4.2 微分中值定理"
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            "content": "4.2.1 三大中值定理"
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            "content": "定理 4.5 (费马定理)"
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            "content": "若 \\(x_0\\) 为 \\(f(x)\\) 极值点，且 \\(f'(x_0)\\) 存在，则 \\(f'(x_0) = 0\\)",
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            "content": "证明 不妨设 \\(x_0\\) 为极小值点，先看右导数 \\(f'(x_0) = f'_+(x_0) = \\lim_{x \\to x_0^+} \\frac{f(x) - f(x_0)}{x - x_0} \\geq 0\\)，同理看左导数 \\(f'(x_0) = f'_-(x_0) = \\lim_{x \\to x_0^-} \\frac{f(x) - f(x_0)}{x - x_0} \\leq 0\\)，由于导数存在，则左右导数相等，故导数 \\(f'(x_0)\\) 只能为 0",
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            "content": "定理 4.6 (Rolle 中值定理)"
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            "content": "若 \\(f(x)\\) 在 \\([a,b]\\) 连续，在 \\((a,b)\\) 可导，\\(f(a) = f(b)\\)，则 \\(\\exists \\xi \\in (a,b)\\) 使得：",
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            "content": "\\[\nf ^ {\\prime} (\\xi) = 0\n\\]"
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            "content": "证明 设 \\(M, m\\) 是 \\(f(x)\\) 在 \\([a, b]\\) 最大、最小值。若 \\(M = m\\)，则 \\(f(x) = c\\)。若 \\(m < M\\)，则 \\(f(x)\\) 在 \\((a, b)\\) 至少有个最值点 (最大 or 最小)，则内部的最值点是极值点，因此由费马定理可知结论成立。",
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            "content": "笔记广义的Rolle中值定理：开区间或无穷区间 \\((a,b)\\)，若 \\(f(x)\\) 在 \\(x\\to a^{+},b^{-}\\) 存在且相等，则Rolle中值定理结论成立。",
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            "content": "推论 4.1 (研究导函数零点情况)"
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            "content": "\\(f(x)\\) 在区间 \\(I\\) 上 \\(k\\) 阶可导，若 \\(f(x)\\) 在 \\(I\\) 上有 \\(n + 1\\) 个互异的实根，则 \\(f^{(k)}(x)\\) 有 \\(n - k\\) 个互异实根。",
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            "content": "练习 4.16Lagrange 中值定理基本应用 (1) 设  \\( f(x) \\)  在  \\( [a, b] \\)  连续， \\( (a, b) \\)  可导，证明： \\( \\exists \\xi \\in (a, b) \\)  使得  \\( \\frac{bf(b) - af(a)}{b - a} = f(\\xi) + \\xi f'(\\xi) \\)",
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            "content": "4.2 微分中值定理"
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            "content": "(2) 设 \\( f(x) \\) 在 \\([a, b]\\) 连续，\\((a, b)\\) 可导，\\( f(b) \\neq f(a) \\)，证明：若 \\( f(x) \\) 非线性函数，则 \\(\\exists \\xi, \\eta \\in (a, b)\\) 使得 \\( f'(\\xi) < \\frac{f(b) - f(a)}{b - a} < f'(\\eta) \\)",
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            "content": "证明 (1) 取  \\( F(x) = xf(x) \\)  即可，用 Lagrange 中值定理",
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            "content": "(2) 只说明左侧：要证明 \\( f'(\\xi) < \\frac{f(b) - f(a)}{b - a} \\)，反设不成立，即 \\(\\forall x\\) 有 \\( f'(x) \\geq \\frac{f(b) - f(a)}{b - a} \\)，可构造 \\( F(x) = f(x) - \\frac{f(b) - f(a)}{b - a} x \\)，且 \\( F'(x) \\geq 0 \\)，而 \\( F(a) = F(b) = 0 \\) 这说明 \\( F'(x) = c \\)，这与 \\( f(x) \\) 非线性矛盾。",
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            "content": "定理 4.8 (Cauchy 中值定理)"
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            "content": "\\(f, g\\) 满足 \\([a, b]\\) 连续，\\((a, b)\\) 可导，\\(f'(x), g'(x)\\) 不同时为 0，\\(g(a) \\neq g(b)\\)，则 \\(\\exists \\xi \\in (a, b)\\) 使得",
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            "content": "\\[\n\\frac {f ^ {\\prime} (\\xi)}{g ^ {\\prime} (\\xi)} = \\frac {f (b) - f (a)}{g (b) - g (a)}\n\\]"
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            "content": "证明 考虑 \\(\\frac{f(b) - f(a)}{g(b) - g(a)} g'(\\xi) = f'(\\xi)\\)，采用积分法构造辅助函数，\\(F(x) = f(x) - \\frac{f(b) - f(a)}{g(b) - g(a)} g(x)\\)，\\(F(b) - F(a) = [f(b) - f(a)] - \\frac{f(b) - f(a)}{g(b) - g(a)} [g(b) - g(a)] = 0\\)，因此 \\(F(a) = F(b)\\)，根据 Rolle 中值定理可知结论成立。",
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            "content": "笔记 一般而言条件都是 \\( g'(x) \\neq 0 \\)，\\( g'(x) \\) 与 \\( f'(x) \\) 不同时为零相对更广泛一点，避开了特殊的情况。",
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            "content": "练习 4.17Cauchy 中值定理基本应用 (1) 重点: 设 \\(f(x)\\) 在 \\([a, b]\\) 连续, \\((a, b)\\) 可导, \\(ab > 0\\), 证明 \\(\\exists \\xi \\in (a, b)\\) 使得 \\(\\frac{af(b) - bf(a)}{a - b} = f(\\xi) - \\xi f'(\\xi)\\)",
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            "content": "(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) \\)",
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            "content": "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：",
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            "content": "\\[\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\\]"
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            "content": "根据 Cauchy 中值定理即可证明。",
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            "content": "(2) 等价于  \\( \\frac{\\frac{e^{b}}{b}-\\frac{e^{a}}{a}}{\\frac{1}{b}-\\frac{1}{a}}=e^{\\theta}-\\theta e^{\\theta} \\)",
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            "content": "4.2.2 中值问题：构造辅助函数"
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            "content": "定理 4.9 (直接积分法)"
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            "content": "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",
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            "content": "定理 4.10 (常用中值定理辅助函数)"
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            "content": "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：",
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            "content": "\\[\nm f (\\xi) + n f ^ {\\prime} (\\xi): f (x) e ^ {\\frac {m}{n} x}\n\\]"
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            "content": "- \\(mf(\\xi) + n\\xi f'(\\xi)\\): \\(x^{m}f^{n}(x)\\) (注意不是 \\(x^{\\frac{m}{n}}f(x)!\\))",
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            "content": "- \\(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",
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            "content": "\\[\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\\]"
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            "content": "4.2 微分中值定理"
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            "content": "\\[\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\\]"
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            "content": "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) \\)",
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            "content": "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。",
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            "content": "\\[\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\\]"
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            "content": "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。",
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            "content": "\\[\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\\]"
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            "content": "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)\\)",
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            "content": "(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) \\)\n\n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\\)",
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            "content": "(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。",
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            "content": "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\\)",
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            "content": "(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) \\)\n\n(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 \\)",
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            "content": "(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。",
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            "content": "(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",
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            "content": "(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。",
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            "content": "(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)\\)",
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            "content": "4.2 微分中值定理"
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            "content": "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：",
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            "content": "\\[\n0 <   F (c) = F (c) - F (a) = F ^ {\\prime} (\\eta) (c - a) \\Rightarrow F ^ {\\prime} (\\eta) > 0\n\\]"
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            "content": "根据导数介值性定理可知结论成立。",
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            "content": "(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\\)",
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            "content": "4.2.3 带积分的微分中值问题"
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            "content": "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。",
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            "content": "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}\\)",
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            "content": "(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",
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            "content": "(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",
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            "content": "\\[\n\\int_ {0} ^ {1} x f (x) d x = \\eta f (\\eta) \\cdot 1 = F (\\eta) = f (1)\n\\]"
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            "content": "而  \\( F(1)=f(1) \\) ，因此根据 Rolle 中值定理可知。",
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            "content": "(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。",
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            "content": "(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。",
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            "content": "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\\)",
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            "content": "4.2.4 多个中值点的中值问题"
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            "content": "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：",
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            "content": "(1) \\( \\exists\\xi\\in(0,1) \\) 使得 \\( f(\\xi)=1-\\xi \\)",
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            "content": "(2) 存在互异的 \\(\\eta_1, \\eta_2 \\in (0,1)\\) 使得 \\(f'(\\eta_1)f'(\\eta_2) = 1\\)",
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            "content": "(3) 存在互异的 \\(\\eta_1, \\eta_2 \\in (0,1)\\) 使得 \\(f'(\\eta_1)(f'(\\eta_2) + 1) = 2\\)",
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            "content": "4.2 微分中值定理"
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            "content": "(3) 提示：先找 \\(\\xi\\) 满足 \\(f(\\xi) = 2(1 - \\xi)\\)，再用两次 Lagrange 中值定理。",
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            "content": "4.2.5 中值等式问题"
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            "content": "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。",
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            "content": "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)\\)",
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            "content": "(2) 变形： \\( f(x) \\)  在  \\( [a, b] \\)  二阶可导，a < c < b，证明  \\( \\exists \\xi \\in (a, b) \\)  使得",
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            "content": "\\[\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\\]"
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            "content": "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。",
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            "content": "(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",
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            "content": "\\[\nF (x) = (x - a) [ f (b) - f (a) ] - (b - a) [ f (x) - f (a) ] - \\frac {k}{2} (b - x) (b - a) (x - a)\n\\]"
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            "content": "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。",
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            "content": "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",
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            "content": "\\[\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\\]"
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            "content": "(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",
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            "content": "\\[\nf (b) - 2 f (\\frac {a + b}{2}) + f (a) = \\frac {1}{4} (b - a) ^ {2} f ^ {\\prime \\prime} (\\xi)\n\\]"
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            "content": "(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",
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            "content": "\\[\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\\]"
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            "content": "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",
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            "content": "\\[\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\\]"
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            "content": "\\( F(a)=F(b)=0 \\) ，根据 Rolle 得到  \\( \\exists\\eta\\in(a,b) \\)  使得  \\( F'(\\eta)=0 \\) ，即",
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            "content": "\\[\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\\]"
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            "content": "而根据  \\( \\eta \\)  在  \\( \\frac{a+\\eta}{2} \\)  的 Taylor 展开， \\( \\exists\\xi\\in\\left(\\frac{a+\\eta}{2},\\eta\\right) \\)",
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            "content": "\\[\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\\]"
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            "content": "因此对比得到  \\( k = f''(\\xi) \\)",
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            "content": "4.2.6 中值不等式"
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            "content": "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)\\)",
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            "content": "(3) 特例 1: 区间 \\([0,2]\\), \\(|f(x)| \\leq 1\\), \\(|f''(x)| \\leq 2\\) 证明 \\(|f'(x)| \\leq 3\\)",
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            "content": "4.3 洛必达法则与 Taylor 公式"
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            "content": "(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\\)",
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            "content": "(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)\\)",
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            "content": "(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} \\)\n\n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。",
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            "content": "(1) 根据 Taylor 展开得到 \\(\\exists \\xi \\in (a, x), \\eta \\in (x, b)\\) 使得",
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            "content": "\\[\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\\]"
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            "content": "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",
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            "content": "\\[\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\\]"
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            "content": "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)\\)",
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            "content": "(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。",
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            "content": "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}\\)",
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            "content": "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",
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            "content": "\\[\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\\]"
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            "content": "因此相加得到  \\( \\left|f(x_{1})-f(x_{2})\\right|\\leq\\frac{b-a}{2} \\)",
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            "content": "4.2.7 导数极限定理与导数介值定理"
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            "content": "定理 4.11 (导数介值定理：达布定理)"
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            "content": "<table><tr><td>定理4.11(等数升值定理:达布定理)</td></tr><tr><td>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</td></tr></table>"
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            "content": "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",
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            "content": "4.3 洛必达法则与 Taylor 公式"
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            "content": "<table><tr><td>f,g满足(1) <eq>\\lim_{x\\to x_0}f(x)=\\lim_{x\\to x_0}g(x)=0</eq> 或者 <eq>\\lim_{x\\to x_0}g(x)=\\infty(2)x_0</eq> 空心邻域中 <eq>f&#x27;(x),g&#x27;(x)</eq> 有定义(非常重要的条件!)则<eq>\\lim_{x\\to x_0}\\frac{f(x)}{g(x)}=\\lim_{x\\to x_0}\\frac{f&#x27;(x)}{g&#x27;(x)}</eq></td></tr></table>"
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            "content": "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！",
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            "content": "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) \\)",
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            "content": "\\[\nf (x) = \\left\\{ \\begin{array}{l l} \\frac {g (x)}{x}, & x \\neq 0 \\\\ 0, & x = 0 \\end{array} \\right.\n\\]"
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            "content": "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，",
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            "content": "\\[\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\\]"
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            "content": "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}\\)",
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            "content": "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。",
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            "content": "定理 4.13 (Peano 余项 Taylor 公式)"
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            "content": "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",
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            "content": "\\[\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\\]"
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            "content": "\\[\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\\]"
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            "content": "定理 4.14 (Lagrange 余项 Taylor 公式)"
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            "content": "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",
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            "content": "\\[\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\\]"
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            "content": "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",
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            "content": "\\[\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\\]"
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            "content": "4.4 凸函数"
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            "content": "4.4 凸函数"
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            "content": "4.4.1 函数的凹凸性"
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            "content": "定义 4.3 (凸函数)"
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            "content": "\\(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",
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            "content": "\\[\nf (\\lambda x _ {1} + (1 - \\lambda) x _ {2}) \\leq \\lambda f (x _ {1}) + (1 - \\lambda) f (x _ {2})\n\\]"
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            "content": "引理 4.1 (定比分点公式)"
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            "content": "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：",
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            "content": "\\[\nx = \\frac {x _ {2} - x}{x _ {2} - x _ {1}} x _ {1} + \\frac {x - x _ {1}}{x _ {2} - x _ {1}} x _ {2}\n\\]"
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            "content": "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",
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            "content": "定理 4.15 (凸函数的等价条件)"
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            "content": "\\(f\\) 是 \\(I\\) 上的凸函数与下面条件等价：",
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            "angle": 0,
            "content": "- 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)\\)。",
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            "content": "- 若不可导，上述命题等价于： \\( f(x) \\geq f(x_0) + k(x - x_0) \\)",
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            "content": "若 \\(f(x)\\) 可导，则等价于 \\(f'(x)\\) 单增",
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            "type": "text",
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            "angle": 0,
            "content": "- 二阶可导：若 \\(f(x)\\) 二阶可导，则等价于 \\(f''(x) \\geq 0\\)",
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            "content": "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",
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            "content": "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",
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            "content": "\\[\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\\]"
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            "content": "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 \\)",
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            "content": "(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",
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            "content": "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。",
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            "content": "4.4 凸函数"
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            "content": "定理 4.16 (凸性质的保持)"
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            "content": "- 凸函数的数乘和加法保持凸性质，差、积、商不一定",
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            "content": "- 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",
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            "content": "- 最值：若 \\(f(x), g(x)\\) 均为 \\(I\\) 上凸函数，则 \\(\\max \\{f(x), g(x)\\}\\) 为凸函数",
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            "content": "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",
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            "content": "\\[\nF (\\lambda x _ {1} + (1 - \\lambda) x _ {2}) \\leq \\lambda F (x _ {1}) + (1 - \\lambda) F (x _ {2})\n\\]"
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            "content": "\\( 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 \\)",
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            "content": "4.4.2 凸函数的基本性质"
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            "content": "定理 4.17 (凸函数基本性质)"
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            "content": "- \\(f(x)\\) 在 \\(I\\) 中任意一点有左右导函数",
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            "content": "4.5 一元极值与导数零点问题"
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            "content": "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：",
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            "content": "\\[\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\\]"
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            "content": "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。",
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            "content": "4.4.3 利用凹凸性证明不等式"
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            "content": "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}\\)",
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            "content": "(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",
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            "content": "\\[\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\\]"
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            "content": "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",
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            "content": "\\[\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\\]"
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            "content": "(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)",
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            "content": "4.5.1 一元极值的概念与判定"
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            "content": "定理 4.20 (一元极值三大充分条件)"
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            "content": "\\( f(x) \\)  可导，且满足必要条件  \\( f'(x_{0})=0 \\) ，则有三大充分条件：",
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            "content": "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",
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            "content": "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",
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            "content": "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：",
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            "type": "equation",
            "bbox": [
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            "angle": 0,
            "content": "\\[\nf (a) = f (x _ {0}) + \\frac {1}{2} f ^ {\\prime \\prime} (\\xi) (a - x _ {0}) ^ {2}\n\\]"
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            "angle": 0,
            "content": "由于  \\( f(a)=0 \\) ，得到  \\( f''(\\xi)<0 \\)",
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            "type": "page_number",
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            "angle": 0,
            "content": "第5章 不定积分"
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            "angle": 0,
            "content": "5.1 常用不定积分公式"
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            "type": "text",
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            "content": "必背部分:",
            "merge_prev": false
        },
        {
            "type": "text",
            "bbox": [
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            "angle": 0,
            "content": "普通三角：",
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        {
            "type": "equation",
            "bbox": [
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                0.215,
                0.402,
                0.244
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            "angle": 0,
            "content": "\\[\n\\int \\sin x \\mathrm{d} x = - \\cos x + c\n\\]"
        },
        {
            "type": "text",
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            "angle": 0,
            "content": "sec、csc:",
            "merge_prev": false
        },
        {
            "type": "equation",
            "bbox": [
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                0.269
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            "angle": 0,
            "content": "\\[\n\\int \\sec^ {2} x \\mathrm{d} x = \\tan x + c\n\\]"
        },
        {
            "type": "equation",
            "bbox": [
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                0.244
            ],
            "angle": 0,
            "content": "\\[\n\\int \\cos x \\mathrm{d} x = \\sin x + c\n\\]"
        },
        {
            "type": "equation",
            "bbox": [
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                0.244,
                0.715,
                0.265
            ],
            "angle": 0,
            "content": "\\[\n\\csc^ {2} x \\mathrm{d} x = - \\cot x + c\n\\]"
        },
        {
            "type": "equation",
            "bbox": [
                0.226,
                0.268,
                0.443,
                0.298
            ],
            "angle": 0,
            "content": "\\[\n\\int_ {a} ^ {b} \\sec x \\mathrm{d} x = \\ln | \\sec x + \\tan x |\n\\]"
        },
        {
            "type": "equation",
            "bbox": [
                0.534,
                0.268,
                0.877,
                0.296
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            "angle": 0,
            "content": "\\[\n\\int \\sec^ {3} x \\mathrm{d} x = \\frac {1}{2} \\tan x \\sec x + \\frac {1}{2} \\ln | \\tan x + \\sec x |\n\\]"
        },
        {
            "type": "text",
            "bbox": [
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                0.2,
                0.316
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            "angle": 0,
            "content": "乘积三角：",
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        {
            "type": "equation",
            "bbox": [
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                0.298,
                0.426,
                0.318
            ],
            "angle": 0,
            "content": "\\[\n\\sec x \\tan x \\mathrm{d} x = \\sec x + c\n\\]"
        },
        {
            "type": "equation",
            "bbox": [
                0.542,
                0.299,
                0.744,
                0.318
            ],
            "angle": 0,
            "content": "\\[\n\\csc x \\cot x \\mathrm{d} x = - \\csc x + c\n\\]"
        },
        {
            "type": "text",
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                0.326,
                0.168,
                0.342
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            "angle": 0,
            "content": "分式:",
            "merge_prev": false
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        {
            "type": "equation",
            "bbox": [
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                0.32,
                0.455,
                0.349
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            "angle": 0,
            "content": "\\[\n\\int {\\frac {1}{x ^ {2} + a ^ {2}}} \\mathrm{d} x = \\frac {1}{a} \\arctan {\\frac {x}{a}} + c\n\\]"
        },
        {
            "type": "equation",
            "bbox": [
                0.536,
                0.32,
                0.774,
                0.35
            ],
            "angle": 0,
            "content": "\\[\n\\int_ {a} \\frac {1}{x ^ {2} - a ^ {2}} \\mathrm{d} x = \\frac {1}{2 a} \\ln | \\frac {x - a}{x + a} | + c\n\\]"
        },
        {
            "type": "equation",
            "bbox": [
                0.224,
                0.347,
                0.516,
                0.378
            ],
            "angle": 0,
            "content": "\\[\n\\int_ {c} \\frac {1}{\\sqrt {x ^ {2} + a ^ {2}}} \\mathrm{d} x = \\ln | x + \\sqrt {x ^ {2} + a ^ {2}} | + c\n\\]"
        },
        {
            "type": "equation",
            "bbox": [
                0.527,
                0.348,
                0.821,
                0.378
            ],
            "angle": 0,
            "content": "\\[\n\\int {\\frac {1}{\\sqrt {x ^ {2} - a ^ {2}}}} \\mathrm{d} x = \\ln | x + \\sqrt {x ^ {2} - a ^ {2}} | + c\n\\]"
        },
        {
            "type": "equation",
            "bbox": [
                0.224,
                0.377,
                0.449,
                0.404
            ],
            "angle": 0,
            "content": "\\[\n\\int_ {a} \\frac {1}{\\sqrt {a ^ {2} - x ^ {2}}} \\mathrm{d} x = \\arcsin {\\frac {x}{a}} + c\n\\]"
