数学常用公式定理1数学常用公式定理2数学常用公式定理4数学常用公式定理5数学常用公式定理6多元函数微分
平面点集
邻域
点$P$的$\delta$邻域
$ $ \begin{align*} \stackrel {}{U} (P_0, \delta ) & = \left \{ P \mid \left|P P_0\right| \lt \delta \right \} \end{align*} $ $
点$P$的去心$\delta$邻域
$ $ \begin{align*} \stackrel {\circ}{U} (P_0, \delta ) & = \left \{ P \mid 0 \lt \left|P P_0\right| \lt \delta \right \} \end{align*} $ $
点和点集之间的关系
内点
存在点$P$的邻域$\stackrel {}{U}(P)$，且$\stackrel {}{U}(P) \subset E$，则称$P$为$E$的内点
外点
存在点$P$的邻域$\stackrel {}{U}(P)$，且$\stackrel {}{U}(P) \cap E = \varnothing$，则称$P$为$E$的外点
边界点
点$P$的任意邻域$\stackrel {}{U}(P)$，既含有属于$E$的点，又含有不属于$E$的点，则称$P$为$E$的边界点
聚点
对于$\forall \delta \gt 0$，点$P$的去心邻域$\stackrel {0}{U}(P,\delta )$内总有$E$中的点，那么称$P$是$E$的聚点
重要平面点集
开集
如果点集$E$的点都是内点，称$E$为开集
闭集
如果点集$E$的边界$\partial E \subset E$，称$E$为闭集
连通集
如果点集$E$内任何两点，都可用折线连接起来，并且折线上的所有点属于$E$，称$E$为连通集
区域
连通的开集$E$，称为区域或开区域
闭区域
开区域连同其边界一起构成的点集称为闭区域
偏导数
定义
设$z=f(x,y)$，$z$在$(x_0,y_0)$处对$x$的偏导数为
$ $ \begin{align*} \left.\frac{\partial z}{\partial x}\right |_{\substack{x=x_{0} \\ y=y_{0}}}= \lim _{\Delta x \rightarrow 0} \frac{f\left(x_{0}+\Delta x, y_{0}\right)-f\left(x_{0}, y_{0}\right)}{\Delta x} \end{align*} $ $
对$y$的偏导数为
$ $ \begin{align*} \left.\frac{\partial z}{\partial y}\right |_{\substack{x=x_{0} \\ y=y_{0}}}= \lim_{\Delta y \to 0} \frac{f(x_0,y_0 + \Delta y)-f(x_0,y_0)}{\Delta y} \end{align*} $ $
如果$z=f(x,y)$在区域$D$每一点$(x,y)$处，对$x$的偏导数存在，称这个偏导数为函数$z$对自变量$x$的偏导函数；对$y$同理
高阶偏导数
设在区域$D$内，${f}’_x(x,y)=\frac{\partial z}{\partial x} $，${f}’_y(x,y)=\frac{\partial z}{\partial y} $，如果二阶偏导数存在，则
$ $ \begin{align*} &\frac{\partial }{\partial x} \left (\frac{\partial z}{\partial x} \right ) = \frac{\partial ^2z}{\partial x^2}={f}''_{xx}(x,y)\\ &\frac{\partial }{\partial y} \left (\frac{\partial z}{\partial y} \right ) = \frac{\partial ^2z}{\partial y^2}={f}''_{yy}(x,y)\\ &\frac{\partial }{\partial x} \left (\frac{\partial z}{\partial y} \right ) = \frac{\partial ^2z}{\partial x \partial y}={f}''_{xy}(x,y)\\ &\frac{\partial }{\partial y} \left (\frac{\partial z}{\partial x} \right ) = \frac{\partial ^2z}{\partial y \partial x}={f}''_{yx}(x,y) \end{align*} $ $
其中，${f}’’_{xy}(x,y)$与${f}’’_{yx}(x,y)$称为混合偏导数，如果二者在区域$D$内连续，那么必定相等
求导法则
一元函数与多元函数复合情形
设$z=f(u,v)$在点$(u,v)$有连续偏导数，$u=\varphi (t)$、$v=\psi (t)$在$t$点可导，则有
$ $ \begin{align*} \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 } \end{align*} $ $
二阶导
$ $ \begin{align*} \frac{\mathrm{d}^2z}{\mathrm{d}t^2} & = \frac{\mathrm{d} }{\mathrm{d} t}\left (\frac{\partial z}{\partial u} \frac{\mathrm{d}u }{\mathrm{d}t } + \frac{\partial z}{\partial v} \frac{\mathrm{d}v }{\mathrm{d}t } \right ) \\ & = \frac{\mathrm{d} }{\mathrm{d} t}\left ( \frac{\partial z}{\partial u} \frac{\mathrm{d}u }{\mathrm{d}t } \right) + \frac{\mathrm{d} }{\mathrm{d} t}\left ( \frac{\partial z}{\partial v} \frac{\mathrm{d}v }{\mathrm{d}t } \right) \\ & = \left [ \frac{\mathrm{d} }{\mathrm{d} t}\left ( \frac{\partial z}{\partial u} \right)\frac{\mathrm{d}u }{\mathrm{d}t } + \frac{\mathrm{d} }{\mathrm{d} t}\left ( \frac{\mathrm{d}u }{\mathrm{d}t } \right)\frac{\partial z}{\partial u} \right ] + \left [ \frac{\mathrm{d} }{\mathrm{d} t}\left ( \frac{\partial z}{\partial v} \right)\frac{\mathrm{d}v }{\mathrm{d}t } + \frac{\mathrm{d} }{\mathrm{d} t}\left ( \frac{\mathrm{d}v }{\mathrm{d}t } \right)\frac{\partial z}{\partial v} \right ] \\ & = \left \{ \left [ \frac{\partial }{\partial u}\left ( \frac{\partial z}{\partial u} \right)\frac{\mathrm{d}u }{\mathrm{d}t } + \frac{\partial }{\partial v}\left ( \frac{\partial z}{\partial u} \right)\frac{\mathrm{d}v }{\mathrm{d}t } \right ] \frac{\mathrm{d}u }{\mathrm{d}t } + \frac{\partial z}{\partial u} \frac{\mathrm{d}^2u }{\mathrm{d}t^2 } \right \} + \left \{ \left [ \frac{\partial }{\partial u}\left ( \frac{\partial z}{\partial v} \right)\frac{\mathrm{d}u }{\mathrm{d}t } + \frac{\partial }{\partial v}\left ( \frac{\partial z}{\partial v} \right)\frac{\mathrm{d}v }{\mathrm{d}t } \right ] \frac{\mathrm{d}v }{\mathrm{d}t } + \frac{\partial z}{\partial v} \frac{\mathrm{d}^2v }{\mathrm{d}t^2 } \right \} \end{align*} $ $
令$\frac {\partial z}{\partial u} = f_1’$，$\frac {\partial z}{\partial v} = f_2’$，则有
$ $ \begin{align*} \frac{\mathrm{d}z }{\mathrm{d}t } = f_1' \cdot \frac{\mathrm{d}u }{\mathrm{d}t } + f_2' \cdot \frac{\mathrm{d}v }{\mathrm{d}t } \end{align*} $ $
二阶导
$ $ \begin{align*} \frac{\mathrm{d}^2z}{\mathrm{d} t^2} = f_1' \cdot \frac{\mathrm{d} ^2u }{\mathrm{d} t^2 } + f_2' \cdot \frac{\mathrm{d} ^2v }{\mathrm{d} t^2 } + f_{11}'' \cdot \left ( \frac{\mathrm{d} u }{\mathrm{d} t } \right)^2 + 2f_{12}'' \cdot \frac{\mathrm{d} v }{\mathrm{d} t } \cdot \frac{\mathrm{d} u }{\mathrm{d} t } + f_{22}'' \cdot \left (\frac{\mathrm{d} v }{\mathrm{d} t } \right )^2 \end{align*} $ $
多元函数与多元函数复合情形
设$z=f(u,v)$在点$(u,v)$有连续偏导数，$u=\varphi (x,y)$、$v=\psi (x,y)$在点$(x,y)$有对$x$、$y$的偏导数，则有
$ $ \begin{align*} \frac{\partial z }{\partial x } = \frac{\partial z}{\partial u} \frac{\partial u }{\partial x } + \frac{\partial z}{\partial v} \frac{\partial v }{\partial x }\\ \frac{\partial z }{\partial y } = \frac{\partial z}{\partial u} \frac{\partial u }{\partial y } + \frac{\partial z}{\partial v} \frac{\partial v }{\partial y } \end{align*} $ $
二阶导
$ $ \begin{align*} \frac{\partial^2z}{\partial x^2} & = \frac{\partial }{\partial x}\left (\frac{\partial z}{\partial u} \frac{\partial u }{\partial x } + \frac{\partial z}{\partial v} \frac{\partial v }{\partial x } \right ) \\ & = \frac{\partial }{\partial x}\left ( \frac{\partial z}{\partial u} \frac{\partial u }{\partial x } \right) + \frac{\partial }{\partial x}\left ( \frac{\partial z}{\partial v} \frac{\partial v }{\partial x } \right) \\ & = \left [ \frac{\partial }{\partial x}\left ( \frac{\partial z}{\partial u} \right)\frac{\partial u }{\partial x } + \frac{\partial }{\partial x}\left ( \frac{\partial u }{\partial x } \right)\frac{\partial z}{\partial u} \right ] + \left [ \frac{\partial }{\partial x}\left ( \frac{\partial z}{\partial v} \right)\frac{\partial v }{\partial x } + \frac{\partial }{\partial x}\left ( \frac{\partial v }{\partial x } \right)\frac{\partial z}{\partial v} \right ] \\ & = \left \{ \left [ \frac{\partial }{\partial u}\left ( \frac{\partial z}{\partial u} \right)\frac{\partial u }{\partial x } + \frac{\partial }{\partial v}\left ( \frac{\partial z}{\partial u} \right)\frac{\partial v }{\partial x } \right ] \frac{\partial u }{\partial x } + \frac{\partial