常用数学公式

lostune2024-09-042024-09-04

常用等价无穷小

$\sin x\sim x\\$ $\tan x\sim x$ $\arcsin x\sim x$ $\arctan x\sim x\\$ $\ln(1+x)\sim x\\$ $e^{x}-1\sim x\\$ $a^{x}-1\sim x\ln a\\$ $1-\cos x\sim\frac{1}{2}x^{2} \\$ $(1+x)^{a}-1\sim ax$

泰勒展开

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

如

$\sin x=\sin0+(\sin x)^{\prime}|_{x=0}\cdot x+\frac{(\sin x)^{\prime\prime}|_{x=0}}{2!}x^{2}+\frac{(\sin x)^{\prime\prime\prime}|_{x=0}}{3!}x^{3}+o(x^{3})$

即

$\sin x=x-\frac16x^{3}+o(x^{3})$

再如

$\sec x=\sec0+(\sec x)^{\prime}|_{x=0}\cdot x+\frac{(\sec x)^{\prime\prime}|_{x=0}}{2!}x^2+\frac{(\sec x)^{\prime\prime\prime}|_{x=0}}{3!}x^3+o(x^3)$

即

$\sec x=1+\frac{x^2}2+o(x^3)$

同理可得如下重要函数的泰勒公式。

$\sin x=x-\frac{x^{3}}{3!}+o\left(x^{3}\right)\\$ $\cos x=1-\frac{x^{2}}{2!}+\frac{x^{4}}{4!}+o\left(x^{4}\right)\\$ $\arcsin x=x+\frac{x^{3}}{3!}+o\left(x^{3}\right)\\$ $\tan x=x+\frac{x^{3}}{3}+o\left(x^{3}\right)\\$ $e^{x}=1+x+\frac{x^{2}}{2!}+\frac{x^{3}}{3!}+o\left(x^{3}\right)\\$ $\quad\left(1+x\right)^{\alpha}=1+\alpha x+\frac{\alpha\left(\alpha-1\right)}{2!}x^{2}+o\left(x^{2}\right)$ $ln(1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}+o(x^3)\\$

基本求导公式

$(x^{a})^{\prime}=\alpha x^{a-1}\left(\alpha\text{ 为常数 }\right)\\$ $\quad(a^{x})^{\prime}=a^{x}\ln a(a>0,a\neq1)\\$ $\quad(e^{x})^{\prime}=e^{x}\\$ $\quad(\log_{a}x)^{\prime}=\frac{1}{x\ln a}\left(a>0,a\neq1\right)\\$ $\left(\ln\left|x\right|\right)^{\prime}=\frac{1}{x}\\$ $\left(\sin x\right)^{\prime}=\cos x\\$ $\left(\cos x\right)^{\prime}=-\sin x \\$ $\left(\arcsin x\right)^{\prime}=\frac{1}{\sqrt{1-x^{2}}} \\$ $(\arccos x)^{\prime}=-\frac{1}{\sqrt{1-x^{2}}}\\$ $\left(\tan x\right)^{\prime}=\sec^{2}x\\$ $\left(\cot x\right)^{\prime}=-\csc^{2}x\\$ $\left(\arctan x\right)^{\prime}=\frac{1}{1+x^{2}} \\$ $\left(arccotx\right)^{\prime}=-\frac{1}{1+x^{2}}\\$ $\left(\sec x\right)^{\prime}=\sec x\tan x \\$ $\left(\csc x\right)^{\prime}=-\csc x\cot x \\$ $\left[\ln\left(x+\sqrt{x^{2}+1}\right)\right]^{\prime}=\frac{1}{\sqrt{x^{2}+1}} \\$ $\left[\ln\left(x+\sqrt{x^{2}-1}\right)\right]^{\prime}=\frac{1}{\sqrt{x^{2}-1}}$ $\int\sqrt{1+x^2}dx=\frac{1}{2}x\sqrt{1+x^2}+\frac{1}{2}\ln|x+\sqrt{1+x^2}|+C$