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Re: Re: 1^Infinity

  • To: mathgroup at smc.vnet.net
  • Subject: [mg38290] Re: [mg38227] Re: 1^Infinity
  • From: Andrzej Kozlowski <akoz at mimuw.edu.pl>
  • Date: Tue, 10 Dec 2002 04:18:47 -0500 (EST)
  • Sender: owner-wri-mathgroup at wolfram.com

Mathematica generally uses indeterminate for expressions which are by 
nature ambiguous, which means they can be represented as limits leading 
to different answers. This is the case with 1^Infinity. For example, 
consider
(2^(1/n))^n and let n tend to Infinity. This can be "thought of as " 
1^Infinity yet it remains all the time equal to 2.
I assume the same is true with other expressions where Indeterminate is 
returned. Of course this is not "mathematically" the only valid 
approach, one adapt various conventions, for after all these are but 
conventions. But personally I find it rather convenient and have got 
programming examples which only work because Indeterminate rather than 
a number are returned in similar cases. I personally think this answer 
is on the whole more flexible than any alternative, and I would vote 
for keeping it.

Andrzej Kozlowski


On Thursday, December 5, 2002, at 05:36 PM, Steven Shippee wrote:

> Isn't this a bit like trying to divide by zero?
>
> Infinity + Infinity = Infinity; however, one can not subtract infinity 
> from
> infinity or or divide by it.
>
> Steven Shippee
> shippee at jcs.mil
>
> "Ersek, Ted R" <ErsekTR at navair.navy.mil> wrote in message
> news:askfsv$l1n$1 at smc.vnet.net...
>> Hello,
>>  Consider this:
>>
>>     In[1]:=
>>       1^Infinity
>>
>>     Out[1]=
>>       Indeterminate
>>
>>  ---------
>>  I agree 1.0^Infinity  is Indeterminate because 1.0 might be a bit 
>> less
>>  than 1 or a bit greater than 1.
>>  but isn't  (1*1*1* ..... *1)  simply the integer 1.
>>
>> ---------
>>  Thanks,
>>
>>     Ted Ersek
>>
>>
>
>
>
>



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