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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. « Next Oldest | Next Newest »

 Messages In This Thread Fractal behavior of tetration - by Kouznetsov - 01/28/2009, 03:38 AM RE: Fractal behavior of tetration - by bo198214 - 02/01/2009, 11:46 AM RE: Fractal behavior of tetration - by Kouznetsov - 02/01/2009, 06:44 PM RE: Fractal behavior of tetration - by andydude - 02/28/2009, 10:16 AM RE: Fractal behavior of tetration - by bo198214 - 02/28/2009, 10:55 AM

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