I’m interested in finding a way (if possible) of expressing this specific value of the regularized hypergeometric function in terms of known constants. How might I use Mathematica to check

this possibility?

Here is the value

Derivative[{0, 0, 0, 0}, {0, 0, 1}, 0][HypergeometricPFQRegularized][{1, 1, 1, 1}, {2, 2, 2}, -1]

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4

I edited your question. Please don’t offer that kind of “rewards”. Upvote all the answers from all users that you think deserve the upvote instead.

– Dr. belisarius

Jun 16 ’14 at 18:03

4

Or, open a bounty if you feel generous enough.

– Dr. belisarius

Jun 16 ’14 at 18:04

4

Both things are independent. If you want to upvote good Q&A, it’s good and go for it. But you have an specific tool for rewarding good answers: a bounty. Also, if you serially upvote 20 answers from the same user, all fraud alarms will be triggered.

– Dr. belisarius

Jun 16 ’14 at 18:46

3

@belisarius I right on all points. Also see: Please don’t stalker-vote!. You would not be helping the person you were attempting to reward.

– Mr.Wizard♦

Jun 16 ’14 at 19:20

1

You’ll want to read this stuff too: mathematica.stackexchange.com/questions/55708/…

– JEP

Jan 20 ’15 at 20:55

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2 Answers

2

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One can expand the hypergeometric function as a series of the last argument and take the derivative

series[Derivative[n__][f_][args__], k_] :=

Module[{vars = {args} /. Except[_List | List] :> Unique[]},

FullSimplify[# (Last@vars)^k /.

Thread[Flatten@vars -> Flatten@{args}],

Assumptions -> {k âˆˆ Integers, k >= 0}] &@

D[SeriesCoefficient[

FunctionExpand[f @@ vars], {Last@vars, 0, k}], ##] & @@

Transpose@{Flatten@vars, Flatten@{n}}];

simplify[expr_] := Sum[series[expr, k], {k, 0, âˆž}]

The series expansion is

expr = Derivative[{0, 0, 0, 0}, {0, 0, 1}, 0][

HypergeometricPFQRegularized][{1, 1, 1, 1}, {2, 2, 2}, -1];

series[expr, k]

However, the summation returns the initial hypergeometric function

simplify[expr]

Probably, there is no simple form for this expression. However, this method works for another arguments

simplify@Derivative[{0, 0, 0, 0}, {0, 0, 1}, 0][

HypergeometricPFQRegularized][{1, 1, 1, 1}, {2, 2, 2}, 1]

Of course, this method have certain limitations (e.g. series convergence) but sometimes it gives interesting results that are impossible to get with other methods.

@Chris’s sis Please, do no upvote my other answers! Let other people decide how good is this answer.

– ybeltukov

Jan 20 ’15 at 21:11

My answer below was aimed at obtaining a closed form solution for the derivative of the HyperGeometricRegularized function that is needed. The answer that was obtained above is given by the first order term in the series expansion f[x_] = FunctionExpand[HypergeometricPFQRegularized[{1, 1, 1, 1}, {2, 2, x}, -1]].

– JEP

Jan 20 ’15 at 21:36

@JEP It contains the derivative of HypergeometricPFQ. Unfortunately there is some mistakes in further expansion. It seems to me that proper expansion can not be summarized…

– ybeltukov

Jan 20 ’15 at 21:41

This isn’t exactly an answer but perhaps it’s a step in the right direction. Actually with the subsequent edits I think it is an answer. You’ll need to do the calculations yourself and check that I didn’t screw anything up but I think this works and gives you a closed form expression.

f[x_] = FunctionExpand[HypergeometricPFQRegularized[{1, 1, 1, 1}, {2, 2, x}, -1]]

Series[f[x], {x, 2, 2}]

as it gives an answer (i.e. the coefficient of (x-2) ) in terms of HypergeometricPFQ plus other stuff. The documentation has a series expansion for HypergeometricPFQ in terms of these things called Pochhammer symbols which are gamma functions (a)_k = Gamma[1+k]/Gamma[k]=k.

You’ll have p=4, a_1=a_2=a_3=a_4=1 and q=3 with b_1=b_2=2 and b_3=x and z=1. That factor in the series with the “a”s is (Gamma[1+k]/Gamma[k])^4 because p=4 and the “a”s are all unity. Then you have to go through the same reasoning with the “b”s remembering that b_3 carries the x dependence. It’s pretty messy but it might end up giving something reasonable.

EDIT: I managed to beat this into a closed form solution. The series expansion above gives you: (-((3 Zeta[3])/4) + 3/4 EulerGamma Zeta[3] + a derivative of HypergeometricPFQ . Use the series expansion of HypergeometricPFQ[{1,1,1,1},{2,2,2+x},1] around x = 0 will give you that derivative. I found -Sum[PolyGamma[0, 2 + k]/(k^2 (1 + k)^2), {k, 1, Infinity}]=1/6 (36 – 18 EulerGamma – [Pi]^2 + 2 EulerGamma [Pi]^2 – 24 Zeta[3]) for the coefficient of x in the series expansion for HypergeometricPFQ[{1,1,1,1},{2,2,2+x},1]. So your answer should be

(-((3 Zeta[3])/4) + 3/4 EulerGamma Zeta[3] + 1/6 (36 – 18 EulerGamma – [Pi]^2 + 2 EulerGamma [Pi]^2 – 24 Zeta[3]) =

1/12 (72 – 2 [Pi]^2 + EulerGamma (4 [Pi]^2 + 9 (-4 + Zeta[3])) – 57 Zeta[3])

which probably has some mistakes but I don’t think there are any insurmountable difficulties in performing the manipulations. Hmmm. The numerical evaluation of the last formula gives:

-0.667003 . The correct numerical answer is -0.338863 so I messed something up but I think if you write it out carefully as you go you can coax Mathematica into giving a closed form symbolic formula.