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Re: A question regarding a hyperbolic geometric function

• To: mathgroup at smc.vnet.net
• Subject: [mg86466] Re: A question regarding a hyperbolic geometric function
• From: "David W.Cantrell" <DWCantrell at sigmaxi.net>
• Date: Wed, 12 Mar 2008 00:10:43 -0500 (EST)
• References: <200803100704.CAA24775@smc.vnet.net> <fr5e09$o5p$1@smc.vnet.net>

"Ali K. Ozdagli" <ozdagli at gmail.com> wrote:
> Hi Bob and all,
>
> I originally figured this problem while I was working with a second
> order ODE of which solution involves Hypergeometric1F1. I have plugged
> the solution back to the ODE and it turned out that the equation is
> not satisfied for all values of the independent variable. I know the
> solution is correct as I have checked it both with Mathematica and
> another software. The mistake in the result becomes apparent when you
> check the accuracy and precision with Accuracy[] and Precision[]. For
> example, Hypergeometric1F1[-26.9, -20.1, 30000] gives
>
> 5.868090710725125*10^13007
>
> However, if you check accuracy and precision of the result you get
>
> Accuracy[Hypergeometric1F1[-26.9, -20.1, 30000]] = -12991.8 (negative is
> bad!!!)
> Precision[Hypergeometric1F1[-26.9, -20.1, 30000]] = 15.9546
>
> Where the latter is also the machine precision in Mathematica. This
> means that the results are precisely wrong although we want them to be
> roughly correct. :)
>
> Any ideas about this?

Both Accuracy and Precision are as one should expect! And they have nothing
to do with the particular function, Hypergeometric1F1. Consider that

In[42]:= Accuracy[6. * 10^13007]

Out[42]= -12991.8

In[43]:= Precision[6. * 10^13007]

Out[43]= 15.9546

give the same values you had.

Now let's look at an example with a smaller third argument:

In[45]:= N[Hypergeometric1F1[-269/10, -201/10, 3000]]

Out[45]= 4.389014940203321*10^1288

Note that this is still a very large number. It should be no surprise that

In[46]:= Accuracy[N[Hypergeometric1F1[-269/10, -201/10, 3000]]]

Out[46]= -1272.69

And it's easy to predict that, if we want Accuracy to be positive, we'll
need to request at least 1289 digits.

In[47]:= Accuracy[N[Hypergeometric1F1[-269/10, -201/10, 3000], 1289]]

Out[47]= 0.357633

BTW, I'm quite skeptical that you actually _need_ such Accuracy. For most
purposes, high Precision is adequate.

David

> On 3/10/08, Bob Hanlon <hanlonr at cox.net> wrote:
> > 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
> >  >
> >
> >

• References:
  • A question regarding a hyperbolic geometric function
    • From: "Ali K. Ozdagli" <ozdagli@gmail.com>