# Zoom out fractals? (A question about selfsimilarity)

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 ".

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 ?

• How do you mean this $g=f(f(\dots))$? – Berci Sep 30 '12 at 14:49
• @Berci : Well let $f$ be $z^2 + 1$ , then you get the Mandelbrot. I think that example is clear. – mick Sep 30 '12 at 16:53
• The Newton fractal probably fits the bill: en.wikipedia.org/wiki/Newton_fractal – lhf Dec 30 '13 at 0:54
• en.wikipedia.org/wiki/… – Dan Rust Dec 30 '13 at 1:36
• No guys , Newton fractal is not an iteration of an entire function $f$ like mandelbrot's $f = z^2 + 1$. Also some tiling is not an entire function $f$. – mick Jan 1 '14 at 21:25

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 $\mathrm{Re}(z)^2 > 1$?