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Re: Simplification with Integers assumption



Sorry, I fooled myself by misstyping Element[x,Integers] instead of 

Element[n,Integers].

With this corrected, I get for: Simplify[Sum[n!/(2*p)!/(n - 2*p)!, {p, 

1, Infinity}], Assumptions -> n > 0 && Element[n, Integers]] zero, what 

is obviously wrong.

Daniel



dh wrote:

> $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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