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 >> >> > > > >