        },
        {
            "type": "text",
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                0.167,
                0.422
            ],
            "angle": 0,
            "content": "对数:",
            "merge_prev": false
        },
        {
            "type": "equation",
            "bbox": [
                0.229,
                0.404,
                0.414,
                0.429
            ],
            "angle": 0,
            "content": "\\[\n\\int \\ln x \\mathrm{d} x = x \\ln x - x + c\n\\]"
        },
        {
            "type": "equation",
            "bbox": [
                0.525,
                0.399,
                0.837,
                0.431
            ],
            "angle": 0,
            "content": "\\[\n\\int \\ln (1 + x) \\mathrm{d} x = (1 + x) \\ln (1 + x) - x + c\n\\]"
        },
        {
            "type": "text",
            "bbox": [
                0.124,
                0.432,
                0.204,
                0.448
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            "angle": 0,
            "content": "会算部分:",
            "merge_prev": false
        },
        {
            "type": "text",
            "bbox": [
                0.123,
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                0.489
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            "angle": 0,
            "content": "三角：",
            "merge_prev": false
        },
        {
            "type": "equation",
            "bbox": [
                0.215,
                0.467,
                0.427,
                0.495
            ],
            "angle": 0,
            "content": "\\[\n\\int \\tan x \\mathrm{d} x = - \\ln | \\cos x | + c\n\\]"
        },
        {
            "type": "equation",
            "bbox": [
                0.67,
                0.466,
                0.861,
                0.495
            ],
            "angle": 0,
            "content": "\\[\n\\int \\cot x \\mathrm{d} x = \\ln | \\sin x | + c\n\\]"
        },
        {
            "type": "text",
            "bbox": [
                0.113,
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                0.193,
                0.515
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            "angle": 0,
            "content": "平方三角：",
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        },
        {
            "type": "equation",
            "bbox": [
                0.215,
                0.494,
                0.43,
                0.523
            ],
            "angle": 0,
            "content": "\\[\n\\int_ {a} ^ {b} \\sin^ {2} x \\mathrm{d} x = \\frac {x}{2} - \\frac {\\sin 2 x}{4} + c\n\\]"
        },
        {
            "type": "equation",
            "bbox": [
                0.67,
                0.494,
                0.885,
                0.522
            ],
            "angle": 0,
            "content": "\\[\n\\int \\cos^ {2} x \\mathrm{d} x = \\frac {x}{2} + \\frac {\\sin 2 x}{4} + c\n\\]"
        },
        {
            "type": "equation",
            "bbox": [
                0.215,
                0.521,
                0.422,
                0.546
            ],
            "angle": 0,
            "content": "\\[\n\\int \\tan^ {2} x \\mathrm{d} x = \\tan x - x + c\n\\]"
        },
        {
            "type": "equation",
            "bbox": [
                0.667,
                0.521,
                0.887,
                0.549
            ],
            "angle": 0,
            "content": "\\[\n\\int \\cot^ {2} x \\mathrm{d} x = - \\cot x - x + c\n\\]"
        },
        {
            "type": "text",
            "bbox": [
                0.114,
                0.552,
                0.193,
                0.568
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            "angle": 0,
            "content": "复合三角：",
            "merge_prev": false
        },
        {
            "type": "equation",
            "bbox": [
                0.215,
                0.546,
                0.572,
                0.575
            ],
            "angle": 0,
            "content": "\\[\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\\]"
        },
        {
            "type": "equation",
            "bbox": [
                0.216,
                0.573,
                0.57,
                0.601
            ],
            "angle": 0,
            "content": "\\[\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\\]"
        },
        {
            "type": "equation",
            "bbox": [
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                0.599,
                0.518,
                0.628
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            "angle": 0,
            "content": "\\[\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\\]"
        },
        {
            "type": "equation",
            "bbox": [
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                0.628,
                0.517,
                0.653
            ],
            "angle": 0,
            "content": "\\[\n\\int x \\sin n x \\mathrm{d} x = \\frac {1}{n ^ {2}} \\sin n x - \\frac {x}{n} \\cos n x + c\n\\]"
        },
        {
            "type": "text",
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                0.16,
                0.673
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            "angle": 0,
            "content": "根式:",
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        },
        {
            "type": "equation",
            "bbox": [
                0.215,
                0.652,
                0.659,
                0.68
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            "angle": 0,
            "content": "\\[\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\\]"
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            "type": "equation",
            "bbox": [
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                0.678,
                0.593,
                0.707
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            "angle": 0,
            "content": "\\[\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\\]"
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            "content": "5.2 不定积分基本方法"
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            "content": "定理 5.1 (分部积分公式)"
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            "angle": 0,
            "content": "函数 \\(u, v\\) 的积分满足：",
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            "content": "\\[\n\\int u \\mathrm{d} v = u v - \\int v \\mathrm{d} u\n\\]"
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            "angle": 0,
            "content": "5.2 不定积分基本方法"
        },
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            "content": "定理 5.2 (表格法)"
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            "content": "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：",
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            "content": "<table><tr><td>u的各阶导数</td><td>u</td><td><eq>u&#x27;</eq></td><td><eq>u&#x27;&#x27;</eq></td><td><eq>u&#x27;&#x27;&#x27;</eq></td><td>...</td><td><eq>u^{(n+1)}</eq></td></tr><tr><td><eq>v^{(n+1)}</eq>的各阶原函数</td><td><eq>v^{(n+1)}</eq></td><td><eq>v^{(n)}</eq></td><td><eq>v^{(n-1)}</eq></td><td><eq>v^{(n-2)}</eq></td><td>...</td><td>v</td></tr></table>"
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            "content": "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\\)",
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            "content": "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",
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            "content": "练习  \\( 5.1P_{n}(x)e^{kx} \\)  型 计算  \\( \\int(x^{3}+2x+6)e^{2x}\\mathrm{d}x \\)",
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            "content": "解 先列表格:",
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            "content": "<table><tr><td><eq>x^{3} + 2x + 6</eq></td><td><eq>3x^{2} + 2</eq></td><td><eq>6x</eq></td><td><eq>6</eq></td><td><eq>0</eq></td></tr><tr><td><eq>e^{2x}</eq></td><td><eq>\\frac{1}{2}e^{2x}</eq></td><td><eq>\\frac{1}{4}e^{2x}</eq></td><td><eq>\\frac{1}{8}e^{2x}</eq></td><td><eq>\\frac{1}{16}e^{2x}</eq></td></tr></table>"
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            "content": "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\\)",
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            "content": "定理 5.3 (积分有理分解)"
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            "content": "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：",
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            "content": "\\[\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\\]"
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            "content": "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。",
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            "content": "定理 5.4 (三角有理式转化为有理式)"
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            "content": "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",
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            "content": "44"
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            "content": "\\[\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\\]"
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            "content": "因此",
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            "content": "\\[\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\\]"
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            "content": "5.3 不定积分表选例"
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            "content": "定理 5.5 (两类无理根式的不定积分)"
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            "content": "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。",
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            "content": "5.3 不定积分表选例"
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            "content": "下面假设必背部分都会背，且不做计算证明",
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            "content": "5.3.1 三角函数：重点"
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            "content": "练习 5.2 核心会算"
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            "content": "\\[\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\\]"
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            "content": "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",
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            "content": "\\[\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\\]"
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            "content": "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",
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            "content": "(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\\)",
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            "content": "(3) 这个经常会用到!",
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            "content": "\\[\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\\]"
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            "content": "因此 \\(I = \\frac{1}{2}\\sec x\\tan x + \\ln |\\sec x + \\tan x| + c\\)",
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            "content": "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}\\)",
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            "content": "(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\\)",
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            "content": "解(1)根据 \\( 1+\\cos x=2\\cos^{2}\\frac{x}{2},1-\\cos x=2\\sin^{2}\\frac{x}{2} \\) ，因此",
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            "content": "\\[\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\\]"
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            "content": "(2) 根据诱导公式 \\(\\sin x = \\cos \\left(\\frac{\\pi}{2} - x\\right)\\)",
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            "content": "\\[\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\\]"
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            "content": "5.3 不定积分表选例"
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            "content": "(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",
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            "content": "\\[\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\\]"
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            "content": "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\\)",
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            "content": "定理 5.6 ( \\( \\frac{1}{a\\sin^{2}x + b\\cos^{2}x} \\) 积分)"
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            "content": "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",
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            "content": "\\[\n\\int {\\frac {\\sec^ {2} x \\mathrm{d} x}{a \\tan^ {2} x + b}}\n\\]"
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            "content": "直接用 \\(\\mathrm{d}(\\tan x) = \\sec^2 x\\mathrm{d}x\\) 即可。",
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            "content": "定理5.7 \\(\\left(\\frac{1}{a\\sin^2x + b\\sin x\\cos x + c\\cos^2x + d}\\right.\\) 积分）"
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            "content": "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",
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            "content": "\\[\n\\int {\\frac {\\sec^ {2} x \\mathrm{d} x}{a \\tan^ {2} x + b \\tan x + c + d \\sec^ {2} x}}\n\\]"
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            "content": "再利用 \\(\\sec^2 x = \\tan^2 x + 1\\) ，全部转换为 \\(\\tan x\\) ，再分母配方即可",
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            "content": "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} \\) \n解(1)同除以 \\( \\cos^{2}x \\) ，得到 \\( \\int\\frac{\\sec^{2}x\\mathrm{d}x}{2+\\tan^{2}x} \\) ，易算",
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            "content": "(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",
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            "content": "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} \\) \n\n(4)  \\( \\int\\sin^{6}x + \\cos^{6}x\\mathrm{d}x \\)  (5)  \\( \\int\\frac{\\mathrm{d}x}{\\sin^{6}x + \\cos^{6}x} \\)",
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            "content": "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)\\)",
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            "content": "(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：",
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            "content": "\\[\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\\]"
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            "content": "(3) 直接降到二次，并转换为 \\(\\tan x, \\sec x\\)：使用 \\(\\sec^2 x = 1 + \\tan^2 x\\)",
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            "content": "\\[\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\\]"
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            "content": "(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：",
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            "content": "5.3 不定积分表选例"
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            "content": "结果为 \\(\\frac{5}{8} x + \\frac{3}{32}\\sin 4x + c\\)",
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            "content": "\\[\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\\]"
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            "content": "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。",
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            "content": "定理 5.8 ( \\( \\frac{a\\sin x + b\\cos x}{c\\sin x + d\\cos x} \\)  型计算技巧)"
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            "content": "47"
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            "angle": 0,
            "content": "\\(\\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",
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            "content": "推论 5.1 ( \\( \\tan x, \\cot x \\)  分式型)"
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            "content": "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",
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            "content": "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} \\)",
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            "content": "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",
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            "content": "\\[\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\\]"
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            "content": "(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",
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            "content": "\\[\nI = \\frac {3}{3 4} x + \\frac {5}{3 4} \\ln | 5 \\sin x + 3 \\cos x | + c\n\\]"
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            "content": "定理 5.9 ( \\( \\frac{1}{a\\sin x + b\\cos x + c} \\)  积分)"
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            "content": "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",
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            "content": "\\[\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\\]"
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            "content": "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。",
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            "content": "练习 5.8 相关训练 (1) \\( \\int\\frac{dx}{\\sin x+2\\cos x+3} \\)",
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            "content": "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",
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            "content": "\\[\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\\]"
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            "content": "推论 5.2 ( \\( \\frac{p\\sin x + q\\cos x + s}{a\\sin x + b\\cos x + c}dx \\) )"
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            "content": "5.3 不定积分表选例"
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            "content": "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\\)",
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            "content": "解(1)写为 \\(\\frac{1}{2}\\int \\sqrt{(a^2 - b^2)\\sin^2x + b^2}\\mathrm{d}(\\sin^2 x)\\) 即可",
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            "content": "5.3.2 对数与指数"
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            "content": "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 \\)",
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            "content": "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\\)",
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            "content": "(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\\)",
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            "content": "(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",
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            "content": "\\[\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\\]"
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            "content": "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\\)",
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            "content": "<table><tr><td>$ \\ln x $</td><td>$ \\frac{1}{x} $</td></tr><tr><td>$ x^{n} $</td><td>$ \\frac{1}{n+1}x^{n+1} $</td></tr></table>"
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            "content": "(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，",
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            "content": "\\[\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\\]"
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            "content": "(3) 同理用表格法一步一步列，",
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            "content": "\\[\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\\]"
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            "content": "48"
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            "content": "(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)\\)",
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            "content": "5.3.3 分式问题"
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            "content": "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}\\)",
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            "content": "(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\\)",
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            "content": "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",
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            "content": "(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。",
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            "content": "(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。",
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            "content": "5.3 不定积分表选例"
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            "content": "(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)",
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            "content": "(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).",
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            "content": "练习 5.13 次方和差 尽量记住下面的分解方法:",
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            "content": "(1) \\(\\int \\frac{\\mathrm{d}x}{x^3 + 1}\\) (2) \\(\\int \\frac{\\mathrm{d}x}{x^3 - 1}\\)",
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            "content": "(3) \\( \\int\\frac{dx}{x^{4}-1} \\)  (4)重点： \\( \\int\\frac{dx}{x^{4}+1} \\) (一定要记住！)",
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            "content": "(5) \\( \\int\\frac{dx}{x^{6}-1} \\)  (6) \\( \\int\\frac{dx}{x^{6}+1} \\)",
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            "content": "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",
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            "content": "(2) 用立方差即可",
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            "content": "(3) 用 \\(\\frac{1}{x^4 - 1} = \\frac{1}{2}\\left(\\frac{1}{x^2 - 1} - \\frac{1}{x^2 + 1}\\right)\\)",
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            "content": "(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",
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            "content": "\\[\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\\]"
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            "content": "因此所求即 \\(\\frac{1}{2} (N - M)\\)",
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            "content": "(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)",
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            "content": "(6) 类似四次方的讨论 + 立方和",
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            "content": "\\[\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\\]"
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            "content": "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}\\)",
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            "content": "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",
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            "content": "(2) 这种有理分解比较常见，最好记下来，",
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            "content": "\\[\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\\]"
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            "content": "5.3.4 根式问题"
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            "content": "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",
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            "content": "5.3 不定积分表选例"
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            "content": "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：",
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            "content": "\\[\nI = \\frac {1}{2} (x \\sqrt {a ^ {2} - x ^ {2}} + a ^ {2} \\arcsin {\\frac {x}{a}}) + C\n\\]"
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            "content": "(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",
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            "content": "\\[\n\\frac {1}{2} (x \\sqrt {x ^ {2} + a ^ {2}} + a ^ {2} \\ln | x + \\sqrt {x ^ {2} + a ^ {2}} |) + C\n\\]"
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            "content": "(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：",
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            "content": "\\[\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\\]"
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            "content": "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\\)",
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            "content": "50"
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            "content": "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\\)",
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            "content": "(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",
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            "content": "\\[\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\\]"
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            "content": "(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",
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            "content": "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\\)",
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            "content": "\\[\n\\int \\frac {\\mathrm{d} x}{(\\sqrt {x} + 1) (x + 3)}\n\\]"
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            "content": "(3) \\(\\int \\frac{\\mathrm{d}x}{x\\sqrt{x^2 - 2x - 3}}\\)",
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            "content": "\\[\nI = \\frac {\\mathrm{d} x}{x + \\sqrt {x ^ {2} - x + 1}}\n\\]"
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            "content": "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",
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            "content": "(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\n\\(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\\)",
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            "content": "(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",
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            "content": "\\[\nI = - \\int {\\frac {2}{t ^ {2} + 3}} \\mathrm{d} t = - \\frac {2}{\\sqrt {3}} \\arctan {\\frac {t}{\\sqrt {3}}} + C\n\\]"
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            "content": "(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",
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            "content": "\\[\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\\]"
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            "content": "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。",
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            "content": "(4) 可以用 Euler 变换或者配方三角替换。",
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            "content": "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\\)",
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            "content": "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：",
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            "content": "\\[\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\\]"
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            "content": "(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\\)",
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            "content": "5.3.6 递推问题"
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            "content": "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 \\)",
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            "content": "\\[\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\\]"
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            "content": "\\[\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\\]"
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            "content": "\\[\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\\]"
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            "content": "\\[\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\\]"
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            "content": "第6章 定积分"
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            "content": "6.1 定积分理论"
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            "content": "6.1.1 可积性"
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            "content": "定义 6.1 (Riemann 积分)"
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            "content": "\\(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\\)",
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            "content": "定义 6.2 (Darboux 上下和)"
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            "content": "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",
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            "content": "\\[\nS (T) = \\sum_ {i = 1} ^ {n} M _ {i} \\Delta x _ {i}, s (T) = \\sum_ {i = 1} ^ {n} m _ {i} \\Delta x _ {i}\n\\]"
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            "content": "定义 \\(f(x)\\) 在 \\([a,b]\\) 的上下积分为：",
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            "content": "\\[\nS = \\inf _ {T} S (T), s = \\sup _ {T} s (T)\n\\]"
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            "content": "定理 6.1 (可积函数的性质)"
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            "content": "- 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)",
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            "content": "\\[\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\\]"
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            "content": "\\(f(x)\\) 在 \\([a,b]\\) 可积当且仅当",
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            "content": "- 上下积分角度： \\( f(x) \\) 在 \\([a, b]\\) 的上积分 \\( s \\) 等于下积分 \\( S \\)",
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            "content": "- 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\\)",
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            "content": "- 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\\)",
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            "content": "- 实变角度：\\(f(x)\\) 在 \\([a, b]\\) 几乎处处连续",
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            "content": "练习6.1Dirichlet函数的不可积性证明Dirichlet函数在 \\([a,b]\\) 不可积",
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            "content": "6.1 定积分理论"
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            "content": "定理 6.3 (Riemann 可积函数类)"
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            "content": "\\(f(x)\\) 定义于 \\([a,b]\\)，常见的有三个Riemann可积函数类：",
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            "content": "•  \\( [a,b] \\)  连续函数",
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            "content": "- 在 \\([a, b]\\) 只有有限个间断点的有界函数",
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            "content": "- \\([a, b]\\) 上单调函数",
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            "content": "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",
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            "content": "(2) 将间断点处的  \\( \\Delta x_{i} \\)  设置得足够小，控制住间断点差距",
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            "content": "(3) \\( \\omega_{i}=f(x_{i})-f(x_{i-1}) \\) ，只需要 \\( \\omega_{i} \\) 取得足够小",
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            "content": "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\\)",
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            "content": "(2) 证明下面的 \\(f(x)\\) 在 \\([0,1]\\) 上可积:",
