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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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