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Re: lisp question


From: Marco Almeida
Subject: Re: lisp question
Date: Tue, 01 May 2007 22:11:42 +0100
User-agent: Wanderlust/2.15.5 (Almost Unreal) SEMI/1.14.6 (Maruoka) FLIM/1.14.8 (Shij┼Ź) APEL/10.7 Emacs/21.4 (powerpc-unknown-linux-gnu) MULE/5.0 (SAKAKI)

At Tue, 01 May 2007 21:10:44 +0200,
Jesper Harder wrote:
> 
> A Soare <address@hidden> writes:
> 
> > What determined the mathematicians to give the definition of x^0 = 1,
> > not defined for x=0 ?
> 
> Because 0/0 is undefined:
> 
>        x^1      x
> x^0 =  ---- =  ---
>        x^1      x
> 
> x=0 ->  0/0.
> 

As far as I know, 0^0 = 0 by convention (pretty much like 0 != 1)
Also, I think that your argument is wrong. 

0^y = 0 for any y
0 is the characteristic of R and you can not divid by it

This means that your first step :

       x^1   
x^0 =  ---- 
       x^1  

is not valid, regardeless of the value you chose as exponent because you will 
be trying a division by zero.


Also, from Concrete Mathematics p.162 (R. Graham, D. Knuth, O. Patashnik): 

Some textbooks leave the quantity 0^0 undefined, because the functions 0^x and 
x^0 have different limiting values when x decreases to 0. But this is a 
mistake. We must define x^0=1 for all x , if the binomial theorem is to be 
valid when x=0 , y=0 , and/or x=-y . The theorem is too important to be 
arbitrarily restricted! By contrast, the function 0^x is quite unimportant.

> 
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