# Tetration Forum

Full Version: Fractal behavior of tetration
You're currently viewing a stripped down version of our content. View the full version with proper formatting.
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 $7\le\Re(z)\le 9$, $0\le\Im(z)\le 2$
[attachment=435]
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<double> variable; these regions are left blanc.

In such a way, tetration gives also a new kind of fractal.
As a reminder there is also a thread about the tetration fractal, which somehow looks similar to your fractal structures.
bo198214 Wrote:tetration fractal
Dear 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?
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
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.