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Re: HypergeometricPFQ

  • To: mathgroup at
  • Subject: Re: HypergeometricPFQ
  • From: victor
  • Date: Thu, 19 Nov 92 12:00:36 CST

This is a solution of your problem:

Unprotect[ HypergeometricPFQ]

 HypergeometricPFQ[{a_, b_}, {c_}, arg_ ] := 

	Hypergeometric2F1[ a, b, c, arg] /;
(And @@ (NumberQ /@ {a,b,c,arg})) && Accuracy[{a,b,c,arg}] < Infinity

Protect[ HypergeometricPFQ]

Begin forwarded message:

Date: Wed, 18 Nov 92 14:56:12 CST
From: lsf at (Sam Finn)
To: mathgroup at
Subject: HypergeometricPFQ

Re: HypergeometricPFQ 

    Mathematica 2.1 on a SPARC2

The evaluation rules for HypergeometricPFQ are messed up: 

In[1]:= Integrate[1/(x^(7/3)(x^2+1.44^2)),{x,1,Infinity}]

                                  5    8
Out[1]= 0.3 HypergeometricPFQ[{1, -}, {-}, -2.0736]
                                  3    3

In[2]:= N[%]

Out[2]= 0.3 HypergeometricPFQ[{1.,1.66667}, {2.66667}, -2.0736]

Now, I contacted support at with this bug, and after a week of
waiting for a reply I phoned. I was told that they don't know why this
is happening, but that it is fixed in 2.2. No, 2.2 is not available: it
is Wolfram internal. The only workaround is to hand substitute into
Hypergeometric2F1. When asked if a bug fix would be available for 2.1,
I was told that it is fixed in 2.2.

Since Wolfram won't help (except when it comes to spending my money), I
thought maybe someone in the mathgroup has been through this and
tracked down the problem (probably in HypergeometricPFQ.m). Regardless,
I thought you might be interested in hearing a testimonial to the
quality technical support that is yours when you become a Mathematica
Plus subscriber. 

Thanks for any help you can offer,

L. S. Finn
Physics and Astronomy
Northwestern University

P.S. And also, will somebody please remind me why I even bother trying
to make productive use of Mathematica?

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