Thank u so much for your attention for my question, the minimum is the value required, the "fmincon" matlab function is used, and the algorithm is default, but there is a problem not addressed, the following is the code:
function responsemum = numresponse(num) %%%%function 1
swith num
case 1
responsenum = @(x,y)x+y;
case 2
responsenum = @(x,y)x.*y; %%%%end of the function 1
function y = innerprod(fun1,fun2) %%%% function 2
y = @(x,y)fun1(x,y).*fun2(x,y); %%%end of function2
metag = innerprod(numresponse(1),numresponse(2));
TL = [-1,-1];
TU = [1,1];
T0 = [0,0];
[tmin,hlmin,exitflag] = fmincon(@(x,y)metag(x,y),T0,[],[],[],[],TL,TU);
there is some error message after run the code:
Error using@(x,y)metag(x,y) not enough input arguments
if u can give me any help, i will be appreciated, Thanks!

2 个评论

It's difficult to read the code. Where do functions end?
First, thank u so much for your attention about my question, i have edit my question, and give some explanation about my question!

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 采纳的回答

[tmin,hlmin,exitflag] = fmincon(@(xy)metag(xy(1),xy(2)), T0, [], [], [], [], TL, TU);

1 个评论

Mr. Walter, Thank u so much for your help, every time you give me a fish, hope i can learn how to fishing, and i can survive by myself in the future!

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