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Re: Series on HypergeometricPFQ

  • To: mathgroup at smc.vnet.net
  • Subject: [mg126373] Re: Series on HypergeometricPFQ
  • From: Alexei Boulbitch <Alexei.Boulbitch at iee.lu>
  • Date: Fri, 4 May 2012 06:27:18 -0400 (EDT)
  • Delivered-to: l-mathgroup@mail-archive0.wolfram.com

Hi, all

I encountered errors when doing the following expansion:

Series[HypergeometricPFQ[{a, a}, {b, c, d}, x], {x, \[Infinity], 0}]

The error message is:

Table::iterb: "Iterator {K3,0,Ceiling[-Re[a]]} does not have appropriate bounds. "

Any idea what's wrong in the above expansion? Many thanks!

--- work around ---

Alternatively, I could first add a small eps factor in the expansion, and eventually take it to zero:

Series[HypergeometricPFQ[{a, a+eps}, {b, c, d}, x], {x, \[Infinity], 0}]

The result looks reasonable. However it is very complicated and unnatural for a large expression.


Hi,

It seems that you simply have a diverging solution. Indeed, making use of your trick with eps one finds

s = Series[
  HypergeometricPFQ[{a, a + eps}, {b, c, d}, x], {x, \[Infinity], 0}]

((-1)^(1/4 (1 + 4 a - 2 b - 2 c - 2 d + 2 eps)) ((-1)^(1/4) E^(
       I a \[Pi] - (I b \[Pi])/2 - (I c \[Pi])/2 - (I d \[Pi])/2 + (
        I eps \[Pi])/2 - 2 Sqrt[x]) - (-1)^(3/4)
        E^(-I a \[Pi] + (I b \[Pi])/2 + (I c \[Pi])/2 + (I d \[Pi])/
        2 - (I eps \[Pi])/2 + 2 Sqrt[x])) x^(
    1/4 (1 + 4 a - 2 b - 2 c - 2 d + 2 eps))
     Gamma[b] Gamma[c] Gamma[d])/(2 Sqrt[\[Pi]]
     Gamma[a] Gamma[a + eps]) + (Gamma[b] Gamma[c] Gamma[
     d] (((-1)^-a x^-a Gamma[a] Gamma[eps])/(
      Gamma[-a + b] Gamma[-a + c] Gamma[-a + d]) + ((-1)^(-a - eps)
        x^(-a - eps) Gamma[-eps] Gamma[a + eps])/(
      Gamma[-a + b - eps] Gamma[-a + c - eps] Gamma[-a + d -
         eps])))/(Gamma[a] Gamma[a + eps])

If I check now its behaviour at eps->0 I find

s /. \[Epsilon] -> 0

Infinity::indet: Indeterminate expression ComplexInfinity+ComplexInfinity encountered. >>

Indeterminate

It is since there are Gamma[eps] and Gamma[-eps] in this expression. Indeed,

Gamma[\[Epsilon]] /. \[Epsilon] -> 0

ComplexInfinity

However, there are only two terms, one containing Gamma[eps] and the other Gamma[-eps], and they are met in combination in the denominator of your expression. This part of in the denominator (let us call it "trm")is:

trm = ((-1)^-a x^-a Gamma[a] Gamma[eps])/(
   Gamma[-a + b] Gamma[-a + c] Gamma[-a + d]) + ((-1)^(-a - eps)
     x^(-a - eps) Gamma[-eps] Gamma[a + eps])/(
   Gamma[-a + b - eps] Gamma[-a + c - eps] Gamma[-a + d - eps]);

If one goes to eps->0 in term, one finds a finite result (let us call it "trmExp"):

trmExp = Series[trm, {eps, 0, 0}] // Normal // FullSimplify


-((E^(-I a \[Pi])
   x^-a Gamma[
   a] (EulerGamma - I \[Pi] + HarmonicNumber[-1 - a + c] - Log[x] +
    PolyGamma[0, a] + PolyGamma[0, -a + b] + PolyGamma[0, -a + d]))/(
 Gamma[-a + b] Gamma[-a + c] Gamma[-a + d]))

Now one may substitute the term trmExp, that we obtained into the place of trm:

res = (((-1)^(
       1/4 (1 + 4 a - 2 b - 2 c - 2 d + 2 eps)) ((-1)^(1/4) E^(
          I a \[Pi] - (I b \[Pi])/2 - (I c \[Pi])/2 - (I d \[Pi])/
           2 + (I eps \[Pi])/2 - 2 Sqrt[x]) - (-1)^(3/4)
           E^(-I a \[Pi] + (I b \[Pi])/2 + (I c \[Pi])/2 + (
           I d \[Pi])/2 - (I eps \[Pi])/2 + 2 Sqrt[x])) x^(
       1/4 (1 + 4 a - 2 b - 2 c - 2 d + 2 eps))
        Gamma[b] Gamma[c] Gamma[d])/(2 Sqrt[\[Pi]]
        Gamma[a] Gamma[a + eps]) + (Gamma[b] Gamma[c] Gamma[
        d] (trmExp))/(Gamma[a] Gamma[a + eps])) /. eps -> 0

((-1)^(1/4 (1 + 4 a - 2 b - 2 c - 2 d)) ((-1)^(1/4) E^(
     I a \[Pi] - (I b \[Pi])/2 - (I c \[Pi])/2 - (I d \[Pi])/2 -
      2 Sqrt[x]) - (-1)^(3/4)
      E^(-I a \[Pi] + (I b \[Pi])/2 + (I c \[Pi])/2 + (I d \[Pi])/2 +
      2 Sqrt[x])) x^(1/4 (1 + 4 a - 2 b - 2 c - 2 d))
   Gamma[b] Gamma[c] Gamma[d])/(
 2 Sqrt[\[Pi]]
   Gamma[a]^2) - (E^(-I a \[Pi])
     x^-a Gamma[b] Gamma[c] Gamma[
     d] (EulerGamma - I \[Pi] + HarmonicNumber[-1 - a + c] - Log[x] +
      PolyGamma[0, a] + PolyGamma[0, -a + b] +
      PolyGamma[0, -a + d]))/(Gamma[
     a] Gamma[-a + b] Gamma[-a + c] Gamma[-a + d])


So, it seems to be possible to get through, but be careful with this solution. I did it on-the-napkin-like,
and if you like this approach, you should check the correctness of each step.

Have fun, Alexei


Alexei BOULBITCH, Dr., habil.
IEE S.A.
ZAE Weiergewan,
11, rue Edmond Reuter,
L-5326 Contern, LUXEMBOURG

Office phone :  +352-2454-2566
Office fax:       +352-2454-3566
mobile phone:  +49 151 52 40 66 44

e-mail: alexei.boulbitch at iee.lu






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