How to minimize a nonlinear function?
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Rafael Zanetti
le 12 Nov 2020
Réponse apportée : Star Strider
le 12 Nov 2020
Hi everyone, I heve tried to solve a nonlinear equation in matlab where x0 = [0;1;2] is the initial value for search
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So I wrote a function :
function f = myfun(x)
f = (1./(1+(x1-x2).^2)) + sin.*(0.5*pi.*x2.*x3) + exp(((-x1+x)./x2)-2).^2)
end
Then:
x1 = 0; x2 = 1; x3 = 2; x0 = [x1;x2;x3];
options = optimoptions('fsolve','Display','iter');
[x,fval] = fsolve(@myfun,x0,options)
... But the following error appears:
"Caused by:
Failure in initial objective function evaluation. FSOLVE cannot continue."
How can i fix that.
I thank you.
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Réponse acceptée
Ameer Hamza
le 12 Nov 2020
Modifié(e) : Ameer Hamza
le 12 Nov 2020
There are issues with the definition of 'myfun'. Compare it with the following to see the errors.
function f = myfun(x)
x1 = x(1);
x2 = x(2);
x3 = x(3);
f = (1./(1+(x1-x2).^2)) + sin(0.5*pi.*x2.*x3) + exp(((-x1+x)./x2)-2).^2;
end
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Star Strider
le 12 Nov 2020
It took a bit to correct the parentheses problems in ‘myfun’.
With that solved:
myfun = @(x1,x2,x3) (1./(1+(x1-x2).^2)) + sin(0.5*pi.*x2.*x3) + exp((-(x1+x3)./x2-2).^2);
options = optimoptions('fsolve','Display','iter')
x1 = 0; x2 = 1; x3 = 2; x0 = [x1;x2;x3];
[x,fval] = fsolve(@(p)myfun(p(1),p(2),p(3)),x0,options)
produces:
x =
-0.823049966296178
1.775543504609500
1.177574125107562
fval =
1.262730886088113e+02
.
.
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