Re: 0^0=Indeterminate quantity right ?

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Article #1041

#1041

From: Kevin Easton
Date: Thu, 02 Jan 2003 13:26

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In comp.os.linux.development.apps Erik Max Francis <max@alcyone.com> wrote:
> Kasper Dupont wrote:
> 
>> Erik Max Francis wrote:
>>
>> > x^0 = 1 holds for all x
>> 
>> OTOH 0^x = 0, and you can probably argue for anything
>> between 0 and 1 by similar arguments.
> 
> This is a very weak argument, because for one thing, 0^x = 0 is simply
> not true; consider x = -1.  (And y^x isn't defined at _all_ for negative
> irrational x, regardless of y!)

Wouldn't that make e^x a noncontinous function? I was sure that e^(-Pi)
is well-defined - certainly e^(-j.Pi) is.

	- Kevin.

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From: Kevin Easton <kevin@-nospam-pcug.org.au>
Newsgroups: alt.linux,comp.os.linux.development.apps,fr.education.entraide.maths
Subject: Re: 0^0=Indeterminate quantity right ?
Date: Thu, 2 Jan 2003 13:26:33 +0000 (UTC)
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