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            "content": "\\[\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\\]"
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            "content": "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。",
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            "content": "(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：",
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            "content": "\\[\n\\sum \\omega_ {i} \\Delta x _ {i} <   \\epsilon\n\\]"
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            "content": "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",
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            "content": "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",
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            "content": "\\[\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\\]"
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            "content": "6.1 定积分理论"
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            "content": "(2)例如原函数如下，其在 \\([0,1]\\) 可导，但是 \\(F^{\\prime}(x)\\) 无界",
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            "content": "\\[\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\\]"
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            "content": "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)\\)",
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            "content": "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) \\) ：",
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            "content": "\\[\nF (x) = \\left\\{ \\begin{array}{l l} 0, & x = 0 \\\\ x ^ {2} \\sin \\frac {1}{x}, & x \\in (0, 1 ] \\end{array} \\right.\n\\]"
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            "content": "\\(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\\)",
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            "content": "定理 6.5 (微积分基本定理)"
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            "content": "\\(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",
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            "content": "- \\(F(x)\\) 是 \\([a, b]\\) 上的连续函数",
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            "content": "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)\\)",
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            "content": "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",
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            "content": "\\[\n\\lim _ {h \\rightarrow 0} \\int_ {a} ^ {b} \\frac {f (x + h) - f (x)}{h} \\mathrm{d} x = f (b) - f (a)\n\\]"
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            "content": "\\[\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\\]"
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            "content": "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。",
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            "content": "6.1 定积分理论"
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            "content": "(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",
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            "content": "练习6.7除法增加条件可积设 \\(f(x)\\) 在 \\([a,b]\\) 上函数，证明",
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            "content": "(1) 若 \\( f(x) \\) 可积，且 \\( |f(x)| \\geq m > 0 \\)，则 \\( \\frac{1}{f(x)} \\) 在 \\([a, b]\\) 可积",
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            "content": "(2) 若 \\( f(x) \\) 在 \\([a, b]\\) 连续且恒不为零，则 \\(\\frac{1}{f(x)}\\) 在 \\([a, b]\\) 可积",
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            "content": "证明 (1) 考虑振幅放缩:",
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            "content": "\\[\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\\]"
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            "content": "(2) 由于连续，能找到最小值，可转换为 (1)",
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            "content": "定理 6.7 (复合 (较难，重点！))"
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            "content": "\\(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",
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            "content": "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",
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            "content": "\\[\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\\]"
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            "content": "6.1.4 定积分的保号性"
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            "content": "定理 6.8 (定积分保号性)"
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            "content": "\\(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\\)",
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            "content": "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",
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            "content": "(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",
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            "content": "\\[\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\\]"
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            "content": "\\[\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\\]"
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            "content": "6.1 定积分理论"
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            "content": "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",
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            "content": "\\[\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\\]"
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            "content": "这与 \\(f(x)\\sin (x - \\xi)\\) 非零矛盾，因此 \\(f(x)\\) 至少有两个零点",
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            "content": "(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。",
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            "content": "6.1.5 积分中值定理"
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            "content": "\\(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\\)",
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            "content": "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）。",
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            "content": "推论 6.1 (第一中值定理的推广)"
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            "content": "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\\)",
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            "content": "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\\)",
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            "content": "6.2 三角函数定积分计算"
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            "content": "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\\)",
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            "content": "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",
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            "content": "\\[\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\\]"
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            "content": "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。",
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            "content": "6.2 三角函数定积分计算"
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            "content": "6.2.1 \\([0, \\frac{\\pi}{2}]\\) 上三角函数积分公式",
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            "content": "定理6.11 \\(\\left[\\frac{0}{2},\\frac{\\pi}{2}\\right]\\) 三角函数积分）"
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            "content": "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\\)",
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            "content": "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\\)",
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            "content": "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}\\)",
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            "content": "(2) 计算 \\(\\int_0^{\\frac{\\pi}{2}}\\ln \\tan x\\mathrm{d}x\\)",
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            "content": "(3) 计算  \\( \\int_{0}^{+\\infty}\\frac{\\mathrm{d}x}{(1+x^{2})(1+x^{2022})} \\)",
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            "content": "(4) 计算 \\(\\int_0^{+\\infty} \\frac{\\ln x}{4 + x^2} \\, \\mathrm{d}x\\)",
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            "content": "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}\\)",
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            "content": "(2) \\(\\int_{0}^{\\frac{\\pi}{2}}\\ln \\sin x\\mathrm{d}x - \\int_{0}^{\\frac{\\pi}{2}}\\ln \\cos x\\mathrm{d}x = 0\\)",
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            "content": "(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}\\)",
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            "content": "(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\\)",
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            "content": "6.2 三角函数定积分计算"
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            "content": "定理 6.12 (Wallis 推广)"
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            "content": "\\[\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\\]"
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            "content": "练习6.12基础练习 \\((1)\\int_{0}^{\\frac{\\pi}{2}}\\cos^{3}x\\sin^{2}x\\mathrm{d}x\\)",
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            "content": "\\[\n(3) \\int_ {0} ^ {\\frac {\\pi}{2}} \\sin^ {3} \\cos^ {5} x \\mathrm{d} x\n\\]"
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            "content": "\\[\n(4) \\int_ {0} ^ {\\frac {\\pi}{2}} \\sin^ {2} x \\cos^ {4} x \\mathrm{d} x\n\\]"
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            "content": "\\[\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\\]"
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            "content": "\\[\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\\]"
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            "content": "\\[\n(3) I (3, 5) = \\frac {2 \\cdot 4 \\cdot 2}{8 \\cdot 6 \\cdot 4 \\cdot 2} = \\frac {1}{2 4}\n\\]"
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            "content": "\\[\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\\]"
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            "content": "6.2.2 \\([0, \\pi], [0, 2\\pi]\\) 上三角函数积分"
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            "content": "定理 6.13 (对称性的积分)"
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            "content": "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\\)",
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            "content": "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。",
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            "content": "定理 6.14 ([0, π] 上 Wallis 公式推广)"
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            "content": "对任意非负整数 \\(m, n\\)，有",
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            "content": "\\[\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\\]"
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            "content": "注意这里 \\(J(m,n)\\neq J(n,m)\\)",
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            "content": "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。",
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            "content": "定理 6.15 ( \\( [0, 2\\pi] \\)  上 Wallis 公式推广)"
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            "content": "\\[\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\\]"
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            "content": "6.2 三角函数定积分计算"
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            "content": "这里 \\(K(m,n) = K(n,m)\\)",
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            "content": "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 \\)",
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            "content": "(2) 偶数幂，因此 \\(2I(4,3) = 2\\frac{3 \\cdot 2}{7 \\cdot 5 \\cdot 3} = \\frac{4}{35}\\)",
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            "content": "(3) 奇数，为 0",
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            "content": "(4) 偶数幂，但 \\(m, n\\) 均为偶数，因此结果为 \\(\\frac{\\pi}{16}\\)",
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            "content": "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} \\)",
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            "content": "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。",
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            "content": "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\\)",
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            "content": "(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",
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            "content": "(3) 偶函数，转换为 \\(\\int_0^\\pi \\cos^3 x\\sin^2 x\\mathrm{d}x = 0\\)",
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            "content": "(4) 奇函数，转换为  \\( -\\int_{0}^{\\pi}\\cos^{4}x\\sin^{2}xdx=-2I(4,3)=-\\frac{4}{35} \\)",
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            "content": "(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}\\)",
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            "content": "(6) 绝对值肯定轴对称，因此 \\(2I(5,0) = \\frac{16}{15}\\)",
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            "content": "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\\)",
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            "content": "(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} \\)",
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            "content": "\\[\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\\]"
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            "content": "(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",
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            "content": "\\[\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\\]"
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            "content": "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。",
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            "content": "6.2.4 Euler 积分"
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            "content": "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 \\)",
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            "content": "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",
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            "content": "\\[\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\\]"
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            "content": "因此得到 \\(I = -\\frac{\\pi}{2}\\ln 2\\)",
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            "content": "笔记要证明Euler积分收敛，只需证明 \\(\\lim_{x\\to 0^{+}}x\\ln \\sin x = 0\\)",
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            "content": "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\\)",
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            "content": "\\[\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\\]"
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            "content": "(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\\)",
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            "content": "(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)",
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            "content": "(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)",
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            "content": "6.2.5 Dirichlet 积分"
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            "content": "定理 6.17 (Dirichlet 积分)"
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            "content": "Dirichlet 积分  \\( \\int_{0}^{\\pi}\\frac{\\sin\\frac{2n+1}{2}x}{2\\sin\\frac{x}{2}}=\\frac{\\pi}{2}dx \\) 。",
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            "content": "\\[\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\\]"
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            "content": "6.3 积分 (不) 等式"
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            "content": "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：\n \\( \\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} \\)",
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            "content": "(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",
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            "content": "(3) 第三个设 \\(I_{n} = \\int_{0}^{\\pi}\\frac{\\sin 2nx}{\\sin x}\\mathrm{d}x,\\)",
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            "content": "\\[\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\\]"
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            "content": "61"
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            "content": "而 \\(I_{1} = 0\\) ，因此 \\(I_{n} = 0\\)",
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            "content": "6.2.6 其他几个常见三角积分"
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            "content": "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}}\\)",
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            "content": "\\[\nf \\left(\\frac {\\pi}{2}\\right) = \\frac {2}{\\pi} \\leq \\frac {\\sin x}{x} <   f (0) = 1\n\\]"
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            "content": "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：",
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            "content": "\\[\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\\]"
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            "content": "(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}\\):",
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            "content": "\\[\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\\]"
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            "content": "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\\)",
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            "content": "(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 \\)",
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            "content": "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",
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            "content": "\\[\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\\]"
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            "content": "6.3 积分 (不) 等式"
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            "content": "同理取 \\([\\frac{1}{2}, 1]\\) 的部分可得到",
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            "content": "\\[\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\\]"
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            "content": "因此两式相加除以 2 即得到结果。",
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            "content": "定理 6.19 (Cauchy-Schwarz 不等式)"
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            "content": "\\[\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\\]"
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            "content": "推论6.2"
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            "content": "\\[\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\\]"
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            "content": "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 \\) \n(2) 二阶导:",
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            "content": "63"
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            "content": "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",
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            "content": "因此 \\(\\int_0^2 f(x)\\mathrm{d}x\\leq 3\\)",
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            "content": "(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\\)",
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            "content": "(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 \\)",
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            "content": "\\[\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\\]"
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            "content": "- 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\\)",
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            "content": "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。",
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            "content": "定理 7.1 (反常积分 Leibniz 公式)"
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            "content": "- 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)\\)",
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            "content": "- 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)\\)",
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            "content": "定理 7.2 (定义判敛)"
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            "content": "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\\)",
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            "content": "- 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\\)",
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            "content": "证明 根据收敛的定义可知。",
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            "content": "定理 7.3 ( \\( \\frac{1}{x^{p}} \\) 的反常积分)"
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            "content": "\\(\\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",
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            "content": "证明 只讨论 \\([1, +\\infty)\\) 上的，直接积分：",
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            "content": "\\[\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\\]"
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            "content": "7.1 反常积分理论"
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            "content": "7.1.2 反常积分比较判别法"
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            "content": "定理 7.4 (Cauchy 收敛准则)"
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            "content": "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",
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            "content": "\\[\n\\left| \\int_ {u _ {1}} ^ {u _ {2}} f (x) \\mathrm{d} x \\right| <   \\epsilon\n\\]"
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            "angle": 0,
            "content": "- 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",
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            "content": "\\[\n\\left| \\int_ {u _ {1}} ^ {u _ {2}} f (x) \\mathrm{d} x \\right| <   \\epsilon\n\\]"
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            "content": "定义 7.2 (绝对收敛与条件收敛)"
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            "content": "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。",
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            "content": "定理 7.5 (绝对收敛)"
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            "content": "- 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",
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            "content": "\\[\n| \\int_ {a} ^ {+ \\infty} f (x) \\mathrm{d} x | \\leq \\int_ {a} ^ {+ \\infty} | f (x) | \\mathrm{d} x\n\\]"
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            "content": "7.1 反常积分理论"
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            "content": "(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\\)",
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            "content": "(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。",
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            "content": "推论 7.2 (对数、指数比较判别法)"
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            "content": "\\[\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\\]"
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            "content": "- \\(x \\to \\infty\\) 时，等价视为 \\(\\ln x \\sim x^{\\epsilon}, e^{x} \\sim x^{+\\infty}\\)",
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            "content": "- \\(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\\)",
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            "content": "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！）",
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            "content": "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",
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            "content": "(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",
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            "content": "(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。",
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            "content": "- 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。",
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            "content": "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",
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            "content": "7.1 反常积分理论"
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            "content": "3. \\(\\int_0^{+\\infty}\\frac{\\cos x}{x^p}\\mathrm{d}x: 1 < p < 2\\) 时绝对收敛，\\(0 < p \\leq 1\\) 时条件收敛，其余发散",
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            "content": "4. 换元： \\(\\int_0^{+\\infty}\\sin x^2\\mathrm{d}x,\\int_0^{+\\infty}\\cos x^2\\mathrm{d}x,\\int_0^{+\\infty}x\\sin x^4\\mathrm{d}x,\\int_0^1\\frac{1}{x^p}\\sin \\frac{1}{x}\\mathrm{d}x\\)",
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            "angle": 0,
            "content": "5. \\(\\int_0^{+\\infty}x^p\\sin x^q\\mathrm{d}x\\)",
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            "angle": 0,
            "content": "6. \\(\\int_0^{+\\infty}\\frac{\\sin x}{x^p + x^q}\\mathrm{d}x\\)",
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            "content": "证明 (1) \\(p > 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",
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            "content": "\\[\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\\]"
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            "content": "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",
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            "content": "\\[\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\\]"
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            "content": "显然发散。",
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            "content": "(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",
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            "type": "text",
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            "content": "(3) \\( \\frac{\\cos x}{x^{p}}\\sim\\frac{1}{x^{p}} \\) ，结合(1)得到0<p<1时条件收敛，其余发散。",
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            "content": "(4) 前三个做 \\( t = x^2 \\) 换元，第一个变成 \\( \\frac{1}{2} \\int_{0}^{+\\infty} \\frac{\\sin t}{\\sqrt{t}} \\, \\mathrm{d}t \\)，因此条件收敛。第二个 \\( \\int_{0}^{+\\infty} \\cos x^2 \\, \\mathrm{d}x \\) 同理条件收敛。第三个换元得到 \\( \\frac{1}{2} \\int_{1}^{+\\infty} \\sin t^2 \\, \\mathrm{d}t \\)，变成第一个，故条件收敛。第四个做 \\( t = \\frac{1}{x} \\) 换元得到 \\( \\int_{+\\infty}^{1} -t^{p-2} \\sin t \\, \\mathrm{d}t = \\int_{1}^{+\\infty} \\frac{\\sin t}{t^{2-p}} \\, \\mathrm{d}t \\)，\\( 2 - p > 1 \\) 时绝对收敛，\\( 0 < 2 - p \\leq 1 \\) 条件收敛。",
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            "content": "(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",
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            "content": "68"
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            "content": "\\[\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\\]"
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            "angle": 0,
            "content": "转换为前面的问题。",
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            "content": "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。",
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            "content": "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):",
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            "content": "(1) 判断 \\(\\int_0^{+\\infty} \\frac{\\sin x}{2x + 3\\sin x} \\mathrm{d}x\\) 的敛散性",
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            "content": "7.1 反常积分理论"
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            "content": "根据 Dirichlet 判别法知前者收敛。后者先进行放缩",
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            "content": "\\[\n0 \\leq \\frac {3 \\sin^ {2} x}{2 x (2 x + 3 \\sin x)} \\leq \\frac {3}{2 x (2 x - 3)}\n\\]"
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            "content": "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",
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            "content": "7.1.4 函数在无穷远处的性质"
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            "content": "一、若极限存在，则必为0"
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            "content": "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",
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            "content": "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)。",
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            "content": "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\\)。",
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            "content": "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"
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            "content": "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}\\)",
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            "content": "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。",
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            "content": "69"
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            "content": "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\\)",
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            "content": "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。"
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            "content": "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\\)",
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            "content": "7.2 反常积分的计算"
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            "content": "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",
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            "content": "\\[\n| f (x) | \\geq | f (x _ {n}) | - | f (x) - f (x _ {n}) | > \\frac {\\epsilon_ {0}}{2}\n\\]"
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            "content": "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。",
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            "content": "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}\\)",
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            "content": "7.2 反常积分的计算"
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            "content": "7.2.1 两个常用的基础反常积分"
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            "content": "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}}\\)",
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            "content": "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}}\\)",
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            "content": "7.2.2 富如兰尼积分"
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            "content": "定理 7.8 (两侧极限存在则无穷积分收敛)"
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            "content": "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)\\)",
