@[TOC]

泰勒公式

$\begin{aligned} & f(x)=f(x_0)+f'(x_0)(x-x_0)+\frac{f''(x_0)}{2!}(x-x_0)^2+\frac{f^{(3)}(x_0)}{2!}(x-x_0)^3+\cdots \\ & f(a)=f(x)+f'(x)(a-x)+\frac{f''(\xi_1)}{2!}(a-x)^2 \\ \end{aligned}$

一元函数泰勒展开

$\begin{aligned} \sin x&=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\cdots\\ \cos x&=1-\frac{x^2}{2!}+\frac{x^4}{4!}+\cdots\\ \tan x&=x+\frac{x^3}{3}+\frac{x^5}{5}+\cdots\\ \arctan x&=x-\frac{x^3}{3}+\frac{x^5}{5}-\cdots\\ \ln(1+x)&=x-\frac{x^2}{2}+\frac{x^3}{3}-\cdots\\ \text{e}^x&=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\cdots\\ \frac{1}{1-x}&=1+x+x^2+x^3+\cdots\\ (1+x)^n&=1+nx+\frac{n(n-1)}{2}x^2\cdots\\ \end{aligned}$

多元函数泰勒展开

$f(\boldsymbol{x})=\sum_{k=0}^{\infty}\frac{1}{k!} (h_1\frac{\partial}{\partial x_1}+\cdots+h_n\frac{\partial}{\partial x_n})^kf(\boldsymbol{x}_0)$

其中$\boldsymbol{h}=\boldsymbol{x}-\boldsymbol{x}_0$。二阶泰勒展开矢量形式

$f(\boldsymbol{x})=f(\boldsymbol{x}_0)+\nabla f(\boldsymbol{x}_0)^T(\boldsymbol{x}-\boldsymbol{x}_0)+ \frac{1}{2}(\boldsymbol{x}-\boldsymbol{x}_0)^T\nabla^2f(\boldsymbol{x}_0)(\boldsymbol{x}-\boldsymbol{x}_0) +o(\Vert\boldsymbol{x}-\boldsymbol{x}_0\Vert^2)$

其中$\nabla f(\boldsymbol{x}_0)=\frac{\partial f(\boldsymbol{x})}{\partial \boldsymbol{x}}\bigg|_{\boldsymbol{x}=\boldsymbol{x}_0}$。
三元二阶泰勒展开，令

$\boldsymbol{v}=\begin{bmatrix}x\\y\\z\end{bmatrix},f=f(\boldsymbol{v}), \frac{\partial f_0}{\partial x}=\frac{\partial f(\boldsymbol{v})}{\partial x}\bigg|_{\boldsymbol{v}=\boldsymbol{v}_0}$

则

$f(\boldsymbol{v})=f(\begin{bmatrix}x_0\\y_0\\z_0\end{bmatrix})+ \begin{bmatrix}\frac{\partial f_0}{\partial x} & \frac{\partial f_0}{\partial y} & \frac{\partial f_0}{\partial z}\end{bmatrix} \begin{bmatrix}h_x\\h_y\\h_z\end{bmatrix}+ \frac{1}{2}\begin{bmatrix}h_x & h_y & h_z\end{bmatrix}\begin{bmatrix} \frac{\partial^2f_0}{\partial x^2} & \frac{\partial^2f_0}{\partial x\partial y}& \frac{\partial^2f_0}{\partial x\partial z} \\ \frac{\partial^2f_0}{\partial y\partial x} & \frac{\partial^2f_0}{\partial y^2}& \frac{\partial^2f_0}{\partial y\partial z} \\ \frac{\partial^2f_0}{\partial z\partial x} & \frac{\partial^2f_0}{\partial x\partial y}& \frac{\partial^2f_0}{\partial z^2} \end{bmatrix}\begin{bmatrix} h_x \\ h_y \\ h_z\end{bmatrix}$

求导

$\frac{\partial f(\boldsymbol{v})}{\partial \boldsymbol{v}}= \begin{bmatrix}\frac{\partial f_0}{\partial x} \\ \frac{\partial f_0}{\partial y} \\ \frac{\partial f_0}{\partial z}\end{bmatrix}+ \begin{bmatrix} \frac{\partial^2f_0}{\partial x^2} & \frac{\partial^2f_0}{\partial x\partial y}& \frac{\partial^2f_0}{\partial x\partial z} \\ \frac{\partial^2f_0}{\partial y\partial x} & \frac{\partial^2f_0}{\partial y^2}& \frac{\partial^2f_0}{\partial y\partial z} \\ \frac{\partial^2f_0}{\partial z\partial x} & \frac{\partial^2f_0}{\partial x\partial y}& \frac{\partial^2f_0}{\partial z^2} \end{bmatrix}\begin{bmatrix}h_x\\h_y\\h_z\end{bmatrix}$ $\frac{\partial f(\boldsymbol{v})}{\partial \boldsymbol{v}}= \nabla f(\boldsymbol{x}_0)+\nabla^2f(\boldsymbol{x}_0)(\boldsymbol{v}-\boldsymbol{v}_0)$

