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Re: Numerical evaluation of HypergeometricPFQ[{a,b,c},{d,e},1]
*To*: mathgroup at christensen.cybernetics.net
*Subject*: [mg512] Re: [mg497] Numerical evaluation of HypergeometricPFQ[{a,b,c},{d,e},1]
*From*: bob Hanlon <hanlon at pafosu2.hq.af.mil>
*Date*: Mon, 6 Mar 1995 19:29:21
In my earlier post I was careless and left off a Sign[u] factor in dealing
with the even integers. This would result in a sign error for negative
even integers. I concluded that the easiest way to deal with the negative
integers is to use the negative symmetry of the Sin function for all
negative values. I also made a few other minor changes.
Bob Hanlon
___________
theIntegral::usage = "theIntegral[u] is the integral of
Sin[u t]/Sin[t] on the interval [0, Pi/2].";
theIntegral[ u_?Negative ] := - theIntegral[ -u ];
(* Sin[u t] has negative symmetry *)
theIntegral[ u_ /; IntegerQ[(u - 1)/2] ] := Pi/2;
(* General expression is indeterminate for odd integers. *)
(* Odd intgers: Gradshteyn & Ryzhik, 3.612.3 *)
theIntegral[ u_ /; IntegerQ[u/2] ] :=
2 Sum[(-1)^(k-1) / (2k - 1), {k, u/2}];
(* Provides exact numeric results for even integers. *)
(* Even integers: Gradshteyn & Ryzhik, 3.612.4 *)
theIntegral[ u_ /; Chop[u - Round[u]] == 0 ] := theIntegral[ Round[u] ];
(* use integers for their real equivalents *)
theIntegral[ u_ ] := Cos[u Pi/2] (- PolyGamma[(1 - u)/4] +
PolyGamma[(3 - u)/4] + PolyGamma[(1 + u)/4] -
PolyGamma[(3 + u)/4]) / 4;
(* PolyGamma[z] = PolyGamma[0, z] is the digamma function, *)
(* that is, logarithmic derivative of the gamma function. *)
On Sun, 5 Mar 1995, bob Hanlon wrote:
> The specific result (Integrate[Sin[u t]/Sin[t], {t, 0, Pi/2}]) is given below.
>
>
> Bob Hanlon
> hanlon at pafosu2.hq.af.mil
> ____________________
>
> Integrate[Sin[u t]/Sin[t], {t, 0, Pi/2}] // Simplify
>
> Pi u 1 3 - u 3 + u
> u Cos[----] HypergeometricPFQ[{-, 1, 1}, {-----, -----}, 1]
> 2 2 2 2
> -----------------------------------------------------------
> 2
> 1 - u
>
> Clear[iss];
>
> iss::usage = "iss[u] evaluates the integral: \n
> Integrate[Sin[u t]/Sin[t], {t, 0, Pi/2}]";
>
> iss[ u_ /; u == 0 ] = 0;
>
> iss[ u_ /; IntegerQ[(u - 1)/2] ] := Pi/2 Sign[u];
> (* general expression is indeterminate for odd integers *)
> (* Gradshteyn & Ryzhik, 3.612.3 *)
>
> iss[ u_ /; IntegerQ[u/2] ] := 2 Sum[(-1)^(k-1) / (2k - 1),
> {k, u/2 Sign[u]}]; (* Gradshteyn & Ryzhik, 3.612.4 *)
>
> iss[ u_ ] := - Cos[u Pi/2] (PolyGamma[(1 - u)/4] -
> PolyGamma[(3 - u)/4] - PolyGamma[(1 + u)/4] +
> PolyGamma[(3 + u)/4]) / 4;
> (* PolyGamma[z] = PolyGamma[0, z] is the digamma function,
> that is, the logarithmic derivative of the gamma function *)
>
> Plot[{iss[u], Pi/2 Sign[u]}, {u, -5, 11}];
>
> _______________
>
> > On Wed, 1 Mar 1995, NELSON M. BLACHMAN wrote:
> >
> > > This is a postscript to my 26 February message (reproduced below)
> > > complaining about Mma's not giving a numerical value for
> > > HypergeometricPFQ[{1/2,1,1},{1.45,1.55},1].
> > >
> > > I've since found that Mma does evaluate things like
> > > HypergeometricPFQ[{1/2,1,1},{1.4,1.6},0.99999], but it takes
> > > something like an hour on my 486DX33 PC--and it takes longer
> > > and longer as the last argument gets closer and closer to 1.
> > > So it's good that Mma quickly announces its inability to
> > > compute HypergeometricPFQ[{1/2,1,1},{1.4,1.6},1].
> > >
> > > A different method is evidently needed when the last argument is
> > > 1--provided that the sum of the components of the middle argument
> > > exceeds the sum of the components of the first argument. (If not,
> > > the hypergeometric function seems likely to be infinite.) So I
> > > continue to hope for a formula for 3F2[{a,b,c},{d,e},1] as a
> > > ratio of products of gamma functions of (e, f, and e + f minus
> > > 0, a, b, c, a + b, etc.), though I've so far been unable to
> > > devise a satisfactory conjecture of this sort.
> > >
> > > Nelson M. Blachman
> > >
> > >
> > > >From: GTEWD::BLACHMAN "NELSON M. BLACHMAN" 26-FEB-1995 23:54:14.57
> > > >To: MX%"mathgroup at christensen.cybernetics.net"
> > > >CC: BLACHMAN
> > > >Subj: Numerical Evaluation of HypergeometricPFQ
> > >
> > > I was pleased to see just now that Mma's able to evaluate
> > > Integrate[Sin[t u]/Sin[t],{t,0,Pi/2}] in terms of HypergeomtricPFQ.
> > > When I asked it to plot the result, however, I found it apparently
> > > unable to determine numerical values for HypergeomtricPFQ.
> > >
> > > Maple can sometimes compute numerical values for HypergeomtricPFQ,
> > > but for the particular denominator indices here it complains of
> > > iteration limits' being exceeded. Does anyone know of a way to
> > > get Mma 2.2 to evaluate HypergeomtricPFQ numerically?
> > >
> > > I suspect there's a simple expression for HypergeomtricPFQ[{ },{ },1]
> > > in terms of gamma functions when P = Q + 1; there is, anyhow, if
> > > P = 2 and Q = 1. Does anyone know if that's true and, if so, what
> > > it is?
> > > Nelson M. Blachman
> > > GTE Government Systems Corp.
> > > Mountain View, California
> > >
> > > Mathematica 2.2 for DOS 387
> > > Copyright 1988-93 Wolfram Research, Inc.
> > >
> > > In[1]:= Integrate[Sin[t u]/Sin[t],{t,0,Pi/2}]
> > >
> > > 3 u Pi u 1 3 - u 3 + u
> > > (- + -) u Cos[----] HypergeometricPFQ[{-, 1, 1}, {-----, -----}, 1]
> > > 2 2 2 2 2 2
> > > Out[1]= -------------------------------------------------------------------
> > > 3 u 1 u 1 u
> > > 4 (-(-) - -) (-(-) - -) (- - -)
> > > 2 2 2 2 2 2
> > >
> > > In[2]:= f[t_]:= %1 /. u -> t Simplify ELIMINATES THE FIRST FACTOR IN
> > > THE NUMERATOR AND IN THE DENOMINATOR.
> > > In[3]:= f[.1] // N
> > >
> > > Out[3]= 0.0997665 HypergeometricPFQ[{0.5, 1., 1.}, {1.45, 1.55}, 1.]
> > >
> > >
> >
> >
>
>
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