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+--- Thread: theta and the Riemann mapping (/showthread.php?tid=700)
theta and the Riemann mapping - sheldonison - 09/21/2011
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
RE: theta and the Riemann mapping - sheldonison - 09/28/2011
(09/21/2011, 05:23 AM)sheldonison Wrote: 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.From my point of view, the biggest obstacle to using the theta(z) in my own program to generate numerical results for sexp(z) is the really nasty singularity that theta(z) has at integer values of z. This is probably also the biggest obstacle to using the Riemann mapping in Kneser's construction to generate numerical results for sexp(z). Both have fascinating very complex singularities, as you superexponentially approach integer values for sexp(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.
\( u(y) = \frac{\log (\text{RiemanMapping}(y)/y)}{2\pi i} \)
....
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
This is the contour for theta(z)+z, going through 5 iterations, from z=-2 to z=3 \( \text{superf}^{-1}(\text{sexp}(z)) \), where we approach within sexp(3.5) or 1.6E-78 of the integer values of z.
This is the corresponding Riemann unit circle mapping, approaching within 1.6E-78 of the singularity. Surprisingly, At this level, there is still a visible discontinuity in the Riemann mapping unit circle! Later in this post, I post the numerical values for 100 terms of the Taylor series of the Kneser Riemann mapping unit circle function.
Going back to the theta(z)+z contour, now I fill in the detail, getting superexponentially closer and closer to the singularity at sexp(z=-1)=0. The earlier plot above approached to within sexp(z)=1.6E-78. In green, the first extension goes 1/sexp(3.5) to 1/sexp(4.5). For comparison, a googolplex is roughly sexp(4.53). But the detail increases as we superexponentially approach zero. The next extension, in red goes from 1/sexp(4.5) to 1/sexp(5.5). Sexp(5.5) is a number too larger to meaningfully describe. The next extension, in green, goes from 1/sexp(5.5) to 1/sexp(6.5). And the next extension, in red goes from 1/sexp(6.5) to 1/sexp(7.5). And finally, in green 1/sexp(7.5) to 1/sexp(8.5). I used superexponential approximations to calculate these plots, which was actually quite a difficult calculation! Underneath, is the exact corresponding theta(z) plot. Notice that theta(z) function will slowly approach +real infinity, and -imag infinity, as the function superexponentially approaches the singularity at zero. Also, notice that the singularity becomes ever more and more complicated, winding and unwinding along repeated paths superexponentially close to paths that it has already followed.
For refeence, this is the chi-star contour plot, through 5 iterations, from z=-2 to z=3 \( \text{Schroeder}(\text{sexp}(z)) \), where we approach within sexp(3.5) or 1.6E-78 of the integer values of z.
I'm not an expert on Riemann mappings, and I've never used any of the kernals that can be used to calculate a Riemann mapping. My own approach is to avoid the singularity, where the functions are more well behaved. For my kneser.gp program, I iteratively calculated the theta(z) mapping at imag(z)=0.12i, instead of at the real axis. This is far enough away from the singularity to avoid all of the numerical difficulties. So, I thought to myself, why not use the exact same idea to iteratively calculate Kneser's Riemann mapping? So, I modified a version of my "kneser.gp" program, to use the Riemann mapping in place of the unit circle theta function, using the equations from the first post in this thread, for sexp(z) base "e". I used the code I developed in kneser.gp, to iteratively generated the Riemann mapping approximations and sexp(z) approximations from each other, using the Riemann mapping unit circle function to give increasingly more accurate approximations for sexp(z) for imag(z)>=0.12i. After fifteen iterations, the result, was a 110 term taylor series for the Riemann mapping unit circle, accurate to 32 decimal digits, for imag(z)>0.12i, along with a taylor series for sexp(z), centered at z=0, and also accurate to 32 decimal digits. This compares with thirteen iterations required for the theta(z) mapping, so this Riemann mapping approach appears to be a little bit less efficient than the theta(z) approach. Here is the resulting Riemann mapping unit circle taylor series that I iteratively generated. The absolute value of the Taylor series terms is slowly decreasing until the 98th term, so the terms beyond that have no numerical value, and are doubling due to random noise in the numerical calculations, since 0.12i corresponds to a radius of 0.47 for the unit circle function.
Now that I know definitively that theta(z) is a completely different function than the Riemann mapping unit circle, I will update the naming conventions used in the next version of my kneser.gp program update, to eliminate any references to the Riemann mapping unit circle. I realize that anyone familiar with Kneser's approach must have been confused by my naming conventions.
- Sheldon Levenstein
Code:
a0= 0
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a109= 945.99666317573730972320340967517 - 411.13177402160055414902246549332iRE: theta and the Riemann mapping - sheldonison - 10/11/2011
I've been playing around with Mike's pari-gp implementation of the HSB graphing system, and I thought I'd add a busy plot, showing the evolution of the superfunction (on top), to the sexp(z) function (on the bottom), with the z+theta(z) contour highlighted in yellow. I repeated the contour between the two graphs, with more details. The superfunction is exactly Periodic, with a period=4.4470 + 1.0579i; below the sexp(z) function converges to the same period as imag(z) increases. I'm using the HSB graphing system unmodified, however, I used some tricks to preserve as much accurate information as possible, when iterating exp(z) to huge numbers, while eventually treat the imaginary component as a random number, to model the chaos as the points randomly go from huge positive, to huge negative.
So this plot, in addition to showing the superfunction, and the sexp(z), also shows one unit length of the z+theta(z) mapping linking the two functions. The highlighted yellow path shows one unit length, developed by taking the inverse superfunction from z=0 to z=1. You can also see the the cyan contour from the previous iteration, going from -infinity to 0. In the HSB graphing system, negative numbers are shown in cyan, and positive numbers are shown in red, with zero being black. The larger the magnitude, the brighter the plot. For larger magnitudes approaching 1E100, the saturation increases to pure white. The rainbow in the sexp(z) function occurs at approximately sexp(2.5 to 3), as the sexp(z) function begins to swings from large positive to large negative numbers. I tried to show how z+theta(z) switches back and forth, turning around infinite number of times as necessary, as it takes an infinitely long path from the inverse_superfunction(0) to the inverse_superfunction(1). I cut off the infinite section going towards superfunction(z)=1. The yellow contour gets mapped to sexp(-1) to sexp(0). If the yellow contour is repeated, then everything above the yellow contour has a 1 to 1 bijection with the upper half of the complex plane for sexp(z). Sometime, I will post the pseudo recursive relationship, governing the pseudo repeating pattern, and the turn arounds in the contour, which occur at 1/sexp(n) for integer values of n>=5; the graph in the middle shows some of the pattern.
- Sheldon