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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 30, 2012 12:01 PM
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Dan Christensen <Dan_Christensen@sympatico.ca> wrote:
> On Apr 29, 8:32=A0am, David W. Cantrell <DWCantr...@sigmaxi.net> wrote:
> > Dan Christensen <Dan_Christen...@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?!

>
> Personally, I would find a result based on the assumption that 0^0=1
> to be less convincing. Compared to, say, the constructivist program,
> this is just a minor speed-bump.
>

> > Good luck finding _any_ textbook that does it that way.
>
> I am quite sure this is something new.


Hmm... It seems that you are implicitly proposing that, in order to be
fully correct, all calculus texts need to be rewritten so that the
derivative of x^n is stated as you did above. Good luck with that.

> > > 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.

>
> No, because g is well-defined everywhere. f(x,y) = x^y is not
> well-defined everywhere -- not unless you assign some arbitrary value to
> 0^0. IMHO, the path-dependent limits make it difficult if not impossible
> to justify any particular value.


Wow, you completely miss my point!

1. It seems that you are happy with g, the floor function, simply because
it is a _fait accompli_. Note that, for integer x, g also has
path-dependent limits. For integer x:

Choosing g(x) = x really _is_ arbitrary. We could just as well have chosen
g(x) = x - 1 or g(x) = x - 1/2 or ...

If the definition of g still doesn't bother you, that's a good thing. The
value of a function _AT_ a point need have no relation to any possible
limit as that point is being _APPROACHED_.

2. Since there is consensus among mathematicians that 0^0 = 1, I consider
that to be a _fait accompli_ as well, and so f(x,y) is well defined at the
origin, just as g(x) is well defined for integer x.

3. If your only complaint about 0^0 = 1 is that it seems arbitrary to you,
I can help you to see that it is not arbitrary. But I refuse to waste any
more time until you clearly acknowledge that limits can have nothing to do
with the value of 0^0.

David W. Cantrell



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