Re: Simplification with Integers assumption
- To: mathgroup at smc.vnet.net
- Subject: [mg74726] Re: Simplification with Integers assumption
- From: dh <dh at metrohm.ch>
- Date: Mon, 2 Apr 2007 06:58:01 -0400 (EDT)
- References: <eunqc2$7ic$1@smc.vnet.net>
$Version 5.1 for Microsoft Windows (October 25, 2004)
Hi,
works o.k. in my version:
Simplify[Sum[n!/(2*p)!/(n - 2*p)!, {p, 1, Infinity}], Assumptions -> n >
0] and Simplify[Sum[n!/(2*p)!/(n - 2*p)!, {p, 1, Infinity}], Assumptions
-> n > 0 && Element[x, Integers]] give:
-(((-2 + 2^n)*n!*Gamma[-n]*Sin[n*Pi])/(2*Pi))
and with FullSimplify both give the simpler:-1 + 2^(-1 + n)
Daniel
did wrote:
> On Mathematica 5.2 Windows, with the 4 similar commands:
>
> Simplify[ Sum[ n! / (2*p)! / (n - 2*p)! , {p, 1, Infinity} ] ,
> Assumptions -> n > 0]
>
> FullSimplify[ Sum[ n! / (2*p)! / (n - 2*p)! , {p, 1, Infinity} ] ,
> Assumptions -> n > 0]
>
> Simplify[ Sum[ n! / (2*p)! / (n - 2*p)! , {p, 1, Infinity} ] ,
> Assumptions -> n > 0 && n =E2=88=88 Integers]
>
> FullSimplify[ Sum[ n! / (2*p)! / (n - 2*p)! , {p, 1, Infinity}] ,
> Assumptions -> n > 0 && n =E2=88=88 Integers]
>
>
> I get the different answers:
>
> Out[1]= -(-2 + 2^n) n! Gamma[n] Sin[n Pi] / (2 Pi)
>
> Out[2]= -1 + 2^(-1+n)
>
> Out[3]= 0
>
> Out[4]= 0
>
> Outputs 1 & 2 look OK, but 3 & 4 are not. It seems that, with the
> assumption n Integer,
> Mathematica simplifies Sin[n Pi] by 0, omitting that Gamma[-n] is infinite.
> Is it the expected behavior?
>
> In this example, the simplest form can be obtained without imposing n
> Integer (I
> presume it's the correct answer), but in other situations it will be
> required. What
> is the safe way to do it?
>
> Thanks
>
>