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Topic: 0^0=1
Replies: 145   Last Post: Jun 5, 2012 1:10 PM

 Messages: [ Previous | Next ]
 Dan Christensen Posts: 8,219 Registered: 7/9/08
Re: 0^0=1
Posted: May 8, 2012 11:54 PM

On May 8, 6:14 pm, William Hale <bill...@yahoo.com> wrote:
> In article
>  Dan Christensen <Dan_Christen...@sympatico.ca> wrote:

> > On May 8, 2:36 pm, Frederick Williams <freddywilli...@btinternet.com>
> > wrote:

> > > Dan Christensen wrote:
>
> > > > 0^3 = 1*0*0*0 = 0
> > > > 0^2 = 1*0*0   = 0
> > > > 0^1 = 1*0     = 0
> > > > 0^0 = 1

>
> > > > Can we formalize this argument?  How do you go from 0^1 = 0 (a direct
> > > > consequence of the usual definition of ^ on N) to 0^0 = 1?

>
> > > The usual definition is
>
> > >   x^0 = 1,   x^{n+1} = x * x^n.
>
> > So, by definition, 0^0 would be 1.
>
> Yes. Is there something wrong with that?
>

I supposed there is nothing inherently wrong with defining your way
out of a controversy. It's just not very intuitively satisfying that
the only apparent motive is convenience. And that it probably isn't
even necessary to define a value for 0^0.

>
>

> > > If, rather, you mean
>
> > >   x^1 = x,   x^{n+1} = x * x^n
>
> So, by definition, 1^1 would be 1.

True.

> Why don't you think there's something
> wrong with this?

To my knowledge, there is nothing very controversial 1^1 = 1. Correct
me if I am wrong, but I don't think there is any weird behaviour of
f(x,y)=x^y around (1,1) in the reals -- no path-dependent limits, etc.

> > Yes. Here is my understanding: This is the form you must use for N not
> > including zero. The definition can be easily extended to N including
> > zero, however.

>
> > If you want the usual rules of exponents to apply, 0^0 could be either
> > left undefined, or defined to be 1 (as above) or 0. Neither
> > alternative should lead to inconsistencies.

>
> > For no other reason than it can be used to produce the required
> > results in some cases, if a value is chosen, it is usually (always?)
> > 1. Is that a good enough reason to choose 0^0 = 1? Many specialists
> > seem to think so -- this, despite path-dependent limits of f(x,y) =
> > x^y toward (0,0) when the definition of ^ is extended to the reals,
> > e.g lim(x->0): x^0 = 1 and lim (y->0): 0^y = 0.

>
> > Perhaps because of these path-dependent limits in the reals, many
> > authors remain unconvinced, and leave 0^0 undefined. In some (all?)
> > cases, results equivalent to those obtained with 0^0=1 can be obtained
> > by considering additional zero-cases with 0^0 being left undefined.

>
> > FWIW, my personal preference would be to leave 0^0 undefined when
> > writing formal proofs.

>

Dan
Also see video demo

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