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superfunctions of eta converge towards each other - Printable Version +- Tetration Forum (https://tetrationforum.org) +-- Forum: Tetration and Related Topics (https://tetrationforum.org/forumdisplay.php?fid=1) +--- Forum: Mathematical and General Discussion (https://tetrationforum.org/forumdisplay.php?fid=3) +--- Thread: superfunctions of eta converge towards each other (/showthread.php?tid=634) Pages:
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RE: superfunctions of eta converge towards each other - sheldonison - 05/27/2011 Continuing with the conjecture, \( \text{sexp}_\eta(z) = \text{cheta(z+k+\theta(z)) \), here are some more graphs, and then some comments. cheta(z) is the upper entire superfunciton of eta. \( \theta(z) \) from z=-1+i to z=1+i, which I previously posted, where theta(z) is well behaved, and decaying as imag(z) increases. Next, lets regenerate this graph with \( \Im(z)=0.001 \), which is closer to the real axis, and closer to the singularity at ineger values of z. To get closer to the real axis, we need to switch to a graph of the contour of \( \text{cheta(z+k) \), by graphing the contour of \( \text{cheta}^{[-1]}(\text{sexp}_\eta(z))-\Re(k) \). The range if this graph is identical to the previous graph, from z=-1+0.001i to 1+0.001i. Only the real part of the constant k has been included, so the imag(z) values represent the contour values of cheta(z). This last cheta(z) contour plot is a graph of the Kneser mapping contour for cheta(z), at the real axis itself, calculating the contour \( \text{cheta}^{[-1]}(z)-k \), with a range of z from \( z=-10^{11}+e\pi i \) to \( z=+10^{11}+e\pi i \). This is equivalent to the range of z from \( \text{sexp}_\eta(-2+\delta_a) \) to \( \text{sexp}_\eta(-3-\delta_b) \), where delta_a and delta_b are very small. Near -2, \( \delta_a \approx 10^{-1634000000} \). I think that the \( \delta_b \approx 10^{-\exp(1634000000)} \)... messy arithmetic. The graph has been modified so that it matches the range from -1 to +1 from above. For base e, I have observed that the graph of the Knser mapping contour continues becoming more and more complex, as we superexponentially approach the singularity. Similar complexity may occur for base eta. I may post more in the future. Tommy wrote: Quote:another question is : how many superfunctions can a function have ? Hey Tommy. There are some restrictions. These Kneser/Riemann mappings involve 1) a theta(z) function quickly decaying to zero at +imag infinity, and 2) a resulting function with singularities at the integer values, where the singularities results in a Schwarz reflection, which allows the function to be defined for imag(z)<0. We already have the regular superfunction for base e, as an example, which is Kneser mapped to produced sexp_e(z). This is another example, where the upper/entire superfunction, cheta(z) is Kneser mapped to produce sexp_eta(z). These two restrictions, limit the kinds of functions involved. For tetration, this works for bases>eta, using the standard Kneser mapping, and for base cheta, as is conjectured. For bases<eta, other theta(z) mappings are possible. I made an entire post about them last year, where I discussed base 2. http://math.eretrandre.org/tetrationforum/showthread.php?tid=515 I have to refresh my memory on what I've posted, but I also derived a new different tetration solution for each base less than eta, using a Kneser mapping. - Sheldon RE: superfunctions of eta converge towards each other - sheldonison - 05/30/2011 (05/24/2011, 02:18 PM)sheldonison Wrote: ..... This time, I calculated the Kneser Riemann mapping from cheta(z), which is the upper super-exponential of eta, calculating \( \text{sexp}_\eta(z) \), and theta(z) using my iterated Kneser/Riemann mapping algorithm \( \text{sexp}_\eta(z) = \text{cheta(z+\theta(z)) \) Here are the first 80 terms of the resulting \( \text{sexp}_\eta \)taylor series, centered at 0, accurate to nearly 50 decimal digits, which is comparable to the accuracy of the cheta series from which it was generated. This required approximately 20 iterations of the iterative Kneser mapping algorithm. Code: a0= 1RE: superfunctions of eta converge towards each other - tommy1729 - 06/06/2011 perhaps not so relevant but the following idea inspires me : sexp(f(z)) = cheta(f(z)) f(z) satisfies f(z) = f(z) + theta(f(z)) + k hence theta(f(z)) = -k but theta is not a constant function , thus there is no f(z) apart from id(z). RE: superfunctions of eta converge towards each other - sheldonison - 12/05/2012 (05/23/2011, 09:01 PM)sheldonison Wrote: ... Here, sexp(z) is the lower superfunction, with sexp(0)=1, and cheta(z) is