08/16/2018, 12:48 PM
(This post was last modified: 08/16/2018, 11:35 PM by sheldonison.)
(08/16/2018, 08:54 AM)Chenjesu Wrote: But that directly contradicts what you said before that we don't actually know what the half-exp really is in terms of functions. So how do they graph a plot function with no explicit representation? Do they use a taylor series? hypergeometric series? bessel series? Or what?
Actually, what I wrote is:
Quote:Unfortunately, the Riemann mapping is hard to compute, so realistically, all that we have is the proof of the existence of a nicely behaved solution. In general, Riemann mappings cannot be expressed in terms of elementary functions or series and therefore Kneser's solution probably isn't expressible in terms of elementary functions.
.. Paulsen's paper gives numerical results just like Trapmann's paper ... There are a few really good programs that can easily calculate sexp for any base, like my pari-gp program, fatou.gp
My program iterates, generating more and more accurate representations for two different functions; one function is a Fourier series involving the Schroder function which is mathematically equivalent to Kneser's Riemann mapping. The 1-cyclic Fourier series is mapping this image (equivalent to Figure 1 from Paulson's paper) to the real axis. The region above is mapped to the upper half of the complex plane. The 1-cyclic Fourier series does not converges near the singularities at the real axis, so near the real axis one can use a 2nd function. The 2nd function involved in the iteration sequence is an approximation for the Taylor series for the slog, centered at a point on the real axis. The two functions combined allow one to generate the slog and sexp for any point in the complex plane. If the two functions match each other (within the each function's error term) where both functions are well behaved, then as the error term goes to zero, the pair of functions converge to Kneser's solution. In the limit, the 1-cyclic Fourier series becomes mathematically equivalent to Kneser's Riemann mapping.
- Sheldon

