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MathGroup Archive 2007

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Re: Simplification with Integers assumption

  • To: mathgroup at smc.vnet.net
  • Subject: [mg74723] Re: Simplification with Integers assumption
  • From: "David W.Cantrell" <DWCantrell at sigmaxi.net>
  • Date: Mon, 2 Apr 2007 06:56:29 -0400 (EDT)
  • References: <eunqc2$7ic$1@smc.vnet.net>

"did" <didier.oslo at hotmail.com> 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)

........................Gamma[-n] is what you intended above, I believe.

> 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?

Well, it certainly does not surprise me.

But your post did lead me to find something surprising. Suppose we fix n as
a positive integer -- say, 3 for example:

In[6]:= Sum[3/((2p)!(3 - 2p)!), {p, 1, Infinity}]

results in nothing but warning messages and gibberish. OTOH, if we replace
Infinity with a large integer -- say, 100 or 1000 -- Mathematica has no
trouble giving the correct answer.

In[7]:= Sum[3/((2p)!(3 - 2p)!), {p, 1, 1000}]

Out[7]= 3/2

Of course, that answer is _also_ the correct one for the infinite sum at
which Mathematica balked. Why couldn't Mathematica do the infinite sum? Is
this problem fixed in the development version perhaps?

David W. Cantrell


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