       Re: Re: approximation for partitial binomial sum

• To: mathgroup at smc.vnet.net
• Subject: [mg36403] Re: [mg36381] Re: [mg36239] approximation for partitial binomial sum
• From: Daniel Lichtblau <danl at wolfram.com>
• Date: Wed, 4 Sep 2002 21:22:17 -0400 (EDT)
• References: <200209040656.CAA28091@smc.vnet.net>
• Sender: owner-wri-mathgroup at wolfram.com

```Constantine wrote:
>
> Hi.
> I want to get some F and R such that:
>
> F[n,p] + R[n,p] =  Sum[Binomial[n,k] p^(n-k) (1-p)^k, {k, 0, Floor[n/2] - 1}],
> when F[n,p] is an approximation to the sum and the R is the remaining error.
>
> Thanks in advance for any hint.
> Constantine.
>
> At 06:34 AM 8/28/2002 -0400, you wrote:
>
> >In a message dated 8/28/02 4:44:13 AM, celster at cs.technion.ac.il writes:
> >
> >
> >>I'm looking for a way of finding the approximation for partitial binomial
> >>sum.
> >>I'll be pleasant for any hint..
> > [...]
> Office: Taub 411
> Tel: +972 4 8294375

You can get a closed form in terms of special functions if you split
into two cases depending on whether n is even or odd.

In:= n = 2*m;

In:= InputForm[Sum[Binomial[n,k]*p^(n-k)*(1-p)^k, {k,0,m-1}]]
Out//InputForm=
p^(2*m)*((p^(-1))^(2*m) - ((-4 + 4/p)^m*Gamma[1/2 + m]*
Hypergeometric2F1[1, -m, 1 + m, (-1 + p)/p])/(Sqrt[Pi]*Gamma[1 +
m]))

In:= n = 2*m+1;

In:= InputForm[Sum[Binomial[n,k]*p^(n-k)*(1-p)^k, {k,0,m-1}]]
Out//InputForm=
p^(1 + 2*m)*((p^(-1))^(1 + 2*m) - (2^(1 + 2*m)*(-1 + p^(-1))^m*Gamma[3/2
+ m]*
Hypergeometric2F1[1, -1 - m, 1 + m, (-1 + p)/p])/(Sqrt[Pi]*Gamma[2 +
m]))

Daniel Lichtblau
Wolfram Research

```

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