z}{\partial u} \frac{\partial ^2u }{\partial x^2 } \right \} + \left \{ \left [ \frac{\partial }{\partial u}\left ( \frac{\partial z}{\partial v} \right)\frac{\partial u }{\partial x } + \frac{\partial }{\partial v}\left ( \frac{\partial z}{\partial v} \right)\frac{\partial v }{\partial x } \right ] \frac{\partial v }{\partial x } + \frac{\partial z}{\partial v} \frac{\partial ^2v }{\partial x^2 } \right \} \\ & = \left \{ \left [ \frac{\partial^2z}{\partial u^2}\frac{\partial u }{\partial x } + \frac{\partial^2z}{\partial u \partial v}\frac{\partial v }{\partial x } \right ] \frac{\partial u }{\partial x } + \frac{\partial z}{\partial u} \frac{\partial ^2u }{\partial x^2 } \right \} + \left \{ \left [ \frac{\partial^2z}{\partial u \partial v}\frac{\partial u }{\partial x } + \frac{\partial^2z}{\partial v^2}\frac{\partial v }{\partial x } \right ] \frac{\partial v }{\partial x } + \frac{\partial z}{\partial v} \frac{\partial ^2v }{\partial x^2 } \right \} \\ \\ \frac{\partial^2z}{\partial y^2} & = \frac{\partial }{\partial y}\left (\frac{\partial z}{\partial u} \frac{\partial u }{\partial y } + \frac{\partial z}{\partial v} \frac{\partial v }{\partial y } \right ) \\ & = \frac{\partial }{\partial y}\left ( \frac{\partial z}{\partial u} \frac{\partial u }{\partial y } \right) + \frac{\partial }{\partial y}\left ( \frac{\partial z}{\partial v} \frac{\partial v }{\partial y } \right) \\ & = \left [ \frac{\partial }{\partial y}\left ( \frac{\partial z}{\partial u} \right)\frac{\partial u }{\partial y } + \frac{\partial }{\partial y}\left ( \frac{\partial u }{\partial y } \right)\frac{\partial z}{\partial u} \right ] + \left [ \frac{\partial }{\partial y}\left ( \frac{\partial z}{\partial v} \right)\frac{\partial v }{\partial y } + \frac{\partial }{\partial y}\left ( \frac{\partial v }{\partial y } \right)\frac{\partial z}{\partial v} \right ] \\ & = \left \{ \left [ \frac{\partial }{\partial u}\left ( \frac{\partial z}{\partial u} \right)\frac{\partial u }{\partial y } + \frac{\partial }{\partial v}\left ( \frac{\partial z}{\partial u} \right)\frac{\partial v }{\partial y } \right ] \frac{\partial u }{\partial y } + \frac{\partial z}{\partial u} \frac{\partial ^2u }{\partial y^2 } \right \} + \left \{ \left [ \frac{\partial }{\partial u}\left ( \frac{\partial z}{\partial v} \right)\frac{\partial u }{\partial y } + \frac{\partial }{\partial v}\left ( \frac{\partial z}{\partial v} \right)\frac{\partial v }{\partial y } \right ] \frac{\partial v }{\partial y } + \frac{\partial z}{\partial v} \frac{\partial ^2v }{\partial y^2 } \right \} \\ & = \left \{ \left [ \frac{\partial^2z}{\partial u^2}\frac{\partial u }{\partial y } + \frac{\partial^2z}{\partial u \partial v}\frac{\partial v }{\partial y } \right ] \frac{\partial u }{\partial y } + \frac{\partial z}{\partial u} \frac{\partial ^2u }{\partial y^2 } \right \} + \left \{ \left [ \frac{\partial^2z}{\partial u \partial v}\frac{\partial u }{\partial y } + \frac{\partial^2z}{\partial v^2}\frac{\partial v }{\partial y } \right ] \frac{\partial v }{\partial y } + \frac{\partial z}{\partial v} \frac{\partial ^2v }{\partial y^2 } \right \} \end{align*} $ $
