Is there something wrong with my anonymous function definition?

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Sumeet
Sumeet 2023-2-12
评论： Sumeet 2023-2-12
A-> fun1 = @(x1,x2) (x1 - 3.67.*10^-6).^2 + (x2-3.67.*10^-7).^2;
B-> fun1 = @(x) ((x(1) - 3.67.*10^-6).^2 + (x(2)-3.67.*10^-7).^2);
I wish to define my function as in A, however I run into errors saying "Failure in initial objective Function evaluation. FMINCON cannot continue.". Only the function in B runs smoothly.
Wished to check if there's something I'm missing out on. Thanks!

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采纳的回答

Torsten
Torsten 2023-2-12
编辑：Torsten 2023-2-12
Fmincon expects fun1 to have one vector of length n of parameter values as input, not n scalar values as for your function fun1. Thus you have to modify your function to fit what fmincon needs.
Use
fun1 = @(x1,x2) (x1 - 3.67e-6).^2 + (x2-3.67e-7).^2;
F = @(x) fun1(x(1),x(2))
and pass F to "fmincon".

更多回答（1 个）

Sulaymon Eshkabilov
Ran in:
If x1 and x2 are scalars, then A and B are equivalent:
% A
fun1 = @(x1,x2) (x1 - 3.67.*10^-6).^2 + (x2-3.67.*10^-7).^2;
x1 = 1;
x2 = 2;
A =fun1(x1, x2)
A = 5.0000
x = [x1, x2];
% B
fun1 = @(x) ((x(1) - 3.67.*10^-6).^2 + (x(2)-3.67.*10^-7).^2);
B = fun1(x)
B = 5.0000
If x1 and x2 are not scalar. x1 and x2 are vectors (col or row) of thte same size. Then A and B are not the same - see:
% A
fun1 = @(x1,x2) (x1 - 3.67.*10^-6).^2 + (x2-3.67.*10^-7).^2;
x1 = 1:3;
x2 = -3:-1;
A = fun1(x1, x2)
A = 1×3
10.0000 8.0000 10.0000
% B
fun1 = @(x) ((x(1) - 3.67.*10^-6).^2 + (x(2)-3.67.*10^-7).^2);
x1 = 1:3;
x2 = -3:-1;
x = [x1, x2];
B = fun1(x)
B = 5.0000
Now, to make both equivalent:
% A
fun1 = @(x1,x2) (x1 - 3.67.*10^-6).^2 + (x2-3.67.*10^-7).^2;
x1 = 1:3;
x2 = -3:-1;
A = fun1(x1, x2)
A = 1×3
10.0000 8.0000 10.0000
% B
fun1 = @(x) ((x(1,:) - 3.67.*10^-6).^2 + (x(2,:)-3.67.*10^-7).^2);
x1 = 1:3;
x2 = -3:-1;
x = [x1; x2];
B = fun1(x)
B = 1×3
10.0000 8.0000 10.0000
Similarly, one can adjust ver B if x1 and x2 are column vectors.

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