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 Equations for Kneser sexp algorithm sheldonison Long Time Fellow Posts: 684 Threads: 24 Joined: Oct 2008 08/15/2010, 10:05 AM (This post was last modified: 08/15/2010, 04:03 PM by sheldonison.) (08/15/2010, 06:34 AM)bo198214 Wrote: I am still a bit confused about the (even basic) steps in your program and about the key relations that make it work. So let me present from my view: In the Kneser construction there is the regular Abel function at the primary fixpoint (in the upper halfplane). You call this function $\text{isuperf}$. It maps the upper halfplane H to some area $\Omega$ which is bounded by the arcs $\text{isuperf}((0,1))+k$, $k\in\mathbb{Z}$. The Riemann mapping in Kneser's construction then is the bijective map that maps $\Omega$ back to the upper halfplane. So that Kneser then defines: $\text{slog}(z) = \rho(\text{isuperf}(z))$, or in super expressions: $\text{sexp}(z) = \text{superf}(\rho^{-1}(z))$. What has this function to do with your $\text{RiemannCircle}$??? Apart from $\rho^{-1}(z)=\text{isuperf}(\text{sexp}(z))=\theta(z)+z$.I guess Kneser is mapping the slog from the Abel function (or as I was referring to it, the inverse superfunction). I never quite finished understanding Kneser's algorithm the last time we exchanged posts online, and blindly went forward anyway, even though I knew there was some kind of serious gap in my understanding of Kneser's Riemann mapping approach. Generating the sexp(z) via a conformal map from the superfunction is probably theoretically equivalent, but it causes a lot of confusion. Apologies. added comment: finally now I think I could unerstand Kneser's solution! I was mentally stuck on conformally mapping the superfunction, with a complex theta. Now it all makes sense. The reason his notation uses the abel function so much is because he's conformally mapping the slog. On well, then the conformal map I'm trying to generate is a slightly different conformal map than Kneser's solution, but mathematically equivalent. You gave the equivalence equation for rho^-1(z)=z+theta(z). Again, sorry for the confusion. Quote:.... And there my questions start. It seems that Riemaprx is not equal to sexp as should: $\text{superf}(z+\text{RiemannCircle}(e^{2\pi i z})) = \text{superf}(z+\theta(z))=\text{sexp}(z)$. I guess this is due to truncation of the negative indices in the Fourier-Series. Can you confirm this? Yes, the Riemaprx (Riemann approximation) was generated from an imperfect sexp(z), so we had to truncate the negative indices. However, the Riemaprx is a true superfunction, and it converges, but the truncation of the negative indices means that it is no longer real valued at the real axis. Typically, there is a small error term. It is tricky to calculate the Riemaprx at the real axis though, because of the number of terms required for convergence, but if you could calculate it exactly to infinite precision, then all the higher harmonics would be gone, and you would be left with mostly a low frequency error component, where the real axis is no longer exact. Quote:So if it is not the same as sexp, you force it into a real-valued sexp (which I guess Riemaprx isnt) by this conjugation trick. (Which really causes me headache seen from a theoretic side, I guess the so constructed function is not even continuous on the unit circle) Would it be equally possible to just define sexp just as the real part of Riemaprx, or doesnt the whole algorithm converge then? I would imagine the complex conjugation trick would cause some really bad theoretical headaches. We're generating the laurent series for a unit circle of the discontinuous sexp(z) approximation, generated from riemaprx(z) and its complex conjugate. Once again, we throw away the negative indices! This time, we're left with an sexp(z) function that is real valued, but is no longer an exact superfunction. But, between sexp(z=-0.5) and sexp(z=0.5), it is a reasonable enough approximation, and is used to generate another slightly better Riemaprx(z) function. I briefly looked at trying to generate the "anti" error term, to cancel out the errors, and push the function back onto the real axis, but abandoned that approach when it turned out that using the sexp(z) generated from the riemaprx and its complex conjugate worked so well. - Sheldon bo198214 Administrator Posts: 1,594 Threads: 101 Joined: Aug 2007 06/10/2011, 08:48 AM (08/08/2010, 07:14 PM)sheldonison Wrote: Theta(z) has a singularity at all integer values of n. Theta(z) is represented by an