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Re: 0^0
Posted:
Oct 19, 2006 2:19 AM
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Proginoskes wrote:
> Akira Bergman wrote:
> > It is double valued;
> > 0^0 = {0,1}
>
> No, it isn't. It's defined to be 1, mainly because it's convenient to
> do it this way when working with polynomials.
>
Definitions can change. Suppressed values will show up eventually.
> In calculus, 0^0 represents a function whose value is approaching 0,
> raised to the power of a function whose value is approaching 0. This is
> an indeterminate form, which is dealt with in Calculus, first semester.
>
This is said by me below with x^y. Regardless of where it is thought,
it does not mean it is a settled issue.
> > It is a special case of x^y. It looks like 0 from x, and like 1 from y.
> >
> >
> > Now consider measuring zero relative to itself;
> > 0! / 0^0 = {1/1,1/0}
> > One value diverges, demonstrating that 0-base counting can not be done.
>
> Well, you can't do 1-base counting either, but only because of lack of
> any possible digits. Same for 0-base. You're right for the wrong
> reason.
There is an extra reason for 0, that is duality. 1 is the trivial case.
>
> > To overcome this measurement difficulty we break the symmetry of x^x
> > form to x^y form and then freeze x to its base. Taking the base as 2;
> > 0! / 2^0 = 1/1 = 1
> > Then 0 acquires one state and enables measurement.
> >
> > Obviously the base can not be 1. The next logical choice is 2. Boolean
> > is the minimum space.
>
> Whatever that's supposed to mean.
You are tough.
>
> --- Christopher Heckman
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