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MathGroup Archive 2005

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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
> 



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