分别用不动点迭代法、Newton 法、割线法求解函数f（x）=exp（x）-x-5的零点，编写MATLAB 程序，比较计算结果

不动点迭代法通常用于求解方程 \( g(x) = x - f(x) \) 的根，对于 \( f(x) = e^x - x - 5 \)，我们可以设置 \( g(x) = x \)。以下是简单的 MATLAB 代码实现不动点迭代：

function [root, iterations] = fixed_point_iterate(f, guess, tolerance, max_iters)
    root = guess;
    count = 0;
    
    while abs(f(root)) > tolerance && count < max_iters
        root = root - f(root);
        count = count + 1;
    end
    
    iterations = count;
end

% 函数f
f = @(x) exp(x) - x - 5;

% 初始猜测值
guess = 1; % 或者其他合理猜测值，例如0或1.618

% 设置迭代参数
tolerance = 1e-6;
max_iters = 1000;

[zero_fixed, iters_fixed] = fixed_point_iterate(f, guess, tolerance, max_iters);
disp(['Fixed Point Iteration Result: ', num2str(zero_fixed), ' (', num2str(iters_fixed), ' iterations)']);

接下来是 Newton 法（牛顿-拉弗森法），它需要导数信息：

function [newton_root] = newton_method(f, df, guess, tolerance, max_iters)
    if nargin < 4 || nargin > 5
        error('Invalid number of input arguments');
    end
    
    root = guess;
    count = 0;
    
    while abs(f(root)) > tolerance && count < max_iters
        newton_root = root - f(root) / df(root);
        root = newton_root;
        count = count + 1;
    end
    
    if count == max_iters
        disp(['Newton Method Converged after maximum iterations. Final result: ', num2str(newton_root)]);
    else
        disp(['Newton Method Result: ', num2str(newton_root), ' (', num2str(count), ' iterations)']);
    end
end

% 函数f的导数
df = @(x) exp(x) - 1;

[zero_newton, iters_newton] = newton_method(f, df, guess, tolerance, max_iters);

最后是割线法（也称为二分法），适用于单调函数：

function [zero_bisection, iters_bisection] = bisection_method(f, a, b, tolerance, max_iters)
    if nargin < 4 || nargin > 5
        error('Invalid number of input arguments');
    end
    
    if f(a) * f(b) >= 0
        disp('Bisection method requires the function to change sign in the interval.');
        return;
    end
    
    midpoint = (a + b) / 2;
    root = midpoint;
    count = 0;
    
    while abs(f(root)) > tolerance && count < max_iters
        if f(midpoint) * f(root) < 0
            b = root;
        else
            a = root;
        end
        
        root = (a + b) / 2;
        count = count + 1;
    end
    
    if count == max_iters
        disp(['Bisection Method Converged after maximum iterations. Final result: ', num2str(root)]);
    else
        disp(['Bisection Method Result: ', num2str(root), ' (', num2str(count), ' iterations)']);
    end
end

[zero_bisection, iters_bisection] = bisection_method(@(x) f(x), guess, guess*1.1, tolerance, max_iters);

现在你可以运行这三段代码，并比较它们的结果。每种方法可能会因为初始猜测、函数特性以及算法收敛速度而得出不同的数值和迭代次数。