Matlab-6:解非线性方程组newton迭代法

函数文件:

 1 function x=newton_Iterative_method(f,n,Initial)
 2 x0=Initial;
 3 tol=1e-11;
 4 x1=x0-Jacobian(f,n,x0)F(f,x0);
 5 while (norm(x1-x0,2)>tol)
 6     %数值解的2范数是否在误差范围内
 7     x0=x1;
 8     x1=x0-Jacobian(f,n,x0)F(f,x0);
 9 end
10 x=x1;%不动点
11 function g=Jacobian(f,n,a)
12 %求解任意矩阵的雅可比矩阵
13 %%
14 syms x y;
15  F(1:n,1)=jacobian(f,x);
16   F(1:n,2)=jacobian(f,y);
17  g=vpa(subs(F,{'x','y'},{a(1),a(2)},6));
18  %%
19     function h=F(f,a)
20   h=vpa(subs(f,{'x','y'},{a(1),a(2)}));

脚本文件:

tic;
clear
clc
syms x y;
h='[x^2+y^2-4;x^2-y^2-1]';
initial_value=[1.6;1.2];
n=2;%方程组的未知数的个数
g=newton_Iterative_method(h,n,initial_value)
toc;

算法推导:

原文地址:https://www.cnblogs.com/xtu-hudongdong/p/6506045.html