MATLAB实例：非线性方程数值解法(迭代解)
作者：凯鲁嘎吉 - 博客园 http://www.cnblogs.com/kailugaji/
很久之前写过一篇关于“MATLAB用二分法、不动点迭代法及Newton迭代（切线）法求非线性方程的根”，本博文相当于之前这一篇的延续与拓展，介绍四种求解一元非线性方程的数值解法(迭代解)，包括：牛顿迭代法，Halley迭代法，Householder迭代法以及预测校正牛顿-哈雷迭代法(Predictor-Corrector Newton-Halley，PCNH)，具体参考文献[1]，来源于这篇文章：THREE-STEP ITERATIVE METHOD WITH EIGHTEENTH ORDER CONVERGENCE FOR SOLVING NONLINEAR EQUATIONS。
1. 迭代更新公式
2. MATLAB程序
newton.m
function [x1, k]=newton(t1,esp,m)
syms x;
fun=x^3+4*(x^2)-10;
for k=1:m
if abs(subs(diff(fun,\'x\'),x,t1))<esp
x1=t1;
break;
else
if subs(diff(fun,\'x\',2),x,t1)==0
break;
disp(\'解题失败！\')
else
t0=t1;
t1=t0-subs(fun,x,t0)/subs(diff(fun,\'x\'),x,t0);
if abs(t1-t0)<esp
x1=t1;
break;
end
end
end
end
% x1=vpa(x1,15);
halley.m
function [x1, k]=halley(t1,esp,m)
syms x;
fun=x^3+4*(x^2)-10;
for k=1:m
if abs(subs(diff(fun,\'x\'),x,t1))<esp
x1=t1;
break;
else
if subs(diff(fun,\'x\',2),x,t1)==0
break;
disp(\'解题失败！\')
else
t0=t1;
t1=t0-(2*subs(fun,x,t0)*subs(diff(fun,\'x\'), x, t0))/(2*(subs(diff(fun,\'x\'), x, t0))^2-subs(fun, x, t0)*subs(diff(fun,\'x\',2),x,t0));
if abs(t1-t0)<esp
x1=t1;
break;
end
end
end
end
% x1=vpa(x1,15);
householder.m
function [x1, k]=householder(t1,esp,m)
syms x;
fun=x^3+4*(x^2)-10;
for k=1:m
if abs(subs(diff(fun,\'x\'),x,t1))<esp
x1=t1;
break;
else
if subs(diff(fun,\'x\',2),x,t1)==0
break;
disp(\'解题失败！\')
else
t0=t1;
t1=t0-(subs(fun, x, t0))/(subs(diff(fun,\'x\'),x,t0))-(((subs(fun, x, t0))^2)*subs(diff(fun,\'x\',2),x,t0))/(2*(subs(diff(fun,\'x\',2),x,t0))^3);
if abs(t1-t0)<esp
x1=t1;
break;
end
end
end
end
% x1=vpa(x1,15);
PCNH.m
function [x1, k]=PCNH(t1,esp,m)
syms x;
fun=x^3+4*(x^2)-10;
for k=1:m
if abs(subs(diff(fun,\'x\'),x,t1))<esp
x1=t1;
break;
else
if subs(diff(fun,\'x\',2),x,t1)==0
break;
disp(\'解题失败！\')
else
t0=t1;
w=t0-subs(fun,x,t0)/subs(diff(fun,\'x\'),x,t0);
y=w-(2*subs(fun,x,w)*subs(diff(fun,\'x\'), x, w))/(2*(subs(diff(fun,\'x\'), x, w))^2-subs(fun, x, w)*subs(diff(fun,\'x\',2),x,w));
t1=y-(subs(fun, x, y))/(subs(diff(fun,\'x\'),x,y))-(((subs(fun, x, y))^2)*subs(diff(fun,\'x\',2),x,y))/(2*(subs(diff(fun,\'x\',2),x,y))^3);
if abs(t1-t0)<esp
x1=t1;
break;
end
end
end
end
% x1=vpa(x1,15);
demo.m
clear
clc
% Input: 初始值，迭代终止条件，最大迭代次数
[x1, k1]=newton(1,1e-4,20); % 牛顿迭代法
[x2, k2]=halley(1,1e-4,20); % Halley迭代法
[x3, k3]=householder(1,1e-4,20); % Householder迭代法
[x4, k4]=PCNH(1,1e-4,20); % 预测校正牛顿-哈雷迭代法(PCNH)
fprintf(\'牛顿迭代法求解得到的方程的根为：%.15f, 实际迭代次数为：%d次\n\', x1, k1);
fprintf(\'Halley迭代法求解得到的方程的根为：%.15f, 实际迭代次数为：%d次\n\', x2, k2);
fprintf(\'Householder迭代法求解得到的方程的根为：%.15f, 实际迭代次数为：%d次\n\', x3, k3);
fprintf(\'预测校正牛顿-哈雷迭代法(PCNH)求解得到的方程的根为：%.15f, 实际迭代次数为：%d次\n\', x4, k4);
%% 函数图像
x=-5:0.01:5;
y=x.^3+4.*(x.^2)-10;
y_0=zeros(length(x));
plot(x, y, \'r-\', x, y_0, \'b-\');
xlabel(\'x\');
ylabel(\'f(x)\');
title(\'f(x)=x^3+4{x^2}-10\');
saveas(gcf,sprintf(\'函数图像.jpg\'),\'bmp\'); %保存图片
3. 数值结果
求解$f(x)=x^3+4{x^2}-10=0$方程在$x_0=1$附近的根。
牛顿迭代法求解得到的方程的根为：1.365230013435367, 实际迭代次数为：4次
Halley迭代法求解得到的方程的根为：1.365230013414097, 实际迭代次数为：3次
Householder迭代法求解得到的方程的根为：1.365230013391664, 实际迭代次数为：3次
预测校正牛顿-哈雷迭代法(PCNH)求解得到的方程的根为：1.365230013414097, 实际迭代次数为：2次
函数图像：
4. 参考文献
[1] Bahgat, Mohamed & Hafiz, Mohammad. (2014). THREE-STEP ITERATIVE METHOD WITH EIGHTEENTH ORDER CONVERGENCE FOR SOLVING NONLINEAR EQUATIONS. International Journal of Pure and Applied Mathematics. 93.