令$\frac {\partial z}{\partial u} = f_1’$，$\frac {\partial z}{\partial v} = f_2’$，则有
$ $ \begin{align*} \frac{\partial z }{\partial x } = f_1' \cdot \frac{\partial u }{\partial x } + f_2' \cdot \frac{\partial v }{\partial x }\\ \frac{\partial z }{\partial y } = f_1' \cdot \frac{\partial u }{\partial y } + f_2' \cdot \frac{\partial v }{\partial y } \end{align*} $ $
二阶导
$ $ \begin{align*} \frac{\partial^2z}{\partial x^2} = f_1' \cdot \frac{\partial ^2u }{\partial x^2 } + f_2' \cdot \frac{\partial ^2v }{\partial x^2 } + f_{11}'' \cdot \left ( \frac{\partial u }{\partial x } \right)^2 + 2f_{12}'' \cdot \frac{\partial v }{\partial x } \cdot \frac{\partial u }{\partial x } + f_{22}'' \cdot \left (\frac{\partial v }{\partial x } \right )^2\\ \\ \frac{\partial^2z}{\partial y^2} = f_1' \cdot \frac{\partial ^2u }{\partial y^2 } + f_2' \cdot \frac{\partial ^2v }{\partial y^2 } + f_{11}'' \cdot \left ( \frac{\partial u }{\partial y } \right)^2 + 2f_{12}'' \cdot \frac{\partial v }{\partial y } \cdot \frac{\partial u }{\partial y } + f_{22}'' \cdot \left (\frac{\partial v }{\partial y } \right )^2 \end{align*} $ $
隐函数
隐函数存在定理1
设函数$F(x,y)$在点$P(x_0,y_0)$的某一邻域内具有连续偏导数，且$F(x_0,y_0)=0$，$F_y(x_0,y_0) \ne 0$，则方程$F(x,y)=0$在点$(x_0,y_0)$的某邻域内恒能唯一确定一个连续且具有连续导数的函数$y=f(x)$，它满足条件$y_0=f(x_0)$，并有
$ $ \begin{align*} \frac {\mathrm{d}y}{\mathrm{d}x} = -\frac {F_x}{F_y} \end{align*} $ $
隐函数存在定理2
设函数$F(x,y,z)$在点$P(x_0,y_0,z_0)$的某一邻域内具有连续偏导数，且$F(x_0,y_0,z_0)=0$，$F_z(x_0,y_0,z_0) \ne 0$，则方程$F(x,y,z)=0$在点$(x_0,y_0,z_0)$的某邻域内恒能唯一确定一个连续且具有连续导数的函数$z=f(x,y)$，它满足条件$z_0=f(x_0,y_0)$，并有
$ $ \begin{align*} \frac {\partial z}{\partial x} = -\frac {F_x}{F_z}，\frac {\partial z}{\partial y} = -\frac {F_y}{F_z} \end{align*} $ $
全微分
定理一
如果$z=f(x,y)$在点$(x,y)$可微分，那么在点$(x,y)$的偏导数$\frac{\partial z}{\partial x} $、$\frac{\partial z}{\partial y} $必定存在，且全微分为
$ $ \begin{align*} \mathrm{d}z = \frac{\partial z}{\partial x} \Delta x + \frac{\partial z}{\partial y} \Delta y \end{align*} $ $
通常写作
$ $ \begin{align*} \mathrm{d}z = \frac{\partial z}{\partial x} \mathrm{d}x + \frac{\partial z}{\partial y} \mathrm{d} y \end{align*} $ $
全增量： $ $ \begin{array}{l} \begin{align*} \Delta z& = f(x+\Delta x,y+\Delta y)-f(x,y)\\ & = A\Delta x+B\Delta y+o(\sqrt[]{(\Delta x)^2+(\Delta y)^2} )\\ \end{align*}\\ \\ 其中，\Delta x \to 0，\Delta y \to 0，A=\frac{\partial z}{\partial x}，B=\frac{\partial z}{\partial y} \end{array} $ $
定理二
如果$z=f(x,y)$的偏导数$\frac{\partial z}{\partial x} $、$\frac{\partial z}{\partial y} $在点$(x,y)$连续，那么在该点可微分
多元函数极值
极值求法
定理一
设$z=f(x,y)$在点$(x_0,y_0)$有偏导数，且有极值，则有
$ $ \begin{align*} {f}'_x(x_0,y_0) = 0\\ {f}'_y(x_0,y_0) = 0 \end{align*} $ $
定理二
设$z=f(x,y)$在点$(x_0,y_0)$的某邻域内有一阶和二阶连续偏导数，又${f}’_x(x_0,y_0) = 0$，${f}’_y(x_0,y_0) = 0$，令
$ $ \begin{align*} {f}''_{xx}(x_0,y_0) = A,\space {f}''_{xy}(x_0,y_0) = B,\space {f}''_{yy}(x_0,y_0) = C \end{align*} $ $
则$f(x,y)$在$(x_0,y_0)$处是否取得极值的条件如下：
- $AC-B^2 \gt 0$时，有极值，且$A \lt 0$时有极大值，$A \gt 0$时有极小值
- $AC-B^2 \lt 0$时，没有极值
- $AC-B^2 = 0$时，需要讨论
如果三元函数$u=f(x,y,z)$在点$(x_0,y_0,z_0)$具有偏导数，那么它在点$(x_0,y_0,z_0)$具有极值的充分必要条件是 $ $ \begin{align*} f'_x(x_0,y_0,z_0)=0,f'_y(x_0,y_0,z_0)=0,f'_z(x_0,y_0,z_0)=0 \end{align*} $ $
条件极值
拉格朗日乘数法
要找$z=f(x,y)$在附加条件$\varphi (x,y)=0$下的可能极值点，先做拉格朗日函数
