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

>

>

```

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