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MathGroup Archive 2007

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Simplification with Integers assumption

  • To: mathgroup at smc.vnet.net
  • Subject: [mg74709] Simplification with Integers assumption
  • From: "did" <didier.oslo at hotmail.com>
  • Date: Sun, 1 Apr 2007 04:19:44 -0400 (EDT)

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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