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.] > > > > > > > > > > > >