the upper superfunction ....I made lots of minor updates and clarifications all over this reply. Apparently, when I posted this last year nobody noticed that the imaginary part of k=1.0471975511965977 is exactly Pi/3, which of course begs for an explanation! I didn't notice it either, until I started to work with the formal Abel series solution for iterates of \( \exp(x)-1 \), which is parabolic with a fixed point of zero. We start by noticing that solutions of \( g(z)=\text{cheta}(z)=\exp^{[z]}_\eta(2e) \) are conjugate to solutions for \( h(z)=(\exp(1)-1)^{[z]} \) so that \( h(z) = \frac{g(z)}{e}-1 \). So the two problems are trivially interchangeable. On mathstack, Will Jagy explained how to generate the formal abel function solution for the parabolic case. There are also papers by Baker on the abel function of exp(z)-1. Here is the formal solution for the abel function of exp(z)-1. I posted more terms below. \( \alpha(\exp(z)-1)=\alpha(z)+1 \) \( \alpha(z) = \frac{-2}{z} + \frac{\log(z)}{3} + \frac{-z}{36} + \frac{z^2}{540} + \frac{z^3}{7776} + \frac {-71z^4}{435456} + \frac{8759z^5}{163296000} + O(z^6) \) \( \alpha(z)=\text{cheta^{-1}(e(z+1))+k \) for \( k\approx -2.025912 \) The formal abel solution is only valid for real(x)>=0. It is also a divergent series, meaning that you need to truncate to some optimal number of terms for any particular value of z, but it is nonetheless very accurate. The 40 term series posted below was generated in pari-gp and is accurate to 32 decimal digits for |z|<=0.15. For larger values of z, iterate log(z+1) until the value is smaller than 0.15, and then evaluate the formal series. There is an analogous abel function for sexpeta. \( \alpha_2(z)=\text{sexpeta^{-1}(e(z+1))+k \) for \( k\approx 3.029297 \) \( \alpha_2(z) = \frac{-2}{z} + \frac{\log(-z)}{3} + \frac{-z}{36} + \frac{z^2}{540} + \frac{z^3}{7776} + \frac {-71z^4}{435456} + \frac{8759z^5}{163296000} + O(z^6) \) This nearly identical \( \alpha_2(z) \) abel function is for the "sexpeta" superfunction of exp(z)-1. \( \alpha_2^{-1}(z) \) approaches zero from the negative real numbers, as z goes to infinity. \( \alpha_2(z) \) is only valid if real(z)<=0 . In this series, the log(z) term was replaced with log(-z), so that the abel function is real valued at the real axis for negative real numbers. For example, if we ignore the fact that \( \alpha(z) \) is not valid for \( \Re(z)<0 \), then \( \alpha(-0.01)\approx198.465+\frac{\pi i}{3} \), whereas \( \alpha_2(-0.01)\approx198.465 \). That \( \pi i/3=\log(-1)/3 \) difference between the two functions is exactly the imaginary part of the \( \theta(z) \) constant term for the two superfunctions of \( \exp_\eta(z) \), that I numerically calculated last year. If z=0.15i, than for the 40 term series posted below, both abel functions are valid, and should be accurate to >30 decimal digits. The two approximations differ by exactly \( \frac{\pi i}{3} \). But, as |z| grows, the formal solution is no longer very accurate, and one must iterate exp(z)-1 for \( \alpha_2(z) \), and iterate log(z+1) for \( \alpha_2(z) \), until each |z| is a smaller number before evaluating the formal solution. This iteration leads to the two inverse abel functions (superfunctions) behaving very differently as z approaches the real axis. But as imaginary of z increases, the inverse of the two functions converge towards each other. \( \lim_{z \to \Im \infty}\hspace{2 mm}\alpha(\alpha_2^{-1}(z))-z = \pi i/3 \), which leads to the \( \theta(z) \) function I calculated. My definition for theta is \( \theta(z)=\alpha(\alpha_2^{-1}(z))-z \). The formal abel series solution may allow one to prove the exponential convergence as \( \Im(z) \) increases, which is conjectured to be: \( \alpha(\alpha^{-1}_2(z))-z=\pi i/3 + \sum_{n=1}^{\infty}a_n \exp(2n\pi zi) \). I've wanted to understand Ecalle cylinders for awhile .... ![]() If anyone wants the pari-gp code for parabolic abel solutions for the general case, for \( f(x)=x+ax^2+... \), I could also post that. Also, I assume there is no equivalent formal solution for the superfunction, \( \alpha^{-1}(z) \), for the parabolic case. The best reasonable approximation I could generate for the reciprocal of the superfunction of exp(z)-1 is: fixed typo, updated approximation with emperical error bounds \( \frac{1}{\alpha^{-1}(z)} \approx \frac{-z}{2}+\frac{-1}{6}\log(\frac{-z}{2})+O(z^{-1}) \) \( \frac{1}{\alpha_2^{-1}(z)} \approx \frac{-z}{2}+\frac{-1}{6}\log(\frac{z}{2})+O(z^{-1}) \) The \( O(z^{-1}) \) term seems to be \( (\frac{\log(\frac{-z}{2})}{18}-\frac{1}{36})\times z^{-1} \) - Sheldon Code: First 30 terms, formal abel series term for exp(z)-1. log(z)/3 term also required |