06/21/2009, 05:18 PM

(06/21/2009, 01:57 PM)Tetratophile Wrote: btw, why does kouznetsov have IMAGINARY infinities and as the values of the super logarithm at the fixed points? The choice seems arbitrary.Infinity can be imagined as a point on the extended complex plane, which is a sphere.

You can regard infinity as any other point on the complex plane.

The tetrational is not holomorphic (not even continuous) at infinity.

However one can approach infinity from different angles/sectors where the limit may be a single value.

E.g. if you approach infinity along the real axis the limit is infinity.

If you however approach infinity along the imaginary axis then the limit is . This is writtten exactly as:

for each real .

for each real .

for each real .

Quote: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);

I guess its about deforming one initial region into the other initial region, and continuing the one Abel function to the other region. Then we have reduced the problem to both Abel function having the same domain, to which we can apply the theorem.

For different fixed point pairs this would not be possible.

Quote: 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,No, they are not analytic in the punctured neighborhood. Or in other words the fixed points are not isolated singularities. They are branching points, we need cut-line ending in the fixed point, to have it holomorphic.

Your drawing however is I also see the situation (except holomorphy at the fixed points).