$f(\boldsymbol{v})$是一个标量，而$\frac{\partial f(\boldsymbol{v})}{\partial \boldsymbol{v}}$是一个矢量。
令$\frac{\partial f(\boldsymbol{v})}{\partial \boldsymbol{v}}=0$可解得

$\boldsymbol{v}-\boldsymbol{v}_0= -(\nabla^2f(\boldsymbol{x}_0))^{-1}\nabla f(\boldsymbol{x}_0)$

代入$f(\boldsymbol{x})$得

$\begin{aligned} f(\boldsymbol{x})&=f(\boldsymbol{x}_0)+\nabla f(\boldsymbol{x}_0)^T(-(\nabla^2f(\boldsymbol{x}_0))^{-1}\nabla f(\boldsymbol{x}_0))\\ &+\frac{1}{2}(\nabla^2f(\boldsymbol{x}_0))^{-1}\nabla f(\boldsymbol{x}_0))^T\nabla^2f(\boldsymbol{x}_0)(\nabla^2f(\boldsymbol{x}_0))^{-1}\nabla f(\boldsymbol{x}_0))\\ &=f_0-(f'_0)^T(f''_0)^{-1}f'_0+\frac{1}{2}((f''_0)^{-1}f'_0)^Tf''_0((f''_0)^{-1}f'_0)\\ &=f_0-(f'_0)^T(f''_0)^{-1}f'_0+\frac{1}{2}(f'_0)^T(f''_0)^{-1}f'_0\\ &=f_0-\frac{1}{2}(f'_0)^T(f''_0)^{-1}f'_0\\ &=f(\boldsymbol{x}_0)+\frac{1}{2}\frac{\partial f(\boldsymbol{x}_0)}{\partial \boldsymbol{x}}^T\boldsymbol{h} \end{aligned}$

空间解析几何

平面

$A(x-x_0)+B(y-y_0)+C(z-z_0)=0 \\ Ax+By+Cz+D=0 \\ \frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1 \\ \vec{n}=(A,B,C) \\$

直线

$\frac{x-x_0}{m}=\frac{y-y_0}{n}=\frac{z-z_0}{p} \\ \vec{s}=(m,n,p)$

质心

$\begin{aligned} & \bar{x}=\frac{\iint x\rho(x,y)\text{d}\sigma}{\iint\rho(x,y)\text{d}\sigma} \\ & \bar{z}=\frac{\iiint z\rho(x,y,z)\text{d}\sigma}{\iiint\rho(x,y,z)\text{d}\sigma} \\ \end{aligned}$

双曲函数

$\cosh x=\frac{\text{e}^x+\text{e}^{-x}}{2},\sinh x=\frac{\text{e}^x-\text{e}^{-x}}{2} \\ \cosh(-x)=\cosh(x),\sinh(-x)=-\sinh(x) \\ \cosh(x+y)=\cosh x\cosh y+\sinh x\sinh y \\ \sinh(x+y)=\sinh x\cosh y+\cosh x\sinh y \\ \cosh^2x-\sinh^2x=1 \\ \cosh 2x=\cosh^2x+\sinh^2x \\ \sinh 2x=2\sinh x\cosh x \\ \cosh'x=\sinh x,\quad\sinh'x=\cosh x$

导数

$(\cot x)'=-\frac{1}{\sin^2x}$

隐函数求导

$F(x,y)=0,\frac{\text{d}y}{\text{d}x}=-\frac{F_x}{F_y} \\ F(x,y,z)=0,\frac{\partial z}{\partial x}=-\frac{F_x}{F_z}, \frac{\partial z}{\partial y}=-\frac{F_y}{F_z}$

不定积分

$\begin{aligned} & \int\frac{1}{\sqrt{x^2+1}}\text{d}x=\ln|x+\sqrt{x^2+1}|+C \\ & \int\text{e}^{ax}\cos bx\text{d}x=\frac{\text{e}^{ax}}{a^2+b^2}(a\cos bx+b\sin bx)+C \\ & \int\text{e}^{ax}\sin bx\text{d}x=\frac{\text{e}^{ax}}{a^2+b^2}(a\sin bx-b\cos bx)+C \\ & \int\text{e}^{ax}\cos bx+\text{j}\text{e}^{ax}\sin bx\text{d}x=\int \text{e}^{(a+\text{j}b)x}\text{d}x=\frac{\text{e}^{(a+\text{j}b)x}}{a+\text{j}b}+C \\ & \int \sec x \mathrm{~d} x=\ln |\sec x+\tan x|+C \\ & \int \csc x \mathrm{~d} x=\ln |\csc x-\cot x|+C \\ \end{aligned}$

