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Re: 0^0=1
Posted:
Apr 23, 2012 11:47 AM
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Dan Christensen writes:
> On Apr 23, 10:18 am, Jussi Piitulainen wrote:
> > You mean first define x^0 = 1 for all non-0 x and then add 0^0 =
> > 19?
>
> Yes.
>
> x^0 = 19 if x=0; 1 otherwise
>
> (Note: I am NOT claiming that 0^0 = 19. I am trying to make the case
> that 0^0 should be left undefined.)
Yes. Why?
And why not show something that works _better_ with 0^0 undefined? Now
you are adding unnecessary conditions to a useful general theorem and
gaining nothing, which makes it _worse_.
> > The binomial theorem gives unusual results with 0^0 = 19:
> >
> > (x + y)^n = sum for k = 0, ..., n of C(n,k) x^k y^(n - k)
>
> [snip]
>
> This theorem could not be formally derived from the usual definition
> of integer exponentiation (above). You could, however, prove:
>
> (x + y)^n = x^n if y = 0; sum for k = 0, ..., n of C(n,k) x^k y^(n -
> k) otherwise
You need to handle separately also the cases where x = 0, and leave
undefined the cases where x + y = 0 and n = 0. For no gain.
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