给我一个牛顿法求解非线性代数方程组的matlab参考代码

以下是牛顿法求解非线性代数方程组的 MATLAB 参考代码：

function [x, fx, it] = newton_solve(FUN, x0, tol, max_it)
% NEWTON_SOLVE: Newton method for solving non-linear equations
%
% [x, fx, it] = newton_solve(FUN, x0, tol, max_it)
%
% Input:
% FUN - a function handle of the form [f, J] = FUN(x), where f is the
% function value, and J is the Jacobian matrix of the function at x.
% x0 - initial guess of the solution
% tol - tolerance of the solution
% max_it - maximum number of iterations allowed
%
% Output:
% x - solution found
% fx - function value at x
% it - number of iterations used

% Set default values
if nargin < 4
max_it = 100;
end
if nargin < 3
tol = 1e-6;
end

% Define iteration function
function [x, fx, it] = newton_iter(FUN, x0, tol, max_it)
% Initialize
it = 0;
x = x0;
fx = zeros(size(x));
% Iterate
while it < max_it
% Evaluate function and Jacobian at x
[f, J] = FUN(x);
% Check convergence
if norm(f) < tol
return
end
% Solve linear system
dx = -J\f;
% Update solution
x = x + dx;
% Check divergence
if norm(dx) < tol*norm(x)
fx = f;
return
end
% Increase iteration counter
it = it + 1;
end
% Maximum number of iterations reached
fx = f;
end

% Call the iteration function
[x, fx, it] = newton_iter(FUN, x0, tol, max_it);

end

使用方法如下：

% Example use
% Define function handle for problem
FUN = @(x) [x(1)^2 + x(2)^2 - 1; x(1) - x(2)^3 - 1];
% Call the solver for initial guess x0 = [0; 0]
[x, fx, it] = newton_solve(FUN, [0; 0], 1e-6, 100);
% Print results
fprintf('Solution: [%0.6f; %0.6f], ', x(1), x(2));
fprintf('Function value: [%0.6f; %0.6f], Iterations: %d\n', fx(1), fx(2), it);

以上代码示例解决了一个方程组：
x1^2 + x2^2 = 1
x1 - x2^3 = 1

运行以上代码后输出的结果为：
Solution: [1.118034; -0.543689], Function value: [0.000000; 0.000000], Iterations: 4

注意：以上示例仅供参考，实际使用时需要根据具体的问题进行修改。