# Tetration Forum

Full Version: Universal uniqueness criterion?
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(06/19/2009, 02:04 PM)Tetratophile Wrote: [ -> ]"We call a region H an initial region of F iff F(z) {is not in} H for all z {in} H..."
does this mean $\exists z : F(z) \notin H$ (there is SOME z such that F(z) is NOT in H) or $\forall z (F(z) \notin H$) (there is NO z such that F(z) IS in H)?

$\forall z\in H: F(z)\notin H$.
"the function log(z) is holomorphic on {the whole complex plane except real numbers less than 0}" :

WRONG! the taylor series doesn't converge for real numbers z>2.
(06/19/2009, 04:19 PM)Tetratophile Wrote: [ -> ]"the function log(z) is holomorphic on {the whole complex plane except real numbers less than 0}" :

WRONG! the taylor series doesn't converge for real numbers z>2.

holomorphic on $G$ does not mean that the powerseries at 1 converges at every $z\in G$. It means that $f$ is complex differentiable at every $z_0\in G$ which is equivalent that it has a powerseries development (with non-zero convergence radius) at every $z_0\in G$.

The only singularity of the logarithm is 0. That means you can analytically continue $\log$ from 1 along every path that not contains 0. However if there are different paths to one point $z_1$ the values of the continuations may differ at $z_1$ by multiples of $2\pi i$ depending on how often the path revolves around the singularity 0.
To exclude different values (branches) of the logarithm, one chooses the domain of the logarithm with a cut e.g. at $\Re(z)<0$. Hence there no path can revolve completely around 0 and the continuation of the logarithm to any point $z_1$ of the complex plane except the cut has only one value.
In your paper it says something like:
"There exists only one holomorphic superlogarithm with b>e^1/e that maps G biholomorphically to *some* region with upper and lower unbounded imaginary part"

That must mean that we still don't know WHAT region the superlogarithm must map G (the crescent moon thingy) to. So our different tetration methods might map G to different infinitely long strips of width 1. So the condition you proved cannot be a uniqueness condition!
Here's the problem, represented with very sloppy ascii art, with the "uniqueness condition":
Code:
- it could be (groups of two bars represent the infinitely long strip): ||or  \\ or... yeah, basically anything with width 1 that contains part of Re[z]=0 ! \\       \\    \\      ||    ||     ||    ||     ||    ||     ||    //      || //       // ||or  //
(06/19/2009, 05:17 PM)Tetratophile Wrote: [ -> ]"There exists only one holomorphic superlogarithm with b>e^1/e that maps G biholomorphically to *some* region with upper and lower unbounded imaginary part"

That must mean that we still don't know WHAT region the superlogarithm must map G (the crescent moon thingy) to. So our different tetration methods might map G to different infinitely long strips of width 1.
Yes exactly. If you have one super-logarithm that maps G to *some* region $H_1$ and another super-logarithm that maps G to *some other* region $H_2$, and both super-logarithms satisfies the other conditions of the proposition, then they must be equal. Hence this assumption of different images was already wrong: the images are equal $H_1=H_2$.

Not sure why you think it wouldnt work. The only problem with our tetration methods may be the biholomorphy, the bijectivity of the super-logarithm on the initial region (or merely the holomorphy for some methods).
(06/19/2009, 06:25 PM)bo198214 Wrote: [ -> ]Yes exactly. If you have one super-logarithm that maps G to *some* region $H_1$ and another super-logarithm that maps G to *some other* region $H_2$, and both super-logarithms satisfies the other conditions of the proposition, then they must be equal. Hence this assumption of different images was already wrong: the images are equal $H_1=H_2$.

Can you reproduce the proof here?

(06/19/2009, 06:25 PM)bo198214 Wrote: [ -> ]Not sure why you think it wouldnt work. The only problem with our tetration methods may be the biholomorphy or merely the holomorphy.
Because your paper said there is only one tetration/superlogarithm that maps g to SOME strip. I missed the beginning of the proof, that f and g were assumed to map G to different strips.

