Re: simplifying inside sum, Mathematica 5.1

*To*: mathgroup at smc.vnet.net*Subject*: [mg53737] Re: simplifying inside sum, Mathematica 5.1*From*: "Jens-Peer Kuska" <kuska at informatik.uni-leipzig.de>*Date*: Tue, 25 Jan 2005 05:04:02 -0500 (EST)*Organization*: Uni Leipzig*References*: <ct4h70$av2$1@smc.vnet.net>*Sender*: owner-wri-mathgroup at wolfram.com

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 >