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Re: 0^0 = 1?

  • To: mathgroup at smc.vnet.net
  • Subject: [mg95644] Re: 0^0 = 1?
  • From: Dave Seaman <dseaman at no.such.host>
  • Date: Fri, 23 Jan 2009 05:09:45 -0500 (EST)
  • Organization: Information Technology at Purdue
  • References: <gl7211$c8r$1@smc.vnet.net> <gl9mua$ajr$1@smc.vnet.net>

On Thu, 22 Jan 2009 11:56:58 +0000 (UTC), dh wrote:


> Hi,

> 0^0 means the limit if both base and exponent go to zero. 

No, that is not how 0^0 is defined.  Does 2+2 mean the limit as both
summands go to 2?  The value may happen to be the same in that case, but
that is not how 2+2 is defined.

The value of x^y for cardinal numbers x and y is the cardinality of the
set of mappings from y into x.  In the case where x and y are the empty
set, there is exactly one such mapping.  Hence, 0^0 = 1.

It's a theorem of ZF (as stated in Suppes, _Axiomatic_Set_Theory_) that
m^0 = 1 for every cardinal number m.

Another way is to notice that 0^0 represents an empty product, whose
value is the identity element in the monoid of the integers (or the
reals).

	In[1]:= Product[0,{k,0}]

	Out[1]= 1

One might also consider the series expansion for Exp[0], which reduces to

	1 = 0^0/0! + (lots of terms that all reduce to zero).

Having x^y be discontinuous at (0,0) does not "cause problems" any more
than having the Sign function be discontinuous at 0 causes problems.
Anyone who works with limits should be aware that you can't just blindly
assume continuity when evaluating limits.  You have to consider the
actual definition of the limit.




-- 
Dave Seaman
Third Circuit ignores precedent in Mumia Abu-Jamal ruling.
<http://www.indybay.org/newsitems/2008/03/29/18489281.php>


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