# 泰勒公式的结构分析与应用Structure Analysis and Application of Taylor Expansion Theorem

DOI: 10.12677/PM.2021.114080, PDF, HTML, XML, 下载: 27  浏览: 95

Abstract: Taylor expansion theorem is an important tool to study the properties of functions with differential theory, and it is also a teaching difficulty. Through the analysis of the structure of Taylor expansion theorem, it is revealed that the core idea of mastering Taylor expansion formula is information mining of special points, which provides help for students to master Taylor expansion formula and its application.

1. 引言

2. 泰勒中值定理及其结构分析

$f\left(x\right)=f\left({x}_{0}\right)+{f}^{\prime }\left({x}_{0}\right)\left(x-{x}_{0}\right)+\frac{{{f}^{\prime }}^{\prime }\left({x}_{0}\right)}{2!}{\left(x-{x}_{0}\right)}^{2}+\cdots +\frac{{f}^{\left(n\right)}\left({x}_{0}\right)}{n!}{\left(x-{x}_{0}\right)}^{n}+{R}_{n}\left(x\right)$ (*)

(*)式中 ${p}_{n}\left(x\right)=f\left({x}_{0}\right)+{f}^{\prime }\left({x}_{0}\right)\left(x-{x}_{0}\right)+\frac{{{f}^{\prime }}^{\prime }\left({x}_{0}\right)}{2!}{\left(x-{x}_{0}\right)}^{2}\text{+}\cdots +\frac{{f}^{\left(n\right)}\left({x}_{0}\right)}{n!}{\left(x-{x}_{0}\right)}^{n}$

$f\left(x\right)=f\left({x}_{0}\right)+{f}^{\prime }\left({x}_{0}\right)\left(x-{x}_{0}\right)+\frac{{{f}^{\prime }}^{\prime }\left({x}_{0}\right)}{2!}{\left(x-{x}_{0}\right)}^{2}\cdots +\frac{{f}^{\left(n\right)}\left({x}_{0}\right)}{n!}{\left(x-{x}_{0}\right)}^{n}+{R}_{n}\left(x\right)$ (**)

3. 泰勒公式的应用分析

3.1. 极限计算

(*)式的典型应用是用于求解由不同函数经过复合和四则运算形成的复杂函数的极限问题，这类极限利用泰勒公式，将不同的函数统一化成多项式的形式，实现了不同函数的形式统一，进而使极限的求解成为可能。

3.2. 中间导数估计

(**)式的典型应用是用于函数的中间导数估计，这种估计理论在现代分析学理论中具有非常重要的作用。下面，通过例子说明中间导数估计。

$f\left(x\right)=f\left(t\right)+{f}^{\prime }\left(t\right)\left(x-t\right)+\frac{1}{2}{f}^{″}\left(t\right){\left(x-t\right)}^{2}+\frac{1}{3!}{f}^{‴}\left(\xi \right){\left(x-t\right)}^{3}$

(先估计 ${f}^{\prime }\left(t\right)$，此时，必须消去 ${f}^{″}\left(t\right)$ 的影响，可以通过选取适当的、相互关联的 $x$，得到两个不同的展开式，也可视为关于 ${f}^{\prime }\left(t\right)$${f}^{″}\left(t\right)$ 的方程组，通过求解方程组达到目的)。

$f\left(t+a\right)=f\left(t\right)+a{f}^{\prime }\left(t\right)+\frac{1}{2}{f}^{″}\left(t\right){a}^{2}+\frac{1}{3!}{f}^{‴}\left({\xi }_{1}\right){a}^{3}$

$f\left(t-a\right)=f\left(t\right)-a{f}^{\prime }\left(t\right)+\frac{1}{2}{f}^{″}\left(t\right){a}^{2}-\frac{1}{3!}{f}^{‴}\left({\xi }_{2}\right){a}^{3}$

