Solve equation with 2 variables being vectors

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Nicole
Nicole el 8 de Mayo de 2014
Comentada: Justin el 12 de Mayo de 2014
Example C=@(A,B) A+B*A A=[1 2 3] B=[4 2 6]
I want C to be a matrix of all possible solutions
Right now I have C as a function and A and B as user inputs
Is there any easier way than writing a loop?
Clarification: C is a very complicated equation with a 2 variables that can have multiple inputs. I want to know all possible solutions. Is there a way to get all possible solutions without nesting loops. I'm making a contour map of 2 inputs and 2 outputs.
  3 comentarios
Nicole
Nicole el 9 de Mayo de 2014
bsxfun does not work because I have 4 variables, c and a are system dependent constants, Ff and Xi are inputs.
Xf=solve(Ff==2*c*a*(Xf/Xi-Xi^2/Xf^2),Xf);
Xf=Xf(1);
Xfs=matlabFunction(Xf);
where Xfs=@(Ff,Xi,a,c)
Justin
Justin el 12 de Mayo de 2014
You should be able to put an array into a symbolic variable and solve for each value of that array. That means you could define Ff and Xi using the meshgrid method in my answer below. You could:
[Xi, Ff] = meshgrid(Xi, Ff)
and then solve with the values of Xi and Ff that are returned which should give you every possible combination.

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Respuestas (2)

Justin
Justin el 8 de Mayo de 2014
Editada: Justin el 8 de Mayo de 2014
Edited for clarification:
This may do the trick. It has a loop but only one for implementing the function. After it is run the a, b, and c should hopefully be everything you need.
A=[1 2 3];
B=[4 2 6];
[a, b] = meshgrid(A, B);
for i = 1:length(p)
c(i) = C(a(i), b(i));
end
Let me know how that works.
_______________________________________________
OLD STUFF:
I'm not sure if I understand the question right but bsxfun may be useful. bsxfun automatically extends an array along a singleton dimension to match the other array's size. You could use it to get the results you want (I think).
A=[1 2 3];
B=[4 2 6];
bsxfun(@plus, A, B')
ans =
5 6 7
3 4 5
7 8 9

Matt J
Matt J el 9 de Mayo de 2014
Editada: Matt J el 9 de Mayo de 2014
C=bsxfun(@(A,B) A+B.*A, A(:),B(:).')

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