06/21/2009, 01:57 PM
(This post was last modified: 06/21/2009, 04:27 PM by Base-Acid Tetration.)
btw, why does kouznetsov have IMAGINARY infinities \( i\infty \) and \( -i\infty \) as the values of the super logarithm at the fixed points? The choice seems arbitrary. I think it should be a kind of negative infinity because of how you get to the fixed points. exp^(some complex number) (1) = i or some other nonreal complex number, take the logarithm of that number an infinite number of times and you get the fixed point. for some complex c, corresponds to a complex iteration plus a negative infinity itterations of exp: \( \exp^{-\infty+c}(1). \) We lose injectivity, but it's a fair price.
no, it isn't finished.... It just lists all i know about the situation. I don't know how to proceed. I just have a hunch that the neighborhood of the fixed points might be important (the only thing the domains D1 and D2 share as part of their boundaries is the fixed point pair, and both D1 and D2 include a subset of the neighborhood); if I can show that the Abel functions are analytic and take the same values in all directions in the deleted neighborhood of the fixed points, I will have established the existence of a unique analytic continuation of both A1 and A2.
I have attached a drawing to summarize the problem and the proposed proof technique.
(06/21/2009, 08:19 AM)bo198214 Wrote: Is the proof already finished?
I dont see where it shows that \( A_1(z)=A_2(z) \) for \( z\in D_1\cap D_2 \).
no, it isn't finished.... It just lists all i know about the situation. I don't know how to proceed. I just have a hunch that the neighborhood of the fixed points might be important (the only thing the domains D1 and D2 share as part of their boundaries is the fixed point pair, and both D1 and D2 include a subset of the neighborhood); if I can show that the Abel functions are analytic and take the same values in all directions in the deleted neighborhood of the fixed points, I will have established the existence of a unique analytic continuation of both A1 and A2.
I have attached a drawing to summarize the problem and the proposed proof technique.

