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Re: Numerical accuracy of Hypergeometric2F1

• To: mathgroup at smc.vnet.net
• Subject: [mg55733] Re: Numerical accuracy of Hypergeometric2F1
• From: Paul Abbott <paul at physics.uwa.edu.au>
• Date: Tue, 5 Apr 2005 03:20:52 -0400 (EDT)
• Organization: The University of Western Australia
• References: <d2qj7t$p0o$1@smc.vnet.net>
• Sender: owner-wri-mathgroup at wolfram.com

In article <d2qj7t$p0o$1 at smc.vnet.net>,
 "Christos Argyropoulos M.D." <chrisarg at fuse.net> wrote:

> Hi,
> Recently I ran into a problem in applied statistics which required 
> evaluation of a specific Hypergeometric2F1 functional i.e. 
> Hypergeometric2F1[k+1/2,1,3/2,x] where Element[k,Integers], k>0 , 
> Element[x, Reals] and  0<=x<=1.
> It appears that for "large" values of k , Mathematica returns the wrong 
> answer .
> If one re-writes the hypergeometric as a polynomial (using the distant 
> neighbor relation  http://functions.wolfram.com/07.23.17.0007.01), get 
> rid of the ratio of Pochhamer functions and then examine the numerical 
> values given by the two formulations diverge.
> \!\(fpol[x_, k_] :=
>     Module[{}, coeff = Table[1, {k + 1}];
>       For[i = 2, i \[LessEqual] k + 1, \(i++\),
>         coeff[\([i]\)] =
>           coeff[\([i -
>                   1]\)]\ \(\(-k\) + i - 1\)\/\(i - k - 1/2\)]; \
> {Hypergeometric2F1[k + 1/2, 1, 3/2,
>           x], \(1\/\(2  k -
>                 1\)\) \(\[Sum]\+\(i = 1\)\%\(k + 1\)\((\((1 - 
> x)\)\^\(-i\)\ \
> coeff[\([i]\)])\)\)}]\)
> 
> In[4]:=
> SetPrecision[fpol[0.7,100],16]
> Out[4]=
> \!\({\(-2.5109638442451095387366941924357`15.9546*^56\),
>     2.0578481017803705921545468593643`15.9546*^51}\)

No! You are applying SetPrecision to an expression that has been 
computed using machine precision. Instead you want to compute

  fpol[SetPrecision[0.7,20], 100]

or

  fpol[0.7`20, 100]

or 

  N[fpol[7/10,100], 20]

This has nothing to do with any specific Hypergeometric2F1. It is a 
general property of the way Mathematica's works with machine and 
arbitrary precision expressions.

Cheers,
Paul

-- 
Paul Abbott                                   Phone: +61 8 6488 2734
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