Create a local system of equations function from a set of anonymous functions

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Fernando
Fernando 2023-7-7
编辑： Dyuman Joshi 2023-7-7
Hello,
I have a set of N anounimous functions which are created using this code
equations = {'@(x1,x2)x1.^2+x2-6','@(x1,x2)-2.5.*x1+x2.^2-2'};%Set your equations
N = numel(equations);
f = cell(1,N);
for i = 1:N
f{i} = str2func(equations{i});
end
The thing is, in a later part of the program I would like to use fsolve to solve the system of equation and fsolve requires the a function in this equivalent form:
function F = root2d(x)
F(1) = x(1).^2+x(2)-6;
F(2) = -2.5.*x(1)+x(2).^2-2;
end
where root2d is the function inputted in fsolve. I was wondering if there was a way to create a function in this form from the upper form or a way to use fsolve using the upper form?
Kind regards

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回答（2 个）

Paul
Paul 2023-7-7
编辑：Paul 2023-7-7
Ran in:
One approach:
equations = {'@(x1,x2)x1.^2+x2-6','@(x1,x2)-2.5.*x1+x2.^2-2'};%Set your equations
N = numel(equations);
f = cell(1,N);
for i = 1:N
f{i} = str2func(equations{i});
end
F = @(x) cellfun(@(func) func(x(1),x(2)),f)
F = function_handle with value:
@(x)cellfun(@(func)func(x(1),x(2)),f)
sol1 = fsolve(F,[2 2])
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.
sol1 = 1×2
1.8509 2.5743
F(sol1)
ans = 1×2
1.0e-13 * -0.0178 0.3109
[f{1}(sol1(1),sol1(2)) , f{2}(sol1(1),sol1(2))]
ans = 1×2
1.0e-13 * -0.0178 0.3109
Dyuman Joshi
Dyuman Joshi 2023-7-7
编辑：Dyuman Joshi 2023-7-7
Ran in:
Another approach (Note - requires Symbolic Toolbox)
equations = {'@(x1,x2) x1.^2+x2-6','@(x1,x2) -2.5.*x1+x2.^2-2'};%Set your equations
f = str2sym(equations);
z = matlabFunction(f,"File", "root2d", "Vars",{symvar(f)});
out = fsolve(z,[2 2])
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.
out = 1×2
1.8509 2.5743
z(out)
ans = 1×2
1.0e-13 * -0.0178 0.3109

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