5月18日

 代码

f = @(x) x.^2 - sin(x);

a = 0;
b = 1;

tol = 1e-6;

x1 = a + 0.382*(b-a);
x2 = a + 0.618*(b-a);

f1 = f(x1);
f2 = f(x2);

% Iterate until the interval [a,b] is smaller than the tolerance
while abs(b-a) > tol
    if f1 < f2
        b = x2;
        x2 = x1;
        f2 = f1;
        x1 = a + 0.382*(b-a);
        f1 = f(x1);
    else
        a = x1;
        x1 = x2;
        f1 = f2;
        x2 = a + 0.618*(b-a);
        f2 = f(x2);
    end
end

min_point = (a+b)/2;
min_value = f(min_point);
fprintf('Minimum point: %f\nMinimum value: %f\n', min_point, min_value);

结果