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Re: 0^0=1
Posted:
May 7, 2012 4:51 AM
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On Apr 21, 4:23 am, "DonH" <donlhumphr...@bigpond.com> wrote:
> "Richard Tobin" <rich...@cogsci.ed.ac.uk> wrote in message
>
> news:jmsvh1$nml$1@matchbox.inf.ed.ac.uk...
>
> > In article
> > <fc51d462-6ab6-4087-97c1-63bda0170...@fv28g2000vbb.googlegroups.com>,
> > Tonico <Tonic...@yahoo.com> wrote:
>
> >>Or, since we're dealing with real or complex numbers, consider lim x^x
> >>= lim e^{x log(x)} , when x --> 0 ...this is a good reason to define
> >>0^0 = 1 .
>
> > But equally one could consider the limit of 0^x.
>
> > For integers the set mapping argument seems compelling, but when
> > dealing with the reals it seems better to leave it undefined or define
> > it explicitly if needed.
>
> > -- Richard
>
> # Yes, 0^0 may be an arbitrary device to accommodate a computer, as Barron's
> Maths Study Dictionary excludes zero(^0) (pg.62) when defining the "zero
> exponent".
> Which still leaves, eg. 5^0 = 1.
> If I have five apples and multiply them by themselves zero times, do I
> end up with one apple?
> Also, does 5^0 = 5 * 1/5?
> Has x^0 = 1 ever been proved? Or, is it merely a convenient dogma?
I'm not a mathematician, and I can't read all 380 posts in this
thread.
But, my understanding of the reason that x^0 = 1 is a consequence of
negative exponents. E.g.
if 2^-2 equals 1/2^2 and x^a * x^b = x^(a+b), then 2^2 * 2^-2 = 2^(2 +
-2) = 2^0.
We can then expand the example to get 2^0 = 2^2 * 1/(2^2) = 1.
However, if x = 0, then the derivation of a^0 = 1 no longer works:
0^2 * 0^-2 = 0*0 / (0*0) = 0/0 which is undefined.
Hence x^0 = 1 is based on a derivation that doesn't work for x=0, and
hence to quote the consequence of that derivation to argue the value
of 0^0 is plain wrong.
I presume that I'm "missing something" and someone will be along to
spank me soon.
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