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Mathematica does not evaluate the derivative of the following MeijerG function.

<< NumericalCalculus`;

ND[
  MeijerG[{{2/3, 2/3, -(7/3), b1}, {13/6}}, {{5/3, 5/3, 7/6}, {5/3, 5/3}}, 1/5]
, {b1, 1}, 5/3]

Why is that?

Notice that the arguments are compatible with the usual rules for the MeijerG function, i.e.

ND[
  MeijerG[{{2/3, 2/3, -(7/3), b1}, {13/6}}, {{5/3, 5/3, 7/6}, {5/3, 5/3}}, 1/5]
, {b1, 0}, 5/3]

makes sense and can be evaluated with Mathematica.

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  • $\begingroup$ Does b1 have an assigned value? $\endgroup$
    – mikado
    Aug 13 at 14:35
  • $\begingroup$ @mikado yes it does, b1 is equal to 5/3 $\endgroup$ Aug 13 at 14:41
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It seems that in evaluating the derivative ND is transforming the function to

MeijerG[{{-(7/3), 2/3, 2/3, 5/3 + 1}, {13/6}}, {{7/6, 5/3, 5/3}, {5/3, 5/3}}, 1/5]

which is ill defined. Perhaps there is such a rule that holds for symbolic derivatives but it's not defined for generic arguments. As it is right now, it looks like a bug and you could report it.

If you need just the numerical derivative, you can code it by hand via the definition. $$ \lim_{n\to\infty} n\big(f\big(x+\tfrac1n\big)-f(x)\big)\,. $$

If you want to get precise results you can even use LinearModelFit to estimate the error, like so

f[b1_] := MeijerG[{{2/3, 2/3, -(7/3), b1}, {13/6}}, {{5/3, 5/3, 7/6}, {5/3, 5/3}}, 1/5];

Table[{nn, (f[SetPrecision[5/3 + 1/nn, 50]] - f[SetPrecision[5/3, 50]]) nn}, {nn, 10^8, 10^9, 10^8}];
ListPlot[%]
fit = LinearModelFit[%%, {1}, x];
{fit["BestFit"], Max[Abs /@ fit["FitResiduals"]]}
(* {numerical derivative, error estimate} *)

This is probably like shooting a fly with a cannon, but it works.

Adjust the ranges of nn so that the plot looks approximately constant and adjust the precision inside SetPrecision so that the plot does not look too noisy.

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  • $\begingroup$ Amazing!! Many thanks! $\endgroup$ Aug 13 at 15:04

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