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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    $\begingroup$ 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. $\endgroup$ Jun 16 '14 at 18:03
  • 4
    $\begingroup$ Or, open a bounty if you feel generous enough. $\endgroup$ Jun 16 '14 at 18:04
  • 4
    $\begingroup$ 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. $\endgroup$ Jun 16 '14 at 18:46
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    $\begingroup$ @belisarius is right on all points. Also see: Please don't stalker-vote!. You would not be helping the person you were attempting to reward. $\endgroup$
    – Mr.Wizard
    Jun 16 '14 at 19:20
  • 1
    $\begingroup$ You'll want to read this stuff too: mathematica.stackexchange.com/questions/55708/… $\endgroup$
    – JEP
    Jan 20 '15 at 20:55

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 ∈ Integers, k >= 0}] &@
        FunctionExpand[f @@ vars], {Last@vars, 0, k}], ##] & @@ 
    Transpose@{Flatten@vars, Flatten@{n}}];

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

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]

enter image description here

However, the summation returns the initial hypergeometric function


enter image description here

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]  

enter image description here

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.

  • $\begingroup$ @Chris's sis Please, do no upvote my other answers! Let other people decide how good is this answer. $\endgroup$
    – ybeltukov
    Jan 20 '15 at 21:11
  • $\begingroup$ 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]]. $\endgroup$
    – JEP
    Jan 20 '15 at 21:36
  • $\begingroup$ @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... $\endgroup$
    – 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.


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