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Re: A question regarding a hyperbolic geometric function

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  • Subject: [mg86450] Re: [mg86396] A question regarding a hyperbolic geometric function
  • From: Bob Hanlon <hanlonr at cox.net>
  • Date: Tue, 11 Mar 2008 03:02:11 -0500 (EST)
  • Reply-to: hanlonr at cox.net

v1 = Hypergeometric1F1[-26.9, -20.1, 30000]

5.86809071072512533577549124209830605005026`\
 15.954589770191005*^13007

Use exact values until evaluating the Hypergeometric1F1 with extended precision

v2 = N[Hypergeometric1F1[-269/10, -201/10, 30000], 100]

5.86809071072500808515263985241332834422719\
 \
6477241041525402740155487068020893691436344\
 \
4849172908349727305685895816448516416585189\
 1451997662`100.*^13007

Although even without extra precision v1 is very close to v2

(v1 - v2)/v2

1.99810515261962031039159`2.2552081099371497\
  *^-14


Bob Hanlon

---- "Ali K. Ozdagli" <ozdagli at gmail.com> wrote: 
> Hi,
> 
> I am working with Mathematica in order to solve an ordinary
> differential equation with several boundary conditions. It turned out
> that the solution is Kummer confluent hypergeometric function,
> HyperGeometric1F1[a,b,x]. My problem is that for the values of a, b
> and x I am interested in, e.g. a=-26.9, b=-20.1, x=300000, the
> numerical accuracy of Mathematica is very poor.
> 
> Can somebody suggest me a way to improve the mathematical accuracy of
> HyperGeometric1F1? I prefer a quick and easy way but also appreciate
> any hard way.
> 
> Best,
> 
> Ali
> 
> -- 
> 
> Ali K. Ozdagli
> Ph.D. Student in Economics
> at University of Chicago
> 



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