Re: Accuracy problem in Mathematica
- To: mathgroup at smc.vnet.net
- Subject: [mg49107] Re: Accuracy problem in Mathematica
- From: "Steve Luttrell" <steve_usenet at _removemefirst_luttrell.org.uk>
- Date: Thu, 1 Jul 2004 05:26:01 -0400 (EDT)
- References: <cbu21o$5ea$1@smc.vnet.net>
- Sender: owner-wri-mathgroup at wolfram.com
Mathematica knows how to explicitly sum this series. In Mathematica 5 the
following input:
1/(1 - a^2)^n +
Sum[((2*n - k - 1)!/((n - 1)!*(n - k)!*2^(2*n - k)))*(1/(1 + a)^k -
1/(1 - a)^k), {k, 1, n}]
gives the following output
(1 - a^2)^(-n) - (2^(1 -
2*n)*Pi*Csc[2*n*Pi]*(a*HypergeometricPFQRegularized[{1, 1 - n}, {2 -
2*n}, -(2/(a - 1))] +
HypergeometricPFQRegularized[{1, 1 - n}, {2 - 2*n}, -(2/(a - 1))] +
a*HypergeometricPFQRegularized[{1, 1 - n}, {2 - 2*n}, 2/(a + 1)] -
HypergeometricPFQRegularized[{1, 1 - n},
{2 - 2*n}, 2/(a + 1)]))/((a - 1)*(a + 1)*(n - 1)!*Gamma[n])
I have not attempted to simplify this further.
Steve Luttrell
"aaaa" <aaa at huji.ac.il> wrote in message news:cbu21o$5ea$1 at smc.vnet.net...
> Hello,
>
> I'm having a problem with a calculation in Mathematica which I can't
> solve. I have an expression which I know to be (from analytical reasons)
> always between 0 and 1. It's a function of a and n ( n being natural and
> a rational) and it looks like this:
>
> 1/(1-a^2)^n +
>
> Sum[((2*n - k - 1)!/((n - 1)!*(n - k)!*2^(2*n - k)))*(1/(1 + a)^k - 1/(1
> - a)^k), {k, 1, n}]
>
>
>
> Let's say a=0.5.
>
> Now, when I try to calculate for small n, it's ok. When calculating for
> large n's (around 400 and above) I'm starting to get wrong results (the
> number not being between 0 and 1). The problem is that the first term
> (the first line before the sum) is VERY VERY close to the negative of
> the second term (the sum), and it's getting closer as n grows. When
> using large n's, Mathematica says they are the same number or even that
> the last term is bigger (which means the whole expression becomes
> negative) - which is wrong. It's a matter of accuracy, and I'm not sure
> how I can fix it.
>
> Can anybody help me?
>
>
>
> Itamar
>