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一、初等数学部分

1、特殊函数值

$\begin{array}{l} \sin 0=0 . \quad \sin \frac{\pi}{6}=\frac{1}{2}, \quad \sin \frac{\pi}{4}=\frac{\sqrt{2}}{2}, \quad \sin \frac{\pi}{3}=\frac{\sqrt{3}}{2}, \\ \sin \frac{\pi}{2}=1, \quad \sin \pi=0, \quad \sin \frac{3 \pi}{2}=-1, \quad \sin 2 \pi=0 . \\ \cos 0=1, \quad \cos \frac{\pi}{6}=\frac{\sqrt{3}}{2}, \quad \cos \frac{\pi}{4}=\frac{\sqrt{2}}{2}, \quad \cos \frac{\pi}{3}=\frac{1}{2} \\ \cos \frac{\pi}{2}=0, \quad \cos \pi=-1, \quad \cos \frac{3 \pi}{2} m 0, \quad \cos 2 \pi=1 \end{array} $

$\tan 0=0 . \quad \tan \frac{\pi}{6}=\frac{\sqrt{3}}{3}, \quad \tan \frac{\pi}{4}=1 . \quad \tan \frac{\pi}{3}=\sqrt{3}$

$\cot \frac{\pi}{2}=0, \quad \lim _{x \rightarrow} \cot x=\infty, \cot \frac{3 \pi}{2}=0 . \quad \lim _{\rightarrow \rightarrow 1} \cot x=\infty \text {, }$

2、倍角公式

$\begin{array}{l} \sin 2_{a}=2 \sin \operatorname{acos} a+\quad \cos 2 g=\cos ^{2} a-\sin ^{2} a=1-2 \sin ^{2} a=2 \cos ^{2} a^{-1}\\ \sin 3 a=-4 \sin ^{3} a+3 \sin a . \quad \cos 3 a=4 \cos ^{2} e-3 \cos a\\ \tan 2 g=\frac{2 \tan \mathrm{c}}{1-\tan ^{2} a}, \quad \cot 2 a=\frac{\cot ^{2} a-1}{2 \cot a} . \end{array}​$

3、半角公式

$\begin{array}{l} \sin ^{2} \frac{\alpha}{2}=\frac{1}{2}(1-\cos \alpha), \quad \cos ^{2} \frac{\alpha}{2}=\frac{1}{2}(1+\cos \alpha),(\text { 降幂公式) }\\ \sin \frac{\alpha}{2}=\pm \sqrt{\frac{1-\cos \alpha}{2}}, \quad \cos \frac{\alpha}{2}=\pm \sqrt{\frac{1+\cos \alpha}{2}},\\ \tan \frac{\alpha}{2}=\frac{1-\cos \alpha}{\sin \alpha}=\frac{\sin \alpha}{1+\cos \alpha}=\pm \sqrt{\frac{1-\cos \alpha}{1+\cos \alpha}},\\ \cot \frac{\alpha}{2}=\frac{\sin \alpha}{1-\cos \alpha}=\frac{1+\cos \alpha}{\sin \alpha}=\pm \sqrt{\frac{1+\cos \alpha}{1-\cos \alpha}} . \end{array}​$

4、和差公式

$\begin{array}{l} \sin (\alpha \pm \beta)=\sin \alpha \cos \beta \pm \cos \alpha \sin \beta, \quad \cos (\alpha \pm \beta)=\cos \alpha \cos \beta \mp \sin \alpha \sin \beta, \\ \tan (\alpha \pm \beta)=\frac{\tan \alpha \pm \tan \beta}{1 \mp \tan \alpha \tan \beta}, \quad \cot (\alpha \pm \beta)=\frac{\cot \alpha \cot \beta \mp 1}{\cot \beta \pm \cot \alpha} \end{array}$

5、积化和差

$\begin{array}{l} \sin \alpha \cos \beta=\frac{1}{2}[\sin (\alpha+\beta)+\sin (\alpha-\beta)], \quad \cos \alpha \sin \beta=\frac{1}{2}[\sin (\alpha+\beta)-\sin (\alpha-\beta)] \\ \cos \alpha \cos \beta=\frac{1}{2}[\cos (\alpha+\beta)+\cos (\alpha-\beta)], \quad \sin \alpha \sin \beta=\frac{1}{2}[\cos (\alpha-\beta)-\cos (\alpha+\beta)] \end{array}$

