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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 (Shijō) 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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