I am still a bit confused about the (even basic) steps in your program and about the key relations that make it work. So let me present from my view:
In the Kneser construction there is the regular Abel function at the primary fixpoint (in the upper halfplane). You call this function \( \text{isuperf} \). It maps the upper halfplane H to some area \( \Omega \) which is bounded by the arcs \( \text{isuperf}((0,1))+k \), \( k\in\mathbb{Z} \).
The Riemann mapping in Kneser's construction then is the bijective map that maps \( \Omega \) back to the upper halfplane. So that Kneser then defines:
\( \text{slog}(z) = \rho(\text{isuperf}(z)) \), or in super expressions:
\( \text{sexp}(z) = \text{superf}(\rho^{-1}(z)) \).
What has this function to do with your \( \text{RiemannCircle} \)???
Apart from
\( \rho^{-1}(z)=\text{isuperf}(\text{sexp}(z))=\theta(z)+z \).
Why is it called RiemannCircle?
Ok, but even if only the naming confuses me here, i am still a bit puzzled what makes your algorithm work. I understand your basic steps as follows:
1. from sexp obtain the Fourier-coefficients (with positive indices) of \( \theta(z)= \text{isuperf}(\text{sexp}(z))-z \) and call the Fourier-Series \( \text{RiemannCircle}(e^{2\pi i z})=\theta(z) \).
2. from RiemannCircle obtain the Powerseries-coefficients of \( \text{Riemaprx}(z)=\text{superf}(z+\text{RiemannCircle}(e^{2\pi i z})) \).
And there my questions start. It seems that Riemaprx is not equal to sexp as should:
\( \text{superf}(z+\text{RiemannCircle}(e^{2\pi i z})) = \text{superf}(z+\theta(z))=\text{sexp}(z) \). I guess this is due to truncation of the negative indices in the Fourier-Series. Can you confirm this?
So if it is not the same as sexp, you force it into a real-valued sexp (which I guess Riemaprx isnt) by this conjugation trick. (Which really causes me headache seen from a theoretic side, I guess the so constructed function is not even continuous on the unit circle) Would it be equally possible to just define sexp just as the real part of Riemaprx, or doesnt the whole algorithm converge then?
In the Kneser construction there is the regular Abel function at the primary fixpoint (in the upper halfplane). You call this function \( \text{isuperf} \). It maps the upper halfplane H to some area \( \Omega \) which is bounded by the arcs \( \text{isuperf}((0,1))+k \), \( k\in\mathbb{Z} \).
The Riemann mapping in Kneser's construction then is the bijective map that maps \( \Omega \) back to the upper halfplane. So that Kneser then defines:
\( \text{slog}(z) = \rho(\text{isuperf}(z)) \), or in super expressions:
\( \text{sexp}(z) = \text{superf}(\rho^{-1}(z)) \).
What has this function to do with your \( \text{RiemannCircle} \)???
Apart from
\( \rho^{-1}(z)=\text{isuperf}(\text{sexp}(z))=\theta(z)+z \).
Why is it called RiemannCircle?
Ok, but even if only the naming confuses me here, i am still a bit puzzled what makes your algorithm work. I understand your basic steps as follows:
1. from sexp obtain the Fourier-coefficients (with positive indices) of \( \theta(z)= \text{isuperf}(\text{sexp}(z))-z \) and call the Fourier-Series \( \text{RiemannCircle}(e^{2\pi i z})=\theta(z) \).
2. from RiemannCircle obtain the Powerseries-coefficients of \( \text{Riemaprx}(z)=\text{superf}(z+\text{RiemannCircle}(e^{2\pi i z})) \).
And there my questions start. It seems that Riemaprx is not equal to sexp as should:
\( \text{superf}(z+\text{RiemannCircle}(e^{2\pi i z})) = \text{superf}(z+\theta(z))=\text{sexp}(z) \). I guess this is due to truncation of the negative indices in the Fourier-Series. Can you confirm this?
So if it is not the same as sexp, you force it into a real-valued sexp (which I guess Riemaprx isnt) by this conjugation trick. (Which really causes me headache seen from a theoretic side, I guess the so constructed function is not even continuous on the unit circle) Would it be equally possible to just define sexp just as the real part of Riemaprx, or doesnt the whole algorithm converge then?