6、和差化积

$\sin \alpha+\sin \beta=2 \sin \frac{\alpha+\beta}{2} \cos \frac{\alpha-\beta}{2}, \quad \sin \alpha-\sin \beta=2 \sin \frac{\alpha-\beta}{2} \cos \frac{\alpha+\beta}{2}$

$\cos \alpha+\cos \beta=2 \cos \frac{\alpha+\beta}{2} \cos \frac{\alpha-\beta}{2}, \quad \cos \alpha-\cos \beta=-2 \sin \frac{\alpha+\beta}{2} \sin \frac{\alpha-\beta}{2}$

7、对数运算

$\log _{a}(M N)=\log _{a} M+\log _{a} N$​(积的对数=对数的积)​

$\log _{a} \frac{M}{N}=\log _{a} M-\log _{a} N()$

二、高等数学部分

1、基本求导公式

$\begin{array}{l} \left(x^{\alpha}\right)^{\prime}=\alpha x^{a-1}(\alpha \text { 为常数 }), \quad\left(a^{x}\right)^{\prime}=a^{x} \ln a(a>0, a \neq 1), \quad\left(\mathrm{e}^{x}\right)^{\prime}=\mathrm{e}^{x}, \quad\left(\log _{a} x\right)^{\prime}=\frac{1}{x \ln a}(a>0, a \neq 1), \\ (\ln x)^{\prime}=\frac{1}{x}, \quad(\sin x)^{\prime}=\cos x, \quad(\cos x)^{\prime}=-\sin x, \quad(\arcsin x)^{\prime}=\frac{1}{\sqrt{1-x^{2}}} \\ (\arccos x)^{\prime}=-\frac{1}{\sqrt{1-x^{2}}}, \quad(\tan x)^{\prime}=\sec ^{2} x, \quad(\cot x)^{\prime}=-\csc ^{2} x, \quad(\arctan x)^{\prime}=\frac{1}{1+x^{2}}, \\ (\operatorname{arccot} x)^{\prime}=-\frac{1}{1+x^{2}}, \quad(\sec x)^{\prime}=\sec x \tan x, \quad(\csc x)^{\prime}=-\csc x \cot x \\ {\left[\ln \left(x+\sqrt{x^{2}+1}\right)\right]^{\prime}=\frac{1}{\sqrt{x^{2}+1}}, \quad\left[\ln \left(x+\sqrt{x^{2}-1}\right)\right]^{\prime}=\frac{1}{\sqrt{x^{2}-1}}} \end{array}$

2、变限积分求导公式

$F^{\prime}(x)=\frac{\mathrm{d}}{\mathrm{d} x}\left[\int_{\varepsilon(x)}^{\varphi_{n}(x)} f(t) \mathrm{d} t\right]=f\left[\varphi_{2}(x)\right] \varphi_{2}^{\prime}(x)-f\left[\varphi_{1}(x)\right] \varphi_{1}^{\prime}(x)​$

3、泰勒公式

当x=0时的泰勒公式被称为麦克劳林公式

$f(x)=f(0)+f^{\prime}(0) x+\frac{f^{\prime \prime}(0)}{2 !} x^{2}+\cdots+\frac{f^{(n)}(0)}{n !} x^{n}+\frac{f^{(n+1)}(\xi)}{(n+1) !} x^{n+1}$

$
f(x)=f(0)+f^{\prime}(0) x+\frac{f^{\prime \prime}(0)}{2 !} x^{2}+\cdots+\frac{f^{(n)}(0)}{n !} x^{n}+o\left(x^{n}\right)
$

