06/05/2014, 06:51 PM
(06/05/2014, 01:53 PM)sheldonison Wrote: There are equations to show the mathematical equivalence of the theta mappings used in kneser.gp for real bases to Kneser's Riemann mapping, assuming convergence; I posted the relevant equations below. ...I feel like Tommy sometimes (in a good way, I hope he takes this as a compliment). I feel like sometimes I can intuitively see what's going on, but I lack the rigorous mathematical tools to either "prove" what I believe intuitively, or to be able to put my intuition to work. (I see Tommy complaining a lot about people going back to Cauchy integrals, and I can understand his frustration. I'm still a newb with Cauchy integrals myself.)
\( \alpha(\text{tet}_b(z)) = z + \theta(z)\;\;\;\theta(z) = \sum_{n=0}^{\infty} a_n (\exp(2\pi i z))^n \)
For real bases, the z+theta(z) mapping is mathematically equivalent to Kneser's Riemann mapping. Kneser's Riemann mapping function would be generated from the \( a_n \) coefficients of the \( \theta(z) \) mapping as follows.
The Riemann mapping starts with, \( z + \theta(z)\; \) which is modified too \( \exp(2\pi i (z + \theta(z))) \) and then we map \( z \mapsto \frac{\ln(z)}{2\pi i} \)
\( \text{RiemannMapping}(z)= \exp(2\pi i \times( \frac{\ln (z)}{2\pi i} + \sum_{n=0}^{\infty} a_n z^n)) \)
\( \text{RiemannMapping}(z)= z \times \exp( \sum_{n=0}^{\infty} 2\pi i a_n z^n)\;\; \) which generates Kneser's Riemann mapping from Sheldon's theta(z) coefficients
Intuitively, I have a good understanding of Kneser's method. But turning that intuition into calculations is the hard part. I don't know how to do Riemann mappings in general.
I followed the steps outlined in the paper by Murid, Ali H. M. and Mohamed, Nurul Akmal (which I discussed in post #10 from this old thread). I tested it with basic functions at first, like x+x^2, just to make sure that I could make it work. Then I tested with functions with simple singularities on the boundary of a disc, like ln(x), or ln(ln(x)), or ln^[3](x), etc. Once I was sure that I had the right set of equations, I proceeded to analyze the Kneser construction.
Even though the math worked, and even though I understood the Riemann mapping in principle, I didn't understand how/why the math worked. It's like knowing that A -> B -> C, and you understand A, and you can confirm C, but you don't understand B. That's where I'm at with the Kneser method. So I still have some basic complex analysis to study and digest and internalize, before I can really analyze your method. I'd love to be able to understand it, so I can compare it to the way I constructed it, and see what the differences are, and see if/how they converge on the same solution.
Sadly, I've lost my Kneser construction code. Once I finish my slog library, that's my next big project, to rebuild the Kneser construction the way I did last time.
~ Jay Daniel Fox

