帕德近似，原理与Matlab实现，指数函数的Pade近似

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基本原理

  泰勒展开以多项式的形式对$f(x)$进行逼近，帕德近似则是以有理分式的形式对目标函数$f(x)$进行逼近。帕德近似的基本形式如下所示（以零点为邻域）：
$$R_{m,n}\left(x\right)=\frac{p\left(x\right)}{q\left(x\right)}=\frac{p_0+p_{1}x+p_2{x}^2+\dots +p_m{x}^m}{1+q_{1}x+q_2{x}^2+\dots +q_n{x}^n}$$

其中，$p(x)$和$q(x)$分别为$m$和$n$阶多项式，$R_{m,n}\left(x\right)$也被称作$f(x)$的$[m,n]$阶帕德近似。注意，$q(x)$的常数项为$1$。

求解方法1

  利用$R_{m,n}(x)$和$f(x)，$在$x=0$处，前$m+n$阶导数与$f(x)$对应项相等，求解帕德近似的系数：
$$\{\begin{array}{c}{R}_{m,n}\left(x_0\right)=f\left(x_0\right)\\ R_{m,n}^{{}^{\prime }}\left(x_0\right)=f^{\prime }\left(x_0\right)\\ R_{m,n}^{{}^{\prime \prime }}\left(x_0\right)=f^{\prime \prime }\left(x_0\right)\\ ⋮\\ R_{m,n}^{\left(m+n\right)}\left(x_0\right)=f^{\left(m+n\right)}\left(x_0\right)\end{array}\to \{\begin{array}{c}{p}_0,p_1,p_2,\dots +p_m\\ q_1,q_2,\dots ,q_n\end{array}$$

  示例：求解$f\left(x\right)=e^x$的$[1,1]$阶帕德近似：
$$\{\begin{array}{l}{R}_{1,1}\left(0\right)=\frac{p_0+p_{1}x}{1+q_{1}x}=\frac{p_0}{1}=1\\ R_{1,1}^{\prime }\left(0\right)=\frac{p_1-p_0{q}_1}{{\left(1+q_{1}x\right)}^2}=\frac{p_1-p_0{q}_1}{{\left(1\right)}^2}=1\\ R_{1,1}^{\prime \prime }\left(0\right)=\frac{-2\left(p_1-p_0{q}_1\right)\left(1+q_{1}x\right)q_1}{{\left(1+q_{1}x\right)}^2}=\frac{-2\left(p_1-p_0{q}_1\right)q_1}{{\left(1\right)}^2}=1\end{array}\to \{\begin{array}{l}{p}_0=1\\ p_1=0.5\\ q_1=-0.5\end{array}$$

  于是，$f\left(x\right)=e^x$的$[1,1]$阶帕德近似表达式为：
$$R_{1,1}\left(x\right)=\frac{1+0.5x}{1-0.5x}$$

求解方法2

  利用$f(x)$在$x=0$处，$m+n$阶泰勒展开表达式，求解帕德近似$R_{m,n}(x)$的系数：
$$\{\begin{array}{l}f\left(x\right)=p\left(x\right)+O\left(x^{m+n}\right)=f\left(0\right)+\frac{f^{\prime }\left(0\right)}{1!}x+\frac{f^{\prime \prime }\left(0\right)}{2!}{x}^2+\dots +\frac{f^{\left(m+n\right)}\left(0\right)}{n!}{x}^{m+n}+O\left(x^{m+n}\right)\\ f\left(x\right)=R_{m,n}\left(x\right)+O\left(x^{m+n}\right)=\frac{p_0+p_{1}x+p_2{x}^2+\dots +p_m{x}^m}{1+q_{1}x+q_2{x}^2+\dots +q_n{x}^n}+O\left(x^{m+n}\right)\end{array}$$

  示例，求解$f\left(x\right)=e^x$的$[1,1]$阶帕德近似：
$$\{\begin{array}{l}{e}^x=1+x+\frac{1}{2}{x}^2+O\left(x^2\right)\\ e^x=\frac{p_0+p_{1}x}{1+q_{1}x}+O\left(x^2\right)\end{array}\to \{\begin{array}{l}1+x+0.5x^2=\frac{p_0+p_{1}x}{1+q_{1}x}+O\left(x^2\right)\\ \to 1+\left(1+q_1\right)x+\left(0.5+q_1\right)x^2=p_0+p_{1}x+O\left(x^2\right)\end{array}\to \{\begin{array}{l}1=p_0\\ 1+q_1=p_1\\ 0.5+q_1=0\end{array}\to \{\begin{array}{l}{p}_0=1\\ p_1=0.5\\ q_1=-0.5\end{array}$$

  于是，$f\left(x\right)=e^x$的$[1,1]$阶帕德近似表达式为：
$$R_{1,1}\left(x\right)=\frac{1+0.5x}{1-0.5x}$$

Matlab实现

  实际应用过程中，可以借助Matlab求解Pade近似表达式。Matlab求解指数函数的$[1,1]$、$[2,2]$、$[3,3]$阶Pade近似，代码如下：

clc; clear; syms x; pade(exp(x),x,0,'Order',[1 1]) pade(exp(x),x,0,'Order',[2 2]) pade(exp(x),x,0,'Order',[3 3]) 

clc;clear;close all; syms x;hold on;fplot(exp(x),'r-*','linewidth', 3) fplot(pade(exp(x),x,0,'Order',[1 1]),'g','linewidth', 1) fplot(pade(exp(x),x,0,'Order',[2 2]),'b','linewidth', 1) fplot(pade(exp(x),x,0,'Order',[3 3]),'k','linewidth', 1) grid on;axis([-2 2 0 8]); legend('epx(x)','[1,1]','[2,2]','[3,3]'); 

参考文献

[1] Matlab文档：Pade approximant.