MathGroup Archive 1995

[Date Index] [Thread Index] [Author Index]

Search the Archive

Re: Numerical evaluation of HypergeometricPFQ[{a,b,c},{d,e},1]

  • To: mathgroup at christensen.cybernetics.net
  • Subject: [mg508] Re: [mg497] Numerical evaluation of HypergeometricPFQ[{a,b,c},{d,e},1]
  • From: bob Hanlon <hanlon at pafosu2.hq.af.mil>
  • Date: Sun, 5 Mar 1995 14:27:27

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.]
> > 
> > 
> 
> 





  • Prev by Date: Re: Numerical evaluation of HypergeometricPFQ[{a,b,c},{d,e},1]
  • Next by Date: Re: any better to apply a function to second column?
  • Previous by thread: Re: Numerical evaluation of HypergeometricPFQ[{a,b,c},{d,e},1]
  • Next by thread: Re: Numerical evaluation of HypergeometricPFQ[{a,b,c},{d,e},1]