
Re: 0^0=1
Posted:
Apr 29, 2012 8:32 AM


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 zerocase here could also be easily handled > separately, e.g. d/dx x^n = n * x^(n1) 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 pathdependent 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

