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Re: simplifying inside sum, Mathematica 5.1
*To*: mathgroup at smc.vnet.net
*Subject*: [mg53726] Re: [mg53709] simplifying inside sum, Mathematica 5.1
*From*: Andrzej Kozlowski <akoz at mimuw.edu.pl>
*Date*: Tue, 25 Jan 2005 05:03:25 -0500 (EST)
*References*: <200501240837.DAA28210@smc.vnet.net> <B0C62081-6E02-11D9-890F-000A95B4967A@mimuw.edu.pl>
*Sender*: owner-wri-mathgroup at wolfram.com
On 24 Jan 2005, at 13:23, Andrzej Kozlowski wrote:
>
> On 24 Jan 2005, at 09:37, Richard Fateman wrote:
>
>> 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
>>
>>
>>
>
> First, note that even if Mathematica did "simplify" 0^i inside the sum
> it would not return what you are (presumably) expecting since:
>
> Sum[0*a[i],{i,0,Infinity}]
>
> Indeterminate expression 0*Infinity encountered.
>
> Indeterminate
>
>
>
> The way to deal with this is to replace Infinity by some symbol m and
> after evaluating replace m by Infinity:
>
>
> Sum[0*a[i],{i,0,m}]/.m->Infinity
>
> 0
>
> Curiously using Hold[Infinity] in place of Infinity does not work:
>
> In[18]:=
> Sum[0*a[i], {i, 0, Hold[Infinity]}]
>
> Indeterminate expression 0*Infinity encountered.
>
> Indeterminate
>
> Nevertheless, this works:
>
> ReleaseHold[Sum[a[i]*x^i, {i, 0, Hold[Infinity]}] /.
> x -> 0 /. 0^(i_) :> KroneckerDelta[i, 0]]
>
>
> a[0]
>
> or
>
>
> Sum[a[i]*x^i, {i, 0, Hold[Infinity]}];
>
>
> ReleaseHold[Block[{Power},
> 0^(i_) := KroneckerDelta[i, 0]; % /. x -> 0]]
>
> a[0]
>
> If you prefer you can use m and m->Infinity instead of Hold and
> ReleaseHold and you could of course globally re-define 0^i as
> KroneckerDelta[i, 0].
>
>
>
> Andrzej Kozlowski
>
Perhaps a slightly more elegant approach is to do everything inside
Block, as follows:
Block[{Power,Infinity},
0^(i_) := KroneckerDelta[i, 0]; Sum[a[i]*x^i, {i, 0, Infinity}]/. x
-> 0]
a(0)
Andrzej Kozlowski
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