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I have a list with values at some points:

list1 = {{-5.4, 0.000513081122}, {-4.8, 0.000840821088}, {-4.0, 
0.001059617102}, {-3.2, 0.001114222604}, {-2.4, 
0.001086309651}, {-1.6, 0.001120999228}, {-0.80, 
0.001128926919}, {0.0, 0.001119484321}, {0.80, 
0.001128926919}, {1.6, 0.001120999228}, {2.4, 
0.001086309651}, {3.2, 0.001114222604}, {4.0, 
0.001059617102}, {4.8, 0.000840821088}, {5.4, 0.000513081122}};

Can I predict the value outside the data set, for eg., at -15, 10, 10, 15? I tried the following:

fit = Fit[list1, Table[x^i, {i, 0, 500}], x];
p1 = ListPlot[list1, Joined -> True, 
  PlotRange -> {{-10, 10}, {0.0015, 0.0001}}];

p2 = Plot[fit, {x, -10, 10}, 
 PlotRange -> {{-10, 10}, {0.0015, 0.0001}}];

Show[p1, p2]

However, the curve blows up outside the data point. I tried using this solution, but still no success. How should I proceed. Any help would be appreciated.

Mark Robinson
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    You have 15 data points and 501 parameters to estimate? I think extrapolating is the least of your problems. If you had maybe 2 or 3 parameters to estimate with 15 data points, then using NonlinearModelFit is necessary as that function will give you confidence bands (as Fit just gives you the fit and nothing else). There must be some theoretical model you could use rather than an arbitrary polynomial. Also (unless this data is not real), you really only have 8 points because of the exact symmetry in the data. – JimB Nov 27 '19 at 14:38
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    A polynomial always "blows up" for "large" values. Also, high order polynomials are subject to precision issues. That is why @JimB mentioned the need for a theoretical model based on your problem domain. – Bob Hanlon Nov 27 '19 at 15:22
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    A quick look at your data suggests that extrapolating even slightly outside the data supplied will require a model exploiting strong prior knowledge. It is far from obvious whether the apparent oscillations have finished or whether the curve is about to turn round and head back up again. Blind extrapolation to two or three times the length of the data set cannot be justified. – mikado Nov 27 '19 at 20:32

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