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Re: 0^0=1
Posted:
May 28, 2012 7:27 AM
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Michael Press <rubrum@pacbell.net> wrote:
> In article <zKpkr.6017$%E2.342@viwinnwfe01.internal.bigpond.com>,
> "DonH" <donlhumphries@bigpond.com> wrote:
>
>> "Richard Tobin" <richard@cogsci.ed.ac.uk> wrote in message
>> news:jmsvh1$nml$1@matchbox.inf.ed.ac.uk...
>>> In article
>>> <fc51d462-6ab6-4087-97c1-63bda017042c@fv28g2000vbb.googlegroups.com>,
>>> Tonico <Tonicopm@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?
>
> You get 1 apple^0 = 1; so you get no apples.
X^0=0?
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