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

Why do you say it's a bug?
Not where the minimum is! look at the plot
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

Sign in to comment.

 Accepted Answer

Walter Roberson
Walter Roberson on 20 Jan 2019

0 votes

No bug. fminbnd is a local minimizer, not a global minimizer.

3 Comments

Hi Roberson,
I have a similar problem with fminbnd, see code below
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])
[xmin, min]=fminbnd(f, x1, x2)
xmin = 1.5712e-06
min = -679.5775
We could see from the plot the minimum value should be around -996 and there should be only one local minimum. But fminbnd returns -679.
Thanks a lot in advance!
Best Regards,
Zhe
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
Hi Torsten, thank you very much!

Sign in to comment.

More Answers (0)

Products

Release

R2018b

Tags

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!