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            "content": "\\[\n\\int_ {- \\infty} ^ {+ \\infty} [ f (a + x) - f (x) ] \\mathrm{d} x\n\\]"
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            "content": "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：",
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            "content": "\\[\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\\]"
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            "content": "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)\\)",
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            "content": "7.2 反常积分的计算"
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            "content": "定理 7.9 (富如兰尼积分)"
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            "angle": 0,
            "content": "\\(f(x)\\) 为 \\([0, +\\infty)\\) 连续函数， \\(a, b > 0\\) ，则",
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            "content": "1. 若 \\(\\lim_{x\\to +\\infty}f(x)\\) 存在，则",
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            "content": "\\[\n\\int_ {0} ^ {+ \\infty} \\frac {f (a x) - f (b x)}{x} \\mathrm{d} x = [ f (0) - f (+ \\infty) ] \\ln \\frac {b}{a}\n\\]"
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            "content": "2. 若 \\(\\exists A > 0\\)，使得 \\(\\int_{A}^{+\\infty} \\frac{f(x)}{x} \\mathrm{d}x\\) 收敛，则",
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            "content": "\\[\n\\int_ {0} ^ {+ \\infty} \\frac {f (a x) - f (b x)}{x} \\mathrm{d} x = f (0) \\ln {\\frac {b}{a}}\n\\]"
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            "content": "证明 下面对 \\(I = \\int_{m}^{M}\\frac{f(ax) - f(bx)}{x}\\mathrm{d}x\\) 进行化简：",
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            "content": "\\[\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\\]"
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            "content": "(1) 根据积分第一中值定理得到",
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            "content": "\\[\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\\]"
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            "content": "71"
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            "content": "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}\\)",
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            "content": "(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}\\)",
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            "content": "练习 7.18 富如兰尼积分的应用 计算以下反常积分",
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            "angle": 0,
            "content": "(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\\)",
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            "content": "解 (1) \\(\\arctan 0 = 0, \\arctan (+\\infty) = \\frac{\\pi}{2}\\)，因此积分为 \\(-\\frac{\\pi}{2} \\ln \\frac{b}{a}\\)",
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            "content": "(2) \\( e^{0}=1,e^{-\\infty}=0 \\) ，因此积分为 \\( \\ln\\frac{b}{a} \\)",
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            "content": "(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}\\)",
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            "content": "7.2.3 Dirichlet 积分"
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            "content": "引理7.1"
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            "content": "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}\\)",
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            "angle": 0,
            "content": "证明 首先  \\( \\frac{\\sin bx - \\sin ax}{x} = \\int_{a}^{b} \\cos xy \\, dy \\) ，因此",
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            "content": "7.2 反常积分的计算"
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            "content": "分顺序得到",
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            "content": "\\[\nI = \\int_ {a} ^ {b} \\mathrm{d} y \\int_ {0} ^ {+ \\infty} e ^ {- p x} \\cos x y \\mathrm{d} x\n\\]"
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        {
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            "content": "根据表格法得到:",
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            "content": "\\[\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\\]"
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            "content": "因此得到 \\(\\frac{y^2 + p^2}{p^2}\\int_0^{+\\infty}\\cos xy\\mathrm{d}y = \\frac{1}{p}\\)",
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            "content": "\\[\nI = \\int_ {a} ^ {b} {\\frac {p}{p ^ {2} + y ^ {2}}} \\mathrm{d} y = \\arctan {\\frac {b}{p}} - \\arctan {\\frac {a}{p}}\n\\]"
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            "content": "定理 7.10 (Dirichlet 积分)"
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            "content": "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)\\)",
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            "content": "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",
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            "content": "\\[\nG (0) = \\lim _ {p \\to 0 ^ {+}} G (p) = \\lim _ {p \\to 0 ^ {+}} \\arctan {\\frac {b}{p}} = \\frac {\\pi}{2} \\mathrm{sgn} (b)\n\\]"
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            "content": "特别地， \\(b = 1\\) 得到 \\(\\int_0^{+\\infty}\\frac{\\sin x}{x}\\mathrm{d}x = \\frac{\\pi}{2}\\)",
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            "content": "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\\)",
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            "content": "\\[\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\\]"
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            "content": "解(1)做 \\(t = x^2\\) ，得到 \\(\\frac{1}{2}\\int_0^{+\\infty}\\frac{\\sin t}{t}\\mathrm{d}t = \\frac{\\pi}{4}\\)",
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            "content": "(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",
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            "content": "\\[\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\\]"
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            "content": "(3) 先分部积分得到",
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            "content": "\\[\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\\]"
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            "content": "根据 \\(4\\sin^3 x\\cos x = \\sin 2x(1 - \\cos 2x) = \\sin 2x - \\frac{1}{2}\\sin 4x\\) 得到",
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            "content": "\\[\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\\]"
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            "content": "8.1 数项级数基本理论"
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            "content": "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。",
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            "content": "定理 8.1 (数项级数收敛的各种条件)"
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            "content": "- 必要条件 (一定能推出): \\(\\lim_{n\\to \\infty}u_n = 0\\)",
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            "content": "- 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\\)",
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            "content": "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"
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            "content": "8.2 正项级数"
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            "content": "8.2 正项级数"
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            "content": "8.2.1 比较原则"
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            "content": "定理 8.4 (比较原则)"
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            "content": "\\(\\sum_{n = 1}^{\\infty}u_n,\\sum_{n = 1}^{\\infty}v_n\\) 是两个正项级数，",
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            "content": "- 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",
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            "content": "- 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",
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            "content": "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。",
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            "content": "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",
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            "content": "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}\\)",
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            "content": "(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)\\)",
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            "content": "(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})\\)",
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            "content": "(1) \\(\\frac{1}{3^{\\ln n}} = \\frac{1}{n^{\\ln 3}}\\) 收敛",
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            "content": "8.2 正项级数"
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            "content": "(6) \\(\\lim_{n\\to \\infty}n^2\\frac{\\ln(n + 1)}{e^n}\\rightarrow 0\\)",
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            "content": "(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",
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            "content": "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\n\n\\( (2)a_{n}=(1-\\frac{p\\ln n}{n})^{n} \\) ，讨论  \\( \\sum_{n=1}^{\\infty}b_{n} \\)  的敛散性",
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            "content": "(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。",
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            "content": "(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。",
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            "content": "(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",
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            "content": "定理 8.5 (比式比较判别法)"
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            "content": "\\(\\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",
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            "content": "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。",
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            "content": "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",
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            "content": "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",
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            "content": "8.2.2 比式判别法与根式判别法"
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            "content": "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。",
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            "content": "定理 8.6 (比式判别法与根式判别法)"
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            "content": "\\(\\sum_{n = 1}^{\\infty}u_n\\) 是正项级数，则",
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            "content": "- 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。",
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            "content": "8.2 正项级数"
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            "content": "- 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)",
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            "content": "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。",
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            "content": "(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。",
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            "content": "(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。",
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            "content": "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} \\)",
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            "content": "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}\\)",
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            "content": "(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\\)",
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            "content": "解(1)极限为 \\(\\frac{1}{2}\\), 收敛",
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            "content": "8.3 任意项级数"
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            "content": "8.2.3 积分判别法和拉贝判别法"
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            "content": "定理 8.7 (积分判别法与拉贝判别法)"
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            "content": "\\(\\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",
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            "content": "- 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",
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            "content": "证明 (1) 拉贝判别法:",
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            "content": "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：",
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            "content": "\\[\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\\]"
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            "content": "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",
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            "content": "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。",
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            "content": "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。"
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            "content": "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!} \\)",
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            "content": "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",
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            "content": "\\[\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\\]"
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            "content": "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",
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            "content": "(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",
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            "content": "8.3 任意项级数"
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            "content": "8.3 任意项级数"
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            "content": "8.3.1 交错项级数"
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            "content": "定理 8.8 (Leibniz 判别法)"
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            "content": "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",
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            "content": "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:",
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            "content": "\\[\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\\]"
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            "content": "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。",
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            "content": "练习 8.12 讨论下面级数的条件收敛与绝对收敛性:",
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            "content": "(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}\\)",
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            "content": "(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}})\\)",
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            "content": "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。",
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            "content": "(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。",
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            "content": "(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",
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            "content": "(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",
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            "content": "(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。",
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            "content": "(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。",
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            "content": "笔记 单调性常常使用求导辅助。",
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            "content": "8.4 数项级数理论"
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            "content": "8.3.2 Dirichlet 与 Abel 判别法"
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            "content": "定理 8.9 (Dirichlet 判别法)"
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            "content": "\\(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",
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            "content": "定理 8.10 (Abel 判别法)"
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            "angle": 0,
            "content": "\\(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",
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            "angle": 0,
            "content": "笔记 Leibniz 判别法本质只是 Dirichlet 和 Abel 判别法的一个特例"
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            "content": "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",
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            "content": "(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",
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            "content": "(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",
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            "content": "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。",
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            "content": "(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。",
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            "content": "下面证明 \\(0 < p \\leq 1\\) 时不绝对收敛：",
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            "content": "\\[\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\\]"
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            "bbox": [
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            "angle": 0,
            "content": "而 \\(\\sum_{n=1}^{\\infty} \\frac{1}{n^p}\\) 发散，因此发散。",
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            "content": "定理 8.11 (加括号收敛)"
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            "content": "\\(\\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)",
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            "content": "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。",
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            "content": "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。",
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            "content": "定理 8.12 (加括号收敛反推级数收敛)"
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            "content": "若 \\(\\sum_{n = 1}^{\\infty}u_n\\) 加括号后 \\(\\sum_{n = 1}^{\\infty}v_{n}\\) 收敛，",
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            "content": "- 若每个括号里符号相同，则 \\(\\sum_{n=1}^{\\infty} u_n\\) 收敛",
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            "content": "定义 8.2 (重排)"
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            "content": "考虑级数 \\(\\sum_{n=1}^{\\infty} u_n\\), \\(f\\) 是 \\([n] \\to [n]\\) 的映射, 则称",
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            "content": "\\[\n\\sum_ {n = 1} ^ {\\infty} v _ {n} = \\sum_ {n = 1} ^ {\\infty} u _ {f (n)}\n\\]"
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            "content": "为 \\(\\sum_{n=1}^{\\infty} u_n\\) 的一个重排",
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            "content": "定理 8.13 (正项级数的重排)"
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            "content": "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",
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            "content": "\\[\n\\sum_ {n = 1} ^ {\\infty} u _ {n} = \\sum_ {n = 1} ^ {\\infty} u _ {f (n)}\n\\]"
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            "content": "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",
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            "content": "\\[\n\\lim _ {n \\to \\infty} S _ {n} ^ {\\prime} \\leq \\lim _ {n \\to \\infty} S _ {n} ^ {\\prime}\n\\]"
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            "content": "8.4 数项级数理论"
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            "content": "定理 8.14 (一般项级数的重排)"
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            "content": "\\(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",
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            "angle": 0,
            "content": "\\[\n\\sum_ {n = 1} ^ {\\infty} u _ {n} = \\sum_ {n = 1} ^ {\\infty} u _ {f (n)}\n\\]"
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            "content": "定理 8.15 (绝对收敛级数的重排)"
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            "content": "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",
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            "content": "\\[\n\\sum_ {n = 1} ^ {\\infty} u _ {n} = \\sum_ {n = 1} ^ {\\infty} u _ {f (n)}\n\\]"
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            "angle": 0,
            "content": "证明 正负部分开即可",
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            "content": "定理 8.16 (Riemann 定理：条件收敛级数的重排)"
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            "content": "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\\)",
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            "content": "8.4.3 正部与负部"
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            "content": "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。",
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            "content": "定理 8.18 (级数相互控制)"
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            "content": "设 \\(a_{n}\\leq c_{n}\\leq b_{n}\\) 则",
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            "content": "- 数列 \\(a_{n}, b_{n}\\) 收敛不能得出 \\(c_{n}\\) 收敛",
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            "content": "- 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",
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            "content": "(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。",
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            "content": "推论 8.1 (控制是相互的)"
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            "content": "\\[\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\\]"
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            "content": "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",
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            "content": "(2) 比如 \\(a_{n} = \\frac{1}{n}, b_{n} = \\frac{(-1)^{n}}{\\sqrt{n}}\\)",
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            "content": "(3) 例如 \\(a_{n} = \\frac{(-1)^{n}}{\\sqrt{n}}\\)",
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            "content": "第9章 函数列与函数项级数"
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            "content": "9.1 函数列的一致收敛"
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            "content": "9.1.1 一致收敛的概念"
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            "content": "\\(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\\)",
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            "content": "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。",
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            "content": "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",
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            "content": "(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",
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            "content": "定理 9.1 (Cauchy 收敛准则)"
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            "content": "\\(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\\)",
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            "content": "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)",
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            "content": "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)\\)",
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            "content": "9.1 函数列的一致收敛"
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            "content": "(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)",
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            "content": "定理 9.3 (确界判别法)"
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            "content": "\\(f_{n}\\) 为 \\(I\\) 上函数列，则",
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            "content": "- \\(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\\)。",
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            "content": "- \\(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)",
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            "content": "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\\)",
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            "content": "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。",
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            "content": "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",
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            "content": "9.1.3 一致收敛判别练习"
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            "content": "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：",
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            "content": "\\[\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\\]"
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            "content": "解(1)根据 \\(|f_n(x)|\\leq \\frac{1}{n}\\to 0\\) ，显然",
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            "content": "(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",
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            "content": "(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。",
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            "content": "(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。",
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            "content": "练习 9.4 分式相关 判断在  \\( [0, +\\infty) \\)  的一致收敛性",
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            "content": "\\[\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\\]"
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            "content": "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",
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            "content": "(2) \\(x = 0\\) 显然，\\(x \\neq 0\\) 时 \\(f_{n}(x) \\leq \\frac{1}{n} \\to 0\\) ，一致收敛",
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            "content": "(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。",
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            "content": "(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\\)",
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            "content": "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) \\)",
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            "content": "9.1 函数列的一致收敛"
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            "content": "(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}}\\)",
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            "content": "(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\\)",
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            "content": "(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",
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            "content": "\\[\n\\left| f _ {n} (x) - f ^ {\\prime} (x) \\right| = \\left| f ^ {\\prime} \\left(x _ {n} + \\xi_ {n}\\right) - f ^ {\\prime} (x) \\right| <   \\epsilon\n\\]"
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            "content": "因此一致收敛",
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            "content": "(2) 即 (1) 的特例",
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            "content": "(3) \\( F(x)=\\sum_{k=1}^{n}\\int_{\\frac{k-1}{n}}^{\\frac{k}{n}}f(x+t)\\mathrm{d}t \\) ，因此",
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            "content": "\\[\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\\]"
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            "content": "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",
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            "content": "\\[\n\\left| \\left(x + \\frac {k}{n}\\right) - (x + t) \\right| = \\left| \\frac {k}{n} - t \\right| \\leq \\frac {1}{n} <   \\delta\n\\]"
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            "content": "综上得到",
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            "content": "\\[\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\\]"
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            "content": "9.1.4 四则运算、复合极限的一致收敛性"
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            "content": "定义 9.3 (一致有界)"
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            "content": "\\(\\exists M, \\forall n, \\forall x \\in I\\) 有 \\(|f_n(x)| \\leq M\\)",
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            "content": "引理 9.1 (极限的有界性)"
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            "content": "设 \\(f_{n}(x)\\) 在区间 \\(I\\) 上点态收敛到 \\(f(x)\\)，则",
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            "content": "- 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",
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            "content": "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",
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            "content": "9.2 函数列极限的性质"
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            "content": "定理 9.4 (四则运算)"
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            "content": "\\(f_{n}, g_{n}\\) 在 \\(I\\) 上分别一致收敛到 \\(f(x), g(x)\\)，则",
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            "content": "- 数乘： \\( kf_{n}(x) \\)  一致收敛到  \\( kf(x) \\)",
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            "content": "- 加法： \\( f_{n}(x) + g_{n}(x) \\)  一致收敛到 \\( f(x) + g(x) \\)",
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            "content": "- 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)\\)",
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            "content": "- 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) \\)",
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            "content": "- 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)}\\)",
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            "content": "(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\\)",
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            "content": "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。",
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            "content": "定理 9.5 (复合)"
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            "content": "\\(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))\\)",
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            "content": "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",
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            "content": "\\[\n\\left| f _ {n} (x) \\right| \\leq M\n\\]"
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            "content": "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",
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            "content": "\\[\n\\left| f _ {n} (x) - f (x) \\right| <   \\delta , \\forall x \\in I\n\\]"
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            "content": "从而 \\(n > N\\) 时， \\(\\forall x\\in I\\) 满足 \\(|g(f_n(x)) - g(f(x))| <   \\epsilon\\)",
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            "content": "9.2 函数列极限的性质"
