Re: One to the power Infinity
- To: mathgroup at smc.vnet.net
- Subject: [mg35943] Re: [mg35911] One to the power Infinity
- From: Andrzej Kozlowski <andrzej at platon.c.u-tokyo.ac.jp>
- Date: Fri, 9 Aug 2002 05:17:51 -0400 (EDT)
- Sender: owner-wri-mathgroup at wolfram.com
Presumably the argument is something like this. Take any positive numbers a and s. Then a = (a^(1/s))^s. Now make s go to infinity so a^(1/s)->1. Thus you can "make" 1^Infinity equal to any positive a, which is exactly the kind of situation in which Indeterminate is used for the answer. Best regards Andrzej Andrzej Kozlowski Toyama International University JAPAN http://platon.c.u-tokyo.ac.jp/andrzej/ On Thursday, August 8, 2002, at 07:06 PM, Matthias.Bode at oppenheim.de wrote: > Dear Colleagues, > > when fooling around with MATHEMATICA I found: > > 0^\[Infinity] => 0, as expected; > 0.9^\[Infinity] => 0, as expected; > 2^\[Infinity] => Infinity, as expected; > 1^\[Infinity] => Indeterminate, unexpected. Naively expected: 1. > > For which reason(s) is 1^\[Infinity] defined as Indeterminate? > > Best regards, > > Matthias Bode. > > >