Hi,
On Fri, 14 Jul 2006, 4:39pm -0400, [log in to unmask] wrote:
> I have to agree that 0**0 is not defined in "conventional
> mathematics". Look at any calculus text. There will be a section
> on evaluating "indeterminate forms" using L'Hospital's rule.
> Indeterminate forms include 0 / 0, infinity / infinity, 0 * infinity,
> 0 ** 0, etc. Note that if you take the logarithm of 0 ** 0 using
> the usual rules, you get 0 * log(0) = 0 * (-infinity). These are
> all situations fitting Lawrie's description below where the limit
> may be undefined because it is path dependent. Only specifying a
> particular path can lead to a specific value.
"May be" is the right term; L'Hopital's rule can be
invoked in some cases to show that the value also "may
not be" path-dependent.
A previous posting (sorry, I didn't save it, so I can't
acknowledge the author) pointed out the following. (I
think I'm remembering this correctly.)
Assume:
A. x**1 = x
B. x**(n+1) = x*(x**n)
Evaluating B for n=0 gives:
x**1 = x*(x**0)
Substituting A for the LHS gives:
x = x*(x**0)
Therefore:
x**0 = x/x
The argument is that this works except when x=0.
But if x is a real number, we can apply L'Hopital's rule
to this expression in the limit x->0, which gives
x**0->1 as x->0.
To me, this is pretty convincing evidence that the only
reasonable definition of 0**0 for real x (at least) is 1.
I forget the conditions that are required for L'Hopital's
rule to apply, but do recall that they are met for reals.
Note that x/x->0 as x->0 from both above and below; so
this evaluation, at least, is path-independent.
-P.
--
Peter S. Shenkin Schrodinger, Inc.
VP, Software Development 120 W. 45th St., 29th Floor
646 366 9555 x111 Tel New York, NY 10036
646 366 9550 FAX USERID: shenkin
http://www.schrodinger.com DOMAIN: schrodinger DOT com
Conf. calls: 650 429 3300 or 866 469 3239 (toll-free); code 36695555#
|