Re: Using Fit to interpolate data

*To*: mathgroup at smc.vnet.net*Subject*: [mg127454] Re: Using Fit to interpolate data*From*: Bill Rowe <readnews at sbcglobal.net>*Date*: Thu, 26 Jul 2012 03:33:25 -0400 (EDT)*Delivered-to*: l-mathgroup@mail-archive0.wolfram.com*Delivered-to*: mathgroup-newout@smc.vnet.net*Delivered-to*: mathgroup-newsend@smc.vnet.net

On 7/25/12 at 2:33 AM, carlsonkw at gmail.com (Kris Carlson) wrote: >On Tue, Jul 24, 2012 at 10:14 AM, jf.alcover <jf.alcover at gmail.com> >wrote: >>Why not use FindFit and an Exp ? >>fbddFit = Exp[-a x + b] /. FindFit[ fiberDataDensitiesFeierabend, >>Exp[-a*x + b], {a, b}, x] >Aha. That is helpful. I had tried using a decay function with Fit >and couldn't get the syntax, but now I see it requires FindFit. >Merci. The power law seems to work better, tho. >In[277]:= fbddFit = Fit[fiberDataDensitiesFeierabend, {x^-13}, x] >Out[277]= 2.4929*10^12/x^13 If your criteria for judging the fit is a plot then it is almost always the case a power law fit will seem to fit non-linear data better. And if the only thing you need to be able to do is find a way to interpolate your data for a new prediction this is likely to be OK. But if the parameters are supposed to have some physical or biological meaning (corresponding to some meaningful model) then the high exponent is a very strong sign this isn't a good model and will quite likely have poor prediction power in terms of any extrapolation.