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Is there something wrong with my anonymous function definition?
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)....
3 years 前 | 2 个回答 | 0
提问
3 years 前 | 2 个回答 | 0