4、麦克劳林公式

$
\begin{aligned} &\mathrm{e}^{x}=1+x+\frac{1}{2 !} x^{2}+\cdots+\frac{1}{n !} x^{n}+o\left(x^{n}\right) . \\ &\sin x=x-\frac{x^{3}}{3 !}+\cdots+(-1)^{n} \frac{x^{2 n+1}}{(2 n+1) !}+o\left(x^{2 n+1}\right) \\ &\cos x=1-\frac{x^{2}}{2 !}+\frac{x^{4}}{4 !}-\cdots+(-1)^{n} \frac{x^{2 n}}{(2 n) !}+o\left(x^{2 n}\right) \\ &\frac{1}{1-x}=1+x+x^{2}+\cdots+x^{n}+o\left(x^{n}\right) \\ &\frac{1}{1+x}=1-x+x^{2}-\cdots+(-1)^{n} x^{n}+o\left(x^{n}\right) \\ &\ln (1+x)=x-\frac{x^{2}}{2}+\frac{x^{3}}{3}-\cdots+(-1)^{n-1} \frac{x^{n}}{n}+o\left(x^{n}\right) \\ &(1+x)^{a}=1+\alpha x+\frac{\alpha(\alpha-1)}{2 !} x^{2}+\cdots+\frac{\alpha(\alpha-1) \cdots(\alpha-n+1)}{n !} x^{n}+o\left(x^{n}\right) \end{aligned}
$

不定积分法基本积分公式

$\int x^{k} \mathrm{~d} x=\frac{1}{k+1} x^{k+1}+C, k \neq-1 ;\left\{\begin{array}{l} \int \frac{1}{x^{2}} \mathrm{~d} x=-\frac{1}{x}+C \\ \int \frac{1}{\sqrt{x}} \mathrm{~d} x=2 \sqrt{x}+C \end{array}\right.$

$\begin{array}{l} \int \frac{1}{x} \mathrm{~d} x=\ln |x|+C \\ \int \mathrm{e}^{x} \mathrm{~d} x=\mathrm{e}^{x}+C ; \int a^{x} \mathrm{~d} x=\frac{a^{x}}{\ln a}+C, a>0 \text { 且 } a \neq 1 . \\ \int \sin x \mathrm{~d} x=-\cos x+C ; \int \cos x \mathrm{~d} x=\sin x+C \end{array}$

$\int \tan x \mathrm{~d} x=-\ln |\cos x|+C ; \int \cot x \mathrm{~d} x=\ln |\sin x|+C$

$\begin{array}{l} \int \frac{\mathrm{d} x}{\cos x}=\int \sec x \mathrm{~d} x=\ln |\sec x+\tan x|+C \\ \int \frac{\mathrm{d} x}{\sin x}=\int \csc x \mathrm{~d} x=\ln |\csc x-\cot x|+C \\ \int \sec ^{2} x \mathrm{~d} x=\tan x+C ; \int \csc ^{2} x \mathrm{~d} x=-\cot x+C \\ \int \sec x \tan x \mathrm{~d} x=\sec x+C ; \int \csc x \cot x \mathrm{~d} x=-\csc x+C \end{array}$

$\begin{array}{l} \left\{\begin{array}{l} \int \frac{1}{1+x^{2}} \mathrm{~d} x=\arctan x+C \\ \int \frac{1}{a^{2}+x^{2}} \mathrm{~d} x=\frac{1}{a} \arctan \frac{x}{a}+C(a>0) \end{array}\right.\\ \left\{\begin{array}{l} \int \frac{1}{\sqrt{1-x^{2}}} \mathrm{~d} x=\arcsin x+C \\ \int \frac{1}{\sqrt{a^{2}-x^{2}}} \mathrm{~d} x=\arcsin \frac{x}{a}+C(a>0) . \end{array}\right.\\ \left\{\begin{array}{l} \int \frac{1}{\sqrt{x^{2}+a^{2}}} \mathrm{~d} x=\ln \left(x+\sqrt{x^{2}+a^{2}}\right)+C(\text { 常见 } a=1) \\ \int \frac{1}{\sqrt{x^{2}-a^{2}}} \mathrm{~d} x=\ln \left|x+\sqrt{x^{2}-a^{2}}\right|+C(|x|>|a|) \end{array}\right.\\ \int \frac{1}{x^{2}-a^{2}} \mathrm{~d} x=\frac{1}{2 a} \ln \left|\frac{x-a}{x+a}\right|+C\left(\int \frac{1}{a^{2}-x^{2}} \mathrm{~d} x=\frac{1}{2 a} \ln \left|\frac{x+a}{x-a}\right|+C\right) \text {. }\\ \int \sqrt{a^{2}-x^{2}} \mathrm{~d} x=\frac{a^{2}}{2} \arcsin \frac{x}{a}+\frac{x}{2} \sqrt{a^{2}-x^{2}}+C(a>|x| \geqslant 0)\\ \int \sin ^{2} x \mathrm{~d} x=\frac{x}{2}-\frac{\sin 2 x}{4}+C\left(\sin ^{2} x=\frac{1-\cos 2 x}{2}\right)\\ \int \cos ^{2} x \mathrm{~d} x=\frac{x}{2}+\frac{\sin 2 x}{4}+C\left(\cos ^{2} x=\frac{1+\cos 2 x}{2}\right)\\ \int \tan ^{2} x \mathrm{~d} x=\tan x-x+C\left(\tan ^{2} x=\sec ^{2} x-1\right)\\ \int \cot ^{2} x \mathrm{~d} x=-\cot x-x+C\left(\cot ^{2} x=\csc ^{2} x-1\right) \end{array}$

