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Hypergeometric and MeijerG

  • To: mathgroup at smc.vnet.net
  • Subject: [mg53107] Hypergeometric and MeijerG
  • From: Maxim <ab_def at prontomail.com>
  • Date: Thu, 23 Dec 2004 07:59:45 -0500 (EST)
  • Sender: owner-wri-mathgroup at wolfram.com

Most of the problems described at
http://forums.wolfram.com/mathgroup/archive/2004/Jul/msg00185.html
still remain in Mathematica 5.1, except for the two examples with  
FunctionExpand of MeijerG function, where the errors have been fixed.  
However, it is easy to construct more examples of the same kind:

In[1]:=
MeijerG[{{1/2, 1/2}, {}}, {{0}, {}}, 1/2] // FunctionExpand

Out[1]=
-3*BesselI[0, 1]*(E*EulerGamma*Sqrt[2*Pi] + E*Sqrt[2*Pi]*Log[2]) -
   2*Sqrt[2*Pi]*Derivative[0, 1, 0][Hypergeometric1F1][1/2, 1, 2]

The correct value is (-Sqrt[2*Pi])*(E*BesselI[0, 1]*(EulerGamma - Log[2])  
+ 2*Derivative[0, 1, 0][Hypergeometric1F1][1/2, 1, 2] + Derivative[1, 0,  
0][Hypergeometric1F1][1/2, 1, 2]). Curiously, sometimes it works the other  
way around, when Mathematica internally uses some incorrect identities to  
numerically evaluate MeijerG, and FunctionExpand avoids that:

In[2]:=
MeijerG[{{1, 2}, {}}, {{2}, {}}, -1] // N
MeijerG[{{1, 2}, {}}, {{2}, {}}, -1] // FunctionExpand // N

Out[2]=
-0.3028251167649334  
+ 1.15572734979092195349743323680174483579`19.98675271785018*I

Out[3]=
-0.30282511676493384 - 1.1557273497909217*I

The second one is correct. I believe that the transformation rules used by  
Mathematica sometimes do not have a corresponding condition which should  
specify the domain of the parameter values where the identity represented  
by the rule holds (e.g., true only if Re[z] > 0, and so on). In such case  
the errors like these are unavoidable.

Even automatic simplifications may give an incorrect result:

In[4]:=
MeijerG[{{1}, {}}, {{-(1/2)}, {}}, -(1/2)]

Out[4]=
-2*I*Sqrt[Pi]

The correct value is 2*I*Sqrt[Pi]. Taking limits of the hypergeometric  
function at the points of discontinuity is still wrong as well:

In[5]:=
Limit[Hypergeometric2F1[n, 2*n, (n + 1)^2 - 3, x], n -> -1]

Out[5]=
1 - (2*x)/3

The correct value is 1 - (8*x)/3 + x^2/3 + (-2 + (4*x)/3)*Log[1 - x].

Maxim Rytin
m.r at inbox.ru


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