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I was experimenting a bit with tetration escape visualization and tried to visualize the following equation on the complex plane:

$$z_{n+1} = z_{0}^{\sin(z_{n})}$$

This sequence visualized looks like this: sin

I noticed between $7.8-0.11i$ and $8.03+0.11i$ there was a Mandelbrot set hidden between all the other mess:

mandelbrot

Also between $-4.81-0.15i$ and $4.61+0.15i$ there is a Mandelbrot-like structure: mandelbrot-like

How comes the Mandelbrot set randomly appears in there?

Aura Lee
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    Notice that $,2.5\pi\approx 7.85,$ which means that $,\sin(x+2.5\pi)=\cos(x)\approx 1-x^2/2,$ – Somos Mar 20 '21 at 19:35
  • I am unable by computing in a loop $Z(n+1)=z_0^{\sin Z(n)}$ with $Z(0)=z_0$ to find something ressembling even partially to Mandelbrot set...with the kind of $z_0$ values you give. – Jean Marie Mar 20 '21 at 22:30
  • Could you place an image on some server that would indicated what you mean by "Mandelbrot set out of a mess"... or insert a program in the langage you use ? – Jean Marie Mar 21 '21 at 00:00
  • I now have enough reputation to post images and links. I made a tool to visualize it, you can find it here: https://github.com/GraxCode/towerpower/ Make sure to select "sin" mode. – Aura Lee Mar 21 '21 at 07:29
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    $z_{0}$ is the point on the complex plane. I visualize the value $z$ as a color after enough iterations that the change is below 0.00000001. – Aura Lee Mar 21 '21 at 07:37
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    That second "Mandelbrot-like" structure is surprisingly adorable. Just saying. – heropup Mar 21 '21 at 07:44
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    Looks like a cute crab, right? – Aura Lee Mar 21 '21 at 07:46
  • Thanks for your answer. So you work exactly as for the usual Mandelbrot set but with a formula which is no longer $z_{n+1}=z_n^2+c.$ – Jean Marie Mar 21 '21 at 13:54
  • exactly. $z_{n+1} = z_{0}^{z_{n}}$ would be the tetration escape sequence. – Aura Lee Mar 21 '21 at 18:21

1 Answers1

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Consider iteration of $z \mapsto c^{\sin(z)}$ near $c$:

$$z \mapsto c^{\sin(c + z)} - c$$

The Taylor series near $z = 0$ is (via Wolfram Alpha):

$$ z \mapsto \left(c^{\sin(c)} - c\right) \\ + \left(\log(c) c^{\sin(c)} \cos(c)\right) z \\ + \left(\frac{1}{2} \log(c) c^{\sin(c)} \left(\log(c) \cos(c)^2 - \sin(c)\right)\right) z^2 \\ + O(z^3) $$

Locally ($z$ small), the non-linearity is dominated by the $z^2$ term (as higher order terms will be much smaller). This means miniature quadratic Mandelbrot set copies may appear. Completing the square to the form $z \mapsto a(z + b)^2 + d$ gives:

$$ ab^2 + d = c^{\sin(c)} - c \\ 2ab = \log(c) c^{\sin(c)} \cos(c) \\ a = \frac{1}{2} \log(c) c^{\sin(c)} \left(\log(c) \cos(c)^2 - \sin(c)\right) $$

Affine conjugation via $z \mapsto z - b$ and $z \mapsto \frac{z}{a}$ gives the form $z \mapsto z^2 + C$ where $C = a (d + b)$ which expands to:

$$ C = \frac{1}{2} \log(c) \left(c^{\sin(c)} - c\right) c^{\sin(c)} \left(\log(c) \cos(c)^2 - \sin(c)\right) \\ - \left(\frac{1}{2} \log(c) c^{\sin(c)} \cos(c)\right)^2 + \frac{1}{2} \log(c) c^{\sin(c)} \cos(c) $$

Solving numerically for $C = 0$ gives (via Wolfram Alpha) a number of solutions including: $$c \approx 7.85398163397448\ldots$$

Solving numerically for $C = 0$ near $c = -4.7$ (via Wolfram Alpha) the solution: $$c \approx -4.71238898038469\ldots$$

These both correspond to mini-Mandelbrot islands with period $1$, there should also be mini-Mandelbrot islands corresponding to higher periods, for example I think there should be a period $3$ near $18.155$ (with $C \approx -1.754\ldots$).

Claude
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