【数值分析】迭代法求方程的根（附matlab代码）

题目

用迭代法求方程 $exp(x)+10\ast x-2=0$ 的根，要求根有3位小数，初值 $x_0=0$

解析

迭代方程：$x_{k+1}=(2-e^{x_k})/10$
当 $x_k\in [0,0.5]$ 时，$\phi (x_k)=(2-e^{x_k})/10\in [0,0.5]$
$|{\phi }^{\prime }(x_k)|=|-e^{x_k}|/10\le L<1，L=e^{0.5}/10=0.165$
故 $x_{k+1}=(2-e^{x_k})/10$ 在[0, 0.5]上整体收敛
由 $|x_k-x\ast |\le |x_k-x_{k+1}|\ast L/(1-L)<10^{-4}/2$
得：$|x_k-x_{k+1}|＜((1-L)/L)\ast 10^{-4}/2<10^{-6}$

Matlab代码

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% 简介：用迭代法求方程exp(x) + 10*x -2=0的根，要求根有3位小数
% 作者：不雨_亦潇潇
% 文件：dichotomy2.m
% 日期：20221013
% 博客：https://blog.csdn.net/weixin_43470383/article/details/127222948
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
clc; clear all;
syms x                              
x0 = 0;
x1 = (2-exp(x0)) / 10;
n = 1;

while abs(x0-x1)>10^(-6)
    x0 = x1;
    x1 = (2-exp(x0)) / 10
    n = n+1
end

x0

运行结果

x1 =
0.089482908192435
n =
2
x1 =
0.090639135859584
n =
3
x1 =
0.090512616674365
n =
4
x1 =
0.090526468052644
n =
5
x1 =
0.090524951682840
n =
6
x1 =
0.090525117687371
n =
7
x0 =
0.090524951682840

$|x_6-x_5|=0.000001517<10^{-6}$
故$x\ast \approx x_6$