__builtin_cpow((0,0),(0,0))

[Ronny Peine]

Maybe i should make it more clearer, why 0^x is not defined for real 
exponents x, and not continual in any way.

Be G a set ("Menge" in german) and op : G x G -> G, (a,b) -> a op b.
If op is associative than (G,op) is called a half-group.
Therefore then exponentiation is defined as:
a from G, n from |N>0:
a^1 = a; a^n = a op a^(n-1)
If a neutral element is in G (mostly called the "1") than a^0 is defined 
as 1.

Example (Z,+) is a half-group (it's even a group). Therefor a^n = a + a 
+ a + ... + a (n times).

For real exponents this is not defined in the above case, therefore
(Example: what would be 2^pi?) a definition which is in accordance to 
the previous one was defined:
For A,X from |R, A>0:
A^X = exp(X*ln(A))

with exp(N*X) = exp(X)^N (which can be proofed by induction) it can
be seen that it is in accordance to the previous definition (if X is 
from |N).

The rule a^(1/n) = n-th root of a comes from the proof:
Be a from |R, a>0 and p from Z, q from |N>1, then:

a^p = exp(p * ln(a)) = exp(q * (p/q) * ln(a)) = exp(p/q * ln(a))^q = 
(a^(p/q))^q => a^(p/q) = q-th root of a^p (remind that this is only true 
for a>0).

For 0^x there is no such definition except of x is from |N. Therefore 
0^0 is defined as according to the first rule as 1 (because we look at
the group (|R,*) with a^n= a*a*a* ... *a (n times) and the neutral 
element 1, therefore a^0 = 1 for every element in |R).

I hope that this make things clearer for some who don't believe 0^0 = 1 
in the real case.

cu, Ronny

Robert Dewar wrote:
> Ronny Peine wrote:
> 
>> Well this article was referenced by 
>> http://grouper.ieee.org/groups/754/, so i don't think it's an 
>> unreliable source.
> 
> 
> Since Kahan is one of the primary movers behind 754 that's not so 
> surprising.
> For me, 754 is authoritative significantly because of this connection.
> If there were a case where Kahan disagreed with 754, I would suspect
> that the standard had made a mistake :-)
> 
>