微分学＜6＞——Taylor公式

Taylor公式

Taylor公式的定性分析

定理6.1 Taylor公式(Peano余项)

若函数$f\left(x\right)$在$x_0$处的$n$阶导数均存在,则存在邻域$U\left(x_0,\delta \right)$使得$f\left(x\right)$满足以下公式:
$f\left(x\right)={\sum }_{k=0}^n\frac{f^{\left(k\right)}\left(x_0\right){\left(x-x_0\right)}^k}{k!}+r_n\left(x\right)$
其中Peano余项$r_n\left(x\right)=o\left({\left(x-x_0\right)}^n\right)(x\to x_0)$,代表$f\left(x\right)$在Taylor近似时产生的误差,余下的$n$次多项式记为$P_n\left(x\right)$,即$x_0$处的$n$次Taylor多项式。

$r_n\left(x\right)=f\left(x\right)-P_n\left(x\right)=f\left(x\right)-{\sum }_{k=0}^n\frac{f^{\left(k\right)}\left(x_0\right){\left(x-x_0\right)}^k}{k!}$,
$$\begin{array}{lc}& r_n^{\left(n-1\right)}\left(x\right)\\ =& f^{\left(n-1\right)}\left(x\right)-{\sum }_{k=n-1}^n\frac{f^{\left(k\right)}\left(x_0\right){\left(x-x_0\right)}^{\left(k-\left(n-1\right)\right)}}{\left(k-\left(n-1\right)\right)!}\\ =& f^{\left(n-1\right)}\left(x\right)-\left(\frac{f^{\left(n-1\right)}\left(x_0\right)}{0!}+\frac{f^{\left(n\right)}\left(x_0\right)\left(x-x_0\right)}{1!}\right)\\ =& f^{\left(n-1\right)}\left(x\right)-f^{\left(n-1\right)}\left(x_0\right)-f^{\left(n\right)}\left(x_0\right)\left(x-x_0\right)\end{array}$$,
因为函数$f\left(x\right)$在$x_0$处的$n$阶导数均存在,所以$f\left(x\right)$的前$n-1$阶导函数一定连续且可导。
连续引用n-1次L’Hospital法则,${\lim }_{x\to x_0}\frac{r_n\left(x\right)}{{\left(x-x_0\right)}^n}={\lim }_{x\to x_0}\frac{r_n^{\left(1\right)}\left(x\right)}{n{\left(x-x_0\right)}^{n-1}}=\cdots ={\lim }_{x\to x_0}\frac{r_n^{\left(n-1\right)}\left(x\right)}{n!\left(x-x_0\right)}$,
展开$r_n^{\left(n-1\right)}\left(x\right)$,${\lim }_{x\to x_0}\frac{r_n^{\left(n-1\right)}\left(x\right)}{n!\left(x-x_0\right)}=\frac{1}{n!}{\lim }_{x\to x_0}\left(\frac{f^{\left(n-1\right)}\left(x\right)-f^{\left(n-1\right)}\left(x_0\right)}{x-x_0}-f^{\left(n\right)}\left(x_0\right)\right)$,
根据高阶导数定义,${\lim }_{x\to x_0}\frac{f^{\left(n-1\right)}\left(x\right)-f^{\left(n-1\right)}\left(x_0\right)}{x-x_0}=f^{\left(n\right)}\left(x_0\right)$,
所以${\lim }_{x\to x_0}\frac{r_n\left(x\right)}{{\left(x-x_0\right)}^n}={\lim }_{x\to x_0}\frac{r_n^{\left(n-1\right)}\left(x\right)}{n!\left(x-x_0\right)}=0$,$r_n\left(x\right)=o\left({\left(x-x_0\right)}^n\right)(x\to x_0)$。

Taylor公式的定量分析

定理6.2 Taylor公式(Lagrange余项)

函数$f\left(x\right)$在闭区间$\left[a,b\right]$上存在$n$阶连续导数,在开区间$\left(a,b\right)$上存在$n+1$阶导数,取一定点$x_0\in \left[a,b\right]$,则$\forall x\in \left[a,b\right]$,$f\left(x\right)={\sum }_{k=0}^n\frac{f^{\left(k\right)}\left(x_0\right){\left(x-x_0\right)}^k}{k!}+r_n\left(x\right)$,其中Lagrange余项$r_n\left(x\right)=\frac{f^{\left(n+1\right)}\left(\xi \right){\left(x-x_0\right)}^{n+1}}{\left(n+1\right)!}$,其中$\xi$位于$x$和$x_0$之间。
当$n=0$时,Taylor公式表现为Lagrange中值定理的形式。

构造函数$G\left(t\right)=f\left(x\right)-{\sum }_{k=0}^n\left(\frac{f^{\left(k\right)}\left(t\right)}{k!}{\left(x-t\right)}^k\right)$和$H\left(t\right)={\left(x-t\right)}^{n+1}$,
不妨设$x<x_0$,
$G\left(x\right)=f\left(x\right)-{\sum }_{k=1}^n\left(\frac{f^{\left(k\right)}\left(t\right)}{k!}{\left(x-x\right)}^k\right)-\frac{f\left(t\right)}{0!}{\left(x-t\right)}^0=f\left(x\right)-f\left(x\right)=0$;
$G\left(x_0\right)=f\left(x\right)-{\sum }_{k=0}^n\left(\frac{f^{\left(k\right)}\left(x_0\right)}{k!}{\left(x-x_0\right)}^k\right)$;
$H\left(x\right)={\left(x-x\right)}^{n+1}=0$;
$H\left(x_0\right)={\left(x-x_0\right)}^{n+1}$;
$$\begin{array}{lc}& G^{\prime }\left(t\right)\\ =& -{\sum }_{k=0}^n{\left(\frac{f^{\left(k\right)}\left(t\right){\left(x-t\right)}^k}{k!}\right)}^{\prime }\\ =& {\sum }_{k=1}^n\frac{f^{\left(k\right)}\left(t\right){\left(x-t\right)}^{k-1}}{\left(k-1\right)!}-{\sum }_{k=1}^{n+1}\frac{f^{\left(k\right)}\left(t\right){\left(x-t\right)}^{\left(k-1\right)}}{\left(k-1\right)!}\\ =& -\frac{f^{\left(n+1\right)}\left(t\right){\left(x-t\right)}^n}{n!}\end{array}$$
$H^{\prime }\left(t\right)=-\left(n+1\right){\left(x-t\right)}^n$;
引用Cauchy中值定理,$\exists \xi \in \left(x,x_0\right)$,$\frac{r_n\left(x\right)}{{\left(x-x_0\right)}^{n+1}}=\frac{G\left(x_0\right)}{H\left(x_0\right)}=\frac{G\left(x_0\right)-G\left(x\right)}{H\left(x_0\right)-H\left(x\right)}=\frac{G^{\prime }\left(\xi \right)}{H^{\prime }\left(\xi \right)}=\frac{f^{\left(n+1\right)}\left(\xi \right)}{\left(n+1\right)!}\Rightarrow r_n\left(x\right)=\frac{f^{\left(n+1\right)}\left(\xi \right){\left(x-x_0\right)}^{n+1}}{\left(n+1\right)!}$。