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 > >