$ $ \begin{align*} L(x,y)=f(x,y)+\lambda \varphi (x,y) \end{align*} $ $
其中$\lambda$为参数；求拉格朗日函数对$x$、$y$的一阶偏导，并使之为零，然后与附加条件方程联立
$ $ \begin{align*} \begin{cases} {f}'_x(x,y) + \lambda {\varphi}'_x(x,y) = 0 \\ {f}'_y(x,y) + \lambda {\varphi}'_y(x,y) = 0 \\ \varphi(x,y) = 0 \end{cases} \end{align*} $ $
解出$x$、$y$、$\lambda$，列出所有$(x,y)$组合，就得到了$z=f(x,y)$在附加条件$\varphi (x,y)=0$下的可能极值点（约束条件如果存在端点，端点处的极值需要额外考虑）
当自变量和条件增加时，例如$u=f(x,y,z,t)$在附加条件$\varphi (x,y,z,t)=0$，$\psi (x,y,z,t)=0$下的极值，可做拉格朗日函数
$ $ \begin{align*} L(x,y,z,t)=f(x,y,z,t)+\lambda \varphi (x,y,z,t)+\mu \psi (x,y,z,t) \end{align*} $ $
其中$\varphi$、$\mu$均为参数；求拉格朗日函数对$x$、$y$、$z$、$t$的一阶偏导，并使之为零，然后与附加条件方程联立，并解出$x$、$y$、$z$、$t$、$\varphi$、$\mu$，并列出所有$(x,y,z,t)$组合，就得到了$u=f(x,y,z,t)$在附加条件$\varphi (x,y,z,t)=0$，$\psi (x,y,z,t)=0$下的可能极值点（约束条件如果存在端点，端点处的极值需要额外考虑）
例：将长为2m的铁丝分成三段，依次围成圆、正方形与正三角形，求三个图形面积之和的最小值 $ $ \begin{array}{l} 设圆形的周长为x，正方形周长为y，正三角形周长为z，则x+y+z-2=0\\ \therefore S_圆=\frac{x^2}{4\pi},S_正=\frac{y}{16},S_三=\frac{\sqrt[]{3} }{36}z^2 \\ 令f(x,y,z)=\frac{x^2}{4\pi}+\frac{y^2}{16}+\frac{\sqrt[]{3} }{36}z^2，\varphi (x,y,z)=x+y+z-2\\ 做拉格朗日函数L(x,y,z)=f(x,y,z)+\lambda \varphi (x,y,z)\\ \therefore L(x,y,z)= \frac{x^2}{4\pi}+\frac{y^2}{16}+\frac{\sqrt[]{3} }{36}z^2+\lambda (x+y+z-2)\\ L'_x=\frac{x}{2\pi}+\lambda，L'_y=\frac{y}{8}+\lambda，L'_z=\frac{\sqrt[]{3}}{18}z+\lambda\\ 联立方程组\begin{cases} \frac{x}{2\pi}+\lambda=0\\ \frac{y}{8}+\lambda=0\\ \frac{\sqrt[]{3}}{18}z+\lambda=0\\ x+y+z-2=0 \end{cases}\space 解得\begin{cases} \lambda = -\frac{1}{\pi + 4 + 3\sqrt[]{3} } \\ x=\frac{2\pi}{\pi + 4 + 3\sqrt[]{3}} \\ y=\frac{8}{\pi + 4 + 3\sqrt[]{3}} \\ z=\frac{6\sqrt[]{3}}{\pi + 4 + 3\sqrt[]{3}} \end{cases}\\ 此时S取最小值为\frac{1}{\pi + 4 + 3\sqrt[]{3}} \end{array} $ $
重积分
二重积分的性质
性质1
设$\alpha $与$\beta $为常数，则
$ $ \begin{align*} \iint\limits_{D}^{} [\alpha f(x,y)+\beta g(x,y)]\mathrm{d} \sigma = \alpha \iint \limits_{D}^{}f(x,y)\mathrm{d}\sigma + \beta \iint \limits_{D}^{}g(x,y)\mathrm{d} \sigma \end{align*} $ $
性质2
如果闭区域$D$被有限条曲线分为有限个部分闭区域，那么在$D$上的二重积分等于在各部分闭区域上的二重积分的和，例如$D$分为两个闭区域$D_1$和$D_2$，则
$ $ \begin{align*} \iint \limits _{D}^{}f(x,y)\mathrm{d} \sigma =\iint \limits _{D_1}^{}f(x,y)\mathrm{d} \sigma+\iint \limits _{D_2}{}f(x,y)\mathrm{d} \sigma \end{align*} $ $
性质3
如果在$D$上，$f(x,y)=1$，$\sigma$为$D$的面积，那么
$ $ \begin{align*} \sigma =\iint \limits _{D}^{}1\cdot \mathrm{d}\sigma =\iint \limits _{D}^{} \mathrm{d} \sigma \end{align*} $ $
性质4 ①
如果在$D$上，$f(x,y) \le g(x,y)$，那么有
$ $ \begin{align*} \iint \limits _{D}^{} f(x,y)\mathrm{d} \sigma \le \iint \limits _{D}^{} g(x,y)\mathrm{d} \sigma \end{align*} $ $
性质4 ②
由于$f(x,y) \le \left|f(x,y)\right|$，那么有
$ $ \begin{align*} \left|\iint \limits _{D}^{} f(x,y)\mathrm{d} \sigma\right| \le \iint \limits _{D}^{} \left|f(x,y)\right|\mathrm{d} \sigma \end{align*} $ $
性质5
设$M$和$m$分别是$f(x,y)$在闭区域$D$上的最大值和最小值，$\sigma$是$D$的面积，则有
$ $ \begin{align*} m\sigma \le \iint \limits _{D}^{} f(x,y)\mathrm{d}\sigma \le M\sigma \end{align*} $ $
性质6 二重积分的中值定理
设$f(x,y)$在闭区域$D$上连续，$\sigma$是$D$的面积，则在$D$上至少存在一点$(\xi ,\eta )$，使得
$ $ \begin{align*} \iint \limits _{D}^{} f(x,y)\mathrm{d}\sigma =f(\xi, \eta)\sigma \end{align*} $ $
直角坐标下二重积分的计算
X-型：若区域$D$表示为
$ $ \begin{align*} D=\{(x,y)|a\le x\le b, \varphi _1(x)\le y \le \varphi _2(x)\} \end{align*} $ $
