Bug fminbnd not working

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Stephen Wilkerson
Stephen Wilkerson on 20 Jan 2019
Edited: Torsten on 17 Jul 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 Comments
Stephen Wilkerson
Stephen Wilkerson on 20 Jan 2019
Not where the minimum is! look at the plot
Stephen Wilkerson
Stephen Wilkerson on 20 Jan 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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Accepted Answer

Walter Roberson
Walter Roberson on 20 Jan 2019
No bug. fminbnd is a local minimizer, not a global minimizer.
  3 Comments
Torsten
Torsten on 17 Jul 2022
Edited: Torsten on 17 Jul 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 on 17 Jul 2022
Hi Torsten, thank you very much!

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