【考研数学高数部分】泰勒展开式

泰勒展开式

$$\begin{gathered} \begin{array}{l}\\ f(x)=f(x_0)+f^{\prime }(x_0)(x-x_0)\\ \\ +\frac{f^{\prime \prime }(x_0)}{2!}(x-x_0{)}^2+...\\ \\ +\frac{f^{(n)}(x_0)}{n!}(x-x_0{)}^n\\ \\ +\frac{f^{(n+1)}(x_0+\theta (x-x_0))}{(n+1)!}(x-x_0{)}^{n+1}(0<\theta <1)\end{array} \\ \text{意义：可用n次多项式来近似表达函数f(x), 且误差是当}x\to x_0\text{ 时比}(x-x_0{)}^n\text{高阶无穷小} \end{gathered}$$

一些常用的泰勒展开式：
$$\begin{gathered} sinx=x-\frac{1}{3!}{x}^3+\frac{1}{5!}{x}^5+\cdots +(-1{)}^n\frac{x^{2n+1}}{(2n+1)!}=\sum _{n=0}^{\infty }(-1{)}^n\frac{x^{2n+1}}{(2n+1)!} \\ cosx=1-\frac{1}{2!}{x}^2+\frac{1}{4!}{x}^4+\cdots +(-1{)}^n\frac{x^{2n}}{(2n)!}=\sum _{n=0}^{\infty }(-1{)}^n\frac{x^{2n}}{(2n)!} \\ arcsinx=x+\frac{1}{3!}{x}^3+o(x^3) \\ tanx=x+\frac{1}{3}{x}^3+o(x^3) \\ arctanx=x-\frac{1}{3}{x}^3+o(x^3) \\ ln(1+x)=x-\frac{1}{2}{x}^2+\frac{1}{3}{x}^3-\cdots +(-1{)}^{n-1}\frac{x^n}{n}=\sum _{n=1}^{\infty }(-1{)}^{n-1}\frac{x^n}{n}(-1<x\le 1) \\ {e}^x=1+\frac{x}{1!}+\frac{x^2}{2!}+\frac{x^3}{3!}+\cdots +\frac{x^n}{n!}=\sum _{n=0}^{\infty }\frac{x^n}{n!} \\ (1+x{)}^a=1+\frac{a}{1!}x+\frac{a(a-1)}{2!}{x}^2+\frac{a(a-1)(a-2)}{3!}{x}^3+o(x^3) \\ \frac{1}{1-x}=1+x+x^2+x^3+\cdots +x^n=\sum _{n=0}^{\infty }{x}^n(-1<x<1) \\ \frac{1}{1+x}=1-x+x^2-x^3+\cdots +(-1{)}^n{x}^n=\sum _{n=0}^{\infty }(-1{)}^n{x}^n(-1<x<1) \\ {a}^x=e^{ln(a^x)}=e^{(xlna)}=\sum \frac{(xlna{)}^n}{n!}{a}^x-1=xlna \end{gathered}$$

等价无穷小

$$\begin{gathered} lnx=ln(1+x-1)\sim x-1 \\ lnu\sim u-1,(u\to 0) \end{gathered}$$

高等数学中的
β=O(α)表示β是α的同阶无穷小
β=o(α)表示β是α的高阶无穷小