Re: 1^Infinity
- To: mathgroup at smc.vnet.net
- Subject: [mg38350] Re: 1^Infinity
- From: Selwyn Hollis <selwynh at earthlink.net>
- Date: Thu, 12 Dec 2002 01:37:25 -0500 (EST)
- References: <20021210175827.578$DZ_-_@newsreader.com>
- Sender: owner-wri-mathgroup at wolfram.com
Okay, finally I get it. I apologize for thinking you're all nitwits. :^) So the question is whether 1^Infinity (and related forms) could/should be assigned some value in a non-limit sense. Thinking back on a previous life (FORTRAN) I see that there may be systems in which it would make sense to have 1^Infinity = 1 while 1.^Infinity = Indeterminant. But in a system that "contains" rationals as well as integers, it seems to me that having 1^Infinity = 1 is just as "wrong" as 1.^Infinity = 1. (The "wrongness" of 1.^Infinity = 1, by the way, must be understood in terms of limits, not machine arithmetic, since 1. is a machine number.) As usual, Mathematica does the right thing, defering to generality and consistency with "traditional mathematics." --- Selwyn Hollis David W. Cantrell wrote: > [Message also posted to: comp.soft-sys.math.mathematica] > > Selwyn Hollis <selwynh at earthlink.net> wrote: > >>It is astonishing that debates like this keep coming up. To anyone who >>disagrees with the notion that 1^Infinity is indeterminate, I suggest >>that you write something called "a new kind of calculus." But you might >>want to learn the old kind first. > > > I taught calculus for many years. Unless I've misunderstood Ted, his > question has nothing to do with calculus. (If I have misunderstood you, > Ted, please let me know!) > > What you're thinking about, Selwyn, cannot be debated -- at least not > by reasonable people: > > Certain _limit_ forms, such as "1^oo" and "0^0", are indeterminate. > <snip> > Just as for the arithmetic expression 0^0, whether the arithmetic > expression 1^Infinity should be defined as 1, or be undefined, is still > open to debate by reasonable people. > > Note: Of course, defining the arithmetic expressions 0^0 and 1^Infinity > to be 1 would in no way alter the fact that f(x,y) = x^y has essential > singularities at (0,0) and (1,Infinity), and so would in no way alter > the fact that the limit forms "0^0 and "1^Infinity" are indeterminate. > > David Cantrell >