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Topic: x^0 = 1
Replies: 19   Last Post: May 15, 2012 7:57 AM

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Horand.Gassmann@googlemail.com

Posts: 554
Registered: 2/4/08
Re: x^0 = 1
Posted: May 15, 2012 7:57 AM
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On May 14, 8:24 pm, Kaba <k...@nowhere.com> wrote:
> Gus Gassmann wrote:
> > Which is pretty much what every mathematician here has been saying:
> > the limit of f(x)^g(x), x->0, is undefined, but there is no reason to
> > define 0^0 as a limit.

>
> As you say, the limit of x^y at the origin does not exist. In other
> words, there is no way to make this function continuous at the origin.
> On the other hand, several restrictions of this function can be made
> continuous at the origin, such as x^0 (by 1) and 0^x (by 0). These are
> the two main candidates for a _useful_ definition of 0^0. If the
> definition goes against what is needed in a specific application, then
> this simply means making special cases, and nothing more. It seems to me
> that defining 0^0 = 1 is especially useful in the previous sense (as
> Knuth writes), and 0^0 = 0 less so (yet to see an example).


Precisely. I have not seen an example, either, where it makes sense to
define 0^0 = 0 other than resorting to a continuity argument.

> > After all, you do not introduce the integer 0
> > as the limit of, say, the function 2^(-x), x->oo.

>
> Not sure what you mean by "introducing as a limit".


This was meant more or less polemically. One does not define numbers
or expressions by resorting to limits. The argument that 0^0 should
not be defined as an expression rests solely on the fact that lim
f(x)^g(x) may take on any value or may not even exist when both f and
g tend to 0. Standing this approach on its head, I was looking for a
limit that does exist and *could* be used to define an integer. But of
course no one in their right mind would do that. (If we are talking
about irrationals, there is of course no other way.)

> I would happily introduce a function f : R* --> R, where R* is the
> affinely extended reals with the usual order topology, and then define f
> as an extension of 2^{-x} where f(-oo) = inf, and f(oo) = 0. It's even a
> continuous function:)
>
> P.S. Concerning the phrase "limit being undefined", I would change that
> to "limit does not exist", since the former sounds like the definition
> of limit does not say anything about this case (which is not true).
>
> --http://kaba.hilvi.org





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