1. Re: Suggestion for 2.5 (correction)
In the same vein, we could say that 0^0 is not determinate, since
(exp(-a^2/x^2))^(x^2) is exp(-a^2) always, whatever a is, and its limit
form is 0^0 as x goes to 0.
Anyway, a powerful argument in favor of 0^0=1 could be that, if x goes
to 0, x^x goes to 1. The principle of economy, which here translates as
"use one variable to resolve the indeterminacy, then 2 if you can't",
then backs up the case of 0^0=1.
True, 0^x is always 0 if x is nonzero. It means that x +-->0^x is not
continuous at 0. Not really shocking IMO.
CChris
> Date: Thu, 20 Feb 2003 12:44:45 +0100
> From: Juergen Luethje <eu.lue at gmx.de>
> Subject: Re: Suggestion for 2.5 (correction)
>
> Sorry, there was a typo in my previous post!
>
> Hi Christian,
>
> you wrote:
>
> > From: Martin Stachon <martin.stachon at worldonline.cz>
>
> [...]
> >> Some other indeterminate forms are :
> [...]
> >> 0^0 (although MS Windows Calculator says 0^0=1, my Casio says "Math error"
> >> 
> >
> > 1 is correct, because epsilon^0 is always 1.
> [...]
>
> I disagree. It also could be said that 0^0 = 0, because 0^x = 0.
> It cannot be decided, which of both rules _generally_ is "better":
>
> "0^0 is undefined. Defining 0^0 = 1 allows some formulas to be expressed
> simply (Knuth 1997, p. 56), although the same could be said for the
> alternate definition 0^0 = 0 (Wells 1986, p. 26)."
> [http://mathworld.wolfram.com/Zero.html]
> corrected --------------^
>
> > CChris
>
> Best regards,
> Juergen
>
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