0^0

[cino hilliard]

>>
>>but logic says that 0^O = 1,
>>as x^y is analogous to x<=y,
>>and 0<=0 is true.
>
>
>It is often most convenient in combinatorics to have 0^0 = 1.  See 
>"Concrete Mathematics", (Graham, Knuth, and Patashnik) section 5.1, page 
>162 in my first edition: "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!"
>

Consider (1/x)^(1/x) = 1/(x^(1/x))
Since Limit(1/x=1..infinity) = 0, we have 0^0 = 1/(x^(1/x))
Then
Limit(x^(1/x),x=1..infinity) = 1
So 1/1 = 0^0 = 1

Empirically,
let x = 1000^y.  y=0..infinity
Then
The first 20 expansions of  (1/x)^(1/x) for y=0 are
1.000000000000000000000000000000000000000000000000000000000000000000000
0.9931160484209337715764260768851547466351916218112333612207382247673452
..
..
0.9999999999999999999999999999999999999999999999999999998687526496993394
0.9999999999999999999999999999999999999999999999999999999998618448944204

Probabilistically, This evidence is so overwhelming that we can conclude 0^0 
= 1  in the same way
we conclude primality with the  ispseudoprime test. Or can we? Is the string 
of 9's 0.99999999999999999..9999.. = 1? Perhaps, if we argue that 1 is a 
better approximation than
any other number. In computer lingo it is 1.

Another point is to define  0 = 1/V. Where V is a Very large number so that 
resistence to  0^0 =
1  would be futile.

Could  0^0 be the creation = Big Bang = nothing to the nothing power?

Have fun,
Cino