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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
> 




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