则有
$ $ \begin{align*} \iint\limits_{D}^{} f(x,y)\mathrm{d}\sigma = \int_{a}^{b}\mathrm{d}x\int_{\varphi_{1}(x)}^{\varphi_{2}(x)}(x,y)\mathrm{d}y \end{align*} $ $
Y-型：若区域$D$表示为
$ $ \begin{align*} D=\{(x,y)|\varphi _1(x)\le x \le \varphi _2(x), c \le y \le d\} \end{align*} $ $
则有
$ $ \begin{align*} \iint\limits_{D}^{} f(x,y)\mathrm{d}\sigma = \int_{c}^{d}\mathrm{d}y\int_{\varphi_{1}(y)}^{\varphi_{2}(y)}(x,y)\mathrm{d}x \end{align*} $ $
极坐标系下二重积分的计算
直角坐标与极坐标的对应关系
令$\begin{cases} x=\rho \cos \theta \\ y=\rho \sin \theta \end{cases} $，
于是$f(x,y)=f(\rho \cos\theta,\rho \sin\theta)$，
其中$\rho =\rho (\theta )$，
若区域$D$表示为
$ $ \begin{align*} D=\{(\rho,\theta)|\alpha \le \theta \le \beta ,\rho_1(\theta) \le \rho \le \rho_2(\theta)\} \end{align*} $ $
则有
$ $ \begin{align*} \iint\limits_{D}^{} f(x,y)\mathrm{d}\sigma = \int_{\alpha }^{\beta }\mathrm{d}\theta \int_{\rho_1(\theta )}^{\rho_2(\theta )} f(\rho \cos\theta,\rho \sin\theta)\rho\mathrm{d}\rho \end{align*} $ $
三种情况
1、若极点在区域$D$内部时
$ $ \begin{align*} D=\{(\theta ,\rho )\, |\, 0 \le \theta \le 2\pi ,0 \le \rho \le \rho (\theta )\} \end{align*} $ $
则有
$ $ \begin{align*} \iint\limits_{D}^{} f(\rho \cos\theta,\rho \sin\theta)\rho \mathrm{d}\rho \mathrm{d}\theta = \int_{0}^{2\pi}\mathrm{d}\theta\int_{0}^{\rho (\theta)}f(\rho \cos\theta,\rho \sin\theta)\rho \mathrm{d}\rho \end{align*} $ $
2、若极点在区域$D$的边界线上
$ $ \begin{align*} D=\{(\theta ,\rho )\, |\, \alpha \le \theta \le \beta ,0 \le \rho \le \rho (\theta )\} \end{align*} $ $
则有
$ $ \begin{align*} \iint\limits_{D}^{} f(\rho \cos\theta,\rho \sin\theta)\rho \mathrm{d}\rho \mathrm{d}\theta = \int_{\alpha }^{\beta}\mathrm{d}\theta\int_{0}^{\rho (\theta)}f(\rho \cos\theta,\rho \sin\theta)\rho \mathrm{d}\rho \end{align*} $ $
3、若极点在区域$D$之外
$ $ \begin{align*} D=\{(\theta ,\rho )\, |\, \alpha \le \theta \le \beta ,\rho _{1}(\theta) \le \rho \le \rho _2(\theta )\} \end{align*} $ $
则有
$ $ \begin{align*} \iint\limits_{D}^{} f(\rho \cos\theta,\rho \sin\theta)\rho \mathrm{d}\rho \mathrm{d}\theta = \int_{\alpha }^{\beta}\mathrm{d}\theta\int_{\rho _{1}(\theta)}^{\rho _{2}(\theta)}f(\rho \cos\theta,\rho \sin\theta)\rho \mathrm{d}\rho \end{align*} $ $
二重积分特性
偶倍奇零对称性
设$f(x,y)$在有界闭区域$D$上连续，当积分区域有如下三种对称方式时
若积分区域$D$关于$y$轴对称，则
(其中$D_1$是$D$的右半部分) $ $ \begin{align*} \iint\limits_{D}^{} f(x,y)\mathrm{d}x\mathrm{d}y = \begin{cases} 2\iint\limits_{D_1}^{}f(x,y)\mathrm{d}x\mathrm{d}y ,& \space f(-x,y)=f(x,y) \\ 0 ,& \space f(-x,y)=-f(x,y) \end{cases} \end{align*} $ $
若积分区域$D$关于$x$轴对称，则
(其中$D_1$是$D$的上半部分) $ $ \begin{align*} \iint\limits_{D}^{} f(x,y)\mathrm{d}x\mathrm{d}y = \begin{cases} 2\iint\limits_{D_1}^{}f(x,y)\mathrm{d}x\mathrm{d}y,&\space f(x,-y)=f(x,y) \\ 0,&\space f(x,-y)=-f(x,y) \end{cases} \end{align*} $ $
若积分区域$D$关于原点对称，则
(其中$D_1$是$D$的上半部分) $ $ \begin{align*} \iint\limits_{D}^{} f(x,y)\mathrm{d}x\mathrm{d}y = \begin{cases} 2\iint\limits_{D_1}^{}f(x,y)\mathrm{d}x\mathrm{d}y,&\space f(-x,-y)=f(x,y) \\ 0,&\space f(-x,-y)=-f(x,y) \end{cases} \end{align*} $ $
轮换对称性
设$f(x,y)$在有界闭区域$D$上连续，积分区域$D$关于$y=x$对称，则
$ $ \begin{align*} \iint \limits _{D}^{}f(x,y)\mathrm{d}x\mathrm{d}y= \iint \limits _{D}^{} f(y,x)\mathrm{d}x\mathrm{d}y=\frac{1}{2} \iint \limits _{D}^{}[f(x,y)+f(y,x)]\mathrm{d}x\mathrm{d}y \end{align*} $ $
