0
$\begingroup$

I begin with

FullSimplify[Series[HypergeometricPFQ[{-k,m/2-1,(m-1)/2,m/2},{1/2,1,3m/2-k-2},t],{t,0,3}],
  Assumptions->k\[Element]Integers\[And]m\[Element]Integers\[And]k>=0\[And]m>0]

and obtain

enter image description here

Now I want to get a formula for the $j$th coefficient. Of course it is easy to guess it but then I should prove it, and also there are more complicated cases. So I try

FullSimplify[SeriesCoefficient[HypergeometricPFQ[{-k,m/2-1,(m-1)/2,m/2},{1/2,1,3m/2-k-2},t],{t,0,j}],
  Assumptions->k\[Element]Integers\[And]m\[Element]Integers\[And]k>=0\[And]m>0]

The result (on 11.0.1.0) is zero. I would understand if the engine would reply that it cannot compute it, but why zero??

$\endgroup$
5
  • 1
    $\begingroup$ It seems like the k>=0 assumption makes it zero. Are you sure that's correct? $\endgroup$ Aug 24, 2021 at 10:07
  • $\begingroup$ @SjoerdSmit Thanks for this observation! I confirm that without the k>=0 I seem to obtain correct result. Which is even more strange since there are definitely nonzero cases with nonnegative k. You can observe this by computing the above series using, say, With[{k=3},...] $\endgroup$ Aug 24, 2021 at 11:42
  • 1
    $\begingroup$ FullSimplify[ Piecewise[{{1/(-1 - k)!, j >= 0}}, 0], Assumptions -> k \[Element] Integers \[And] m \[Element] Integers \[And] k >= 0 \[And] m > 0] returns zero. So does Table[1/(-1 - k)!, {k, 100}]. $\endgroup$
    – Michael E2
    Aug 24, 2021 at 12:15
  • $\begingroup$ @MichaelE2 Well this is understandable. But in these coefficients, all negative factorials occasionally cancel out, how to deal with these cases? $\endgroup$ Aug 24, 2021 at 12:20
  • 1
    $\begingroup$ Take the limits as k approaches an integer? However, Limit[(Gamma[j - k] Gamma[-2 - k + (3 m)/2])/(Gamma[-k] Gamma[-2 + j - k + (3 m)/2]), k -> k0, Assumptions -> k0 \[Element] Integers \[And] m \[Element] Integers \[And] m > 0 && j >= 0 && k0 >= 3] returns 0 even though the limits at specific values of j and m are generally nonzero. I'd be inclined to call the zero limit a bug, since it's not even generically true. $\endgroup$
    – Michael E2
    Aug 24, 2021 at 14:41

1 Answer 1

2
$\begingroup$

A good solution is to use the following code for the nth term of the series

HypergeometricPFQn[a_List, b_List, z_, n_] := z^n/n!
  Product[Pochhammer[ai, n], {ai, a}]/Product[Pochhammer[bi, n], {bi, b}];
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.