LudovicoVan
Posts:
2,971
From:
London
Registered:
2/8/08
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Re: 0^0=1
Posted:
May 7, 2012 9:59 AM
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"Ross" <rossclement@gmail.com> wrote in message
news:f71781c8-8b65-406b-a3c6-d1277de56b7f@cl4g2000vbb.googlegroups.com...
> 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.
>>
>> # 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.
Dilution is the easiest strategy to make any discussion and knowledge
impossible.
> But, my understanding of the reason that x^0 = 1 is a consequence of
> negative exponents.
Your understanding is correct, don't get fooled by the spammers.
-LV
> 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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