若被积函数$f(x,y)$满足$f(x,y)=f(y,x)$，则称$f(x,y)$关于$x$和$y$轮换对称，那么
$ $ \begin{align*} \iint \limits _{D}^{}f(x,y)\mathrm{d}x\mathrm{d}y=2 \iint \limits _{D_1}^{}f(x,y)\mathrm{d}x\mathrm{d}y \end{align*} $ $
若被积函数$f(x,y)$满足$f(x,y)=-f(y,x)$，那么
$ $ \begin{align*} \iint \limits _{D}^{}f(x,y)\mathrm{d}x\mathrm{d}y=0 \end{align*} $ $
二重积分换元法
雅可比行列式
$ $ \begin{align*} J = \begin{vmatrix} \frac{\partial (f_1 ,\cdots, f_n)}{\partial (x_1, \cdots, x_n)} \end{vmatrix} = \begin{vmatrix} \frac{\partial f_1}{\partial x_1} & \cdots & \frac{\partial f_1}{\partial x_n} \\ \vdots & \ddots & \vdots \\ \frac{\partial f_n}{\partial x_1} & \cdots & \frac{\partial f_n}{\partial x_n} \end{vmatrix} \end{align*} $ $
换元法
$ $ \begin{align*} &有直角坐标系下的二重积分\iint\limits_{D_{xy}}^{} f(x,y)\mathrm{d} x\mathrm{d} y\\ &令\begin{cases} x=x(u,v)\\ y=y(u,v) \end{cases}，可推出\begin{cases} u=u(x,y)\\ v=v(x,y) \end{cases}\\ \\ &则有：\\ &1. f(x,y) \to f[x(u,v),y(u,v)]，\\ &2. \iint\limits_{D_{xy}}^{} \to \iint\limits_{D_{uv}}^{}，\\ &3. \mathrm{d}x\mathrm{d}y \to \begin{vmatrix} \frac{\partial (x,y)}{\partial (u,v)} \end{vmatrix}\mathrm{d}u\mathrm{d}v \to J\mathrm{d}u\mathrm{d}v，\\ \\ &对J取绝对值，即有：\\ &\iint\limits_{D_{xy}}^{} f(x,y)\mathrm{d} x\mathrm{d} y = \iint\limits_{D_{uv}}^{}f[x(u,v),y(u,v)]\left|J\right|\mathrm{d}u\mathrm{d}v \end{align*} $ $
无穷级数
三大参照级数
几何级数（等比级数）
其中$a \ne 0$，$q$是级数的公比
$ $ \begin{align*} \sum_{n=1}^{\infty} aq^{n-1}=a+aq+aq^2+···+aq^{n-1}+··· \end{align*} $ $
当$\left| q \right| \lt 1$时，级数收敛；当$\left|q \right|\ge1$时，级数发散
调和级数
调和级数$\sum_{n=1}^{\infty } \frac{1}{n} $是发散的
$p$级数
$ $ \begin{align*} \sum_{n=1}^{\infty } \frac{1}{n^p} \end{align*} $ $
$p$级数当$0 \lt p \le 1$时是发散的，当$p\gt 1$时是收敛的
数项级数
如果级数的部分和数列$ \{S_n\} $有极限S（级数的和），则称级数是收敛的，如果没有极限，则称级数是发散的
收敛级数的性质
性质1
若级数$\sum_{n=1}^{\infty } u_n$收敛于$S$，则级数$\sum_{n=1}^{\infty } ku_n$也收敛，且收敛于$kS$
性质2
若级数$\sum_{n=1}^{\infty } u_n$与级数$\sum_{n=1}^{\infty } v_n$分别收敛于$S_1,S_2$，则级数$\sum_{n=1}^{\infty } (u_n \pm v_n)$收敛于$S_1 \pm S_2$；但级数$\sum_{n=1}^{\infty } u_n$与级数$\sum_{n=1}^{\infty } v_n$都发散时，级数$\sum_{n=1}^{\infty } (u_n \pm v_n)$不一定发散
性质3
在级数中去掉、加上或改变有限项，不会改变该级数的敛散性
性质4
如果级数收敛，则该级数的任意项加括号所成的级数仍收敛，且其和不变；如果加括号后所成的级数收敛，则不能断定原级数也收敛；如果加括号后的级数发散，则原级数也发散
级数收敛的必要条件
定理1
若级数$\sum_{n=1}^{\infty } u_n$收敛，则$\lim_{n \to \infty } u_n=0$
推论
若$\lim_{n \to \infty } u_n \ne 0$或不存在，
则级数$\sum_{n=1}^{\infty } u_n$发散
正项级数
正项级数的部分和数列$\{S_n\}$是单调增加的，所以定理2
正项级数$\sum_{n=1}^{\infty } u_n$收敛的充分必要条件是它的部分和数列$\{S_n\}$有界
定理3（比较判别法）
设$\sum_{n=1}^{\infty } u_n$和$\sum_{n=1}^{\infty } v_n$均为正项级数，且$u_n \le v_n$，
- 如果级数$\sum_{n=1}^{\infty } v_n$收敛，则级数$\sum_{n=1}^{\infty } u_n$收敛
- 如果级数$\sum_{n=1}^{\infty } u_n$发散，则级数$\sum_{n=1}^{\infty } v_n$发散
定理4（比较判别法的极限形式）
设$\sum_{n=1}^{\infty } u_n$和$\sum_{n=1}^{\infty } v_n$均为正项级数，且$\lim_{n \to \infty } \frac{u_n}{v_n} = A$
- 若$0 \lt A \lt +\infty$，则$\sum_{n=1}^{\infty } u_n$和$\sum_{n=1}^{\infty } v_n$同时收敛或发散
- 若$A=0$，且$\sum_{n=1}^{\infty } v_n$收敛，则$\sum_{n=1}^{\infty } u_n$收敛
- 若$A=+\infty$且$\sum_{n=1}^{\infty } v_n$发散，则$\sum_{n=1}^{\infty } u_n$发散
定理5（比值判别法）
设$\sum_{n=1}^{\infty } u_n$为正项级数，如果$\lim_{n \to \infty } \frac{u_n+1}{u_n} = \rho $
