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I am working with interpolation function with some non-numerical arguments. A simplified version of the function I am working with is:

FUNC = Interpolation[{{0, a}, {1, b}, {2, c}}]

I would like to compute the derivative of FUNC with respect to a, b and c (for a general function value)

Potential solution: obtain the piecewise function that Interpolation generates. Problem: I do not know how to extract this piecewise function from FUNC.

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  • $\begingroup$ Is not it the same as computing a derivative of the interpolating function? $\endgroup$
    – yarchik
    Commented Sep 27, 2016 at 8:52
  • $\begingroup$ No. Typically, the derivative would be FUNC'[x]. However, I do not want dFUNC/dx, but dFUNC/da, where a is one of the parameters specified above. $\endgroup$
    – Breugem
    Commented Sep 27, 2016 at 8:56
  • $\begingroup$ The piecewise function that the interpolating function represents depends on the interpolation method and the data. Here is one way: mathematica.stackexchange.com/a/59963/4999 $\endgroup$
    – Michael E2
    Commented Sep 27, 2016 at 10:23
  • $\begingroup$ This makes no sense because this derivative is not defined by your data unless it contains {0,a} and {0,a+some value}. $\endgroup$
    – Vsevolod A.
    Commented Mar 25, 2018 at 17:49

1 Answer 1

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f[a_, b_, c_] = Interpolation[
   {{0, a}, {1, b}, {2, c}},
   InterpolationOrder -> 2];

xValues = Range[0, 2, 1/8];

fValues = f[a, b, c] /@ xValues // Simplify;

dfda[x_] = InterpolatingPolynomial[
   {xValues, D[#, a] & /@ fValues} // Transpose, x] //
  Simplify

(*  1/2 (2 - 3 x + x^2)  *)

Plot[dfda[x], {x, 0, 2},
 Frame -> True,
 FrameLabel -> (Style[#, 14, Bold] & /@
    {"x", "df / da"})]

enter image description here

EDIT: As Breugem suggested in his comment to this answer, working directly with the InterpolatingPolynomial is more straightforward.

fp[a_, b_, c_, x_] = 
 InterpolatingPolynomial[{{0, a}, {1, b}, {2, c}}, x] // Simplify

(*  a + (-a + b + 1/2 (a - 2 b + c) (-1 + x)) x  *)

D[fp[a, b, c, x], a] // Simplify

(*  1/2 (2 - 3 x + x^2)  *)
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  • $\begingroup$ This is very good! The interpolating polynomial will become complex for large grids, in which a piecewise polynomial needs to be estimated, which brings us back to the original problem. I guess the most efficient way would be to generate an interpolating polynomial from the start $\endgroup$
    – Breugem
    Commented Sep 27, 2016 at 12:15
  • 1
    $\begingroup$ @Breugem Using an interpolating polynomial on more than a few points is usually a bad idea, unless your data is exact and you know a polynomial is an appropriate model. Piecewise interpolation is usually more appropriate. $\endgroup$
    – Michael E2
    Commented Sep 27, 2016 at 13:12
  • $\begingroup$ I agree:) This was my (potentially unclear) comment (i.e., we cannot use the InterpolatingPolynomial from the answer above) $\endgroup$
    – Breugem
    Commented Sep 27, 2016 at 13:50

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