taking jacobion of function
2 Ansichten (letzte 30 Tage)
Ältere Kommentare anzeigen
Benjamin
am 19 Mär. 2019
Beantwortet: Walter Roberson
am 19 Mär. 2019
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 Kommentare
Akzeptierte Antwort
Walter Roberson
am 19 Mär. 2019
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.
0 Kommentare
Weitere Antworten (0)
Siehe auch
Kategorien
Mehr zu Get Started with Curve Fitting Toolbox finden Sie in Help Center und File Exchange
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!