Detecting an error in Muller method.

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llucia
llucia 2023-4-7
评论： Torsten 2023-4-7
Hi! I am programming Muller method as I wan to find the complex roots of an ecuation.
However, it does not work properly and I cannot detect why.
If someone could help me, I would really appreciate it.
Thanks in advance.
Here is the code:
%Muller, mejora del método de la secante, nos permite sacar tanto reales
%como complejas.
f = @(x) x.^3 + 2*x.^2 + 10*x -20;
x0 = -1;
x1 = 0;
x2 = 1;
eps_x = 10e-10;
eps_f = 10e-10;
maxits = 100;
[x0,x1,x2,n,error] = muller(f,x0,x1,x2,eps_x, eps_f,maxits)
function [x0,x1,x2, n, error] = muller(f, x0, x1, x2, eps_x, eps_f, maxits)
x0 = -1;
x1 = 0;
x2 = 1;
n = 0;
error(1) = eps_x + 1;
while n < maxits && (error(n+1) > eps_x || abs(f(x0)) > eps_f)
c = f(x2);
b = ((x0 - x2)^2*(f(x1)-f(x2))-(x1-x2)^2*(f(x0)-f(x2)))/((x0-x2)*(x1-x2)*(x0-x1));
a = ((x1-x2)*(f(x0)-f(x2))-(x0-x2)*(f(x1)-f(x2)))/((x0-x2)*(x1-x2)*(x0-x1));
x3 = x2 - (2*c)/(b+sign(b)*sqrt(b*b-4*a*c));
x0 = x1;
x1 = x2;
x2 = x3;
n = n+1;
error(n+1) = abs(x2-x1);
end
fprintf ('Las raíces son %.6f %.6f %.6f', x0, x1,x2)
end
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llucia
llucia 2023-4-7
I have written an fprintf and I just get the real root but not the complex ones.
VBBV
VBBV 2023-4-7
That is due to usage of abs in this line
error(n+1) = abs(x2-x1);
%abs

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采纳的回答

Torsten
Torsten 2023-4-7
The argument of the root in the calculation of x2 must become negative.
Or simply start with complex values for x1,x2 and/or x3, e.g.
x0 = -1;
x1 = 0;
x2 = 3*1i;
  4 个评论
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llucia
llucia 2023-4-7
perfect. Thanks a lot. It works perfectly now.
Torsten
Torsten 2023-4-7
In numerical computations, you will never automatically get complex numbers if you don't start with complex numbers and if there are no expressions that can generate complex numbers (like the sqrt here).

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