taking jacobion of function

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Benjamin
Benjamin 2019-3-19
回答： Walter Roberson 2019-3-19
I am using the lsqcurvefit function to fit a function that is dependent on two variables. Let's call these two variables, X, and Y.
If my function looks like:
f(x,y) = a*X + b*Y + c
can someone explain why the Jacobian to this is:
Jacobian=[X(:), Y(:), ones(size(X(:)))];
If I take df/dX, I would get "a", df/dY = "b". Why is the answer what I show it to be?
I am confused and I need to understand what is going on, since I need to change this function to something else. I wanted to start with this one, which is an example I was given, and the Jacobian is known.
  2 个评论
Walter Roberson
Walter Roberson 2019-3-19
You appear to be taking the jacobian with respect to [a b c]
Benjamin
Benjamin 2019-3-19
编辑：Benjamin 2019-3-19
Since my function is a function of x and y, shouldn't the jacobian be taken with respect to x and y? a and b are just fitted coefficients. Also, if I were taking the jacobian wrt to [a b c], wouldnt the partial with respect to c by 0? Why do they have it as a column vector of 1's?
Edit: When I try to take Jacobian with respect to x and y, I get that it expexted another number in that matrix. So it appears that taking wrt a, b, c, is right. I just don't know why.

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Walter Roberson
Walter Roberson 2019-3-19
If you look carefully at https://www.mathworks.com/help/optim/ug/lsqcurvefit.html#buuhcjo-fun then the x it is talking about are the model parameter values, which are distinct from xdata . Your model parameters appear to be a, b, and c so you take the jacobian with respect to those.
Your function contains a + c term. The derivative of c with respect to c is 1, not 0.

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