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MathGroup Archive 2007

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Re: Simplification with Integers assumption

  • To: mathgroup at smc.vnet.net
  • Subject: [mg74737] Re: Simplification with Integers assumption
  • From: "Sem" <sarner2006-sem at yahoo.it>
  • Date: Tue, 3 Apr 2007 00:25:53 -0400 (EDT)
  • References: <eunqc2$7ic$1@smc.vnet.net>

Hi,
I tried in Math 4 (WinXP) your expression:

    In[1]:= FullSimplify[ Sum[ n! / (2*p)! / (n - 2*p)! , {p, 1, Infinity}] 
, {n > 0 && n \[Element] Integers}]
    Out[1]= -1 + 2^(-1+n)

    In[2]:= Simplify[ Sum[ n! / (2*p)! / (n - 2*p)! , {p, 1, Infinity}] , {n 
 > 0 &&
      n \[Element] Integers}]
    Out[2]= (-2+2^n) n! / (2 Gamma[1+n])

Math 5.2 makes me puzzled!

Regards, Sem.

"did"
> 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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