(从上述两个表达式中可以看出，在估计某个中间导数时，如何消去另一个中间导数的影响)。将两式相减得

$2a{f}^{\prime }\left(t\right)=f\left(t+a\right)-f\left(t\right)\left(t-a\right)+\frac{{a}^{3}}{3!}\left[{f}^{‴}\left({\xi }_{2}\right)-{f}^{‴}\left({\xi }_{1}\right)\right]$

${a}^{2}{f}^{″}\left(t\right)=f\left(t+a\right)+f\left(t-a\right)-2f\left(t\right)+\frac{{a}^{3}}{3!}\left[{f}^{‴}\left({\xi }_{1}\right)-{f}^{‴}\left({\xi }_{2}\right)\right]$

$a=1$ 即得 $|{f}^{″}\left(t\right)|\le 4{M}_{0}+\frac{1}{6}{M}_{1}$

3.3. 高阶导数中值估计

$f\left(x\right)=-1+\frac{1}{2}{f}^{″}\left(\xi \right){\left(x-{x}_{0}\right)}^{2}$

$x=0$，则存在 ${\xi }_{1}\in \left(0,{x}_{0}\right)$，使得

$0=f\left(0\right)=-1+\frac{1}{2}{f}^{″}\left({\xi }_{1}\right){\left(0-{x}_{0}\right)}^{2}$

$x=1$，则存在 ${\xi }_{2}\in \left({x}_{0},1\right)$，使得

$0=f\left(1\right)=-1+\frac{1}{2}{f}^{″}\left({\xi }_{1}\right){\left(1-{x}_{0}\right)}^{2}$

$f\left(x\right)=f\left(a\right)+\frac{1}{2}{f}^{″}\left(\xi \right){\left(x-a\right)}^{2}$

$f\left(x\right)=f\left(b\right)+\frac{1}{2}{f}^{″}\left(\xi \right){\left(x-b\right)}^{2}$

$f\left(\frac{a+b}{2}\right)=f\left(a\right)+{f}^{\prime }\left(a\right)\left(\frac{a+b}{2}-a\right)+\frac{1}{2}{f}^{″}\left({\xi }_{1}\right){\left(\frac{a+b}{2}-a\right)}^{2}$

$f\left(\frac{a+b}{2}\right)=f\left(b\right)+{f}^{\prime }\left(b\right)\left(\frac{a+b}{2}-b\right)+\frac{1}{2}{f}^{″}\left({\xi }_{2}\right){\left(\frac{a+b}{2}-b\right)}^{2}$

$f\left(b\right)-f\left(a\right)=\frac{1}{2}\left[{f}^{″}\left({\xi }_{2}\right)-{f}^{″}\left({\xi }_{1}\right)\right]{\left(\frac{b-a}{2}\right)}^{2}$

${f}^{″}\left({\xi }_{2}\right)-{f}^{″}\left({\xi }_{1}\right)=\frac{8}{{\left(b-a\right)}^{2}}\left[f\left(b\right)-f\left(a\right)\right]$

$|{f}^{″}\left(\xi \right)|=\mathrm{max}\left\{|{f}^{″}\left({\xi }_{1}\right)|,|{f}^{″}\left({\xi }_{2}\right)|\right\}$，则

$2|{f}^{″}\left(\xi \right)|\ge |{f}^{″}\left({\xi }_{2}\right)-{f}^{″}\left({\xi }_{1}\right)|=\frac{8}{{\left(b-a\right)}^{2}}|f\left(b\right)-f\left(a\right)|.$

4. 简单小结

 [1] 同济大学数学系. 《高等数学》[M]. 北京: 高等教育出版社, 2014. [2] 王耀革, 崔国忠, 郭从洲. 利用“结构分析-形式统一法”求解数学题目[J]. 理论数学, 2020(5): 524-529. https://doi.org/10.12677/PM.2020.105064 [3] 崔国忠, 石金娥, 郭从洲. 数学分析[M]. 北京: 科学出版社, 2018.