fminbnd finding wrong minimum point for no reason, please help

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Emre
Emre le 19 Oct 2011
f = @(x)(-(1.0/((x-0.3).^2+0.01)+1.0/((x-0.9).^2+0.04)));
x = fminbnd(f, -1, 2);
x
y=f(x);
y
x = -1:0.3:2
y = -(1./((x-0.3).^2+0.01)+1./((x-0.9).^2+0.04))
plot(x,y)
here, minimum of the function should be smt between -50 and -60, but it finds -102.5014!! Why is that? please help.

Réponse acceptée

Andrei Bobrov
Andrei Bobrov le 19 Oct 2011
f = @(x)(-(1.0./((x-0.3).^2+0.01)+1.0./((x-0.9).^2+0.04)));
x1 = fminbnd(f, -1, 2);
x1
y=f(x1);
y
x = sort([-1:0.3:2,x1])
y = -(1./((x-0.3).^2+0.01)+1./((x-0.9).^2+0.04))
plot(x,y)
  3 commentaires
Andrei Bobrov
Andrei Bobrov le 19 Oct 2011
due to the choice of x
Andrei Bobrov
Andrei Bobrov le 19 Oct 2011
use
ezplot(f,[-1,2,f(x1)-.2,0])

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Plus de réponses (1)

Walter Roberson
Walter Roberson le 19 Oct 2011
As was indicated more than once in resposne to your previous question, you are trying to apply fminbnd() to a problem with multiple minima, but fminbnd() is not designed to handle such situations.
If you want to handle functions like the one above, you need a global minimizer.

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