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Re: 1^Infinity

  • To: mathgroup at smc.vnet.net
  • Subject: [mg38322] Re: 1^Infinity
  • From: "David W. Cantrell" <DWCantrell at sigmaxi.org>
  • Date: Thu, 12 Dec 2002 01:34:54 -0500 (EST)
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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.

That is simply a fact. It cannot be altered (unless we were to change
the power function x^y substantially, which surely we don't want to do).
But, when we say that a certain limit form is indeterminate, it is
important to know not only what that _does_ mean, but also what it does
_not_ mean! Let's take the limit form "0^0" as an example. Saying that it
is indeterminate means that, as x and y approach 0, x^y may approach
any of many different possible values (or the limit may not exist).
[The word "indeterminate" is appropriate in that, merely knowing that
both base and power approach 0, we do not have enough information
to be able to determine the limit, if it exists.] This indeterminacy is
due to the fact that f(x,y) = x^y has an essential singularity at (0,0).
But saying that the limit form "0^0" is indeterminate does _not_ mean that
the simple arithmetic expression 0^0 need be undefined. In the arithmetic
expression 0^0, both base and power _are_ 0, they are constant. No limits
are involved. Whether the arithmetic expression 0^0 should be defined as
1, as many prominent mathematicians (including Euler, Knuth, Graham, and
Kahan) have suggested, or should be undefined, is apparently still open to
debate by reasonable people.

Now the OP had asked about 1^Infinity. If he was asking about the limit
form "1^Infinity", then, just as for "0^0", there can be no reasonable
debate. That limit form is indeterminate, period. But, since he mentioned
no limits, I had naturally assumed that he had in mind the simple
arithmetic expression 1^Infinity, in which the base does not merely
approach 1, but rather _is_ 1, and the power does not merely approach
Infinity, by rather _is_ Infinity. Both base and power are constants.
(In case you're balking at the notion of the power being the constant
Infinity: Of course, there is no such constant in the real number
system. But such a constant does exist in an extension of the reals.)
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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