Fractal behavior of tetration

[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<double> variable; these regions are left blanc.

In such a way, tetration gives also a new kind of fractal.

[bo198214]

As a reminder there is also a thread about the tetration fractal, which somehow looks similar to your fractal structures.

[Kouznetsov]

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?

[andydude]

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]

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

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.