Tell me more ×
Mathematica Stack Exchange is a question and answer site for users of Mathematica. It's 100% free, no registration required.
up vote 10 down vote favorite

I would expect all of the following to give the same answer (2.12467) but only half of them give this answer. The others seem to be using the default NormFunction:>(Norm[#,2]&). Can anyone explain this?

α = 3;
n /. FindFit[{1, 3, 9, 20}, x^n, n, x, NormFunction :> ((Null; Norm[#, α]) &)]
n /. FindFit[{1, 3, 9, 20}, x^n, n, x, NormFunction :> ((Norm[#, α]) &)]
n /. FindFit[{1, 3, 9, 20}, x^n, n, x, NormFunction :> (Norm[#, 3] &)]

Block[{α = 3},  n /. FindFit[{1, 3, 9, 20}, x^n, n, x, NormFunction :> (Norm[#, α] &)]]
With[{α = 3}, n /. FindFit[{1, 3, 9, 20}, x^n, n, x, NormFunction :> (Norm[#, α] &)]]
Module[{α = 3}, n /. FindFit[{1, 3, 9, 20}, x^n, n, x, NormFunction :> (Norm[#, α] &)]]

(* Outputs: 2.12467, 2.13284, 2.12467, 2.13284, 2.12467, 2.13284 *)
share|improve this question
Actually, the last three cases can be reduced to the second and third: Block just has the effect of temporarily setting a to 3 therefore the expression is exactly equivalent to the second case (indeed, if executed in that order, the Block is even redundant because a already has the value 3). The With replaces the a by the literal 3, therefore it's exactly equivalent to the third case. And Module replaces a by a temporary variable which is set to 3, but of course the name (and life time) of the variable doesn't matter, therefore it's again equivalent to case 2. – celtschk May 1 '12 at 19:20

2 Answers

up vote 8 down vote

This must have to do with the symbolic preprocessing, happening in FindFit. In all cases when you get 2.13284, this was a result of symbolic preprocessing, which was possible because the norm function could be evaluated on symbolic arguments. The subsequent result is likely explained by the mechanism described by @Searke.

But if you define your own norm as

ClearAll[norm];
norm[vec_?(VectorQ[#, NumericQ] &), alpha_] := Norm[vec, alpha];

and replace Norm with norm in all your examples, you always get 2.12467. You can gain more insight into this by using Trace with TraceInternal -> True.

share|improve this answer
3  
In fact it seems to be enough to write norm=Norm, so the symbolic preprocessing must be doing something very fragile. – Lev Bishop May 1 '12 at 19:36
@LevBishop Good point. Alas, I don't have enough time right now to dig deeper into it. – Leonid Shifrin May 1 '12 at 20:08
up vote 5 down vote

Compare your results to:

FindFit[{1, 3, 9, 20}, x^n, n, x, NormFunction :> ((Norm[#, thisisnotavariable]) &)]

When the NormFunction fails and is not well designed, FindFit just quietly chooses the 2-norm.

This is effectively what is happening in your example above.

share|improve this answer
What, in this case, makes Norm[#, α]& not well designed? – rcollyer May 1 '12 at 19:10
5  
It's not so simple: FindFit[{1, 3, 9, 20}, x^n, n, x, NormFunction :> (notavariable &)] gives a straightforward error message ("not a real number"). And why does adding a redundant compoundexpression (Null;) change my NormFunction from "not well designed" to "well designed"? – Lev Bishop May 1 '12 at 19:11

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.