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It is well known that if we zoom in on the Mandelbrot set we get selfsimilarity.

So I wonder if $g$ is a fractal (in the complex plane) generated by a nonperiodic nonpolynomial entire function $f$

$g:= f(f(…))$

Is it possible that the fractal is infinite in size and that when we zoom out , we get selfsimilarity too ? Lets call such a fractal a ” zoom out fractal “.

- Mandelbrot set: periodicity of secondary and subsequent bulbs as multiples of their parent bulbs
- Supremum of all y-coordinates of the Mandelbrot set
- Finding the all roots of a polynomial by using Newton-Raphson method.
- About fractional iterations and improper integrals
- Why does the mandelbrot set and its different variants follow similar patterns to epi/hypo trochodis and circular multiplication tables?
- Help locating mini mandelbrots

As example Mandelbrot : $f(z)=z^2+1$, g$(z) [= f(f(…))]$ is the fractal.

The fractal has finite size (area or length ) since it diverges for $Re(z)>2$.

$f(z)$ is a nonperiodic nonpolynomial entire function. But when we zoom out we get no selfsimilarity. ( divergence is not considered valid as selfsimilar ) So Mandelbrot is NOT a zoom out fractal.

Does the existance of zoom out fractals require that the fractal is also a zoom in fractal ?

What is the formal way or term to express ‘ zoom out selfsimilarity ‘ or ‘ zoom out fractal ‘ , if any ?

- Do Integrals over Fractals Exist?
- variant on Sierpinski carpet: rescue the tablecloth!
- Calculate moment of inertia of Koch snowflake
- Mandelbrot boundary
- Do there exist periodic fractals $A_f$ of this type?
- What exactly are fractals
- Are the vertices of a Voronoi diagram obtained from a Sierpinski attractor also a kind of attractor?
- How to draw a Mandelbrot Set with the connecting filaments visible?
- What is this pattern found in the first occurrence of each $k \in \{0,1,2,3,4,5,6,7,8,9\}$ in the values of $f(n)=\sqrt{n}-\lfloor \sqrt{n} \rfloor$?
- Fractal derivative of complex order and beyond

A serious candidate for $f(z)$ is the following function : $\displaystyle f(z)=\lim_{\epsilon\to0}\int_\epsilon^{\infty} \dfrac{e^{zt}}{t^t} \, dt $.

The reason is that the entire function $f(z)$ is bounded for $\mathrm{Im}(z)^2 > \pi$

This was given here : Entire function $f(z)$ bounded for $\mathrm{Re}(z)^2 > 1$?

See also : How fast does the function $\displaystyle f(x)=\lim_{\epsilon\to0}\int_\epsilon^{\infty} \dfrac{e^{xt}}{t^t} \, dt $ grow?

It would be nice to see a plot.

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