凑微分法

基本思想

$\int f[g(x)] g^{\prime}(x) \mathrm{d} x=\int f[g(x)] \mathrm{d}[g(x)]=\int f(u) \mathrm{d} u$

常用的幂级数展开公式

$e^{x}=\sum_{z=0}^{\infty} \frac{1}{n !} x^{u}=1+x+\frac{1}{2 !} x^{2}+\cdots+\frac{1}{n !} x^{x}+\cdots, x \in(-\infty,+\infty)$

$\sin x=\sum_{n=0}^{\infty} \frac{(-1)^{n}}{(2 n+1) !} x^{2 n+1}=x-\frac{1}{3 !} x^{3}+\frac{1}{5 !} x^{5}-\cdots+\frac{(-1)^{\pi}}{(2 n+1) !} x^{2 y+1}+\cdots, x \in(-\infty,+\infty)$

$\cos x=\sum_{n=0}^{\infty} \frac{(-1)^{n}}{(2 n) !}^{2 x}=1-\frac{1}{2 !} x^{2}+\frac{1}{4 \mid} x^{4}-\cdots+\frac{(-1)^{n}}{(2 n) !} x^{2 \pi}+\cdots, x \in(-\infty,+\infty)$


$\ln (1+x)=\sum_{y=0}^{\infty} \frac{(-1)^{y}}{n+1} x^{n+1}=x-\frac{1}{2} x^{2}+\frac{1}{3} x^{3}-\cdots+\frac{(-1)^{n}}{n+1} x^{n+1}+\cdots, x \in(-1,1]$

$\frac{1}{+x}=\sum_{x=0}^{\infty}(-1)^{n} x^{x}=1-x+x^{2}-x^{3}+\cdots+(-1)^{x} x^{2}+\cdots, x \in(-1,1)$


$\arcsin x=\sum_{n=0}^{\infty} \frac{(2 n) !}{4^{x}(n \mid)^{2}(2 n+1)} x^{2 n+1}=x+\frac{1}{6} x^{3}+\frac{3}{40} x^{5}+\cdots, x \in(-1,1)$


$\tan x=\sum_{n=1}^{\infty} \frac{B_{2}(-4)^{n}\left(1-4^{x}\right)}{(2 n) !} x^{2 x-1}=x+\frac{1}{3} x^{3}+\frac{2}{15} x^{5}+\frac{17}{315} x^{\top}+\frac{62}{2835} x^{2}+\frac{1382}{155925} x^{11}+\cdots x \in(-1,1)$


$\sec x=\sum_{n=0}^{\infty} \frac{(-1)^{n} E_{2 z} x^{2 n}}{(2 n) !}=1+\frac{1}{2} x^{2}+\frac{5}{24} x^{4}+\frac{61}{720} x^{6}+\cdots, x \in\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$


$\csc x=\sum_{n=0}^{\infty} \frac{(-1)^{n+1} 2\left(2^{2 \pi-1}-1\right) B_{2 n}}{(2 n) !} x^{2 x-1}=\frac{1}{x}+\frac{1}{6} x+\frac{7}{360} x^{3}+\frac{31}{15120} x^{5}+\cdots, x \in(0, \pi)$


