Re: simplifying inside sum, Mathematica 5.1

*To*: mathgroup at smc.vnet.net*Subject*: [mg53749] Re: simplifying inside sum, Mathematica 5.1*From*: Richard Fateman <fateman at cs.berkeley.edu>*Date*: Wed, 26 Jan 2005 04:36:30 -0500 (EST)*Organization*: University of California, Berkeley*References*: <ct4h70$av2$1@smc.vnet.net> <ct56p1$eca$1@smc.vnet.net>*Sender*: owner-wri-mathgroup at wolfram.com

Since the answer given by this process is 0 instead of the arguably correct a[0], the question remains. (Similarly for Dr. Bob and Andrzej). RJF > Jens-Peer Kuska wrote: Hi, > > since 0^0 is indefined *and* i run over i=0,1,... Mathematica will not > simplify this expression. In other cases you may remove the HoldAll > attribut from Sum[] to force the evaluation or you can try > > s = Sum[a[i]*x^i, {i, 0, Infinity}]; > > (s /. x -> 0) /. Verbatim[Sum][s_, iter__] :> > > (Sum @@ {Simplify[s, Element[i, Integers] && i > 0], > iter}) > > clearly the assumption in the Simplify[] call contradicts > > the bounds of the iterator. > > > > Regards > > Jens > > > "Richard Fateman" <fateman at cs.berkeley.edu> schrieb im Newsbeitrag > news:ct4h70$av2$1 at smc.vnet.net... > >>Sum[a[i]*x^i,{i,0,Infinity}] >> >>%/. x->0 >> >> >>leaves 0^i inside the sum, unsimplified. >> >>So does the sum from 1 to Inf. >> >>The correct answers are presumably a[0] and 0. >> >>Is there a way to get Mathematica to do this? >> >>RJF >> > > >

**Follow-Ups**:**Re: Re: simplifying inside sum, Mathematica 5.1***From:*DrBob <drbob@bigfoot.com>