不定积分列表法

$\int x^3\text{e}^{2x}\text{d}x=x^3\frac{1}{2}\text{e}^{2x}-3x^2\frac{1}{4}\text{e}^{2x}+6x\frac{1}{8}\text{e}^{2x}-6\frac{1}{16}\text{e}^{2x} \\$


$\frac{1}{2}\text{e}^{2x}$	$\frac{1}{4}\text{e}^{2x}$	$\frac{1}{8}\text{e}^{2x}$	$\frac{1}{16}\text{e}^{2x}$
+	-	+	-
$x^3$	$3x^2$	$6x$	$6$ \\

没有初等函数的原函数

$\int\text{e}^{x^2}\text{d}x\quad \int\frac{\sin x}{x}\text{d}x\quad \int\frac{1}{\ln x}\text{d}x\quad \int\frac{1}{\sqrt{1+x^4}}\text{d}x\quad \int\sqrt{1-k^2\sin^2x}\text{d}x \\ \int\text{e}^{\frac{1}{x}}\text{d}x\quad$

定积分

$\begin{aligned} \int_0^{+\infty}\text{e}^{-x^2}\text{d}x &\xlongequal[t=\sqrt{x}]{x=t^2}\frac{1}{2}\int_0^{+\infty}\frac{\text{e}^{-t}}{\sqrt{t}}\text{d}t =\frac{1}{2}\Gamma(\frac{1}{2}) =\frac{\sqrt{\pi}}{2} \\ \int_0^{+\infty}\text{e}^{-x^2}\text{d}x &= \sqrt{\int_0^{+\infty}\int_0^{+\infty}\text{e}^{-x^2-y^2}\text{d}x\text{d}y} \\ &= \sqrt{\int_0^{\frac{\pi}{2}}\int_0^{+\infty}\text{e}^{-r^2}r\text{d}r\text{d}\theta} \\ &= \sqrt{\int_0^{\frac{\pi}{2}}\frac{1}{2}\text{d}\theta}=\frac{\sqrt{\pi}}{2} \\ \int_0^{\frac{\pi}{2}}\sin x\text{d}x &= \int_0^{\frac{\pi}{2}}\sin 2x\text{d}x \\ \int_a^bf(x)\text{d}x &= \int_a^bf(a+b-x)\text{d}x \\ \end{aligned}$

$\Gamma$函数

$\begin{aligned} & \Gamma(s)=\int_0^{+\infty}\text{e}^{-x}x^{s-1}\text{d}x \\ & \Gamma(s+1)=s\Gamma(s)=s! \\ \end{aligned}$

余元公式

$\Gamma(s)\Gamma(1-s)=\frac{\pi}{\sin\pi s}$

罗巴切夫斯基(Lobachevsky)积分

若$f(x)$在$[0,+\infty)$有$f(\pi-x)=f(x)$以及$f(x+\pi)=f(x)$，则有

$\int_0^{+\infty} f(x)\frac{\sin x}{x}=\int_0^{\frac{\pi}{2}}f(x)\text{d}x$

微分方程

常用变量代换

$\begin{aligned} & z=\frac{y}{x},z=\frac{1}{y},z=y' \\ \end{aligned}$

二阶线性微分方程解的结构

$\begin{aligned} & y''+P(x)y'+Q(x)y=0 \\ & y=C_1y_1(x)+C_2y_2(x) \\ & y''+P(x)y'+Q(x)y=f(x) \\ & y=C_1y_1(x)+C_2y_2(x)+f^*(x) \\ \end{aligned}$

二阶常系数齐次线性微分方程

特征方程$x^2+px+q=0$，单根$r_1,r_2$，复根$r$，共轭复根$\alpha\pm \text{j}\beta$。

$\begin{aligned} & y''+py'+qy=0 \\ & y=C_1\text{e}^{r_1x}+C_2\text{e}^{r_2x} \\ & y=(C_1+C_2x)\text{e}^{rx} \\ &y=\text{e}^{\alpha x}(C_1\cos\beta x+C_2\sin\beta x) \\ \end{aligned}$

