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Re: Re: 0^0 = 1?

  • To: mathgroup at smc.vnet.net
  • Subject: [mg95655] Re: [mg95644] Re: 0^0 = 1?
  • From: Daniel Lichtblau <danl at wolfram.com>
  • Date: Sat, 24 Jan 2009 06:17:43 -0500 (EST)
  • References: <gl7211$c8r$1@smc.vnet.net> <gl9mua$ajr$1@smc.vnet.net> <200901231009.FAA03550@smc.vnet.net>

Dave Seaman wrote:
> 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.

That's a definition from set theory. I doubt it plays nice in the 
complex plane. The definition of Power we go by is
ower[a,b] == Exp[Log[a]*b]
Among other advantages, it means branch cuts for Power are inherited 
from Log.


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

I think the setting of cardinal numbers is not really a good choice for 
symbolic computation.


> 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

This and the ZF result are arguments for making 0^0 equal to one. They 
are not in any sense "proofs" that it must be one, given that Power 
lives in the setting of functions of complex variables.


> 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).

This point is questionable, since one does not in general encounter the 
formula with a term x0^0 (where x0 is the point of expansion).


> 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.

That's a matter of opinion. Clearly we do not at this time agree with 
you on this.


> 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.

True, but I don't see any relevance to this particular issue. The 
question at hand is whether or not 0^0 should be assigned a concrete 
value. If it is given a value, then since limits of x^y depend on how 
x,y respectively approach 0, they might or might not equal that value 
(so there is no underlying assumption of continuity). But that is 
already the case, and the question at hand is whether the undefined 
value should in fact be defined.


Daniel Lichtblau
Wolfram Research


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