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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>