How to obtained matris vector solution using fminunc function?

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Alexi
Alexi 2023-3-25
编辑： Torsten 2023-3-25
c_1=[3;5;8];
c_2=[2;6;8];
fun = @(x) (x(1)-c_1).^2 + (x(2)-c_2).^2; % cost function
x0 = [[0.1;0.1;0.1],[0.1;0.1;0.1]]; % İnitial Gues
[x,fval] = fminunc(fun,x0) % Results

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Torsten
Torsten 2023-3-25
编辑：Torsten 2023-3-25
Ran in:
You will have to use fminunc three times:
c_1=[3;5;8];
c_2=[2;6;8];
x1_sol = zeros(size(c_1));
x2_sol = zeros(size(c_1));
for i=1:numel(c_1)
fun = @(x) (x(1)-c_1(i)).^2 + (x(2)-c_2(i)).^2; % cost function
x0 = [0.1,0.1]; % İnitial Guess
[x,fval] = fminunc(fun,x0); % Results
x1_sol(i) = x(1);
x2_sol(i) = x(2);
end
Local minimum found. Optimization completed because the size of the gradient is less than the value of the optimality tolerance. Local minimum found. Optimization completed because the size of the gradient is less than the value of the optimality tolerance. Local minimum found. Optimization completed because the size of the gradient is less than the value of the optimality tolerance.
x1_sol
x1_sol = 3×1
3.0000 5.0000 8.0000
x2_sol
x2_sol = 3×1
2.0000 6.0000 8.0000
Or do you try to solve a different problem ?
  2 个评论
Alexi
Alexi 2023-3-25
Thank you for your answer I think I can adapt this approach to my original problem.
Torsten
Torsten 2023-3-25
编辑：Torsten 2023-3-25
Ran in:
Maybe you mean
c_1=[3;5;8];
c_2=[2;6;8];
fun = @(x)sum((x(1)-c_1).^2 + (x(2)-c_2).^2); % cost function
x0 = [0.1,0.1]; % İnitial Guess
[x,fval] = fminunc(fun,x0) % Results
Local minimum found. Optimization completed because the size of the gradient is less than the value of the optimality tolerance.
x = 1×2
5.3333 5.3333
fval = 31.3333
?

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