- 当$\rho \lt 1$时，级数收敛
- 当$\rho \gt 1$时，级数发散
- 当$\rho = 1$时，级数可能收敛，也可能发散
定理6（根值判别法）
设$\sum_{n=1}^{\infty } u_n$为正项级数，如果$\lim_{n \to \infty } \sqrt[n]{u_n}=p$
- 当$p \lt 1$时，级数收敛
- 当$p \gt 1$时，级数发散
- 当$p=1$时，可能收敛，也可能发散
定理7（极限判别法）
设$\sum_{n=1}^{\infty } u_n$为正项级数
- 如果$\lim_{n \to \infty} nu_n=l \gt 0$，则级数发散
- 如果$p \gt 1$，而$\lim_{n \to \infty} n^pu_n=l(0\le l \lt +\infty)$，则级数收敛
交错级数
定理8（莱布尼兹判别法）
如果交错级数$\sum_{n=1}^{\infty } (-1)^{n-1}u_n$满足如下条件，则级数收敛，且其和$S \le u_1$，其余项$r_n$的绝对值为$\left|r_n\right| \le u_{n+1}$
- $u_n \ge u_{n+1}$
- $\lim_{n \to \infty} u_n=0$
任意项级数
定义
如果级数$\sum_{n=1}^{\infty } \left|u_n\right|$收敛，则级数$\sum_{n=1}^{\infty } u_n$也收敛，此时称级数$\sum_{n=1}^{\infty } u_n$绝对收敛
如果级数$\sum_{n=1}^{\infty } \left|u_n\right|$发散，而级数$\sum_{n=1}^{\infty } u_n$也收敛，此时称级数$\sum_{n=1}^{\infty } u_n$条件收敛
定理9
如果级数$\sum_{n=1}^{\infty } \left|u_n\right|$收敛，则级数$\sum_{n=1}^{\infty } u_n$一定收敛
推论
- 如果级数$\sum_{n=1}^{\infty } u_n$发散，则级数$\sum_{n=1}^{\infty } \left|u_n\right|$一定发散
- 当$\sum_{n=1}^{\infty } u_n$条件收敛时，级数$\sum_{n=1}^{\infty} \frac{\left|u_n\right|+u_n}{2}$和$\sum_{n=1}^{\infty} \frac{\left|u_n\right|-u_n}{2} $均发散
幂级数
求幂级数的方法
先求幂级数的收敛区间$(-R,R)$，然后讨论幂级数在$x=R$与$x=-R$处的敛散性，从而得出幂级数的收敛域
定理1（阿贝尔定理）
如果幂级数$\sum_{n=0}^{\infty}a_nx^n$当$x=x_0(x_0\ne0)$时收敛，则对于所有满足$\left|x\right| \lt \left|x_0\right|$的点$x$，幂级数$\sum_{n=0}^{\infty}a_nx^n$绝对收敛，
反之，如果幂级数$\sum_{n=0}^{\infty}a_nx^n$当$x=x_0(x_0\ne0)$时发散，则对于所有满足$\left|x\right| \gt \left|x_0\right|$的点$x$，幂级数$\sum_{n=0}^{\infty}a_nx^n$发散
定理2
设幂级数$\sum_{n=0}^{\infty}a_nx^n$，
有$\lim_{n \to \infty} \left|\frac{a_n+1}{a_n} \right| = \rho$，则
- 若$\rho \ne 0$，则收敛半径$R=\frac{1}{\rho}$
- 若$\rho = 0$，则收敛半径$R = +\infty$
- 若$\rho = +\infty$，则收敛半径$R=0$
幂级数四则运算
设幂级数$\sum_{n=0}^{\infty}a_nx^n$在其收敛区间$(-R_1,R_1)(R_1 \gt 0)$内的和函数为$S_1(x)$，幂级数$\sum_{n=0}^{\infty}b_nx^n$在其收敛区间$(-R_2,R_2)(R_2 \gt 0)$内的和函数为$S_2(x)$，取$R=\mathrm{min}\{R_1,R_2\} $，则有
1.加减运算$\eqref {eq27}$，其收敛区间为$(-R,R)$
$ $ \begin{align} \label {eq27} \sum_{n=0}^{\infty}a_nx^n \pm \sum_{n=0}^{\infty}b_nx^n=\sum_{n=0}^{\infty}(a_n \pm b_n)x^n = S_1(x) \pm S_2(x) \end{align} $ $
2.乘法运算$\eqref {eq28}$，其收敛区间为$(-R,R)$
$ $ \begin{align} \label {eq28} \sum_{n=0}^{\infty}a_nx^n \cdot \sum_{n=0}^{\infty}b_nx^n = a_0b_0 + (a_0b_1+a_1b_0)x+(a_0b_2+a_1b_1+a_2b_0)x^2+ \cdot \cdot \cdot + (a_0b_n+a_1b_{n-1}+a_2b_{n-2}+\cdot \cdot \cdot +a_nb_0)x^n+\cdot \cdot \cdot = S_1(x)S_2(x) \end{align} $ $
幂函数的和函数的性质
幂级数在其收敛区间内可以逐项求导或逐项积分，并且逐项求导或者逐项积分后所得幂级数的收敛区间不变，但在其收敛区间的端点处，级数的敛散性可能会改变
设幂级数$\sum_{n=0}^{\infty}a_nx^n$的和函数为$S(x)$，则
1.$S(x)$在其收敛域上连续，当$x_0 \in (-R,R)$时，有
$ $ \begin{align*} \lim_{x \to x_0 } (\sum_{n=0}^{\infty } a_nx^n) = \sum _{n=0}^{\infty}(\lim_{x \to x_0 }a_nx^n) = \sum _{n=0}^{\infty}a_nx^n_0 = S(x_0) \end{align*} $ $
2.微分运算，$S(x)$在其收敛区间$(-R,R)$内可导，且
$ $ \begin{align*} S'(x)=(\sum_{n=0}^{\infty}a_nx^n)'=\sum_{n=0}^{\infty}(a_nx^n)'=\sum_{n=0}^{\infty}na_nx^{n-1} \end{align*} $ $
其收敛区间为$(-R,R)$
3.积分运算，$S(x)$在其收敛域上可积，且
$ $ \begin{align*} \int_{0}^{x}S(t)\mathrm{d}t=\int_{0}^{x} (\sum_{n=0}^{\infty}a_nt^n)\mathrm{d}t=\sum_{n=0}^{\infty}(\int_{0}^{x}a_nt^n\mathrm{d}t)=\sum_{n=0}^{\infty} \frac{a_n}{n+1}x^{n+1} \end{align*} $ $
其收敛区间为$(-R,R)$
数学常用公式定理1数学常用公式定理2数学常用公式定理4数学常用公式定理5数学常用公式定理6