infinite sequence of fourier terms. The fourier series for theta(z) can be developed from any arbitrary unit length on the real axis of sexp(z), where z>-2. Only terms with positive values of n are included, and all terms a_n for negative values of n are zero. I have to ask here again. "Only terms with positive values of n are included.": Why did you put that restriction? Is there inherent reason, or is it just that you think this is the most natural and simplest way? sheldonison Long Time Fellow Posts: 684 Threads: 24 Joined: Oct 2008 06/10/2011, 01:43 PM (This post was last modified: 06/10/2011, 05:25 PM by sheldonison.) (06/10/2011, 08:48 AM)bo198214 Wrote: (08/08/2010, 07:14 PM)sheldonison Wrote: Theta(z) has a singularity at all integer values of n. Theta(z) is represented by an infinite sequence of fourier terms. The fourier series for theta(z) can be developed from any arbitrary unit length on the real axis of sexp(z), where z>-2. Only terms with positive values of n are included, and all terms a_n for negative values of n are zero. I have to ask here again. "Only terms with positive values of n are included.": Why did you put that restriction? Is there inherent reason, or is it just that you think this is the most natural and simplest way?For clarity, this is the integral, to generate the individual a_n terms for the $\theta(z)$ Fourier series approximation, generated from the sexp(z) approximation. $a_n =\int_{\;-0.5+0.12i}^{\;0.5+0.12i} (\text{superf}^{-1}(\text{sexp}(z))-z) \times \exp(-2n\pi i z) \; \mathrm{d}z$ $\theta(z)=\sum_{n=0}^{\infty}a_n\times \e^{(2n\pi i z)}$ $\text{Riemaprx}(z)=\text{superf}(z+\theta(z))$, Kneser Riemann mapping approximation, for $\Im(z)>=0.12i$ Because sexp(z) here, is only an approximation, including terms with negative values of n would mean that theta(z) would not decay as z goes to $\Im\infty$. It would also mean that the theta(z) function would only be defined at one and only one value of $\Im(z)=0.12i$, since an infinite Fourier series typically only converges where the series is sampled. For this algorithm, I have chosen to sample at imag(z)=0.12i, see Jan 11th 2011 post. By throwing out the a_n terms, with n<0, we have a function which is defined for $\Im(z)>=0.12i$, with singularities at integers. The singularities are because the sexp(z) approximation function we are generating the Fourier series of is only approximate. So $\text{superf^{-1}(\text{sexp}(z))-z$ is only an approximately 1-cyclic function. Since the solution we are ultimately looking for has a_n terms with negative values of n all zero, this allows the iterated sequence of functions to converge. Again, I think the picture in the previous post I linked to helps a lot. After doing the Fourier analysis, the function on this circle is used to generate the next sexp(z) approximation, as described in the algorithm, where f(x) is the previous sexp(z) approximation. The sexp(z) would also require a full Laurent series to converge on the unit circle (and then only converge on the unit circle), but we throw out the coefficients for the z^-n terms. If we had the exact sexp(z) function, and we did an exact infinite Fourier analysis at the real axis, or at any other value of $\Im(z)$, then all of the a_n terms with negative n would be zero, and the infinite sequence of positive a_n terms gives a function which converges for $\Im(z)$>=0, as long as z is not an integer. - Sheldon bo198214 Administrator Posts: 1,594 Threads: 101 Joined: Aug 2007 06/13/2011, 01:12 PM (This post was last modified: 06/13/2011, 01:14 PM by bo198214.) (06/10/2011, 01:43 PM)sheldonison Wrote: $\theta(z)=\sum_{n=0}^{\infty}a_n\times \e^{(2n\pi i z)}$ Because sexp(z) here, is only an approximation, including terms with negative values of n would mean that theta(z) would not decay as z goes to $\Im\infty$. Ahhh! Now I understand $\lim_{z\to i\infty} e^{2n\pi i z} = 0$ you choose *the* $\theta$ that that decays towards ioo. Is this a uniqueness criterion for $\theta$? I mean that $\text{superf}(z+\theta(z))$ is a real analytic superfunction and $\theta$ decays towards $i\infty$. Could be, ha? sheldonison Long Time Fellow Posts: 684 Threads: 24 Joined: Oct 2008 06/14/2011, 03:00 PM (This post was last modified: 06/14/2011, 03:02 PM by sheldonison.) (06/13/2011, 01:12 PM)bo198214 Wrote: Ahhh! Now I understand $\lim_{z\to i\infty} e^{2n\pi i z} = 0$ you choose *the* $\theta$ that that decays towards ioo. Is this a uniqueness criterion for $\theta$? I mean that $\text{superf}(z+\theta(z))$ is a real analytic superfunction