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Re: Using Fit to interpolate data
*To*: mathgroup at smc.vnet.net
*Subject*: [mg127437] Re: Using Fit to interpolate data
*From*: Bill Rowe <readnews at sbcglobal.net>
*Date*: Wed, 25 Jul 2012 02:31:15 -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 7/24/12 at 4:15 AM, carlsonkw at gmail.com (Kris Carlson) wrote:
>Can someone enlighten me about how to fit a curve to data? These
>data are of the density of axons of given diameters in the spinal
>cord. The density of smaller fibers is dramatically larger than that
>of larger fibers. There cannot be density < 0 of any diameter, so
>heuristically to prevent a fit yielding an equation that dips below
>0 I added an end point with fiber diameter = 16 that is 0. Then
>using Fit I try to increase the exponent of x until the curve
>doesn't yield negative values. The fit looks good in large scale but
>when I plot the region of greatest interest, 8 < x < 14, it no
>longer looks so good. Maybe I am simply ignorant about fitting a
>curve to somewhat irregular data? Or can it be done?
>fiberDataDensitiesFeierabend = {{16, 0}, {10.7, 0.11}, {10.4,
>0.19}, {9.77, 0.41}, {8.29, 3.05}, {7.14, 19.86}};
>fbddPlot =
>ListPlot[fiberDataDensitiesFeierabend, PlotMarkers -> {Automatic,
>Medium}]
The plot above suggests an exponential model. And doing
ListLogPlot[fiberDataDensitiesFeierabend]
Seems to confirm this in that the data points seem to line along
a line in log space.
>fbddFit = Fit[fiberDataDensitiesFeierabend, {1, x, x^-13}, x]
Better to use FindFit rather than Fit. Among other things,
FindFit allows you to specify constraints on the model
parameters. Also, using high powers of x as basis functions is
not a good idea. If you absolutely must get a polynomial fit,
choose Chebyshev polynomials or Legendre polynomials as your
basis functions rather than powers of x.
Using FindFit and assuming an exponential model I get:
In[12]:= params=FindFit[fiberDataDensitiesFeierabend, a Exp[b
x], {a, b}, x]
Out[12]= {a->2.06978*10^6,b->-1.61827}
and
Plot[a Exp[b x] /. params, {x, 7, 16.5},
Epilog -> {PointSize[.02], Point[fiberDataDensitiesFeierabend]}]
suggests this is a reasonable fit to the data.
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