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 Fractal behavior of tetration Kouznetsov Fellow Posts: 151 Threads: 9 Joined: Apr 2008 01/28/2009, 03:38 AM (This post was last modified: 02/01/2009, 06:13 PM by Kouznetsov.) Let sexp be holomorphic tetration. Let $F= \{ z\in \mathbb{C}: \exists n\in \mathbb{Z} : \Im(\mathrm{sexp}(z+n))=0 \}$. This $F$ has fractal structure. This structure is dence everywhere, so, if we put a black pixel in vicinity of each element, the resulting picture will be the "Black Square" by Malevich; it is already painted and there is no need to reproduce it again. Therefore, consider the approximation. Let $F_n= \{ z\in \mathbb{C}: \Im(\mathrm{sexp}(z+n))=0 \}$. While $\mathrm{sexp}(z+1)=exp(\mathrm{sexp}(z))$, $F_n\subset F_{n+1}$ id est, all the points of the approximation are also elements of the fractal (although only Malevich could paint all the points of the fractal). As an illustration of $F_8$, centered in point 8+i, I suggest the plot of function $\Im(\mathrm{sexp}(z))$ in the complex $z$ plane, in the range $7\le\Re(z)\le 9$, $0\le\Im(z)\le 2$     Levels $\Im(\mathrm{sexp}(z))=0$ are drawn. Due to more than $10^{100}$ lines in the field of view, not all of them are plotted. Instead, the regions where $|\Im(\mathrm\sexp(z))|<10^{-4}$ are shaded. In some regions, the value of $\Im(\mathrm{sexp}(z))$ is huge and cannot be stored in a complex variable; these regions are left blanc. In such a way, tetration gives also a new kind of fractal. bo198214 Administrator Posts: 1,389 Threads: 90 Joined: Aug 2007 02/01/2009, 11:46 AM (This post was last modified: 02/01/2009, 11:46 AM by bo198214.) As a reminder there is also a thread about the tetration fractal, which somehow looks similar to your fractal structures. Kouznetsov Fellow Posts: 151 Threads: 9 Joined: Apr 2008 02/01/2009, 06:44 PM bo198214 Wrote:tetration fractalDear Bo: 1. Thank you for the link, beautigul pics. However, those pics do not correspond to tetration or superexponential, althouch, they are related to the reiterated exponential. As I understand, in our notations, they correspond to the play with base b. In my case, b=e is fixed. Formally, my set F is periodic, while the structures you refer are not. Perhaps, the pics you mention should be cited. How is it better to cite those pics? 2. I have the expression for the set of branchpoints of the modified slog, while the primary cutlines go to the right hand side direction. The formula is: $S \subset \{L,L^*\}$ $S=S \cup \{ \log(z)+2\pi \mathrm{i} m,~ z\in S,~ m\in \mathbb{N}~: |\Im(\log(z))+2\pi m| > \Im(L)~ \}$ The resulting set S is not dense, as the F above, the set S is countable and its measure is zero. How do you like it? andydude Long Time Fellow Posts: 509 Threads: 44 Joined: Aug 2007 02/28/2009, 10:16 AM Sweet. I like it. This reminds me, there are so many types of fractals: by-period, by-escape, Julia sets, Fatou sets, Mandelbrot sets, and so on. I suppose the Julia set of exponentiation would be the convergence region of ${}^{\infty}z$, right? But what would the Julia set of tetration look like? Andrew Robbins bo198214 Administrator Posts: 1,389 Threads: 90 Joined: Aug 2007 02/28/2009, 10:55 AM (This post was last modified: 02/28/2009, 11:01 AM by bo198214.) andydude Wrote:I suppose the Julia set of exponentiation would be the convergence region of ${}^{\infty}z$, right? By Wikipedia the Julia set of an entire function is Quote:the boundary of the set of points which converge to infinity under iteration For $\exp_b$ this would be the boundary of all points $x$ with $\lim_{n\to\infty} {\exp_b}^{\circ n}(x)=\infty$. I think for $b>e^{1/e}$ this is the whole complex plane, because the points that go to infinity are next to points that doent. For tetration the Julia set are is the boundary of points $x$ such that $\text{sexp}_b^{\circ n}(x)\to\infty$. I guess this depends on which tetration we choose. So maybe Dmitrii can draw a picture . Another option is the Mandelbrot set, which is the set of parameter $b$ for which $f_b^{\circ n}(0)\not\to\infty$, i.e. is bounded. « Next Oldest | Next Newest »

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