and $\theta$ decays towards $i\infty$. Could be, ha?Its definitely a uniqueness criterion. Another way to think about it is from the point of view that $\theta(z)$ is connected to Kneser's unique Riemann mapping, since $\rho^{-1}(z)=\theta(z)+z$. But yes, any sexp(z) solution would either be the unique solution, with $\theta(z)$ exponentially decaying to a constant as $z\to+i\infty$, or else if it is any other solution, than $\theta(z)$ grows exponentially as $z\to+i\infty$. - Sheldon bo198214 Administrator Posts: 1,594 Threads: 101 Joined: Aug 2007 06/14/2011, 05:07 PM (06/14/2011, 03:00 PM)sheldonison Wrote: Its definitely a uniqueness criterion. Another way to think about it is from the point of view that $\theta(z)$ is connected to Kneser's unique Riemann mapping, since $\rho^{-1}(z)=\theta(z)+z$. But yes, any sexp(z) solution would either be the unique solution, with $\theta(z)$ exponentially decaying to a constant as $z\to+i\infty$, or else if it is any other solution, than $\theta(z)$ grows exponentially as $z\to+i\infty$. That screams for a proof, does it? sheldonison Long Time Fellow Posts: 684 Threads: 24 Joined: Oct 2008 06/19/2011, 03:14 AM (This post was last modified: 06/21/2011, 01:40 AM by sheldonison.) (06/14/2011, 05:07 PM)bo198214 Wrote: (06/14/2011, 03:00 PM)sheldonison Wrote: Its definitely a uniqueness criterion... any sexp(z) solution would either be the unique solution, with $\theta(z)$ exponentially decaying to a constant as $z\to+i\infty$, or else if it is any other solution, than $\theta(z)$ grows exponentially as $z\to+i\infty$. That screams for a proof, does it?For purposes of this proof, we assume we have two different real valued sexp solutions. It is assumed that the first sexp_a solution follows the uniqueness criteria, and then a proof is given that the second solution does not follow the uniqueness criteria. $\text{sexp}_a(z)=\text{superf}(z+\theta_a(z))$ $\text{sexp}_b(z)=\text{superf}(z+\theta_b(z))$ The uniqueness criteria is that the 1-periodic function $\theta_a(z)$ exponentially decays to a constant as $z\to+i\infty$, meaning that there all of the individual terms for n>=1 in $\theta_a(z)$ exponentially decay as $z\to+i\infty$. It will be shown that $\theta_b(z)$ does not decay to a constant as $z\to+i\infty$. So there cannot be two different solutions, where $\theta_a(z)$ and $\theta_b(z)$ which both decay to a constant as $z\to+i\infty$, and therefore, sexp_a(z) is a unique solution. $\theta_a(z) = a_0 + \sum_{n=1}^{\infty} a_n \exp(2n\pi i z)$ It can be trivially proven that terms of the form $a_n \exp(2n\pi i z)$ decay to zero as $z\to+i\infty$. Since sexp_a and sexp_b are both real valued, then there exists another real valued 1-periodic function, linking sexp_b to sexp_a, which will be called $\theta_c(z)$. $\text{sexp_b}(z) = \text{sexp_a(z+\theta_c(z))$ Because $\theta_c(z)$ is real valued at the real axis, it must be represented as a sum of this form, where at the real axis, the 2nd set of exponential summation terms are the complex conjugate of the first set of exponential summation terms, so that their sum is a real number. Any real valued 1-periodic function can be represented in this form. $\theta_c(z) = c_0 + \sum_{n=1}^{\infty} c_n \exp(2n\pi i z) + \sum_{n=1}^{\infty} \overline{c_n} \exp(-2n\pi i z)$ For $\theta_c(z)$, there are two sets of exponential terms. Terms of the form $c_n \exp(2n\pi i z)$ decay to zero as $z\to+i\infty$, and likewise it can be trivally shown that terms of the form $\overline{c_n} \exp(-2n\pi i z)$ grow exponentially as $z\to+i\infty$. So, as $z\to+i\infty$, the behavior of $\theta_c(z)$ is determined only by the terms $\overline{c_n} \exp(-2n\pi i z)$. The next step is to generate an equation for $\theta_b(z)$, by using the equation for sexp_b(z) in terms of sexp_a(z) and $\theta_c(z)$. Then substitute the equation for sexp_a(z) in terms of the superfunction, and take the inverse superfunction, which gives an equation $\theta_b(z)$. $\text{sexp}_b(z) = \text{sexp}_a(z+\theta_c(z))$ $\text{sexp}_a(z) = \text{superf}(z+\theta_a(z))$ Then substituting $z+\theta_c(z)$ from the first equation into the second equation in place of (z) to get: $\text{sexp}_b(z) = \text{sexp}_a(z+\theta_c(z)) = \text{superf}(z+\theta_c(z)+\theta_a(z+\theta_c(z)))$ Then, notice that this equation can be compared to the other equation for sexp_b(z), which allows us to get an equation for $\theta_b(z)$. $\text{sexp}_b(z) = \text{superf}(z+\theta_c(z)+\theta_a(z+\theta_c(z))) = \text{superf}(z+\theta_b(z))$ Taking the inverse superfunction