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Topic: 0^0=1
Replies: 19   Last Post: May 1, 2012 4:02 PM

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David W. Cantrell

Posts: 3,395
Registered: 12/3/04
Re: 0^0=1
Posted: Apr 29, 2012 8:32 AM
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Dan Christensen <Dan_Christensen@sympatico.ca> wrote:
<snip>
> It didn't occur to me when first asked, but as with the Binomial
> Theorem, you don't really need to define 0^0=1 to arrive at an
> equivalent result in this case either. It may seem like a stretch, but
> I would think that the zero-case here could also be easily handled
> separately, e.g. d/dx x^n = n * x^(n-1) for n=/=1; 1 otherwise.


That could be done. But why, oh why, would anyone _want_ to be so perverse?!

Good luck finding _any_ textbook that does it that way.

> Sorry, but my concern about path-dependent limits remains.

I said long ago in this thread that limits are irrelevant. Yes, the
function f(x,y) = x^y has an essential singularity at the origin. But
whether 0^0 is or is not defined is unrelated to that fact! Defining 0^0 to
be 1 assigns a value to f _AT_ the point (0,0). By contrast, those limits
which "concern" you deal with f as (x,y) _APPROACHES_ (0,0) instead.

For a moment, let's think about a different function: g(x) = floor(x)
a.k.a. the "greatest integer" function. It has discontinuities at every
integer x. Does it also bother you that, say, g(1) is defined to be 1 even
though the limit of g as x approaches 1 does not exist? If so, why? (It
shouldn't.) If not, then please explain why defining f(0,0) to be 1 does
bother you even though defining g(1) to be 1 doesn't.

Until you realize that limits are a red herring in the matter of defining
0^0, there's nothing more to discuss.

David W. Cantrell



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