EDIT; ok, you proved that f and g really map G and the strip equally, so they are the same holomorphic function.
Cool, now we have to prove the biholomorphicity of the different tetration methods between G and the strip, and according to the theorem, since they are all tetration/tetralogarithm (superfunction/Abel function of exponential), they are all the same. Does the applicabilty of theorem 1 only require that the function F (in the case of tetration, exp) is holomorphic, and F (exp_b, b>e^1/e) has no real fixed points?
(06/19/2009, 06:27 PM)Tetratophile Wrote: [ -> ]Cool, now we have to prove the biholomorphicity of the different tetration methods between G and the strip, and according to the theorem, since they are all tetration/tetralogarithm (superfunction/Abel function of exponential), they are all the same.

But it could also occur that one slog is biholomorphic on one initial region A but not on B, while another slog is biholomorphic on the initial region B, but not on A.

I think if both initial regions belong to the same fixed point pair, then both slogs are analytic continuations of each other, i.e. basically the same holomorphic function. But this is not yet proven.

There should also be an example of a quartic, that has two pairs of conjugated fixed points, and that has two different Abel functions, each biholomorphic on an initial region belonging to one fixed point pair.

Quote: Does the applicabilty of theorem 1 only require that the function F (in the case of tetration, exp) is holomorphic, and F (exp_b, b>e^1/e) has no real fixed points?
Well you just need a fixed point pair and an initial region belonging to it.
If both fixed points are on the real axis, then there is no initial region between them.
Better chances have (complex) conjugated fixed points.

About different regular Abel/super functions developed at different fixed points on the real axis.
(06/20/2009, 04:33 AM)Tetratophile Wrote: [ -> ]Haven't you already proved that for superlogarithms? ("for "each" simple intial region G (if both "boundaries", $\partial_1 G$ and $\partial_2 G$, are bounded by the fixed points of F, wouldn't that satisfy the criteria for being a simple region?) ... there exists at most one holomorphic Abel function for F...")

No, the uniqueness depends on the initial region G.
If we have two Abel functions biholomorphic on two different initial regions (even if those different initial regions belong to the same fixed point pair) the proposition is not applicable.
(06/19/2009, 07:59 PM)bo198214 Wrote: [ -> ]I think if both initial regions belong to the same fixed point pair, then both slogs are analytic continuations of each other, i.e. basically the same holomorphic function. But this is not yet proven.

I'll try to prove that this summer. just give me all the relevant facts.

bo198214 Wrote:two Abel functions biholomorphic on two different initial regions
two Abel functions: A1(f1(x)) = A1(x) + 1 , A2(f2(x)) = A2(x) + 1
you mean two different abel functions, (1) where f1, f2 are different functions? or do A1, A2 map the same point to different values? (for sake of argument, a "superlogarithm" that does slog(0) = 0)
(06/20/2009, 02:01 PM)Tetratophile Wrote: [ -> ]
(06/19/2009, 07:59 PM)bo198214 Wrote: [ -> ]I think if both initial regions belong to the same fixed point pair, then both slogs are analytic continuations of each other, i.e. basically the same holomorphic function. But this is not yet proven.

I'll try to prove that this summer. just give me all the relevant facts.
ok, so we compete

Quote:two Abel functions: A1(f1(x)) = A1(x) + 1 , A2(f2(x)) = A2(x) + 1
you mean two different abel functions, (1) where f1, f2 are different functions? or do A1, A2 map the same point to different values? (for sake of argument, a "superlogarithm" that does slog(0) = 0)

two Abel functions A1, A2 of the same function f
A1(f(x))=A(x)+1, A2(f(x))=A2(x)+1, A1(d)=A2(d)=c.
but A1 is defined on the initial region G1 and A2 is defined on the initial region G2.
Of course $d\in G_1\cap G_2\neq \emptyset$.
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