Re: 0^0 = 1?

*To*: mathgroup at smc.vnet.net*Subject*: [mg95683] Re: 0^0 = 1?*From*: Dave Seaman <dseaman at no.such.host>*Date*: Sun, 25 Jan 2009 06:48:24 -0500 (EST)*References*: <gl7211$c8r$1@smc.vnet.net> <gl9mua$ajr$1@smc.vnet.net>

On Sat, 24 Jan 2009 11:17:16 +0000 (UTC), Daniel Lichtblau wrote: > 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. Branch cuts at 0 are not particularly relevant here, since Exp[-oo*0] = Exp[anything*0] = Exp[0] = 1. >> 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. Because it doesn't support your preferred answer? We define other operations (addition and multiplication, for example) by starting with the cardinals, then extending to the integers, the rationals, the reals, and then the complex numbers. At each stage, we want the new definition to be consistent with the old. >> 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. The ZF result is indeed a proof in the context of the cardinal numbers. The empty product result applies to any monoid. The complex numbers are a monoid. The symbol "0" in Mathematica represents the complex number 0, and as you can see, Mathematica agrees that the empty product yields the complex number 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). > 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). It is not questionable. The series expansion for Exp[x] is Sum[x^k/k!,{k,0,Infinity}] whether one normally encounters it in that form or not. >> 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. But no one has presented an example of an actual "problem" that is caused by the definition. Discontinuous functions exist. That's the basic reason that we need to study limits in the first place. >> 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. We don't "assign" a value to 0^0. We simply look at the definition and notice what it says for the case of 0^0. The definition makes no statement at all about limits, and therefore thinking that the definition has implications regarding limits is a misconception. -- Dave Seaman Third Circuit ignores precedent in Mumia Abu-Jamal ruling. <http://www.indybay.org/newsitems/2008/03/29/18489281.php>

**Follow-Ups**:**Re: Re: 0^0 = 1?***From:*Andrzej Kozlowski <akoz@mimuw.edu.pl>

**Re: Re: 0^0 = 1?***From:*Murray Eisenberg <murray@math.umass.edu>