Kneser method question

[sheldonison]

... Step one : We use the first upper fixpoint of exp : L.

From L we compute the solution to 

Exp(f(z))  = f(L z)

By using the koenigs function.

Step 2 :

We find the value 1. 
From this 1 we trace the positive reals.

We notice the reals make a corner at every rotation L or equivalently at 1,e,e^e,....

First question : why at these values and not at say pi,exp(pi),exp(exp(pi)),... ?

Second question : this indicates there are singularities at 1,e,e^e,....
What kind of singularities are they ?? 
The corners seem to suggest log or sqrt or such, but I am not sure.
...
Question 8 :
The big question :

How does the riemann mapping remove those singularities at 1,e,e^e,... ??

step1:
Koenigs at the primary fixed point=L, generate the Schroeder functional equation 
\( \lambda=L;\;\; \)  lambda is the derivative of exp(L+z) at the fixed point L; for base(e) Lambda=L
\( \Psi_\circ \exp(z)=\lambda\cdot\Psi(z);\;\;\; \Psi(L)=0; \)

step2: 
we take \( \Psi(\Re) \) Kneser calls this his Chi* function
But \( \Psi(0) \) is a really nasty singularity.  
see the pictures in mathstack description of Kneser to see what the Chi* function looks like.  One difficulty is that this singularity is pretty complicated; but for the open set \( (0,1)=\{z\in\mathbb{R} : 0 < z < 1\} \) then the Chi* is defined and multplication by L gives you the open set \( (1,e) \).

Next, Kneser's Riemann mapping puts all of these open sets (-inf,0); (0,1); (1,e); (e,e^e); (e^e,e^e^e); on top of each other.  I would personally proceed by generating a complex valued Abel \( \alpha \) function first, \( \alpha=\frac{\ln\Psi}{\ln \lambda} \).  But then \( \alpha \) also has a singularity at the fixed point of L, in addition to the singarities at 0,1,e,e^e.... so there is that additional complexity which would have to be formally dealt with, though Kneser has to deal with an equivalent problem.  I'm going to cheat and assume we have the desired Tet(z), then we can take \( \exp(2\pi i)\circ\alpha\circ\text{Tet}\circ\frac{\ln(z)}{2\pi i} \); and this is Kneser's region Riemann mapped to a unit circle, and the singularity is at U(1).  The Riemann mapping is a 1to1 mapping of Kneser's region to the unit circle, and the inverse of the Riemann mapping function is analytic inside the unit circle, where the singularities at 0,1,e now correspond to z=1.

So then Kneser unwraps the Riemann mapping, to get an slog/tet which is real valued at the real axis.  What about the singularities at 0,1,e,e^e?  Kneser uses the Schwarz reflection principle, and then to show that the singularity is gone, we have to show that we can now walk around what was the singularity at z=0, and get back to where we started, and that this is bounded.  btw; I admit I'm not able to do a rigorous Kneser proof, but hopefully this helps.

First question : why at these values and not at say pi,exp(pi),exp(exp(pi)),... ?
Second question : this indicates there are singularities at 1,e,e^e,....
What kind of singularities are they ?? 
The corners seem to suggest log or sqrt or such, but I am not sure. 

I don't understand the first question; Kneser's Chi* is the \( \Psi\circ\Re \) complex valued Schroeder function of the real number line, but the singularities are at -infinity,0,1,e ....  The sequence can also be extended backwards by taking the logarithm of \( (-\infty,0) \) to include \( (\pi i+\infty,\pi i-\infty) \).  

The singularity of the Schroeder function at \( \Psi(0) \) is far more complicated than a simple logarithmic branch type singularity, and weaves in and out as it makes its way slowly towards infinity, following subsets of path similar to the chistar itself.  The ChiStar is generated by iterating logarithms; consider the iterated logarithm of the limit as \( \ln^n\circ\lim_{\delta \to 0}0+\delta \).  In particular, when \( \delta\approx\frac{1}{e\uparrow\uparrow{n}};\;n\ge 3 \) interesting behavior occurs in the iterated logarithm of the chistar function, where if you take the logarithm twice you get \( (e\uparrow\uparrow{n-2})+\pi i \) and if you take the logarithm n times the value once again gets very close to zero, so there is a pseudo recursive structure to the ChiStar.  I don't think enough attention has been given to this wonderfully complicated singularity, or the complexity the singularity gives to the region that needs to be Riemann mapped.  JayDFox had a post on the singularity, and I had some posts too

So I guess that makes it question 4.  We continue... First we take a log base L to solve 

Yeah, the log base L gives you a complex valued Abel function with an additional logarithmic singularity at L which is a more straightforward logarithmic singularity.  This singularity at L gives the complex valued superfunction is periodicity of \( \frac{2\pi i}{L}\approx 4.447+1.058i \)  After the Riemann mapping is converted to a 1-cyclic theta mapping.   There is still a singularity in the slog at L; \( \text{slog}(z)=\alpha(z)+\theta(\alpha(z)) \)  but each time you loop around that singularity L, \( \alpha(z)\mapsto \alpha(z)+\frac{2\pi i}{L} \), and the 1-cyclic mapping decays very quickly as imag(z) gets larger, so the magnitude of theta is about 0.0013x smaller for each loop around L.  btw; Kneser used \( \text{slog}(z)=\tau\circ\alpha(z) \) where \( \tau(z)=z+\theta(z) \) and tau is the Riemann mapping to a unit unit circle via \( f(z)=\exp(2\pi i)\circ\tau\circ\frac{\ln(z)}{2\pi i};\;\;f^{-1}(z) \) maps \( \mathbb{U}(z) \) to the real axis.

Question 7 :  How does the riemann mapping not destroy the functional equation ?

The singularity at L does get more complicated after the Riemann mapping, but I would look at it from the point of view of the slog.  \( \text{slog}(z)=\alpha(z)+\theta(\alpha(z)) \); and theta is a 1-cyclic function which does decay to zero as Imag(z) increases to infinity by the way it was defined.  Start wihere alpha(z) is well defined and show that \( \alpha(-0.5)=\alpha(\exp(-0.5))-1 \) if you connect the two staying in the upper half of the complex plane.  So alpha still clearly has the functional equation alpha(exp(z))=alpha(z)+1. Then the question is resolved by showing that the theta 1-cyclic mapping also keeps the functional equation intact.  One can start by analyzing and/or doing numerical approximations of Kneser's slog in a region between the fixed points, L;L*, and extending it to the complex plane.  Hope this helps, and thanks for the questions.