二阶常系数非齐次线性微分方程

$y''+py'+qy=f(x)$

对于微分方程

$f(x)=\text{e}^{\lambda x}P_m(x)$

特解满足形式

$y^*=x^kR_m(x)\text{e}^{\lambda x}$

其中$R_m(x)$是与$P_m(x)$同次的多项式，而$k$按$\lambda$不是特征方程的根、是特征方程的单根、复根，依次取0、1、2。n阶常系数非齐次线性微分方程中，$k$是$\lambda$的重复次数。
对于微分方程

$f(x)=\text{e}^{\lambda x}[P_l(x)\cos\omega x+Q_n(x)\sin\omega x]$

特解满足形式

$y^*=x^k\text{e}^{\lambda x}[R_m^{(1)}(x)\cos\omega x+R_m^{(2)}(x)\sin\omega x]$

其中$R_m^{(1)}(x)$，$R_m^{(2)}(x)$是m次多项式，$m=\max\{l,n\}$，$k$按$\lambda+\omega i$(或$\lambda-\omega i$)不是特征方程的根是特征方程的单根分别取0或1。

欧拉方程

令$x=\text{e}^t$，$t=\ln x$

$\sum_{i=1}^nx^iy^{(i)}=f(x) \\ \frac{\text{d}y}{\text{d}x}=\frac{1}{x}\frac{\text{d}y}{\text{d}t} \\ \frac{\text{d}^2y}{\text{d}x^2}=\frac{1}{x^2}(\frac{\text{d}^2y}{\text{d}t^2}-\frac{\text{d}y}{\text{d}t}) \\ \frac{\text{d}^3y}{\text{d}x^3}=\frac{1}{x^3}(\frac{\text{d}^3y}{\text{d}t^3}-3\frac{\text{d}^2y}{\text{d}t^2}+2\frac{\text{d}y}{\text{d}t}) \\ xy'=\text{D}y \\ x^2y''=\text{D}(\text{D}-1)y \\ x^3y'''=\text{D}(\text{D}-1)(\text{D}-2)y$

级数

正项级数

若$a_n\le b_n$且$b_n$收敛，则$a_n$收敛。
若$a_n\ge b_n$且$b_n$发散，则$a_n$发散。
令$l=\displaystyle\lim_{n\to\infty}\frac{a_n}{b_n}$，
若$01$时发散。
令$\rho=\sqrt[n]{a_n}$，$\rho<1$时收敛，$\rho>1$时发散。

p级数

$1+\frac{1}{2^p}+\frac{1}{3^p}+\frac{1}{4^p}+\cdots+\frac{1}{n^p}$

其中$p>0$，$p\le1$时发散。

交错级数

$\sum_{n=0}^{\infty}(-1)^{n} u_{n}$

若$u_n\ge u_{n+1}$且$\displaystyle\lim_{n\to\infty}u_n=0$则交错级数收敛。
对于一般的级数

$u_1+u_2+\cdots+u_n$

如果级数$|u_n|$收敛则称$u_n$绝对收敛，如果级数$|u_n|$发散但$u_n$收敛则称$u_n$条件收敛。$u_n$绝对收敛是$u_n$收敛的充分不必要条件。

幂级数

$\sum_{n=0}^{\infty} a_{n} x^{n}=a_{0}+a_{1} x+a_{2} x^{2}+\cdots+a_{n} x^{n}$

信号与系统

傅里叶级数

傅里叶级数(FS)将连续周期信号分解为非周期离散频谱。

$f(t)=\frac{a_0}{2}+\sum_{n=1}^{\infty}a_n\cos n\Omega t+\sum_{n=1}^{\infty}b_n\sin n\Omega t$ $a_n=\frac{2}{T}\int_{-\frac{T}{2}}^{\frac{T}{2}}f(t)\cos n\Omega t\text{d}t$ $b_n=\frac{2}{T}\int_{-\frac{T}{2}}^{\frac{T}{2}}f(t)\sin n\Omega t\text{d}t$

其中$T$为周期，$\Omega=\frac{2\pi}{T}$。复数形式

$f(t)=\sum_{n=-\infty}^{\infty}F_n\text{e}^{jn\Omega t}$ $F_n=\frac{1}{T}\int_{-\frac{T}{2}}^{\frac{T}{2}}f(t)\text{e}^{-jn\Omega t}\text{d}t$

傅里叶变换

连续傅里叶变换(FT)将连续非周期信号分解为非周期连续频谱。

拉普拉斯变换

$\frac{\text{d}y}{\text{d}t}f(t)\longrightarrow sF(s)-f(0)$ $\frac{\text{d}^2}{\text{d}t^2}f(t)\longrightarrow s^2F(s)-sf(0)-f'(0)$