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 theta and the Riemann mapping sheldonison Long Time Fellow Posts: 641 Threads: 22 Joined: Oct 2008 09/21/2011, 05:23 AM (This post was last modified: 09/21/2011, 10:54 AM by sheldonison.) The equation to convert the theta(z) unit disk to Kneser's Riemann mapping unit disk is as follows, where u(z) is $\theta$ wrapped around the unit disk. $\text{RiemannMapping}(z)=z \times \exp(u(z) \times 2\pi i)$ I will derive this equation, but first some background. A year or two ago, I came up with the idea to generate sexp(z) from the superfunction(z), via the equation: $\text{sexp}(z)=\text{superfunction}(z+\theta(z))$, where the superfunction is developed from the inverse Schroeder function of exp(z). My assumption at that time, is that Kneser's construction also involved theta(z), which isn't exactly correct. I tried to understand Henryk's post on Kneser's construction, as well as Jay Daniel's post and graphs on the Kneser construction. However, my predisposition for believing Kneser's construction involved a Riemann mapping of theta(z), along with my limited background probably combined to prevent me from fully understanding Kneser's construction. My lack of understanding didn't stop me from deriving my own algorithm to iteratively generate Kneser's sexp(z) solution from $\theta(z)$, which turned out to be a big improvement in numerically calculating the sexp(z) function. Showing how the $\theta(z)$ function is connected to the Riemann mapping helps put my algorithm on firmer theoretical grounds. Kneser iterated logarithms of the upper half of the complex plane, via the Schroeder equation, to generate the Chi-Star. The limit equation for the Schroeder function is for exp(z) is: $S(z) = (\lim_{n\to\infty} (\log^{[n]}(z+L)-L)\times\exp(n L)$ where the Schroeder equation here is developed centered around the fixed point, $L\approx 0.318 + i1.337$. I was also iterating logarithms, but the difference is I was iterating logarithms to generate the inverse of the superfunction, and the superfunction and to calculate theta(z). The inverse of the superfunction, can be developed from the Schroeder equation as follows. $\text{superfunction}^{-1}(z) = \log(S(z-L))/L$ The limit equation to generate the inverse schroder equation is $S^{-1}(z) = \lim_{n\to\infty} \exp^{[n]}(L+z\times\exp(-n L))$ If you have the inverse Schroeder equation, then $S^{-1}(L z) = \exp(S^{-1}(z))$. The inverse Schroder equation can trivially be turned into the complex valued superfunction, which has superfunction(z+1)=exp(superfunction(z)). $\text{superfunction}(z) = S^{-1} (\exp(z L))$ As noted earlier, there is a simple exact mathematical connection between theta(z) and the Riemann mapping unit circle in Kneser Riemann mapping. Start by representing theta(z) as a Fourier series, and then equivalently, as a unit disk. $\theta(z)=\sum_{n=0}^{\infty}a_n\times \e^{(2\pi n i z)}$ Equivalently, one can wrap $\theta(z)$ around a unit disk, with the transformation $y=\exp(2\pi i z)$. Let us call this unit disk version of theta(z), "u(y)", where u stands for the unit disk. It is simply theta(z) wrapped around a unit circle, which is a little bit confusing since the theta(z) unit disk is not the same as the Riemann mapping unit disk in Kneser's construction. It does have some similarities, in that u(0) corresponds to $+\Im \infty$, since the theta(z) function was defined so that only positive values of n are used in the a_n terms of the fourier series. The u(y) function uses exactly the same a_n coefficients as the $\theta(z)$ function, but it is a taylor series instead of a fourier series. $u(y)=\sum_{n=0}^{\infty}a_n\times y^n$ The u(y) function is analytic in the unit disk, except for a singularity at u(1), which corresponds to the singularities in $\theta(z)$ for integer values of z. As stated earlier, the equation to convert the theta(z) unit disk to the Riemann mapping unit disk is: $\text{RiemannMapping}(y) = y \times \exp(u(y) \times 2\pi i)$ To derive this equation, one can start with the solution for sexp(z), and use that to generate $f(z)=z+\theta(z)$. $\text{sexp}(z)=\text{superfunction}(z+\theta(z))$ $f(z)= z+\theta(z)= \text{superfunction}^{-1}(\text{sexp}(z))$ f(z+1)=f(z)+1 by definition, since both the superfunction(z) and sexp(z) are superfunctions of exp(z). f(z) cannot be wrapped around the unit circle directly. But according to Jay's post, Kneser's trick is to multiply f(z) by $2\pi i$ and then exponentiate. call this function g(z). $g(z)=\exp(f(z)\times 2\pi i)=\exp((z+\theta(z))\times 2\pi i)$ Now, g(z+1)=g(z), since $\exp(2\pi i)=1$. So we have succeeded in creating a 1-cyclic function g(z). At the real axis, a unit length of g(z) can be wrapped a unit circle. So g(z), starting from z=0, to z=1, can be wrapped around a unit circle, with the interior uniquely defined by the Riemann mapping. A unit length of g(z) can be calculated from [0..1], by starting with any approximation, and then doing a Riemann mapping. Next, recall that $f(z)= z+\theta(z)$, so $g(z)=\exp((z+\theta(z))\times 2\pi i)=\exp(2z\pi i) \times \exp(\theta(z)\times 2\pi i)$ Next, recall the substitution connecting the u(y) function, which is exactly the substition necessary to wrap g(z) around the unit circle $y= \exp(2\pi i z)$ and $z = \frac {log(y)}{2\pi i}$ $\text{RiemanMapping}(y)=g( \frac {log(y)}{2\pi i})$, which is a Riemann mapping of a unit length of g(z), wrapped around the unit disk. And finally, with trivial algebra, $g(\frac {\log(y)}{2\pi i}) = y \times \exp(u(y)\times 2\pi i)$, which derives the formula for Kneser's Riemann mapping in terms of the unit circle of theta(z). $\text{RiemanMapping}(y) = y \times \exp(u(y)\times 2\pi i)$ This can also be reversed, to derive the unit circle theta(z) function from the Riemann mapping. $u(y) = \frac{\log (\text{RiemanMapping}(y)/y)}{2\pi i}$ The Riemann mapping uniqueness criteria required is RiemannMapping(0)=0. Otherwise dividing by y creates a singularity at the origin. The first derivative of the Riemann mapping taylor series at the origin is equal to the $\exp(2a_0\pi i)$ of the a0 taylor series term of the unit circle function, or the theta(z) function. The a0 term shows how in the limit, as imag(z) goes to infinity sexp(z) converges exponentially towards superfunction(z+a0). So, in summary, Kneser's Riemann mapping construction corresponds exactly equivalent to the theta(z) equations developed in my kneser.gp iterative generation of the sexp(z) function. Both the unit circle version of theta(z) and the Riemann mapping unit circle are defined everywhere in the unit circle, except for the singularity at z=1. Using these equations, I have succeeded in recreating all of the graphs in Jay's post, as well as some new ones of my own which I will post later. - Sheldon Levenstein « Next Oldest | Next Newest »

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