Borrar filtros
Borrar filtros

fsolve for 2 equation with 2 variables

1 visualización (últimos 30 días)
Miroslav Mitev
Miroslav Mitev el 13 de Dic. de 2017
Comentada: Star Strider el 13 de Dic. de 2017
Here is my function, in order to shorten the expression here I add A and B, but in my original function they are inside F(1) and F(2) (i.e. that is not the problem), this is the error I get: Objective function is returning undefined values at initial point.
function F = myfun(x)
N=32;
K=5;
s=1;
b=0.1;
P=5;
A=-2*s^4*x(1)*N+2*s^4*x(1)*K+3*s^2*x(2)*log(2);
B=sqrt(8)*x(2)*s^2*log(2);
g=exprnd(1,1,N-K);
F(1) = sum((g.*(1-b*x(1))/(x(2)*log(2)))-1./g)+(N-K)*((A+sqrt(A^2-B^2))/((4*s^2/sqrt(8))*B))-N*P;
F(2) = b*(sum(log2((g*(1-b*x(1)/x(2)*log(2))))))-(N-K)*log2(1+(2*A^2+2*A*sqrt(A^2-B^2)-B^2)/((4/sqrt(8))*B*(A+(4/sqrt(8))*B+sqrt(A^2-B^2))));
end

Respuesta aceptada

Star Strider
Star Strider el 13 de Dic. de 2017
What is the initial point you chose? Your function returns finite values for random non-zero arguments. It returns [NaN NaN] for [0 0] as an input.
The solution is most likely to use an initial point other than [0 0]. I would use rand(2,1).
  7 comentarios
Matt J
Matt J el 13 de Dic. de 2017
Since it is only a function of 2 variables, you could do a coarse surf() plot of norm(F) and find visually where the roots approximately lie. This would give you a better initial guess than simply randomizing.
Star Strider
Star Strider el 13 de Dic. de 2017
My pleasure.
It might be useful for you to post the symbolic expression you are coding as well as your code for it as a new Question, since that seems to be the problem.

Iniciar sesión para comentar.

Más respuestas (0)

Categorías

Más información sobre Solver Outputs and Iterative Display en Help Center y File Exchange.

Community Treasure Hunt

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

Start Hunting!

Translated by