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            "content": "9.2.1 极限函数的连续性"
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            "content": "定理 9.6 (极限顺序交换)"
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            "content": "\\(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",
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            "content": "\\[\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\\]"
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            "content": "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",
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            "content": "\\[\n| f (x) - A | \\leq | f (x) - f _ {n} (x) | + | f _ {n} (x) - a _ {n} | + | a _ {n} - A | <   3 \\epsilon\n\\]"
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            "content": "9.2 函数列极限的性质"
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            "content": "推论 9.1 (局部连续性)"
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            "content": "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",
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            "content": "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)\\)",
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            "content": "推论 9.2 (整体连续性)"
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            "content": "若连续函数列在 \\(I\\) 上内闭一致收敛于 \\(f\\) ，则 \\(f\\) 在 \\(I\\) 上连续",
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            "content": "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",
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            "content": "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。",
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            "content": "定理 9.7 (一致连续性)"
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            "content": "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",
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            "content": "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。",
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            "content": "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})\\)",
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            "content": "(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)\\)",
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            "content": "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。",
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            "content": "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",
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            "content": "\\[\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\\]"
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            "content": "(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。",
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            "content": "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。",
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            "content": "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\\)",
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            "content": "9.2.2 极限函数的可积性"
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            "content": "定理 9.8 (可积性)"
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            "content": "\\(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",
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            "content": "\\[\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\\]"
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            "content": "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:",
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            "content": "\\[\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\\]"
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            "content": "9.2 函数列极限的性质"
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            "content": "得到 \\(f(x)\\) 的振幅 \\(\\omega_{i}^{f}\\leq \\frac{2\\epsilon}{b - a} +\\omega_{i}^{f_{N}}\\) ，因此",
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            "content": "\\[\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\\]"
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            "content": "(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\\)",
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            "angle": 0,
            "content": "笔记 该定理在 ZJU2020 中出现过。",
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            "content": "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}\\)。",
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            "content": "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。",
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            "content": "推论 9.3 (可积性推广)"
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            "content": "89"
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            "content": "\\(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\\)",
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            "content": "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 \\) \n\n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 \\)",
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            "content": "9.2.3 极限函数的可微性"
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            "content": "定理 9.9 (可微性)"
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            "content": "\\(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",
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            "content": "\\[\n(\\lim _ {n \\to \\infty} f _ {n} (x)) ^ {\\prime} = \\lim _ {n \\to \\infty} f _ {n} ^ {\\prime} (x)\n\\]"
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            "content": "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 \\)",
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            "content": "(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) \\)。",
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            "content": "9.3 函数项级数的一致收敛性"
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            "content": "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}}\\)",
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            "content": "定理 9.10 (关于 x 单调)"
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            "content": "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)\\)",
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            "content": "定理 9.11 (Dini 定理：关于 n 单调)"
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            "content": "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)\\)",
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            "content": "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",
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            "content": "\\[\n\\left| f _ {n} (x) - f (x) \\right| \\leq \\left| f _ {N} (x) - f (x) \\right| <   \\epsilon\n\\]"
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            "content": "(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",
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            "content": "90"
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            "content": "\\[\n\\forall n > N _ {i}, \\forall x \\in U (x _ {i}, \\delta_ {i}), | f _ {n} (x) - f (x) | <   \\epsilon\n\\]"
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            "content": "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 \\) \n笔记 ZJU2020 默写了 Dini 定理的证明",
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            "content": "9.3 函数项级数的一致收敛性"
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            "content": "9.3.1 一致收敛的基本性质与判别"
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            "content": "\\( \\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。",
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            "content": "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",
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            "content": "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",
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            "content": "定理 9.13 (一致收敛充要条件)"
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            "content": "9.3 函数项级数的一致收敛性"
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            "content": "- 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\\)",
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            "content": "- 确界： \\(\\lim_{n\\to \\infty}\\sup_{x\\in D}|\\sum_{k = n + 1}^{\\infty}u_k(x)| = 0\\)",
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            "content": "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",
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            "content": "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 \\) 。",
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            "content": "定理 9.14 (M 判别法)"
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            "content": "\\(\\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",
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            "content": "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\\)",
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            "content": "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",
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            "content": "9.3.2 Dirichlet 与 Abel 判别法"
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            "content": "定理 9.15 (Dirichlet 与 Abel 判别法)"
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            "content": "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",
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            "content": "笔记 单调都是关于 \\(n\\) ，一致都是关于 \\(x\\)",
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            "content": "9.4 函数项级数极限函数的性质"
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            "content": "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：",
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            "content": "\\[\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\\]"
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            "content": "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。",
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            "content": "\\(\\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",
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            "content": "\\[\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\\]"
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            "content": "第 10 章 幂级数"
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            "content": "10.1 幂级数概念及其收敛区间"
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            "content": "<table><tr><td>定义10.1(幂级数)</td></tr><tr><td><eq>\\sum_{n=0}^{\\infty}a_n(x-x_0)^n</eq>称为幂级数,但一般只研究<eq>\\sum_{n=0}^{\\infty}a_nx^n</eq></td></tr></table>"
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            "content": "10.1.1 收敛区间与收敛半径"
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            "content": "定理 10.1 (Abel 定理)"
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            "content": "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。",
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            "content": "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",
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            "content": "定义 10.2 (收敛半径)"
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            "content": "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。",
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            "content": "定理 10.2 (收敛半径的计算)"
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            "content": "幂级数收敛半径 \\(R\\) (可以为 \\(0, \\infty\\)) 的计算有以下几种方式：",
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            "content": "- 根式 (柯西-阿达马): \\(\\lim_{n\\to \\infty}\\sqrt[n]{|a_n|} = \\frac{1}{R}\\) (可以为 \\(0,\\infty\\) )",
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            "content": "- 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}\\)",
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            "content": "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。",
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            "content": "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\\)",
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            "content": "(2) \\(\\lim_{n\\to \\infty}\\sqrt[n]{\\frac{1}{n}} = 1\\) ，收敛半径1，收敛域[-1,1)",
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            "content": "10.1 幂级数概念及其收敛区间"
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            "content": "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}} \\)",
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            "content": "(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",
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            "content": "(3) \\(\\sum_{n=1}^{\\infty} \\frac{x^{2n}}{n-3^{2n}}, \\sum_{n=1}^{\\infty} \\frac{x^{n^2}}{2^n}\\) 收敛域",
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            "content": "(4) \\( \\sum_{n=1}^{\\infty}\\frac{3^{n}+(-2)^{n}}{n}(x+1)^{n} \\)",
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            "content": "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] \\)",
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            "content": "(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",
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            "content": "(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",
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            "content": "(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})\\)",
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            "content": "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！",
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            "content": "94"
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            "content": "(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) \\)\n\n(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}] \\)",
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            "content": "10.1.2 幂级数的一致收敛性"
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            "content": "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。",
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            "content": "定理 10.3 (Abel 第二定理)"
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            "content": "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)",
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            "content": "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。",
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            "content": "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！",
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            "content": "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",
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            "content": "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)\\)",
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            "content": "设 \\((-R, R)\\) 上 \\(f(x) = \\sum_{n=0}^{\\infty} a_n x^n\\)，则对 \\(\\forall x \\in (-R, R)\\) 有：",
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            "content": "\\[\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\\]"
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            "content": "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！",
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            "content": "\\[\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\\]"
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            "content": "笔记 具体用哪个主要看  \\( (-1) \\)  的指数次和分母的关系。",
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            "content": "10.2.1 等比级数"
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            "content": "定理 10.8 (等比级数)"
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            "content": "根据 \\(f(x) = \\sum_{n=0}^{\\infty} x^n = \\frac{1}{1 - x}\\) 可以推出：",
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            "content": "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\\)",
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            "content": "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\\)",
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            "content": "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)}\\)",
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            "content": "解(1)这里展示正推, 进行公式推导:",
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            "content": "\\[\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\\]"
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            "content": "因此 \\(I = f_{3}\\left(\\frac{1}{2}\\right) + 3f_{2}\\left(\\frac{1}{2}\\right) = 6 + 6 = 12\\)",
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            "content": "(2) 这里展示反推:",
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            "content": "\\[\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\\]"
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            "content": "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\\)",
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            "content": "10.2.2 \\(\\arctan x\\) 级数"
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            "content": "练习 10.6arctan x (1) 重点：求  \\( \\sum_{n=1}^{\\infty}\\frac{(-1)^{n-1}}{n(2n+1)} \\)",
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            "content": "10.3 幂级数求和式展开"
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            "content": "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",
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            "content": "\\[\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\\]"
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            "content": "因此结果为 \\(\\ln 2 + \\frac{\\pi}{2} - 2\\)",
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            "content": "10.2.3 \\(e^x\\) 级数"
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            "content": "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] \\)",
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            "content": "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",
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            "content": "\\[\n\\frac {e ^ {x} + e ^ {- x}}{2} = \\sum_ {n = 0} ^ {\\infty} \\frac {x ^ {2 n}}{(2 n) !}\n\\]"
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            "content": "97"
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            "content": "取 \\(x = 1\\) ，得到 \\(\\sum_{n=0}^{\\infty}\\frac{1}{(2n)!} = \\frac{e + e^{-1}}{2}\\)",
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            "content": "10.3 幂级数求和式展开"
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            "content": "10.3.1 基本理论与直接展开"
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            "content": "定理 10.9 (幂级数展开)"
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            "content": "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",
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            "content": "\\[\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\\]"
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            "content": "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",
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            "content": "(2) \\( f(x)=\\frac{1}{x(x+1)} \\)  在 x=1 处幂级数展开",
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            "content": "(3) \\( f(x)=\\frac{1}{1-3x+2x^{2}} \\)  在 x=0 处展开",
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            "content": "(4) \\(f(x) = \\frac{x}{1 + x - 2x^2}\\) 在 \\(x = 0\\) 处展开",
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            "content": "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",
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            "content": "\\[\nf (x) = 1 - \\sum_ {n = 0} ^ {\\infty} \\left(\\frac {- t}{2}\\right) ^ {n}\n\\]"
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            "content": "(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\\)",
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            "content": "10.3 幂级数求和式展开"
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            "content": "10.3.2 求导与求积"
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            "content": "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",
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            "content": "(2) 计算 \\( f(x) = \\arcsin x \\) 在 \\( x = 0 \\) 处的幂级数展开",
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            "content": "(3) 计算 \\( f(x) = \\int_{0}^{x} \\frac{\\sin t}{t} t \\) 在 \\( x = 0 \\) 处的幂级数展开",
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            "content": "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}\\)",
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            "content": "\\[\n(2) f ^ {\\prime} (x) = \\frac {1}{\\sqrt {1 - x ^ {2}}}, \\text { 而 }\n\\]"
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            "content": "\\[\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\\]"
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            "content": "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}\\)",
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            "content": "练习 10.10arctan 经典题目 求幂级数展开:",
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            "content": "(1) \\( f(x)=\\arctan x \\)",
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            "content": "(2) 重点 (ZJU2021): \\( g(x) = \\arctan \\frac{1 - kx}{1 + kx} \\)",
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            "content": "\\[\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\\]"
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            "content": "\\[\nh (x) = 2 \\sum_ {n = 0} ^ {\\infty} \\frac {(- 1) ^ {n}}{2 n + 1} x ^ {2 n + 1}\n\\]"
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            "content": "第 11 章 傅里叶级数"
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            "content": "11.1 以 \\(2\\pi\\) 为周期的傅里叶级数"
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            "content": "定义 11.1 (2π 周期傅里叶展开)"
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            "content": "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",
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            "content": "\\[\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\\]"
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            "content": "\\[\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\\]"
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            "content": "笔记 计算时一定要注意第一项是 \\(\\frac{a_0}{2}\\)!",
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            "content": "定理 11.1 (收敛定理)"
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            "content": "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。",
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            "content": "推论 11.1 (连续特殊情况)"
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            "content": "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\\)",
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            "content": "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}}\\)",
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            "content": "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",
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            "content": "\\[\nx ^ {2} = \\frac {1}{3} \\pi^ {2} + \\sum_ {n = 1} ^ {\\infty} (- 1) ^ {n} \\frac {4}{n ^ {2}}\n\\]"
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            "content": "令 \\(x = \\pi\\) 得到",
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            "content": "\\[\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\\]"
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            "content": "笔记 注意 \\(\\cos n\\pi = (-1)^n\\)。",
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            "content": "11.2 以 \\(2l\\) 为周期的傅里叶级数"
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            "content": "\\(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",
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            "content": "\\[\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\\]"
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            "content": "第 12 章 多元极限与连续"
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            "content": "12.1 累次极限与重极限"
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            "content": "12.1.1 基本概念"
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            "content": "定义 12.1 (重极限)"
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            "content": "\\(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\\)",
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            "content": "笔记 计算重极限的方法:",
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            "content": "- 放缩 + 迫敛：例如 \\(\\left|(x + y)\\sin \\frac{1}{x}\\sin \\frac{1}{y}\\right| \\leq |x + y| = 0\\)",
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            "content": "- 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\\)",
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            "content": "- 两个累次极限存在但不相等",
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            "content": "- 找特殊的趋近方式，如 \\( y = kx \\)，使得获得的重极限不同",
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            "content": "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)\\)",
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            "content": "\\[\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\\]"
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            "content": "解 (1) 设 \\(x = r\\cos \\theta, y = r\\sin \\theta\\)，则",
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            "content": "\\[\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\\]"
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            "content": "因此极限为0",
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            "content": "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。",
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            "content": "12.2 多元函数连续性"
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            "content": "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。",
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            "content": "练习 12.2 常用反例 讨论  \\( (0,0) \\)  处的重极限与累次极限",
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            "content": "(1) 重点: \\(f(x, y) = \\frac{xy}{x^2 + y^2}\\)",
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            "content": "(2) \\( f(x,y)=\\frac{x-y}{x+y} \\)",
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            "content": "(3) \\( f(x,y)=x\\sin\\frac{1}{y} \\)",
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            "content": "102"
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            "content": "(4) \\( f(x,y)=x\\sin\\frac{1}{y}+y\\sin\\frac{1}{x} \\)",
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            "content": "(5) 重点: \\( f(x, y) = \\begin{cases} 1, & 0 < y < x^2, -\\infty < x < +\\infty \\\\ 0, & \\text{其他} \\end{cases} \\)",
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            "content": "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。",
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            "content": "(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",
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            "content": "(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",
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            "content": "(4) 累次极限不存在。而 \\(|f(x, y)| \\leq |x| + |y| \\to 0\\)，重极限为 0。",
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            "content": "(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。",
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            "content": "练习 12.3 反例应用 上述反例可作为以下问题的解答:",
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            "content": "1. 累次极限存在但不相等: \\(\\frac{x - y}{x + y}\\)",
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            "content": "2. 累次极限存在且相等，但重极限不存在： \\(\\frac{xy}{x^2 + y^2}\\)",
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            "content": "3. 累次极限不存在，但重极限存在： \\( x \\sin \\frac{1}{y} + y \\sin \\frac{1}{x} \\)",
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            "content": "4. 重极限与双累次极限都不存在: \\(\\frac{1}{x} + \\frac{1}{y}\\)",
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            "content": "12.1.2 重极限的计算"
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            "content": "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}} \\) \n(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}} \\)",
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            "content": "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",
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            "content": "(3) 齐次，\\(y = kx\\)，不存在",
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            "content": "(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",
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            "content": "12.2 多元函数连续性"
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            "content": "12.2 多元函数连续性"
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            "content": "12.2.1 多元函数连续性"
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            "content": "定义 12.3 (多元函数连续)"
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            "content": "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。",
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            "content": "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。",
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            "content": "\\[\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\\]"
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            "content": "\\(f(x,y)\\) 关于 \\(x,y\\) 均连续，但是 \\(\\lim_{(x,y)\\to (0,0)}f(x,y)\\) 不存在。",
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            "content": "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：",
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            "content": "1. 一致连续条件：\\(y\\) 的连续是关于 \\(x\\) 一致的",
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            "content": "2. Lip 条件： \\( f(x,y) \\)  关于 x 满足 Lip 条件，则  \\( f(x,y) \\)  连续",
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            "content": "3. 单调条件：关于 \\(x\\) 或 \\(y\\) 单调",
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            "content": "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！",
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            "content": "(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",
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            "content": "\\[\nf (x, y) \\leq f (x, y _ {0}) + \\epsilon \\leq f (x _ {0}, y _ {0}) + 2 \\epsilon\n\\]"
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            "content": "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\\) 。",
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            "content": "12.2.2 多元函数的一致连续性"
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            "content": "引理 12.1 (多元三角不等式)"
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            "content": "对 \\(\\forall \\mathbf{x},\\mathbf{y}\\in \\mathbb{R}^n\\) ，有 \\(||\\mathbf{x} - |\\mathbf{y}||\\leq |\\mathbf{x} - \\mathbf{y}|\\)",
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            "content": "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}|\\)",
