08/15/2010, 10:05 AM
(This post was last modified: 08/15/2010, 04:03 PM by sheldonison.)

(08/15/2010, 06:34 AM)bo198214 Wrote: 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:I guess Kneser is mapping the slog from the Abel function (or as I was referring to it, the inverse superfunction). I never quite finished understanding Kneser's algorithm the last time we exchanged posts online, and blindly went forward anyway, even though I knew there was some kind of serious gap in my understanding of Kneser's Riemann mapping approach. Generating the sexp(z) via a conformal map from the superfunction is probably theoretically equivalent, but it causes a lot of confusion. Apologies.

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

\( \rho^{-1}(z)=\text{isuperf}(\text{sexp}(z))=\theta(z)+z \).

added comment: finally now I think I could unerstand Kneser's solution! I was mentally stuck on conformally mapping the superfunction, with a complex theta. Now it all makes sense. The reason his notation uses the abel function so much is because he's conformally mapping the slog. On well, then the conformal map I'm trying to generate is a slightly different conformal map than Kneser's solution, but mathematically equivalent. You gave the equivalence equation for rho^-1(z)=z+theta(z). Again, sorry for the confusion.

Quote:....Yes, the Riemaprx (Riemann approximation) was generated from an imperfect sexp(z), so we had to truncate the negative indices. However, the Riemaprx is a true superfunction, and it converges, but the truncation of the negative indices means that it is no longer real valued at the real axis. Typically, there is a small error term. It is tricky to calculate the Riemaprx at the real axis though, because of the number of terms required for convergence, but if you could calculate it exactly to infinite precision, then all the higher harmonics would be gone, and you would be left with mostly a low frequency error component, where the real axis is no longer exact.

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?

Quote: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?I would imagine the complex conjugation trick would cause some really bad theoretical headaches. We're generating the laurent series for a unit circle of the discontinuous sexp(z) approximation, generated from riemaprx(z) and its complex conjugate. Once again, we throw away the negative indices! This time, we're left with an sexp(z) function that is real valued, but is no longer an exact superfunction. But, between sexp(z=-0.5) and sexp(z=0.5), it is a reasonable enough approximation, and is used to generate another slightly better Riemaprx(z) function.

I briefly looked at trying to generate the "anti" error term, to cancel out the errors, and push the function back onto the real axis, but abandoned that approach when it turned out that using the sexp(z) generated from the riemaprx and its complex conjugate worked so well.

- Sheldon