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 Kneser's Super Logarithm sheldonison Long Time Fellow Posts: 664 Threads: 23 Joined: Oct 2008 01/23/2010, 01:01 PM (This post was last modified: 01/27/2010, 08:36 PM by sheldonison.) (corrections and additions, corrected notation) It occured to me that complex fourier analysis is at the heart of a lot of key topics for tetration. I finally have gotten to the point where I understand Kouznetsov's Cauchy integral, and implemented parts of it. I also think I might understand Kneser's approach, based on the Riemann mapping, starting with the entire analytic super-exponential, developed from the fixed point, c~=0.318+i*1.337. This is the inverse Abel function, where the Abel function is the slog developed from the fixed point c. $\psi^{-1}(z)=\lim_{n\to\infty}( \exp^{\circ n}(c+c^{(z-n)})$, at least I think that is what is meant by $\psi^{-1}(z)$. I tried really hard to understand Jay's post, and this post, but there are gaps.... I tried to fill in the gaps, but I have a feeling I'm going down a slightly different approach here. Is the real valued sexp at the real axis connected to the Schroeder equation, chi, by a complex 1-cyclic function, $\theta(z)-z$ $\text{sexp}_e(z)=\psi^{-1}(\theta(z))$ Where $\theta(z)$ is the Riemann mapping of contour of the limit of iterating the natural logarithm, starting with the real interval, 0..1 (which corresponds to sexp(z) from z=0 to z=1). Then subtracting c, which is the fixed point, ~= 0.318+i*1.337? Then the Riemann mapping unit circle is scaled by 2*pi and converted to a repeating unit length at the real axis. Is this equation correct, and is this Kneser's approach? $\psi(z)=\lim_{n\to\infty}( \log_c(\ln^{\circ n}(z)-c)+{n})$ $ \theta(z)=\operatorname{RiemannMapping} \lim_{n\to\infty}( \log_c(\ln^{\circ n}[0..1]-c)+{n})$ This equation is developed from the inverse sexp function, or the Abel function, $\psi(z)$. Assuming the equation is correct, here is my rough calculation of the resulting limit, whose contour is what I think is to be Riemann mapped, and then used to generate $\theta(z)$. For double precision numbers >0 and <1, the real part of the contour never exceeds 1.78, so most of the contour graph is only for values of z super-exponentially close to zero or one, but I thought I would include the entire contour for completeness. Also, I removed the scaling, and added the next iteration, which is developed from sexp(z), from z=1 to z=2, which shows that $\theta(z+1)=\theta(z)+1$, and that $\theta(z)-z$ is 1-cyclic. Contour values from this graph should have imaginary values of zero for $\psi^{-1}(z)$. Also, the outermost $\log_c$ for the contour is multi-valued, and as n increases, sometimes multiples of -2pi*i have to be added so that the contour equation converges. The other values correspond to other identical parts of the $\psi{-1}(z)$ function, which repeats infinitely many times in the complex plane. Once the Riemann mapping is generated, and converted to a unit length at the real axis, to generate $\theta(z)$, I assume that the $\theta(z)$ goes to a small constant (the center of the Riemann mapping unit circle), as the imaginary axis grows, and has a singularities at unit intervals at the real axis. Presumably, the singularity in the Riemann mapping is at one point on the edge of the unit circle, but not inside the unit circle. The contour is valid before and after the Riemann mapping. But after the Riemann mapping, it would seem that $\theta(z)-z$ is a 1-cyclic function, and it is the only analytic 1-cyclic function that converts from $\psi^{-1}(z)$ to the sexp_e such that $\theta(z)-z$ will be a constant plus the sum of a complex fourier series, with complex coefficients, of $A_n*e^{(+2z\pi n {i})}$. Because of the unique Riemann mapping, all Fourier terms of the form $B_n*e^{(-2z\pi n {i})}$ are zero. Because the Riemann mapping to drive these terms to zero is unique, there is only one sexp solution approaching the fixed point as i increases, with the complex fourier series decaying to zero as i increases. Given that Kouznetsov's solution also assumes the limiting value of f as imaginary increases, it would seem that complex Fourier analysis would allow one to prove why the solution is unique, why Kouznetsov's solution is equal to Kneser's solution, and perhaps even why Kouznetsov's iterated Cauchy integral converges. - Sheldon « Next Oldest | Next Newest »

 Messages In This Thread Kneser's Super Logarithm - by bo198214 - 11/19/2008, 02:20 PM RE: Kneser's Super Logarithm - by bo198214 - 11/19/2008, 03:25 PM RE: Kneser's Super Logarithm - by sheldonison - 01/23/2010, 01:01 PM RE: Kneser's Super Logarithm - by mike3 - 01/25/2010, 06:35 AM RE: Kneser's Super Logarithm - by sheldonison - 01/25/2010, 07:42 AM RE: Kneser's Super Logarithm - by mike3 - 01/26/2010, 06:24 AM RE: Kneser's Super Logarithm - by sheldonison - 01/26/2010, 01:22 PM RE: Kneser's Super Logarithm - by mike3 - 01/27/2010, 06:28 PM RE: Kneser's Super Logarithm - by sheldonison - 01/27/2010, 08:30 PM RE: Kneser's Super Logarithm - by mike3 - 01/28/2010, 08:52 PM RE: Kneser's Super Logarithm - by sheldonison - 01/28/2010, 10:08 PM RE: Kneser's Super Logarithm - by mike3 - 01/29/2010, 06:43 AM RE: Kneser's Super Logarithm - by bo198214 - 01/26/2010, 11:19 PM RE: Kneser's Super Logarithm - by sheldonison - 01/27/2010, 07:51 PM RE: Kneser's Super Logarithm - by bo198214 - 11/22/2008, 06:11 PM RE: Kneser's Super Logarithm - by bo198214 - 11/23/2008, 01:00 PM

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