$\cot x=\sum_{n=0}^{\infty} \frac{(-1)^{n} 2^{2 \pi} B_{2 \pi}}{(2 n) !} x^{2 x-1}=\frac{1}{x}-\frac{1}{3} x-\frac{1}{45} x^{3}-\frac{2}{945} x^{5}-\cdots, x \in(0, \pi)$


$\operatorname{sh} x=\sum_{n=0}^{\infty} \frac{x^{2 n+1}}{(2 n+1) !}=x+\frac{x^{3}}{3 !}+\frac{x^{5}}{5 !}+\frac{x^{\top}}{7 !}+\cdots+\frac{x^{2 n+1}}{(2 n+1) !}+\cdots$


$\operatorname{archx}=\ln 2 x-\sum_{n=1}^{\infty}\left(\frac{(-1)^{x}(2 n) !}{2^{2 x}(n !)^{2}}\right) \frac{x^{-2 x}}{2 n}=\ln 2 x-\left(\left(\frac{1}{2}\right) \frac{x^{-2}}{2}+\left(\frac{1 \times 3}{2 \times 4}\right) \frac{x^{-4}}{4}+\left(\frac{1 \times 3 \times 5}{2 \times 4 \times 6}\right) \frac{x^{-6}}{6}+\cdots\right),|x|>1$

$\operatorname{arth} x=\sum_{n=0}^{\infty} \frac{x^{2 n+1}}{2 n+1}=x+\frac{x^{3}}{3}+\frac{x^{5}}{5}+\frac{x^{7}}{7}+\cdots+\frac{x^{2 n+1}}{2 n+1}+\forall,|x|<1$

二阶常系数齐次线性微分方程的通解

对于 $y^{\prime \prime}+p y^{\prime}+q y=0$, 其对应的特征方程为 $\lambda^{2}+p \lambda+q=0$, 求其特征根,有以下三种情况请大家牢记 (其中 $C_{1}, C_{2}$ 为任意常数).
(1)若 $p^{2}-4 q>0$, 设 $\lambda_{1}, \lambda_{2}$ 是特征方程的两个不等实根, 即 $\lambda_{1} \neq \lambda_{2}$, 可得其通解为
$$
y=C_{1} \mathrm{e}^{2, x}+C_{2} \mathrm{e}^{\lambda_{2} x}
$$

(2)若 $p^{2}-4 q=0$, 设 $\lambda_{1}, \lambda_{2}$ 是特征方程的两个相等的实根, 即二重根,令 $\lambda_{1}=\lambda_{2}=\lambda$, 可得其通解为
$$
y=\left(C_{1}+C_{2} x\right) \mathrm{e}^{\mathrm{k} x}
$$
(3)若 $p^{2}-4 q<0$, 设 $\alpha \pm \beta \mathrm{i}$ 是特征方程的一对共轭复根,可得其通解为
$$
y=\mathrm{e}^{\mathrm{ax}}\left(C_{1} \cos \beta x+C_{2} \sin \beta x\right)
$$

二阶常系数非齐次线性微分方程的特解

对于 $y^{\prime \prime}+p y^{\prime}+q y=f(x), 《$ 考试大纲》规定我们需要会解以下两种情况. 设 $P_{n}(x), P_{m}(x)$ 分别为 $x$ 的 $n$ 次, $m$ 次多项式.
(1) 当自由项 $f(x)=P_{n}(x) \mathrm{e}^{\mathrm{ar}}$ 时,特解要设为 $y^{·}=\mathrm{e}^{a x} Q_{n}(x) x^{k}$,

其中

(2) 当自由项, $f(x)=\operatorname{ear}\left[P_{m}(x) \cos \beta x+P_{n}(\dot{x}) \sin \beta x\right]$​ 时,特解要设为
其中
$y^{·}=\mathrm{e}^{\mathrm{ax}}\left[Q_{i}^{(1)}(x) \cos \beta x+Q_{i}^{(2)}(x) \sin \beta x\right] x^{k}$​

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