fminbnd finding wrong minimum point for no reason, please help

2 visualizaciones (últimos 30 días)
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.

Respuesta aceptada

Andrei Bobrov
Andrei Bobrov el 19 de Oct. de 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 comentarios
Andrei Bobrov
Andrei Bobrov el 19 de Oct. de 2011
due to the choice of x
Andrei Bobrov
Andrei Bobrov el 19 de Oct. de 2011
use
ezplot(f,[-1,2,f(x1)-.2,0])

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Más respuestas (1)

Walter Roberson
Walter Roberson el 19 de Oct. de 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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