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Bug fminbnd not working

3 visualizaciones (últimos 30 días)
Stephen Wilkerson
Stephen Wilkerson el 20 de En. de 2019
Editada: Torsten el 17 de Jul. de 2022
fplot(@(x) x*(sin(x))^2*cos(x),[-2*pi 2*pi]);
[xMin1 fvalmin1] = fminbnd('-x*(sin(x))^2*cos(x)', -6, 6)
returns xMin1 = 1.0954
fvalmin1 = -0.3963
How is this possible, look at the plot?
  3 comentarios
Stephen Wilkerson
Stephen Wilkerson el 20 de En. de 2019
Not where the minimum is! look at the plot
Stephen Wilkerson
Stephen Wilkerson el 20 de En. de 2019
Solved, The function doesn't really do much, it give you the nearest point that's a minimum to it's starting point. In my case it woud be 0 as the starting point. Typical

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Respuesta aceptada

Walter Roberson
Walter Roberson el 20 de En. de 2019
No bug. fminbnd is a local minimizer, not a global minimizer.
  3 comentarios
Torsten
Torsten el 17 de Jul. de 2022
Editada: Torsten el 17 de Jul. de 2022
Since the changes in the x-values are in the order of 1e-6, you must choose a smaller value for TolX:
a=-200; b=-979.8997; c=7.1833e+05; d=24.4232;e=-6.6083;
x1=0; x2=4.1135e-06;
f=@(x)a-(a-b)*cos(c*(x-x1)) + d*e*sin(c*(x-x1))
f = function_handle with value:
@(x)a-(a-b)*cos(c*(x-x1))+d*e*sin(c*(x-x1))
fplot(f, [x1 x2])
options = optimset('TolX',1e-8);
[xmin, min]=fminbnd(f, x1, x2, options)
xmin = 2.8422e-07
min = -996.4246
Zhe Yu
Zhe Yu el 17 de Jul. de 2022
Hi Torsten, thank you very much!

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