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            "content": "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。",
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            "content": "(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。",
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            "content": "定义 13.1 (偏导数)"
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            "content": "\\(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}\\)",
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            "content": "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。",
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            "content": "定义 13.2 (可微)"
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            "content": "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",
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            "content": "\\[\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\\]"
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            "content": "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\\)",
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            "content": "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。",
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            "content": "推论 13.1 (偏导与可微)"
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            "content": "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",
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            "content": "笔记 判断是否可微分两步："
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            "content": "1. 看看 \\(f_{x}, f_{y}\\) 是否都存在，不存在则不可微",
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            "content": "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",
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            "content": "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",
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            "content": "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\\)",
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            "content": "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",
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            "content": "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！)",
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            "content": "\\[\nf _ {x} (0, 0) = \\frac {f (\\Delta x , 0) - f (0 , 0)}{\\Delta x} = 0\n\\]"
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            "content": "同理可知 \\(f_{y}(0,0) = 0\\) ，再考虑极限",
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            "content": "\\[\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\\]"
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            "content": "极限不存在，因此不可微。",
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            "content": "13.1.2 可微性的证明"
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            "content": "13.1 多变量导数与微分"
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            "content": "(2) \\(\\alpha > 0\\)，讨论 \\(f(x, y) = |xy|^{\\alpha}\\) 在原点的可微性",
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            "content": "(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",
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            "content": "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",
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            "content": "(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：",
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            "content": "\\[\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\\]"
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            "content": "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",
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            "content": "(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})\\) 。",
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            "content": "13.1.3 梯度与方向导数"
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            "content": "定义 13.3 (梯度)"
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            "content": "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} \\)",
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            "content": "定义 13.4 (方向导数)"
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            "content": "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}\\)",
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            "content": "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) \\)",
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            "content": "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",
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            "content": "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 \\)",
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            "content": "定理 13.1 (可微函数方向导数的梯度表达)"
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            "content": "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",
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            "content": "\\[\nf _ {l} (P _ {0}) = f _ {x} (P _ {0}) \\cos \\alpha + f _ {y} (P _ {0}) \\cos \\beta = \\nabla f \\cdot \\mathbf {n}\n\\]"
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            "content": "106"
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            "content": "这里 \\(\\mathbf{n} = (\\cos \\alpha, \\cos \\beta)\\) 是 \\(l\\) 方向的单位向量",
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            "content": "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：",
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            "content": "\\[\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\\]"
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            "content": "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。",
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            "content": "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",
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        {
            "type": "header",
            "bbox": [
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            "angle": 0,
            "content": "13.1 多变量导数与微分"
        },
        {
            "type": "text",
            "bbox": [
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                0.906,
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            "content": "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}\\)",
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        },
        {
            "type": "text",
            "bbox": [
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            "content": "(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) \\)",
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        },
        {
            "type": "text",
            "bbox": [
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            "angle": 0,
            "content": "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。",
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        },
        {
            "type": "text",
            "bbox": [
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            "content": "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",
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        },
        {
            "type": "equation",
            "bbox": [
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            "content": "\\[\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\\]"
        },
        {
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            "bbox": [
                0.091,
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            "angle": 0,
            "content": "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。",
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        },
        {
            "type": "paragraph_title",
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            "content": "13.1.4 可微、偏导、方向导数、连续关系总结"
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            "content": "定理 13.2 (可微的性质)"
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            "angle": 0,
            "content": "若 \\(f\\) 在 \\(\\mathbf{x}_0\\) 可微，则",
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        {
            "type": "text",
            "bbox": [
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            "angle": 0,
            "content": "- \\(f\\) 在 \\(\\mathbf{x}_0\\) 处连续",
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            "bbox": [
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            "content": "107"
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            "type": "text",
            "bbox": [
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            "angle": 0,
            "content": "- \\(f\\) 在 \\(\\mathbf{x}_0\\) 处存在偏导数和任意方向导数",
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        {
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            "bbox": [
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            "content": "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",
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        },
        {
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            "content": "定理 13.3 (可微充分条件)"
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            "type": "text",
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            "content": "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",
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            "content": "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",
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            "content": "\\[\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\\]"
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            "content": "根据偏导连续，\\(\\exists \\alpha, \\beta\\) 在 \\((\\Delta x, \\Delta y) \\to (0, 0)\\) 时 \\(\\alpha, \\beta \\to 0\\)，满足",
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            "type": "equation",
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            "content": "\\[\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\\]"
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            "content": "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。",
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        },
        {
            "type": "text",
            "bbox": [
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            "angle": 0,
            "content": "笔记 若 \\(f_{x}, f_{y}\\) 连续，则一般称 \\(f(x, y)\\) 连续可微。",
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            "content": "定理 13.4 (偏导与方向导的性质)"
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            "angle": 0,
            "content": "偏导与方向导数相关性质：",
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            "content": "- 若 \\(f\\) 偏导数在 \\((x_0, y_0)\\) 邻域有界，则 \\(f\\) 在 \\((x_0, y_0)\\) 连续",
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            "angle": 0,
            "content": "- \\(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",
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            "content": "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",
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            "content": "\\[\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\\]"
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            "content": "根据偏导有界条件，因此 \\(|f(x,y) - f(x_0,y_0)|\\to 0\\) ，因此连续。",
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            "content": "(2) 均见下面的重要反例",
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            "content": "13.1 多变量导数与微分"
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            "content": "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 \\) ，",
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            "content": "\\[\nf (x, y) = \\left\\{ \\begin{array}{l l} 1, & 0 <   y <   x ^ {2}, - \\infty <   x <   \\infty \\\\ 0, & \\text {其他} \\end{array} \\right.\n\\]"
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            "content": "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。",
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            "content": "13.1.5 复合求导：链式法则"
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            "content": "定理 13.5 (链式法则)"
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            "content": "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",
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            "content": "\\[\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\\]"
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            "content": "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} \\)",
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            "content": "108"
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            "content": "(2) 计算 \\(x^{x}\\) 的导数",
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            "content": "解(1)展开后如下:",
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            "content": "\\[\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\\]"
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            "content": "(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) \\)",
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            "content": "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}\\)",
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            "content": "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",
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            "content": "(1) \\(\\frac{\\partial z}{\\partial x} = f_{1}(x, \\frac{x}{y}) + f_{2}(x, \\frac{x}{y})\\frac{1}{y}\\)，因此",
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            "content": "\\[\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\\]"
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            "content": "13.1.6 混合偏导"
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            "content": "定理 13.6 (混合偏导定理)"
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            "content": "若 \\(f_{xy}(x,y), f_{yx}(x,y)\\) 均在 \\(x_0, y_0\\) 邻域连续，则",
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            "content": "\\[\nf _ {x y} (x _ {0}, y _ {0}) = f _ {y x} (x _ {0}, y _ {0})\n\\]"
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            "content": "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",
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            "content": "\\[\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\\]"
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            "content": "13.2 多元中值定理与泰勒展开"
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            "content": "再根据 Lagrange 中值定理:",
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            "content": "109"
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            "content": "\\[\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\\]"
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            "content": "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。",
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            "content": "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) \\) :",
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            "content": "\\[\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\\]"
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            "content": "解 直接计算可以得到  \\( f_{xy}(0,0) = -1, f_{yx}(0,0) = 1 \\)",
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            "content": "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)\\)",
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            "content": "13.2 多元中值定理与泰勒展开"
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            "content": "13.2.1 多元中值定理"
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            "content": "定义 13.5 (凸区域)"
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            "content": "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",
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            "content": "\\[\n\\left(x _ {1} + \\lambda \\left(x _ {2} - x _ {1}\\right), y _ {1} + \\lambda \\left(y _ {2} - y _ {1}\\right)\\right) \\in D\n\\]"
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            "content": "即任意两点的连线都含于 D，则称 D 为凸区域。",
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            "content": "凸区域"
        },
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            "content": "非凸区域"
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            "content": "图 13.1: 凸区域"
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            "content": "定理 13.7 (多元中值定理)"
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            "content": "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",
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            "content": "\\[\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\\]"
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            "content": "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",
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            "content": "\\[\n\\Phi^ {\\prime} (\\theta) = f _ {x} (a + \\theta h, b + \\theta k) h + f _ {y} (a + \\theta h, b + \\theta k) k\n\\]"
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            "content": "由于 \\(D\\) 是凸区域，因此 \\((a + \\theta h, b + \\theta k) \\in D\\) ，从而结论成立。",
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            "content": "13.3 隐函数定理"
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            "content": "13.2.2 多元 Taylor 展开"
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            "content": "定理 13.8 (多元 Taylor 展开)"
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            "content": "多元泰勒 \\(f(x + \\Delta x, y + \\Delta y)\\) 展开的公式如下：",
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            "content": "\\[\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\\]"
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            "content": "证明 设  \\( \\Phi(t)=f(x_{0}+th,y_{0}+tk) \\) ，用一元 Taylor 公式证明。",
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            "content": "13.3 隐函数定理"
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            "content": "13.3.1 隐函数存在定理与可微定理"
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            "content": "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。",
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            "content": "定义 13.6 (隐函数)"
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            "content": "\\(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。",
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            "content": "定理 13.9 (隐函数存在唯一性定理)"
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            "content": "若 \\(F(x,y)\\) 满足"
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            "content": "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",
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            "content": "2. \\(F(x_0,y_0) = 0\\) ，且 \\(F_{y}(x_{0},y_{0})\\neq 0\\) 则：",
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            "content": "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)\\)",
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            "content": "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！）",
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            "content": "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}} \\)",
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            "content": "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)} \\)",
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            "content": "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。",
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            "content": "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。",
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            "content": "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}}\\)",
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            "content": "(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}\\)",
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            "content": "(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 \\)",
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            "content": "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：",
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            "content": "\\[\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\\]"
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            "content": "13.3 隐函数定理"
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            "content": "(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：",
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            "content": "\\[\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\\]"
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            "content": "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",
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            "content": "\\[\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\\]"
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            "content": "(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",
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            "content": "\\[\n- \\frac {F _ {y}}{F _ {x}} \\cdot - \\frac {F _ {z}}{F _ {y}} \\cdot - \\frac {F _ {x}}{F _ {z}} = - 1\n\\]"
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            "content": "13.3.2 隐函数组存在定理与微分定理"
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            "content": "定理 13.10 (隐函数组定理)"
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            "content": "若 \\(F(x,y,u,v),G(x,y,u,v)\\) 满足：",
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            "content": "- 在 \\(P_0(x_0, y_0, u_0, v_0)\\) 邻域中对所有变量有连续偏导",
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            "content": "\\[\nF (P _ {0}) = G (P _ {0}) = 0, J = \\left. \\frac {\\partial (F , G)}{\\partial (u , v)} \\right| _ {P _ {0}} \\neq 0\n\\]"
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            "content": "则方程组",
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            "content": "\\[\n\\left\\{ \\begin{array}{l} F (x, y, u, v) = 0 \\\\ G (x, y, u, v) = 0 \\end{array} \\right.\n\\]"
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            "content": "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",
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            "content": "\\[\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\\]"
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            "content": "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)}\\)",
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            "content": "证明 求导结论使用了 Cramer 法则",
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            "content": "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。",
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            "content": "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：",
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            "content": "\\[\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\\]"
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            "content": "练习 13.14 隐函数组求导 (1) 证明",
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            "content": "\\[\n\\left\\{ \\begin{array}{l} x ^ {2} + y ^ {2} + u ^ {2} - v ^ {2} = 2 \\\\ x + y + u + v = 0 \\end{array} \\right.\n\\]"
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            "content": "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}\\)",
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            "content": "13.4 多元无条件极值问题"
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            "content": "(2) 设 \\(u = u(x, y), v = v(x, y)\\) 由方程组",
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            "content": "\\[\n\\left\\{ \\begin{array}{l} u = f (u x, v + y) \\\\ v = g (u - x, v ^ {2} y) \\end{array} \\right.\n\\]"
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            "content": "给出的隐函数，求 \\(u_{x}, v_{x}\\)",
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            "content": "(3) 已知如下方程组，求  \\( z_{x} \\) :",
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            "content": "\\[\n\\left\\{ \\begin{array}{l} x = u + v \\\\ y = u ^ {2} + v ^ {2} \\\\ z = u ^ {3} + v ^ {3} \\end{array} \\right.\n\\]"
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            "angle": 0,
            "content": "(4) 令 \\( u = f(x - ut, y - ut, z - ut), g(x, y, z) = 0 \\)，求 \\( u_x, u_y \\)",
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            "content": "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：",
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            "content": "\\[\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\\]"
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            "content": "因此 \\(J(P_0) \\neq 0\\)，隐函数组存在，根据隐函数组求导：",
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            "content": "\\[\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\\]"
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            "content": "(2) 直接两侧求导，得到",
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            "content": "\\[\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\\]"
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            "content": "解线性方程组得到",
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            "content": "\\[\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\\]"
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            "content": "(3) 令 \\(x, y\\) 为自变量，\\(u, v, z\\) 为因变量，则",
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            "content": "\\[\nz _ {x} = - \\frac {\\frac {\\partial (F , G , H)}{\\partial (x , u , v)}}{\\frac {\\partial (F , G , H)}{\\partial (z , u , v)}}\n\\]"
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            "content": "13.4 多元无条件极值问题"
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            "content": "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。",
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            "content": "13.4 多元无条件极值问题"
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            "content": "定义 13.9 (Hesse 矩阵)"
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            "content": "\\(f:\\mathbb{R}^n\\to \\mathbb{R}\\) ，则其Hesse矩阵为：",
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            "content": "\\[\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\\]"
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            "content": "定义 13.10 (正定、负定、不定)"
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            "content": "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：",
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            "angle": 0,
            "content": "- 正(负)定: 对 \\(\\forall \\mathbf{x}, Q(\\mathbf{x}) \\geq (\\leq)0\\)",
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            "content": "- 不定：\\(\\exists \\mathbf{p}, \\mathbf{q} \\in \\mathbb{R}^n\\)，\\(Q(\\mathbf{p}) \\leq 0 \\leq Q(\\mathbf{p})\\)",
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            "content": "定理 13.11 (正定的判别法)"
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            "content": "\\(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",
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            "content": "13.4.1 无条件极值"
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            "content": "定理 13.12 (极值点必要条件)"
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            "content": "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\\)",
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            "content": "证明 用反证法即可。",
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            "content": "定理 13.13 (极值点充分条件)"
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            "content": "\\(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，",
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            "content": "- 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",
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            "content": "- 若Hesse矩阵 \\(H\\) 可逆但为不定方阵，则 \\(\\mathbf{x}_0\\) 不是 \\(f\\) 的极值点",
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            "content": "- 若 Hesse 矩阵 \\(H\\) 不可逆，则不能判别",
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            "content": "证明 根据 Taylor 公式:",
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            "content": "\\[\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\\]"
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            "content": "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",
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            "content": "推论 13.2 (二元极值判别法)"
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            "content": "\\((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",
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            "content": "\\[\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\\]"
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            "content": "- \\(ac - b^{2} > 0\\) 且 \\(a > 0\\): 极小值",
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            "content": "- \\(ac - b^{2} > 0\\) 且 \\(a < 0\\): 极大值",
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            "content": "113"
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            "content": "- \\(ac - b^{2} < 0\\) ：无极值",
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            "content": "- \\(ac - b^{2} = 0\\) ：不确定",
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            "content": "13.5 条件极值"
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            "content": "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。",
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            "content": "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\\)",
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            "content": "(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。",
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            "content": "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",
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            "content": "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",
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            "content": "\\[\nH = \\left[ \\begin{array}{r r r} 2 & 0 & - 1 \\\\ 0 & 2 & - 1 \\\\ - 1 & - 1 & 2 \\end{array} \\right]\n\\]"
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            "content": "13.5 条件极值"
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            "content": "13.5 条件极值"
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            "content": "13.5.1 Lagrange 乘数法"
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            "content": "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) \\)",
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            "content": "定义 13.11 (Lagrange 函数)"
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            "content": "针对上述目标函数与条件组，定义 Lagrange 函数",
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            "content": "\\[\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\\]"
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            "content": "定理 13.14 (Lagrange 乘数法)"
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            "content": "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：",
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            "content": "\\[\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\\]"
