FSolve with Matrix Input

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Drew
Drew 2013-10-18
评论： Drew 2013-10-18
fsolve is really confusing me. All the examples find show a system of equations being input in character form. Mine are in matrix form. As I understand it x0 is meant to define a boundary condition for solving the system, so that the solver doesn't fly off into infinity and crash if some of the equations are bad. I don't understand the point of function f though. Shouldn't this be the array which contains the system of equations to be solved?
I've seen several examples of fsolve in matlab, but can't seem to find any that show how to pass parameters in matrix form.
In this example I'm using only two equations but in practice I actually have hundreds to solve simultaneously so it's a rather large matrix.
Thanks!
Here is my code.
[A,b] = equationsToMatrix(eq1,eq2)
X0 = [0 0]
fsolve([A,b], X0)
Here is the output
eq1 = - sx - sy/2 == 5
eq2 = - (3*sx)/2 - (3*sy)/2 == 9
A =
[ -1, -1/2]
[ -3/2, -3/2]
b =
5
9
X0 =
0 0
Error using lsqfcnchk (line 109)
If FUN is a MATLAB object, it must have an feval method.
Error in fsolve (line 198)
funfcn = lsqfcnchk(FUN,'fsolve',length(varargin),funValCheck,gradflag);
Error in SolveTesting (line 70)
fsolve([A,b], X0)

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采纳的回答

Yatin
Yatin 2013-10-18
Hi,
I see that your equations are linear. You can therefore use matrix left division ("\") for better speed and accuracy instead of " fsolve ". So as per the above case:
A = [-1, -1/2; -3/2,-3/2];
b = [5;9];|
then
x = A\b;
This should give you your desired result.
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Drew
Drew 2013-10-18
For good measure I have also tried the following.
function f = matrixfun(z,A,b)
%syms x y;
f = A * [z(1);z(2)] - b;
end
Same problem
>> fsolve(@matrixfun,[-4 -5],[],A,b)
Error using fsolve (line 257)
FSOLVE requires all values returned by user functions to be of data type double.
btw, I love the utterly useless error messages that matlab provides.
Drew
Drew 2013-10-18
Solved it, ahhhhhh!
>> fsolve(@(z)double(A)*[z(1);z(2)]-double(b),[-5 -5])
Equation solved.
fsolve completed because the vector of function values is near zero
as measured by the default value of the function tolerance, and
the problem appears regular as measured by the gradient.
<stopping criteria details>
ans =
-4.0000 -2.0000

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