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MathGroup Archive 2005

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Re: Re: simplifying inside sum, Mathematica 5.1

  • To: mathgroup at smc.vnet.net
  • Subject: [mg53773] Re: [mg53749] Re: simplifying inside sum, Mathematica 5.1
  • From: Andrzej Kozlowski <akoz at mimuw.edu.pl>
  • Date: Thu, 27 Jan 2005 05:41:08 -0500 (EST)
  • Sender: owner-wri-mathgroup at wolfram.com

I know it is much to expect,  but how about trying to read postings 
before replying to them or commenting on them? Just to save you the 
trouble here it it again:

Block[{Power,Infinity},
     0^(i_) := KroneckerDelta[i, 0]; Sum[a[i]*x^i, {i, 0, Infinity}]/. x
-> 0]

a(0)


Andrzej




On 26 Jan 2005, at 09:36, Richard Fateman wrote:

>
> 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
>>>
>>
>>
>>
>
>


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