Kneser method question

[tommy1729]

It has been a while since I considered kneser’s method.

So I will try to describe it and then ask my question.
Correct me if I make mistake.

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.

Third question : do we pick branches ? I thought we did not. 

I can imagine that statistically such singularities or corners are expected because we need to respect the rotation. But why it is necessary, I have no formal explaination.

So I guess that makes it question 4.

We continue.

First we take a log base L to solve 

Exp(g(z)) = g(z) + 1

Notice we added a log on top of the singularities , right ?

So we arrive at question 5 and 6 :

Question 5 : did the log remove or simplify the singularities ?
This is only possible with powers like log x^a = a log x I think.

Question 6 : all singularities are still there right ??

Ok, so now we have a kind of Abel function g(z) but it does not map the positive reals to the positive reals.

Notice also the positive reals values are now arranged 1-periodic within the function g(x).

But often stacked on top of each other, hence unfortunately not describable by a Fourier series.

So we need to map the positive reals to the positive reals kinda.

We know that is possible with a somewhat unique analytic function from riemann’s mapping theorem.

However the riemann mapping is mysterious to many.

Apart from how it is done and closed form issues , error terms etc there is also this :

Question 7 :

How does the riemann mapping not destroy the functional equation ?

Question 8 :
The big question :

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

Question 9 : what happened to the other singularities ?

I have the feeling I’m not the only wondering about these things.

Regards

Tommy1729

[Gottfried]

It has been a while since I considered kneser’s method.

So I will try to describe it and then ask my question.
(...)
I have the feeling I’m not the only wondering about these things.

Regards

Tommy1729

Sorry, I've nothing to contribute to the contents currently.    But ----      
--- I think this concise list should be also asked in mathoverflow.net (by someone who can reproduce & discuss it better than me)

Gottfried

[Daniel]

Hello Tommy. I'm very impressed with your analysis and it's base of mathematical concepts. I have some feedback that hopefully will be of use. Questions 1 & 2: When I submitted a paper to the Annals of Mathematics the referee felt I should have discussed the role of germs in the study of iterated functions. Germs are sets of points with the same local properties, so they for equivalence classes. In iteration of functions the point zero is a different germ than it's immediate surrounding points. 

1&2: I don't think {0,1,e,e^e,...} and their forward orbits are singularities, I think they are a different germ from their surrounding points. 

3: A branch was picked when L was selected. In my research I consider the entire countable infinite set of branches indirectly when I deal with the fixed points on the different branches. For example I was able to compute the position and Lyapunov multiplier of a neighboring fixed point.

NOTE: I was wrong about {0,1,e,e^e,...} not being singularities. If they are at corners then there is a singularity in the second derivative.

[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.

[Ember Edison]

Hi. 

How to get the photo like this https://math.eretrandre.org/tetrationfor...p?tid=1172 in fatou.gp?

[sheldonison]

Hi. 

How to get the photo like this https://math.eretrandre.org/tetrationfor...p?tid=1172 in fatou.gp?

I export the set of points to a .csv file; .csv is a comma separated file for spreadsheets.  Then I import the .csv file into excel, and use excel's chart functions, "x-y scatter with smooth lines".   chiStar1_excel

Then sometimes I overlay the excel output with fatou.gp MakeGraph color plot.  This requires manually matching the size of the excel output so that it closely matches the size of the fatou.gp MakeGraph plot.  chistar_contour.png

The tricky thing about making a picture of the chi-star is that it nearly overlaps itself an infinite number of times as each singularity goes to infinity an infinite number of different ways.  Its not an easy thing to image. Here is the excel file for reference.  chistar.xls

The chi-star is a beautiful object that forms the basis of Kneser's tetration/slog.

[Ember Edison]

Hi. 

How to get the photo like this https://math.eretrandre.org/tetrationfor...p?tid=1172 in fatou.gp?

I export the set of points to a .csv file; .csv is a comma separated file for spreadsheets.  Then I import the .csv file into excel, and use excel's chart functions, "x-y scatter with smooth lines".   chiStar1_excel

Then sometimes I overlay the excel output with fatou.gp MakeGraph color plot.  This requires manually matching the size of the excel output so that it closely matches the size of the fatou.gp MakeGraph plot.  chistar_contour.png

The tricky thing about making a picture of the chi-star is that it nearly overlaps itself an infinite number of times as each singularity goes to infinity an infinite number of different ways.  Its not an easy thing to image. Here is the excel file for reference.  chistar.xls

The chi-star is a beautiful object that forms the basis of Kneser's tetration/slog.

......Sorry? So how to use chistar.xls get the chistar_contour.png, I can't reproduce. My import is not working.

[sheldonison]

......Sorry? So how to use chistar.xls get the chistar_contour.png, I can't reproduce. My import is not working.

I can't help you if you don't have a way to load an excel file. Once you can load an excel file, then use the screenshot function to create the image files   Searching on the internet, I see that google sheets also supports x-y scatter charts; however, I have not used google sheets.
https://www.howtogeek.com/226280/how-to-...indows-10/

[Ember Edison]

......Sorry? So how to use chistar.xls get the chistar_contour.png, I can't reproduce. My import is not working.

I can't help you if you don't have a way to load an excel file. Once you can load an excel file, then use the screenshot function to create the image files   Searching on the internet, I see that google sheets also supports x-y scatter charts; however, I have not used google sheets.
https://www.howtogeek.com/226280/how-to-...indows-10/

Thanks. It work when I reinstall the office.

I have some questions.
(1)Can we use \( \Psi \)Schröder and inverse Schröder functions, rewrite the Kneser’s superroot?
(2)Why fatou.gp can't "recursion" well-behaved region to fix the ill-region? sexp can work well when small Imag(z), and pent/hex can work well when small Real(z).

If you have time plaese fix the bug report. It feels so bad for pent(3) can't evaluate, It just a simple imput

[sheldonison]

Thanks. It work when I reinstall the office.

I have some questions.
(1)Can we use \( \Psi \)Schröder and inverse Schröder functions, rewrite the Kneser’s superroot?
(2)Why fatou.gp can't "recursion" well-behaved region to fix the ill-region? sexp can work well when small Imag(z), and pent/hex can work well when small Real(z).

If you have time plaese fix the bug report. It feels so bad for pent(3) can't evaluate, It just a simple imput

I'm focused on rigorously mathematically proving a simplified version of fatou.gp converges to Kneser's slog for real base e, so I'm not working on fixing any other boundary conditions for fatou.gp

You can get an Abel function from the Psi function, by taking the log_lambda
of the Psi function, where \( \lambda \) is the derivative at the fixed point; this
works for any function for which you have Psi.

\( \alpha(z)=\frac{\ln(\Psi(z))}{\ln(\lambda)} \)
But then to get Kneser's slog, you would need Kneser's Tau function; where theta (z) is a 1-cyclic function;  \( \tau(z)=z+\theta(z) \); otherwise as you are well aware, it would be a completely different function from Kneser's slog, and would give different results.  For example, \( \alpha \) would not be real valued at the real axis for real bases>exp(1/e).  

\( \text{slog}(z)=\tau(\alpha(z)) = \alpha(z)+\theta(\alpha(z)) \)
Generating \( \Psi(z) \) is fairly straight forward, but generating \( \tau(z) \) is equivalent to generate Kneser's Riemann mapping.