考研数学基础汇总

Author:张一极

个人博客:极度空间

公众号:视觉迷航

本文适用于有基础,但不够全面作补充用,如有问题,欢迎斧正:zk@likedge.top

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数列:

常见数列求和

$$\begin{array}{l}\sum_{k=1}^{n} k=1+2+3+\cdots+n=\frac{n(n+1)}{2} \\ \sum_{k=1}^{n} k^{2}=1^{2}+2^{2}+3^{2}+\cdots+n^{2}=\frac{n(n+1)(2 n+1)}{6} \\ \sum_{k=1}^{n} \frac{1}{k(k+1)}=\frac{1}{1 \times 2}+\frac{1}{2 \times 3}+\frac{1}{3 \times 4}+\cdots+\frac{1}{n(n+1)}=\frac{n}{n+1}\end{array}$$

三角函数:

$$\begin{array}{l}\arctan 0=0, \quad \arctan \frac{\sqrt{3}}{3}=\frac{\pi}{6}, \quad \arctan 1=\frac{\pi}{4}, \quad \arctan \sqrt{3}=\frac{\pi}{3} \\ \operatorname{arccot} 0=\frac{\pi}{2}, \quad \operatorname{arccot} \sqrt{3}=\frac{\pi}{6}, \quad \operatorname{arccot} 1=\frac{\pi}{4}, \quad \operatorname{arccot} \frac{\sqrt{3}}{3}=\frac{\pi}{3}\end{array}$$

三角变换:

倍角公式和三角函数基本变换

$$\begin{array}{l}\csc \alpha=\frac{1}{\sin \alpha}, \quad \sec \alpha=\frac{1}{\cos \alpha}, \quad \cot \alpha=\frac{1}{\tan \alpha}, \quad \tan \alpha=\frac{\sin \alpha}{\cos \alpha}, \quad \cot \alpha=\frac{\cos \alpha}{\sin \alpha} \\ \sin ^{2} \alpha+\cos ^{2} \alpha=1, \quad 1+\tan ^{2} \alpha=\sec ^{2} \alpha, \quad 1+\cot ^{2} \alpha=\csc ^{2} \alpha\end{array}$$

$$\begin{array}{l}\sin 2 \alpha=2 \sin \alpha \cos \alpha, \quad \cos 2 \alpha=\cos ^{2} \alpha-\sin ^{2} \alpha=1-2 \sin ^{2} \alpha=2 \cos ^{2} \alpha-1 \\ \sin 3 \alpha=-4 \sin ^{3} \alpha+3 \sin \alpha, \quad \cos 3 \alpha=4 \cos ^{3} \alpha-3 \cos \alpha \\ \tan 2 \alpha=\frac{2 \tan \alpha}{1-\tan ^{2} \alpha}, \quad \cot 2 \alpha=\frac{\cot ^{2} \alpha-1}{2 \cot \alpha}\end{array}$$

tan变换

$$u=\tan \frac{x}{2}(-\pi<x<\pi), \text { 则 } \sin x=\frac{2 u}{1+u^{2}}, \cos x=\frac{1-u^{2}}{1+u^{2}}$$

因式分解

$$\begin{array}{ll}(a+b)^{2}=a^{2}+2 a b+b^{2} & \left(2(a-b)^{2}=a^{2}-2 a b+b^{2}\right. \\ (a+b)^{3}=a^{3}+3 a^{2} b+3 a b^{2}+b^{3} & (a)(a-b)^{3}=a^{3}-3 a^{2} b+3 a b^{2}-b^{3}\end{array}$$

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$$\begin{array}{ll}a^{2}-b^{2}=(a+b)(a-b) & (6) a^{3}-b^{3}=(a-b)\left(a^{2}+a b+b^{2}\right) \\ a^{3}+b^{3}=(a+b)\left(a^{2}-a b+b^{2}\right)\end{array}$$

$$\begin{array}{l}a^{\prime \prime}-b^{n}=(a-b)\left(a^{n-1}+a^{n-2} b+\cdots+a b^{n-2}+b^{n-1}\right)(n \text { 是正整数 }) \cdot \beta \\ n \text { 是正偶数时 }, a^{n}-b^{n}=(a+b)\left(a^{n-1}-a^{n-2} b+\cdots+a b^{n-2}-b^{n-1}\right) \\ n \text { 是正奇数时 }, a^{n}+b^{n}=(a+b)\left(a^{n-1}-a^{n-2} b+\cdots-a b^{n-2}+b^{n-1}\right)\end{array}$$

二项式定理

$$\begin{aligned}(a+b)^{n}=\sum_{k=0}^{n} C_{n}^{k} a^{n-k} b^{k}=& a^{n}+n a^{n-1} b+\frac{n(n-1)}{2 !} a^{n-2} b^{2}+\cdots+\\ & \frac{n(n-1) \cdots(n-k+1)}{k !} a^{n-k} b^{k}+\cdots+n a b^{n-1}+b^{n} \end{aligned}$$

阶乘变换

$$\begin{array}{l}(2 n) ! !=2 \cdot 4 \cdot 6 \cdot \cdots \cdot(2 n)=2^{n} \cdot n ! \\ (2 n-1) ! !=1 \cdot 3 \cdot 5 \cdot \cdots \cdot(2 n-1)\end{array}$$

不等式变换

$$\begin{array}{l}|a \pm b| \leqslant|a|+|b| ;<br>\\|| a|-| b|| \leqslant|a-b|;\\ \sqrt{a b} \leqslant \frac{a+b}{2} \leqslant \sqrt{\frac{a^{2}+b^{2}}{2}}(a, b>0) \\ \text { ( } \sqrt[3]{a b c} \leqslant \frac{a+b+c}{3} \leqslant \sqrt{\frac{a^{2}+b^{2}+c^{2}}{3}}(a, b, c>0) \\ 设 a>b>0 \text { ,则 }\left\{\begin{array}{l}\text { 当 } n>0 \text { 时 }, a^{n}>b^{n} \\ \text { 当 } n<0 \text { 时 }, a^{n}<b^{n}\end{array}\right. \\ 0<a<x<b, 0<c<y<d \text { ,则 } \frac{c}{b}<\frac{y}{x}<\frac{d}{a} \text { . } \\ \sin x<x<\tan x\left(0<x<\frac{\pi}{2}\right) \\ \sin x<x(x>0) \\ \arctan x \leqslant x \leqslant \arcsin x(0 \leqslant x \leqslant 1) \\ \mathrm{e}^{x} \geqslant x+1(\forall x) \\ x-1 \geqslant \ln x(x>0)\\\frac{1}{1+x}<\ln \left(1+\frac{1}{x}\right)<\frac{1}{x}(x>0)\end{array}$$