【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 e x p ( x ) + 10 ∗ x − 2 = 0 exp(x) + 10*x -2=0exp(x)+10∗x−2=0 的根，要求根有3位小数，初值 x 0 = 0 x_0 =0x0=0

解析

迭代方程：x k + 1 = ( 2 − e x k ) / 10 x_{k+1}=(2-e^{x_{k}})/10xk+1=(2−exk)/10

当 x k ∈ [ 0 , 0.5 ] x_{k}∈[0, 0.5]xk∈[0,0.5] 时，φ ( x k ) = ( 2 − e x k ) / 10 ∈ [ 0 , 0.5 ] φ(x_{k})=(2-e^{x_{k}})/10 ∈[0, 0.5]φ(xk)=(2−exk)/10∈[0,0.5]

∣ φ ′ ( x k ) ∣ = ∣ − e x k ∣ / 10 ≤ L < 1 ， L = e 0.5 / 10 = 0.165 |φ'(x_{k})| =|-e^{x_k}|/10≤L<1，L=e^{0.5}/10=0.165∣φ′(xk)∣=∣−exk∣/10≤L<1，L=e0.5/10=0.165

故 x k + 1 = ( 2 − e x k ) / 10 x_{k+1}=(2-e^{x_{k}})/10xk+1=(2−exk)/10 在[0, 0.5]上整体收敛

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)>5*10^(-4)
    x0 = x1;
    x1 = (2-exp(x0)) / 10
    n = n+1
end
x1

运行结果

x1 =

0.089482908192435

n =

2

x1 =

0.090639135859584

n =

3

x1 =

0.090512616674365

n =

4

x1 =

0.090512616674365

x ∗ ≈ x 4 = 0.090512616674365 x*≈x_{4}=0.090512616674365x∗≈x4=0.090512616674365