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 > >