How do I use fsolve n*m times in a 3-dimensional array without messing up my dimensions?
1 回表示 (過去 30 日間)
古いコメントを表示
I am searching for solutions to a system of three equations, three unknowns. The system is called eqs and is a function of a(1), a(2), a(3), and the parameter S. For n values of the parameter S, I want to solve eqs m times with m random guesses. Here is the function I am trying to solve m*n times
function F=eqs(a,S)
F(1)=(S-2*(a(3))^2)*(pi-acos(a(1)/a(3))-acos(a(2)/a(3))-acos(((1-2*a(1))^2+...
(1-2*a(2))^2)^0.5/((2*a(3)))))-a(1)*((a(3))^2-(a(1))^2)^0.5-a(2)*...
((a(3))^2-(a(2))^2)^0.5-1/4*((1-2*a(1))^2+(1-2*a(2))^2)^0.5*...
(4*(a(3))^2-(1-2*a(1))^2-(1-2*a(2))^2 )^0.5;
F(2)=(4*a(2)*(a(3))^2+a(2)-4*(a(3))^2)/(4*a(2)-4*(a(3))^2-1)-a(1);
F(3)=(1-2*a(2))^2/(1-2*a(1))^2*(4*(a(3))^2-4*(a(1))^2 )-4*(a(1))^2+...
(1-2*a(1))^2+(1-2*a(2))^2;
end
Here is the code that is not working. I have some problem with my matrix dimensions on the solutions line, but I cannot figure it out. please help.
%Preallocations
n=5;
m=10;
draw=100;
S=zeros(n,1);
guess=zeros(n,m,3);
b=zeros(n,m,1);
c=zeros(n,m,1);
d=zeros(n,m,1);
solutions=zeros(n,m,3);
for t=1:n;
S(t)=(t+n/(3+2^1.5))/n; %this is the parameter that varies. n values.
for k=1:m;
%for every value of S I want to fsolve m times with m random guesses
%generating the guess matrix
b(t,k)=randsample(draw,1);
c(t,k)=randsample(draw,1);
d(t,k)=randsample(draw,1);
guess(t,k,1)=b(t)/(2*draw+1);
guess(t,k,2)=c(t)/(2*draw+1);
guess(t,k,3)=d(t)*1.2/(2*draw+1);
%solving the off-diagonal system
system = @(a)eqs(a,S);
solutions(t,k,:)=fsolve(system,guess(t,k,:));%=[a(1) a(2) a(3)]
end
end
0 件のコメント
採用された回答
Star Strider
2015 年 12 月 29 日
You need to subscript ‘S’ in your function call to pass it to your function as a scalar:
system = @(a)eqs(a,S(t));
and your code works. Otherwise, it is a vector and it will throw an error. I leave it to you to determine if it gives the results you want.
その他の回答 (0 件)
参考
カテゴリ
Help Center および File Exchange で Get Started with MATLAB についてさらに検索
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!