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How do I compute the absolute error n = |xn − xr| for n = 0, 1, . . . , 50, where we take the output xr from MATLAB’s fzero function with initial guess xinit = 1 to be the “true” root, given this.
xn = bisectionMethod(f, a, b, numiter)
f1= @(x) cos(x)-x;
f2= @(x) exp(-x^2)-x;
f3= @(x) (x^3)-(1/2);
x1= bisectionMethod(f1, 0, 1, 50);
x2= bisectionMethod(f2, 0, 1, 50);
x3= bisectionMethod(f3, 0, 1, 50);

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Ameer Hamza
Ameer Hamza 2020-10-8
编辑：Ameer Hamza 2020-10-8

0 个投票

Consider one function
f1 = @(x) cos(x)-x;
x1r = fzero(f1, 0);
You can do it like this
n = 0:50;
x1n = zeros(size(n)); % all solutions for f1
for i = 1:numel(x1n)
x1n(i) = bisectionMethod(f1, 0, 1, n(i));
end
err = abs(x1n-z1r);
A more efficient approach is to modify bisectionMethod() function such that it returns a complete vector in a single call. The above code is repeating the same calculations several times.

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