【数学】泰勒公式推导(佩亚诺余项)

每次忘了都要查半天，自己推导一遍记录下来

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Step 1

设$f(x)$存在n阶导数，根据一阶导数定义：

$\underset{x\to x_0}{\lim }\frac{f(x)-f(x_0)}{x-x_0}=f^{\prime }(x)$

等式脱帽
$\frac{f(x)-f(x_0)}{x-x_0}=f^{\prime }(x_0)+o$
$f(x)-f(x_0)=(x-x_0)f^{\prime }(x_0)+o(x-x_0)$

化简得
$\{\begin{array}{l}f(x)=f(x_0)+(x-x_0)f^{\prime }(x_0)+o(x-x_0)（1）\\ o(x-x_0)=f(x)-f(x_0)-(x-x_0)f^{\prime }(x_0)（2）\end{array}$

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Step 2

根据式$(2)$
$\underset{x\to x_0}{\lim }\frac{o(x-x_0)}{(x-x_0{)}^2}=\underset{x\to x_0}{\lim }\frac{f(x)-f(x_0)-(x-x_0)f^{\prime }(x_0)}{(x-x_0{)}^2}$ $\left(\frac{0}{0}型\right)$

使用洛必达法则
$\underset{x\to x_0}{\lim }\frac{o(x-x_0)}{(x-x_0{)}^2}=\underset{x\to x_0}{\lim }\frac{f^{\prime }(x)-f^{\prime }(x_0)}{2(x-x_0)}=\underset{x\to x_0}{\lim }\frac{f^{\prime \prime }(x)}{2}=\frac{f^{\prime \prime }(x_0)}{2}(常数)$

所以得出$o(x-x_0)$和$(x-x_0{)}^2$是等价无穷小
$\frac{o(x-x_0)}{(x-x_0{)}^2}=\frac{1}{2}{f}^{\prime \prime }(x_0)+o$
$o(x-x_0)=\frac{1}{2}(x-x_0{)}^2{f}^{\prime \prime }(x_0)+o(x-x_0{)}^2$

带入式$(1)$中
$\{\begin{array}{l}f(x)=f(x_0)+(x-x_0)f^{\prime }(x_0)+\frac{1}{2}(x-x_0{)}^2{f}^{\prime \prime }(x_0)+o(x-x_0{)}^2（3）\\ o(x-x_0{)}^2=f(x)-f(x_0)-(x-x_0)f^{\prime }(x_0)-\frac{1}{2}(x-x_0{)}^2{f}^{\prime \prime }(x_0)（4）\end{array}$

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Step 3

根据式$(4)$
$\underset{x\to x_0}{\lim }\frac{o(x-x_0{)}^2}{(x-x_0{)}^3}=\underset{x\to x_0}{\lim }\frac{f(x)-f(x_0)-(x-x_0)f^{\prime }(x_0)-\frac{1}{2}(x-x_0{)}^2{f}^{\prime \prime }(x_0)}{(x-x_0{)}^3}$ $\left(\frac{0}{0}型\right)$

使用洛必达法则
$\underset{x\to x_0}{\lim }\frac{o(x-x_0{)}^2}{(x-x_0{)}^3}\stackrel{2次}{}\underset{x\to x_0}{\lim }\frac{f^{\prime \prime }(x)-f^{\prime \prime }(x_0)}{3\cdot 2(x-x_0)}=\underset{x\to x_0}{\lim }\frac{f^{\prime \prime \prime }(x)}{6}=\frac{f^{\prime \prime \prime }(x_0)}{6}(常数)$

所以得出$o(x-x_0{)}^2$和$(x-x_0{)}^3$是等价无穷小
$\frac{o(x-x_0{)}^2}{(x-x_0{)}^3}=\frac{1}{6}{f}^{\prime \prime \prime }(x_0)+o$
$o(x-x_0{)}^2=\frac{1}{6}(x-x_0{)}^3{f}^{\prime \prime \prime }(x_0)+o(x-x_0{)}^3$

带入式$(1)$中
$\{\begin{array}{l}f(x)=f(x_0)+(x-x_0)f^{\prime }(x_0)+\frac{1}{2}(x-x_0{)}^2{f}^{\prime \prime }(x_0)+\frac{1}{6}(x-x_0{)}^3{f}^{\prime \prime \prime }(x_0)+o(x-x_0{)}^3（5）\\ o(x-x_0{)}^3=f(x)-f(x_0)-(x-x_0)f^{\prime }(x_0)-\frac{1}{2}(x-x_0{)}^2{f}^{\prime \prime }(x_0)-\frac{1}{6}(x-x_0{)}^3{f}^{\prime \prime \prime }(x_0)（6）\end{array}$

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结论

以此类推
$o(x-x_0{)}^{n-1}=\frac{1}{n!}(x-x_0{)}^n{f}^{(n)}(x_0)+o(x-x_0{)}^n$
$f(x)=f(x_0)+(x-x_0)f^{\prime }(x_0)+\frac{1}{2}(x-x_0{)}^2{f}^{\prime \prime }(x_0)+\frac{1}{6}(x-x_0{)}^3{f}^{\prime \prime \prime }(x_0)+\cdots +\frac{1}{n!}(x-x_0{)}^n{f}^{(n)}(x_0)+o(x-x_0{)}^n$

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常用麦克劳林展开式

$（1）e^x=1+x+\frac{x^2}{2!}+\cdots +\frac{x^n}{n!}+o(x^n)=\sum _{n=0}^{\infty }\frac{x^n}{n!}$
$（2）\sin x=x-\frac{x^3}{3!}+\cdots +(-1{)}^n\frac{x^{2n+1}}{(2n+1)!}+o(x^{2n+1})=\sum _{n=0}^{\infty }(-1{)}^n\frac{x^{2n+1}}{(2n+1)!}$
$（3）\cos x=1-\frac{x^2}{2!}+\cdots +(-1{)}^n\frac{x^{2n}}{(2n)!}+o(x^{2n})=\sum _{n=0}^{\infty }(-1{)}^n\frac{x^{2n}}{(2n)!}$
$（4）\ln (1+x)=x-\frac{x^2}{2}+\cdots +(-1{)}^{n-1}\frac{x^n}{(n)}+o(x^n)=\sum _{n=1}^{\infty }(-1{)}^{n-1}\frac{x^n}{n}.(x>-1)$ [注]: 从n=1开始
$（5）\frac{1}{1-x}=1+x+x^2+\cdots +x^n+o(x^n)=\sum _{n=0}^{\infty }{x}^n$
$（6）\frac{1}{1+x}=1-x+x^2-\cdots +(-1{)}^n{x}^n+o(x^n)=\sum _{n=0}^{\infty }(-1{)}^n{x}^n$
$（7）(1+x{)}^{\alpha }=1+\alpha x+\frac{\alpha (\alpha -1)}{2}{x}^2+o(x^2)(x\to 0)$
$（8）\tan x=x+\frac{1}{3}{x}^3+o(x^3)(x\to 0)$
$（9）\arcsin x=x+\frac{1}{6}{x}^3+o(x^3)(x\to 0)$
$（10）\arctan x=x-\frac{1}{3}{x}^3+o(x^3)(x\to 0)$