Nonscalar arrays of function handles are not allowed; use cell arrays instead.

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Tevel Hedi
Tevel Hedi 2022-4-9
编辑： Torsten 2022-4-9
syms o p
fun1=o+p;
fun2=o*p+5;
eq1=matlabFunction(fun1);
eq2=matlabFunction(fun2);
bbb=fsolve(@(o,p) [eq1;eq2],[0,1]);
What am I doing wrong?

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采纳的回答

Steven Lord
Steven Lord 2022-4-9
Ran in:
You need to evaluate the function handles in your fsolve call. Alternately you could skip converting the symbolic expressions into function handles and use solve.
syms o p
fun1=o+p;
fun2=o*p+5;
eq1=matlabFunction(fun1);
eq2=matlabFunction(fun2);
bbb=fsolve(@(op) [eq1(op(1), op(2));eq2(op(1), op(2))],[0,1]) % or
Equation solved. fsolve completed because the vector of function values is near zero as measured by the value of the function tolerance, and the problem appears regular as measured by the gradient.
bbb = 1×2
-2.2361 2.2361
bbb2 = solve(eq1, eq2, o, p)
bbb2 = struct with fields:
o: [2×1 sym] p: [2×1 sym]
vpa(bbb2.o, 5)
ans = 
vpa(bbb2.p, 5)
ans = 

更多回答（2 个）

David Hill
David Hill 2022-4-9
Why use symbolic and convert?
fun=@(x)[x(1)+x(2);x(1)*x(2)+5];
x=fsolve(fun,[0,1]);
Torsten
Torsten 2022-4-9
编辑：Torsten 2022-4-9
syms o p
fun1 = o+p;
fun2 = o*p+5;
fun = [fun1,fun2];
eq = matlabFunction(fun);
bbb = fsolve(@(x)eq(x(1),x(2)),[0,1]);

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