Re: correction to my last post
- To: mathgroup at smc.vnet.net
- Subject: [mg36033] Re: correction to my last post
- From: Andrzej Kozlowski <andrzej at tuins.ac.jp>
- Date: Tue, 13 Aug 2002 05:22:56 -0400 (EDT)
- Sender: owner-wri-mathgroup at wolfram.com
Infinity! == Product[k, {k, 1, Infinity}] == Infinity! == Product[k, {k,
1, Infinity}] == E^-(D[Sum[k^(-s), {k, 1, Infinity}], s] /. s -> 0) ==
E^-Derivative[1][Zeta][0] == Sqrt[2*Pi]
In fact Mathematica can evaluate the last three equalities:
In[12]:=
E^(-(D[Sum[k^(-s), {k, 1, Infinity}], s] /. s -> 0)) ==
E^(-Derivative[1][Zeta][0]) == Sqrt[2*Pi]
Out[12]=
True
while the first four are obvious, aren't they :)
Andrzej
On Tuesday, August 13, 2002, at 06:13 AM, David W. Cantrell wrote:
> [Message also posted to: comp.soft-sys.math.mathematica]
>
> Andrzej Kozlowski <andrzej at tuins.ac.jp> wrote:
>> In[2]:=
>> Limit[(1 + y/Abs[x])^Abs[x], x -> Infinity]
>>
>> Out[2]=
>> E^(I*Im[y] + Re[y])
>>
>> The latter can of course be made equal to any number, real or complex
>> except 0.
>
> Even 0 is possible. Take y to be -Infinity.
>
>> So the whole original confusion relates to the meaning of
>> 1^Infinity. If one interpreted it in the first sense (or simply as
>> Limit[1^x,x->Infinity]) than the answer would have been 1. However,
>> Mathematica adopts the more general approach, considering an expression
>> involving Infinity to be Indeterminate unless all ways of representing
>> it as a limit lead to the same answer. Of course one should not take
>> this too strictly, for Mathematica gives:
>>
>> In[3]:=
>> Infinity!
>>
>> Out[3]=
>> Infinity
>
> This seems to be obviously correct.
>
>> although a pretty good case can be made for Sqrt[2Pi], see
>> <http://functions.wolfram.com/10.01.06.0013> :)
>
> Could you please explain "how a pretty good case can be made for
> Sqrt[2Pi]"? I looked at the link, but could not see any way to make
> a case for Infinity! being Sqrt[2Pi].
>
> David
>
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Andrzej Kozlowski
Toyama International University
JAPAN
http://platon.c.u-tokyo.ac.jp/andrzej/