of both sides, results in this equation $z+\theta_c(z)+\theta_a(z+\theta_c(z)) = z+\theta_b(z)$ $\theta_b(z) = \theta_c(z)+\theta_a(z+\theta_c(z))$ So, now there is an equation for the 1-periodic $\theta_b(z)$ in terms of the 1-periodic $\theta_c(z)$ and $\theta_a(z)$. The properties of $\theta_c(z)$ and $\theta_a(z)$ can be used to prove that $\theta_b(z)$ does not decay to a constant, but rather becomes a function whose amplitude grows arbitrarily large as $z\to+i\infty$, which means that sexp_b(z) does not meet the uniqueness criteria that defines sexp_a(z). First of all, assume that at least some of the c_n terms (n>=1) in $\theta_c(z)$ are non-zero. Otherwise theta_c(z) would be the identity, and sexp_b(z) would be equivalent to sexp_a(z). Now, we go back to the equation for $\theta_b(z)$ $\theta_b(z) = \theta_c(z)+\theta_a(z+\theta_c(z))$ We know that $\theta_c(z)$ grows exponentially as $z\to i\infty$ since the $\overline{c_n} \exp(-2n\pi i z)$ terms in the 1-periodic function $\theta_c(z)$ all grow exponentially. But what about $\theta_a(z+\theta_c(z))$? We need to show that $\theta_a(z+\theta_c(z))$ does not somehow cancel out the exponential growth of $\theta_c(z)$. We take advantage of the fact that as $z\to i\infty \;\; \theta_a(z) \approx a_0$, where the amplitude of the 1-periodic $\theta_z(a)$ terms all decay exponentially as $z\to i\infty$. Where $\Im(z+\theta_c(z))$ is large enough positive, then $\theta_a(z+\theta_c(z)) \approx a_0$ and $\theta_b(z) \approx a_0 + \theta_c(z)$. Where $\Im(z+\theta_c(z))$ approaches the real axis, which happens when $\Im(\theta_c(z))$ is negative enough, then the equations are less clear, since $\theta_a(z)$ has a singularity at the real axis for integer values of z, so $\theta_a(z)$ is not predictable when $z+i\Im(\theta_c(z))$ approaches the real axis. But the overall function, $\theta_b(z)$ for some arbitrarily large value of $\Im(z)\to+i\infty$, can be shown to have an arbitrarily large amplitude, tracking $\theta_b(z) \approx a_0 + \theta_c(z)$, as long as $\Im(z+\theta_c(z))$ is sufficiently large. So $\theta_b(z)$ cannot be converging arbitrarily closely to a constant as $\Im(z)\to+\infty$. That is to say, $\theta_b(z)$ will cover the range of values for $\approx a_0 + \theta_c(z)$, as opposed to converging to a constant, as we would expect if $\theta_b(z)$ were expressible in the same form as $\theta_a(z)$. Therefore, $\theta_b(z)$ does not match the assumption, that sexp_b(z) as an alternative solution which also meets the uniqueness criteria. It seems to me that the last paragraph is hard to follow, and that it probably needs to be rewritten and formalized. But, at the moment, I'm not quite sure how to do that. But at the very least, I think one can compare the complex somewhat unpredictable arbitrarily large amplitude non-converging behavior of $\theta_b(z)$ as $z\to+i\infty$ with the exponential decay of $\theta_a(z)$ to a constant as $z\to+i\infty$, and see the contradiction in assuming that sexp_b(z) is an alternative solution which meets the uniqueness critera. Therefore, there aren't multiple solutions which meet the uniqueness criteria. - Sheldon bo198214 Administrator Posts: 1,594 Threads: 101 Joined: Aug 2007 06/20/2011, 09:22 PM I didnt dive into the details yet, but the approach looks very promising. I am really excited, this seems something that wasnt yet discovered (though one never knows in holomorphic dynamics). Sheldon, you are our man! sheldonison Long Time Fellow Posts: 684 Threads: 24 Joined: Oct 2008 06/21/2011, 01:48 AM (This post was last modified: 06/21/2011, 04:30 PM by sheldonison.) (06/20/2011, 09:22 PM)bo198214 Wrote: I didnt dive into the details yet, but the approach looks very promising. I am really excited, this seems something that wasnt yet discovered (though one never knows in holomorphic dynamics). Sheldon, you are our man!Thanks Henryk. I look forward to your feedback on how to tighten up this proof. Also, I didn't realize that the uniqueness criteria for Kneser's solution had not been proven! I think I may need to separate the more complicated case where $\theta_c(z)$ has singularities as $\Im(z)\to\infty$, since at the singularity, $\theta_c(z)$ may only be growing arbitrarily large in the real direction, and not the imaginary direction. Ideally, it would be possible to show that a nearly identical singularity will always show up in $\theta_b(z)$, but this is clearly a more complicated case. - Sheldon « Next Oldest | Next Newest »

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