Thread Rating:
  • 0 Vote(s) - 0 Average
  • 1
  • 2
  • 3
  • 4
  • 5
Attempting to compute the kslog numerically (i.e., Kneser's construction)
(09/25/2009, 07:56 AM)bo198214 Wrote:
(09/25/2009, 12:45 AM)jaydfox Wrote: And now, of course, we map this region back to the unit disk. Details tomorrow!

Its a pleasure to look at this quality work, partiularly the extension of the chi star. But I am even more curious how you find the Riemann-mapping!

Okay, now for the Riemann mapping itself. Here, I will admit that I was stumped for a while, but after doing fair amount of research, I saw that there were a couple of existing computational approaches, relying on so-called "kernels". Various kernels of interest were the Cauchy kernel, the Bergman kernel, and the Szego kernel. But implementation details were hard to find, until I stumbled across the following paper [1] after much googling:

Note that this only gets us the mapping to the unit disk. After that, it's a simple matter of taking the logarithm and dividing by 2*pi*i to get back a unit interval. An appropriate constant is added as needed to place 0 and 1 at -1 and 0, respectively, and we now have the half kslog! Mirroring across the real line completes the puzzle.

Armed with the referenced paper, I began experimenting with the first phase, which involves creating a parameterized closed curve around a point of interest, and then solving an integration problem numerically, via matrix methods. I initially performed the test with functions I could compute exactly, in order to verify the results against a theoretically exact answer. The method is quite stable: indeed, in my tests, I can solve a 400x400 matrix with merely 53-bit (double) precision, and retain 46 bits (measured versus a solution using 1024 bits). That means I lose only 7 bits during the matrix solution!

This in turn means that a very large system could be solved with rather low precision math, which is good because convergence isn't terribly fast in the case of the kslog (because of the singularities). However, it should be good enough to confirm whether this construction gives (probably) the same results as the intuitive method or the Cauchy integral method. So far, the results are promising.

If you read the referenced paper (which I've also attached, as the website seems to be down at the moment), you will see that the matrix solution is the solution to a special type of integral equation. Because it's merely an integral, we can use polynomial interpolations to improve accuracy, and I've confirmed this numerically (empirical, not theoretical evidence). Note that the singularities at the end of the boundary make unnecessary the "trapezoidal rule" used by the authors. Abandoning the trapezoidal rule and treating the boundary as if it were not cyclic, actually improved convergence (and, hypothetically, the accuracy). As I mentioned earlier, Fejer quadrature actually failed to converge, and produced horribly inaccurate results.

[1] Murid, Ali H. M. and Mohamed, Nurul Akmal
Computation of the Riemann Map using Integral Equation Method and Cauchy’s Integral Formula
In: Simposium Sains Matematik, 2004, Pulai Springs, Skudai. (Unpublished)

Attached Files
.pdf   AliHassanMohamedMurid2004_.pdf (Size: 95.99 KB / Downloads: 684)
~ Jay Daniel Fox

Messages In This Thread
RE: Attempting to compute the kslog numerically (i.e., Kneser's construction) - by jaydfox - 09/25/2009, 03:24 PM

Possibly Related Threads...
Thread Author Replies Views Last Post
  Kneser-iteration on n-periodic-points (base say \sqrt(2)) Gottfried 11 4,089 05/05/2021, 04:53 AM
Last Post: Gottfried
  fast accurate Kneser sexp algorithm sheldonison 38 114,433 01/14/2016, 05:05 AM
Last Post: sheldonison
  "Kneser"/Riemann mapping method code for *complex* bases mike3 2 10,034 08/15/2011, 03:14 PM
Last Post: Gottfried
  Attempt to make own implementation of "Kneser" algorithm: trouble mike3 9 24,206 06/16/2011, 11:48 AM
Last Post: mike3
  An incremental method to compute (Abel) matrix inverses bo198214 3 13,078 07/20/2010, 12:13 PM
Last Post: Gottfried

Users browsing this thread: 1 Guest(s)