Take the 2-minute tour ×
Mathematica Stack Exchange is a question and answer site for users of Mathematica. It's 100% free, no registration required.

Using FindFit, Mathematica selects an optimal algorithm when Automatic is selected. I'd like to know the name of the method applied to the FindFit computation. Is there any option to explain that?

share|improve this question
1  
I don't think you can hijack FindFit[] to say exactly what method it uses. However, the docs say that by default it uses SVD for linear least squares, and Levenberg-Marquardt for nonlinear least squares problems. –  J. M. Apr 17 '13 at 2:50
    
I think we should explain the name of algorithm in academic papers. How can researchers clear this issue when they use FindFit? –  S.Orii Apr 17 '13 at 3:50
    
You can use an explicit setting of Method (e.g. Method -> "LevenbergMarquardt") when invoking FindFit[]. –  J. M. Apr 17 '13 at 4:28
    
A merit to use FindFit is of automatic selection of algorithm when it executes under a condition like "a>0". We may find the algorithm selected by an explicit setting mentioned above. Is it the only way to find the selected algorithm? –  S.Orii Apr 17 '13 at 7:30
    
Well, since FindFit[] is effectively a black box, there's not much hope in figuring out what's going on under the hood. If you're going to be writing about this in some paper, just say you used the Automatic setting, and mention the claim in the docs about the methods internally used if need be. –  J. M. Apr 17 '13 at 8:09
show 1 more comment

1 Answer

As noted by @J.M., the docs say that the default for FindFit[] is SVD for linear least squares and Levenberg-Marquardt for nonlinear least squares. If it is important for the end-user to know explicitly the method used, then the user should explicitly define the method.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.