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            "content": "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。",
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            "content": "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}\\)",
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            "content": "(2)x,y>0，证明 \\( \\frac{x^{n}+y^{n}}{2}\\geq\\left(\\frac{x+y}{2}\\right)^{n} \\)",
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            "content": "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：",
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            "content": "\\[\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\\]"
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            "content": "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",
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            "content": "\\[\nx _ {1} \\dots x _ {n} \\leq \\left(\\frac {a}{n}\\right) ^ {n} = \\left(\\frac {x _ {1} + \\cdots + x _ {n}}{n}\\right) ^ {n}\n\\]"
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            "content": "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",
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            "content": "\\[\nL (x, y, z, \\lambda , \\mu) = x ^ {2} + y ^ {2} + z ^ {2} + \\lambda (x ^ {2} + y ^ {2} - z) + \\mu (x + y + z - 1)\n\\]"
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            "content": "\\[\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\\]"
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            "content": "最大值为 \\(9 + 5\\sqrt{3}\\)，最小值为 \\(9 - 5\\sqrt{3}\\)",
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            "content": "第 14 章 含参积分"
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            "content": "14.1 含参正常积分"
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            "content": "14.1.1 含参正常积分基本概念与定理"
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            "content": "定义 14.1 (含参正常积分)"
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            "content": "\\(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：",
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            "content": "\\[\nF (t) = \\int_ {\\alpha (t)} ^ {\\beta (t)} f (x, t) \\mathrm{d} x, t \\in [ a, b ]\n\\]"
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            "content": "定理 14.1 (规则区域含参正常积分性质)"
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            "content": "若 \\(f(x,t)\\) 在 \\([a,b]\\times [c,d]\\) 连续，则",
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            "content": "- 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\\)",
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            "content": "- 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\\)",
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            "content": "- 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\\)",
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            "content": "定理 14.2 (不规则区域的连续与可微性)"
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            "content": "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",
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            "content": "- 连续性： \\(\\varphi (t) = \\int_{\\alpha (t)}^{\\beta (t)}f(x,t)\\mathrm{d}x\\) 在 \\([c,d]\\) 上连续，即",
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            "content": "\\[\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\\]"
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            "content": "- 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)",
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            "content": "\\[\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\\]"
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            "content": "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)\\)",
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            "content": "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",
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            "content": "\\[\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\\]"
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            "content": "进而得到  \\( F''(y) = f(y) + f(y) = 2f(y) \\)",
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            "content": "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):",
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            "content": "\\[\n\\lim _ {x \\to 0} \\frac {2 x \\sqrt {1 + x ^ {4}}}{2 x} = \\lim _ {x \\to 0} \\sqrt {1 + x ^ {4}} = 1\n\\]"
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            "content": "14.2 含参反常积分与一致收敛"
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            "content": "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：",
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            "content": "\\[\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\\]"
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            "content": "14.1.2 常用含参正常积分的计算"
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            "content": "引理14.1"
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            "content": "\\(|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}}\\)",
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            "content": "证明 使用万能公式，在前面三角函数不定积分中涉及过。",
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            "content": "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\\)",
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            "content": "(2) 计算 \\(\\int_0^1 \\sin \\left(\\ln \\frac{1}{x}\\right) \\frac{x^b - x^a}{\\ln x} \\, \\mathrm{d}x\\)",
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            "content": "(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) \\)",
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            "content": "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",
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            "content": "\\[\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\\]"
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            "content": "(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",
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            "content": "\\[\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\\]"
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            "content": "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)\\)",
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            "content": "(3) 注意到",
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            "content": "\\[\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\\]"
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            "content": "而 \\(\\frac{y}{1 + y\\cos x}\\) 连续，因此",
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            "content": "\\[\nF (a) = \\int_ {- a} ^ {a} \\mathrm{d} y \\int_ {0} ^ {\\frac {\\pi}{2}} \\frac {1}{1 + y \\cos x} \\mathrm{d} x\n\\]"
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            "content": "最终积分结果为 \\(\\pi \\arcsin a\\)",
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            "content": "14.2 含参反常积分与一致收敛"
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            "content": "14.2 含参反常积分与一致收敛"
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            "content": "14.2.1 基本概念"
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            "content": "\\(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。",
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            "content": "定义 14.3 (含参反常积分一致收敛)"
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            "content": "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)\\)",
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            "content": "定理 14.3 (一致收敛基本充要条件)"
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            "content": "- 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\\)",
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            "content": "- 上确界判别法： \\(\\lim_{A\\to +\\infty}\\sup_{t\\in I}|\\int_{A}^{+\\infty}f(x,t)\\mathrm{d}x| = 0\\)。",
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            "content": "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"
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            "content": "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",
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            "content": "\\[\n\\left| \\int_ {A ^ {\\prime}} ^ {+ \\infty} \\frac {\\sin u}{u} \\mathrm{d} u \\right| <   \\epsilon\n\\]"
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            "content": "\\[\n\\left| \\int_ {A} ^ {+ \\infty} \\frac {\\sin x y}{y} \\mathrm{d} y \\right| <   \\epsilon\n\\]"
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            "content": "\\[\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\\]"
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            "content": "14.2 含参反常积分与一致收敛"
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            "content": "定理 14.4 (M 判别法)"
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            "content": "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。",
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            "content": "定理 14.5 (比较判别法)"
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            "content": "\\(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",
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            "content": "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 \\) 。",
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            "content": "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",
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            "content": "(2) 证明 \\(\\int_{0}^{+\\infty} e^{-x^2 y} \\, \\mathrm{d}y\\) 在 \\(x \\in [a, b]\\) 一致收敛",
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            "content": "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。",
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            "content": "(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",
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            "content": "笔记 即使含参反常积分是常数，也不代表其一致收敛。",
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            "content": "14.2.2 Dirichlet 与 Abel 判别法"
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            "content": "定理 14.6 (Dirichlet 与 Abel 判别法)"
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            "content": "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",
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            "content": "- 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",
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            "content": "- 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",
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            "content": "笔记 注意这里单调都是关于 x 的，一致都是关于 t 的。",
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            "content": "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",
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            "content": "(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",
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            "content": "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。",
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            "content": "120"
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            "content": "(2) 对于 \\([a, b] \\subset (0, +\\infty)\\)，对 \\(\\forall x \\in [a, b]\\)，有",
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            "content": "而 \\(\\left|\\frac{y}{1 + y^2}\\right| \\leq \\frac{1}{2}\\) 一致有界，且",
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            "content": "14.3 含参反常积分的性质"
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            "content": "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",
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            "content": "(2) \\(\\varphi(y) = \\int_{0}^{+\\infty}\\frac{\\sin x^2}{1 + xy}\\mathrm{d}x\\) 在 \\(y\\in [0, + \\infty)\\) 一致收敛",
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            "content": "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<e^{-xy}\\leq1 \\)  一致有界，由 Abel 判别法可知\n(2) 用 Abel",
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            "content": "14.3.1 含参反常积分的极限与连续性"
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            "content": "定理 14.7 (连续性)"
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            "content": "设 \\(f(x,t)\\) 在 \\(I\\times [c, + \\infty)\\) 上连续，若 \\(\\Phi (t) = \\int_{c}^{+\\infty}f(x,t)\\mathrm{d}x\\) 在 \\(I\\) 上一致收敛，则 \\(\\Phi (t)\\) 在 \\(I\\) 上连续，即\\(\\lim_{t\\to t_0}\\int_c^{+\\infty}f(x,t)\\mathrm{d}x = \\int_c{^+\\infty}f(x,t_0)\\mathrm{d}x\\)",
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            "content": "14.3.2 含参反常积分的可微性与可积性"
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            "content": "若 \\(f(x,t)\\) 与 \\(f_{t}(x,t)\\) 在 \\([c, +\\infty) \\times I\\) 上连续，\\(\\Phi(t) = \\int_{c}^{+\\infty} f(x,t) \\mathrm{d}x\\) 在 \\(I\\) 上收敛，\\(\\int_{c}^{+\\infty} f_{t}(x,t) \\mathrm{d}x\\) 在 \\(I\\) 上一致收敛，则",
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            "content": "\\[\n\\frac {\\mathrm{d}}{\\mathrm{d} t} \\int_ {c} ^ {+ \\infty} f (x, t) \\mathrm{d} x = \\int_ {c} ^ {+ \\infty} \\frac {\\mathrm{d}}{\\mathrm{d} t} f (x, t) \\mathrm{d} x\n\\]"
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            "content": "定理 14.9 (可积性)"
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            "content": "14.3 含参反常积分的性质"
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            "content": "分顺序得到",
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            "content": "\\[\nI = \\int_ {a} ^ {b} \\mathrm{d} y \\int_ {0} ^ {+ \\infty} e ^ {- p x} \\cos x y \\mathrm{d} x\n\\]"
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            "content": "根据表格法得到:",
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            "content": "\\[\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\\]"
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            "content": "因此得到 \\(\\frac{y^2 + p^2}{p^2}\\int_0^{+\\infty}\\cos xy\\mathrm{d}y = \\frac{1}{p}\\)",
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            "content": "\\[\nI = \\int_ {a} ^ {b} \\frac {p}{p ^ {2} + y ^ {2}} \\mathrm{d} y = \\arctan {\\frac {b}{p}} - \\arctan {\\frac {a}{p}}\n\\]"
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            "content": "定理 14.11 (Dirichlet 积分)"
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            "content": "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)\\)",
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            "content": "定理 14.12 (正态积分公式)"
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            "content": "\\[\n\\int_ {0} ^ {+ \\infty} e ^ {- x ^ {2}} \\mathrm{d} x = \\frac {\\sqrt {\\pi}}{2}\n\\]"
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            "content": "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\\)",
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            "content": "14.3.3 含参反常积分与函数项级数的关系"
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            "content": "定理 14.13 (一致收敛级数表示法 (ZJU 考研考过))"
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            "content": "\\(\\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",
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            "content": "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",
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            "content": "14.3 含参反常积分的性质"
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            "content": "(因为一致收敛要求所有 x 都趋于 0)。",
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            "content": "第 15 章 多重积分"
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            "content": "15.1 二重积分"
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            "content": "15.1.1 二重积分的概念与计算"
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            "content": "定义 15.1 (二重积分)"
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            "content": "\\( 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",
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            "content": "\\[\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\\]"
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            "content": "定理 15.1 (二重积分与累次积分)"
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            "content": "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",
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            "content": "\\[\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\\]"
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            "content": "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\\)",
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            "content": "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 \\)",
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            "content": "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",
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            "content": "(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",
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            "content": "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}\\)",
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            "content": "(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",
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            "content": "\\[\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\\]"
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            "content": "15.1 二重积分"
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            "content": "(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! \\)",
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            "content": "15.1.2 二重积分可积性"
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            "content": "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",
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            "content": "\\[\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\\]"
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            "content": "(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",
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            "content": "\\[\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\\]"
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            "content": "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",
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            "content": "(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。",
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            "content": "15.1.3 二重积分换元法"
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            "content": "定理 15.2 (二重积分换元法)"
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            "content": "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",
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            "content": "\\[\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\\]"
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            "content": "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。",
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            "content": "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",
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            "content": "(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",
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            "content": "(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",
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            "content": "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：",
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            "content": "\\[\nu = a, v = 2 x + a\n\\]"
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            "content": "15.1 二重积分"
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            "content": "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})\\)",
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            "content": "(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}\\)",
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            "content": "(3) 做 \\(xy = u, \\frac{y^2}{x} = v\\)",
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            "content": "15.1.4 极坐标换元"
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            "content": "定理 15.3 (二重积分极坐标换元)"
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            "content": "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",
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            "content": "图 15.1: 二重积分极坐标换元"
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            "content": "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\\)",
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            "content": "(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",
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            "content": "(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",
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            "content": "\\[\nD = \\left\\{(x, y): 0 \\leq y \\leq b \\sqrt {1 - \\frac {x ^ {2}}{a ^ {2}}}, 0 \\leq x \\leq a \\right\\}\n\\]"
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            "content": "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",
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            "content": "\\[\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\\]"
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            "content": "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]\\)",
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            "content": "15.2 三重积分"
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            "content": "\\[\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\\]"
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            "content": "(2)一般都是先 \\(r\\) 后 \\(\\theta\\) ，将 \\(D\\) 分为左上，右下两个三角形，则",
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            "content": "\\[\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\\]"
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            "content": "先 \\(\\theta\\) 后 \\(r\\) 会比较麻烦",
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            "content": "三类特殊的圆"
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            "content": "定理 15.4 (三类特殊圆极坐标)"
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            "content": "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：",
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            "content": "- \\((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}]\\)",
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            "content": "- \\(x^{2} + (y - a)^{2} = a^{2}\\)，即 \\(x^{2} + y^{2} = 2ay\\)，则 \\(r = 2a\\sin \\theta, \\theta \\in [0, \\pi]\\)",
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            "content": "- \\((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}]\\)",
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            "content": "图 15.2: 三个特殊的圆"
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            "content": "15.2 三重积分"
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            "content": "15.2.1 投影法"
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            "content": "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",
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            "content": "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：",
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            "content": "\\[\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\\]"
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            "content": "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",
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            "content": "(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",
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            "content": "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}\\)",
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            "content": "15.2 三重积分"
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            "content": "(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",
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            "content": "\\[\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\\]"
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            "content": "15.2.2 截面法"
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            "content": "定理 15.6 (截面法)"
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            "content": "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",
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            "content": "\\[\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\\]"
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            "content": "15.2.3 三重积分换元法"
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            "content": "定理 15.7 (三重积分换元)"
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            "content": "若坐标替换 \\(x = x(u,v,w),y = y(u,v,w),z = z(u,v,w)\\) ，则",
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            "content": "\\[\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\\]"
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            "content": "\\[\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\\]"
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            "content": "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|\\)",
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            "content": "15.2.4 柱面坐标变换与球坐标变换"
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            "content": "定理 15.8 (柱面变换)"
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            "content": "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：",
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            "content": "\\(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\\)",
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            "content": "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",
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            "content": "\\[\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\\]"
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            "content": "15.2 三重积分"
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            "content": "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}\\)",
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            "content": "定理 15.9 (球坐标变换)"
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            "content": "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",
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            "content": "图 15.3: 三重积分球坐标换元"
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            "content": "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",
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            "content": "定理 15.10 (广义球坐标变换)"
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            "content": "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：",
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            "content": "\\[\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\\]"
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            "content": "第 16 章 曲线积分"
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            "content": "16.1 曲线的参数化"
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            "content": "16.2 第一型曲线积分"
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            "content": "16.2.1 第一型曲线积分的基本概念"
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            "content": "定义 16.1 (第一型曲线积分)"
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            "content": "\\( \\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",
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            "content": "\\[\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\\]"
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            "content": "定理 16.1 (第一型曲线积分的计算)"
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            "content": "\\(\\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",
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            "content": "\\[\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\\]"
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            "content": "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。",
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            "content": "推论 16.1 (二维第一型曲线积分的计算)"
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            "content": "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",
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            "content": "\\[\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\\]"
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            "content": "16.2 第一型曲线积分"
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            "content": "(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= \\) \n \\( \\sqrt{a^{2}+b^{2}}(2\\pi a^{2}+\\frac{8}{3}\\pi^{3}b^{2}) \\)",
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            "content": "(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：",
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            "content": "\\[\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\\]"
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            "content": "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",
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            "content": "(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",
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            "content": "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)\\)",
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            "content": "(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}\\)",
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            "content": "16.2.2 使用对称性"
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            "content": "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) \\)",
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            "content": "(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})\\)",
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            "content": "(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}}\\)",
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            "content": "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\\)",
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            "content": "(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：",
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            "content": "\\[\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\\]"
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            "content": "(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：",
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            "content": "\\[\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\\]"
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            "content": "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",
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            "content": "(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\\)",
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            "content": "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",
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            "content": "\\[\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\\]"
