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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
> > >
> >
> >
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