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How do I extrapolate from these non linear values?


I'm looking for a formula that'll give me more values down the graph following the trend of these numbers.

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closed as not a real question by tomasz, William, rschwieb, Michael Greinecker, J. M. Sep 19 '12 at 9:28

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That would depend on what you want to obtain... –  tomasz Sep 6 '12 at 20:31
@tomasz thanks for your comment, i'm looking for a formula that'll give me more values down the graph following the trend of these numbers. Thanks. –  Harry Sep 6 '12 at 20:39

3 Answers 3

up vote 2 down vote accepted

There seems to be a very good fit to an exponential: $$L_j \approx \exp(11.9095801316190 + .157003890964286\; j), \ j = 1 \ldots 15 $$

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As a worst-case scenario option, you could use the Lagrange polynomial equation http://mathworld.wolfram.com/LagrangeInterpolatingPolynomial.html .It takes a while even with Wolfram, though, so only use it only if approximations fail you.

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You can try various things. I would first divide them all by $10^6$ to avoid overflow, then try polynomial fits of increasing order and see how they do. You could also take logs and try a linear fit to see if they fit an exponential.

What do you know about where they came from? Are they measured data that might have some noise? Are they results of an integer calculation, exact, and maybe factoring will teach you something? Again, if they are integers, they might fit a polynomial exactly and taking successive differences will find that out.

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Polynomial fit for extrapolation?? I suppose that could work if you stuck to a low-order polynomial, and had a domain for extrapolation that was well-suited. –  Arkamis Sep 6 '12 at 20:49
A professor of mine once gave a problem where based on a 7th degree polynomial fit of some population data we were supposed to determine when the human race would go extinct. –  axblount Sep 6 '12 at 20:56
@EdGorcenski: sure, but it can do funny things. Excel's quadratic fit doesn't look too bad, though the cubic looks like it is having roundoff problems already. –  Ross Millikan Sep 6 '12 at 20:56
@RossMillikan Sure, a quadratic might fit well, even a cubic if you restrict one of the coefficients to be zero. But unless the data were definitely quadratic, you won't get an exact fit that way. –  Arkamis Sep 6 '12 at 21:02
@Ed G: I didn't think an exact fit was required. Robert Israel's exponential is terrific –  Ross Millikan Sep 6 '12 at 21:10

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