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            "content": "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,\\)",
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            "content": "16.3 第二型曲线积分"
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            "content": "即 \\(xy + yz + xz = -\\frac{1}{2}\\)，因此：",
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            "content": "\\[\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\\]"
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            "content": "(2) 和第一题本质一样",
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            "content": "16.2.3 曲线积分中值问题"
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            "content": "定理 16.2 (曲线积分中值定理)"
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            "content": "\\(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",
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            "content": "\\[\n\\int_ {L} f (x, y) \\mathrm{d} s = f (x _ {0}, y _ {0}) \\Delta L\n\\]"
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            "content": "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))\\)",
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            "content": "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",
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            "content": "\\[\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\\]"
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            "content": "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：",
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            "content": "\\[\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\\]"
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            "content": "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。",
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            "content": "16.3 第二型曲线积分"
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            "content": "16.3.1 基本概念与参数计算"
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            "content": "定义 16.2 (第二型曲线积分)"
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            "content": "\\(\\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：",
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            "content": "\\[\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\\]"
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            "content": "定理 16.3 (第二型曲线积分的计算)"
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            "content": "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",
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            "content": "\\[\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\\]"
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            "content": "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",
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            "content": "16.3 第二型曲线积分"
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            "content": "(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。",
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            "content": "(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\\)",
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            "content": "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",
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            "content": "(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",
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            "content": "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\\)",
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            "content": "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",
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            "content": "(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。",
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            "content": "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",
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            "content": "\\[\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\\]"
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            "content": "(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：",
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            "content": "\\[\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\\]"
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            "content": "\\[\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\\]"
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            "content": "16.3.2 使用对称性"
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            "content": "16.3 第二型曲线积分"
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            "content": "定理 16.4 (dx 的符号)"
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            "content": "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。",
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            "content": "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。",
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            "content": "(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。",
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            "content": "(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。",
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            "content": "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]\\)",
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            "content": "可以算 \\(\\int_{L} y \\, \\mathrm{d}y\\) 比较简单，因此最终结果为 0。",
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            "content": "(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\\)。",
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            "content": "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：",
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            "content": "16.4 Green 公式"
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            "content": "16.3.3 两类曲线积分的关系"
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            "content": "定理 16.5 (光滑曲线切向量)"
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            "content": "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",
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            "content": "\\[\n\\mathrm{d} x = \\cos (\\mathbf {t}, x) \\mathrm{d} s, \\quad \\mathrm{d} y = \\cos (\\mathbf {t}, y) \\mathrm{d} s\n\\]"
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            "content": "定理 16.6 (两类曲线积分的关系)"
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            "content": "根据光滑曲线切向量的结论，可知：",
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            "content": "\\[\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\\]"
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            "content": "16.4 Green 公式"
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            "content": "16.4.1 Green 公式及其基本应用"
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            "content": "定理 16.7 (Green 公式)"
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            "content": "\\(\\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",
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            "content": "\\[\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\\]"
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            "content": "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：",
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            "content": "\\[\n\\left| \\begin{array}{c c} \\frac {\\partial}{\\partial x} & \\frac {\\partial}{\\partial y} \\\\ P & Q \\end{array} \\right|\n\\]"
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            "content": "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",
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            "content": "\\[\nI = \\int_ {L} [ e ^ {x} \\sin y - b (x + y) ] \\mathrm{d} x + (e ^ {x} \\cos y - a x) \\mathrm{d} y\n\\]"
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            "content": "(2)L 为  \\( \\sin x \\)  从  \\( (0,0) \\)  到  \\( (\\pi,0) \\)  的一段，计算：",
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            "content": "\\[\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\\]"
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            "content": "(3) \\(L\\) 为曲线 \\(y = \\sin x\\) 从 \\((0,0)\\) 到 \\((\\pi ,0)\\) 的一段，计算",
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            "content": "\\[\nI = \\int_ {L} e ^ {x} \\left[ (1 - \\cos y) \\mathrm{d} x - (y - \\sin y) \\mathrm{d} y \\right]\n\\]"
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            "content": "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\\)",
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            "content": "(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 = \\)",
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            "content": "16.4 Green 公式"
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            "content": "\\(\\int_{\\pi}^{0}x\\mathrm{d}x = -\\frac{1}{2}\\pi^2\\) ，因此：",
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            "content": "\\[\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\\]"
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            "content": "(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 \\)",
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            "content": "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",
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            "content": "定理 16.8 (区域面积转换为曲线积分)"
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            "content": "平面区域 \\(D\\) 的面积为：",
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            "content": "\\[\nS _ {D} = \\iint_ {D} \\mathrm{d} S = \\frac {1}{2} \\oint_ {L} x \\mathrm{d} y - y \\mathrm{d} x\n\\]"
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            "content": "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",
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            "content": "16.4.2 内部有瑕点"
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            "content": "定理 16.9 (瑕点 Green 公式)"
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            "content": "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：",
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            "content": "\\[\n\\int_ {L} P \\mathrm{d} x + Q \\mathrm{d} y = \\int_ {L _ {1}} P \\mathrm{d} x + Q \\mathrm{d} y\n\\]"
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            "content": "保证  \\( L_{1} \\)  曲线表达式的特殊性即可快速计算出结果。",
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            "content": "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}} \\) 。",
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            "content": "推论 16.2 (常用函数 1)"
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            "content": "若 \\(P_0(x_0,y_0)\\) 则：",
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            "content": "\\[\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\\]"
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            "type": "header",
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            "content": "16.4 Green 公式"
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            "content": "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\\)",
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            "content": "推论 16.3 (常用函数 2)"
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            "content": "\\( c_{1},c_{2}>0 \\)  时， \\( a_{2}=-b_{1},\\frac{a_{1}}{c_{1}}=\\frac{b_{2}}{c_{2}} \\)  有：",
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            "content": "\\[\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\\]"
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            "content": "引理 16.1 (推广椭圆面积)"
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            "content": "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}}\\)",
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            "content": "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。",
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            "content": "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}}\\)",
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            "content": "推论 16.4 (常用函数 3)"
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            "content": "137"
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            "content": "若 \\(a, b, c\\) 满足 \\(a, c > 0, ac - b^2 > 0\\)，则",
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            "content": "\\[\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\\]"
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            "content": "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：",
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            "content": "\\[\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\\]"
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            "content": "16.4.3 曲线积分与路径无关性"
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            "content": "定理 16.10 (路径无关性)"
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            "content": "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:",
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            "content": "- \\(D\\) 内处处有 \\(P_{y} = Q_{x}\\)",
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            "content": "- \\(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\\)",
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            "content": "- \\(D\\) 内存在原函数 \\(u(x,y)\\) 使得 \\(\\mathrm{du} = P\\mathrm{dx} + Q\\mathrm{dy}\\)",
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            "content": "第 17 章 曲面积分"
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            "content": "17.1 曲面的面积"
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            "content": "定理 17.1 (曲面面积公式)"
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            "content": "\\(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：",
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            "content": "\\[\n\\Delta S = \\iint_ {D} \\sqrt {1 + f _ {x} ^ {2} + f _ {y} ^ {2}} \\mathrm{d} x \\mathrm{d} y\n\\]"
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            "content": "定理 17.2 (参数方程曲面面积)"
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            "content": "\\(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：",
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            "content": "\\[\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\\]"
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            "content": "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",
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            "content": "\\[\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\\]"
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            "content": "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！",
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            "content": "17.2 第一型曲面积分"
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            "content": "17.2.1 基本概念与计算公式"
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            "content": "\\(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",
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            "content": "定理 17.3 (投影法)"
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            "content": "17.2 第一型曲面积分"
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            "content": "定理 17.4 (球坐标变换结论)"
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            "content": "对于球面 \\(x^{2} + y^{2} + z^{2} = a^{2}\\) ，令 \\(z = \\sqrt{a^2 - x^2 - y^2}\\) ，则得到",
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            "content": "\\[\n\\sqrt {1 + z _ {x} ^ {2} + z _ {y} ^ {2}} = \\frac {a}{z}\n\\]"
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            "content": "证明 直接计算可验证",
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            "content": "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。",
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            "content": "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。",
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            "content": "(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",
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            "content": "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",
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            "content": "\\[\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\\]"
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            "content": "(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：",
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            "content": "\\[\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\\]"
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            "content": "定理 17.5 (参数方程法)"
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            "content": "若 \\(S\\) 由参数方程 \\(x(u,v),y(u,v),z(u,v)\\) 给出， \\((u,v)\\in D\\) ，则",
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            "content": "\\[\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\\]"
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            "content": "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",
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            "content": "\\[\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\\]"
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            "content": "17.3 第二型曲面积分"
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            "content": "(2) \\(\\iint_{S} (x + y + z)^{3} \\mathrm{d} S\\)，其中 \\(S\\) 为 \\(|x| + |y| + |z| = 1\\)",
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            "content": "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。",
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            "content": "(2) 根据关于原点中心对称性得到为 0。",
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            "content": "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\\) 。",
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            "content": "17.3 第二型曲面积分"
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            "content": "17.3.1 基本概念、投影法、参数方程法"
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            "content": "定义 17.2 (第二型曲面积分)"
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            "content": "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):",
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            "content": "\\[\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\\]"
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            "content": "定理 17.7 (投影法)"
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            "content": "若曲面 \\(S\\) 的方程为 \\(z = z(x,y)\\)，在 \\(xy\\) 平面投影区域为 \\(D_{xy}\\)，则",
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            "content": "\\[\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\\]"
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            "content": "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。",
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            "content": "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",
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            "content": "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：",
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            "content": "140"
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            "content": "\\[\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\\]"
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            "content": "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}\\)",
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            "content": "(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。",
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            "content": "定理 17.8 (参数方程法)"
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            "content": "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",
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            "content": "17.3 第二型曲面积分"
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            "content": "若向量 \\((A, B, C)\\) 与 \\(S\\) 法向量方向一致则取正，否则取负。",
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            "content": "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",
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            "content": "17.3.2 利用对称性"
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            "content": "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。",
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            "content": "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:",
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            "content": "(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\\)",
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            "content": "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",
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            "content": "(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",
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            "content": "(3) 无论怎么考虑都非零",
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            "content": "(4) 考虑 yz， \\( \\sin x \\)  相反，dxdy 为正，因此积分为 0。",
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            "content": "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",
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            "content": "\\[\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\\]"
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            "content": "17.3.3 两类曲面积分的联系"
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            "content": "定理 17.9 (两类曲面积分的联系)"
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            "content": "\\(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",
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            "content": "\\[\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\\]"
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            "content": "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：",
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            "content": "\\[\n\\mathbf {n} = \\pm \\frac {1}{\\sqrt {F _ {x} ^ {2} + F _ {y} ^ {2} + F _ {z} ^ {2}}} (F _ {x}, F _ {y}, F _ {z})\n\\]"
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            "content": "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：",
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            "content": "\\[\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\\]"
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            "content": "这里乘上2的原因是关于 \\(z\\) 正负有两块。",
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            "content": "17.4 Gauss 公式"
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            "content": "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：",
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            "content": "\\[\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\\]"
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            "content": "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 \\)",
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            "content": "17.4 Gauss 公式"
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            "content": "17.4.1 Gauss 公式与基本使用"
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            "content": "定理 17.10 (Gauss 公式)"
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            "content": "\\(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",
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            "content": "\\[\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\\]"
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            "content": "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。",
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            "content": "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} \\) :",
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            "content": "\\[\nI _ {1} = - \\iiint_ {V} (0 + 1 + 0) \\mathrm{d} x \\mathrm{d} y \\mathrm{d} z = - 1 8 \\pi\n\\]"
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            "content": "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，：",
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            "content": "\\[\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\\]"
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            "content": "因此结果为 \\(I = \\frac{1}{9} (-18\\pi - 9\\pi) = -3\\pi\\)",
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            "content": "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 \\)",
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            "content": "(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",
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            "content": "\\[\n\\vert x - y + z \\vert + \\vert y - z + x \\vert + \\vert z - x + y \\vert = 1\n\\]"
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            "content": "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。",
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            "content": "(2) 使用 Gauss 公式即 \\(I = \\iiint_{V} 6 \\mathrm{~d} V\\)，使用坐标变换",
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            "content": "\\[\nu = x + z - y, v = y + x - z, w = y + z - x\n\\]"
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            "content": "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",
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            "content": "17.5 Stokes 公式"
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            "angle": 0,
            "content": "此最终结果为:",
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            "bbox": [
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            "angle": 0,
            "content": "\\[\nI = \\frac {3}{2} \\times 8 \\times \\frac {1}{6} = 2\n\\]"
        },
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            "content": "17.4.2 内部有瑕点"
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            "content": "练习 17.14 经典题目 S 为不经过原点的光滑封闭曲面，计算",
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            "content": "\\[\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\\]"
        },
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            "content": "解记 \\(r^2 = ax^2 + by^2 + cz^2\\)，则两侧对 \\(x\\) 求导可以得到",
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            "content": "\\[\nr _ {x} = \\frac {a x}{r}\n\\]"
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            "content": "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",
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            "angle": 0,
            "content": "\\[\nP _ {x} + Q _ {y} + R _ {z} = 0\n\\]"
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            "content": "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",
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            "content": "\\[\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\\]"
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            "content": "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.\\)",
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            "content": "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",
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            "bbox": [
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            "content": "\\[\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\\]"
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            "content": "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",
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            "content": "\\[\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\\]"
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            "content": "\\(\\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\\)",
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            "content": "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",
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            "content": "17.5 Stokes 公式"
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            "content": "17.5.1 Stokes 公式与基本使用"
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            "content": "定理 17.11 (Stokes 公式)"
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            "content": "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",
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            "content": "\\[\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\\]"
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            "content": "17.5 Stokes 公式"
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            "type": "text",
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            "content": "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。",
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            "type": "text",
            "bbox": [
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            "content": "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。",
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            "bbox": [
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            "content": "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。",
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            "content": "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:",
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            "content": "(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。",
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            "type": "text",
            "bbox": [
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            "content": "(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。",
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            "content": "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：",
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            "bbox": [
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            "content": "\\[\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\\]"
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            "content": "(2) \\(\\mathbf{n} = (x, y, z)\\)，因此根据Stokes公式得到",
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            "bbox": [
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            "content": "\\[\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\\]"
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            "content": "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：",
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            "bbox": [
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            "content": "\\[\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\\]"
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            "content": "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：",
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            "content": "\\(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：",
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            "content": "- 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\\)",
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            "content": "- \\(V\\) 内处处成立: \\(P_{y} = Q_{x}, Q_{z} = R_{y}, R_{x} = P_{z}\\)",
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            "content": "- 确界定理：\\(S\\) 是非空实数集，若 \\(S\\) 有上界，则 \\(S\\) 必有上确界",
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