Is there a function $f:\mathbb N\to \mathbb R$ such that $\lim_{n\to\infty} f(n) = 0$ and for every $c\in\mathbb C$:

If $z_0=0$, $z_{n+1}=z_n^2+c$ and $|z_k|<2$, then there exists a point $c'$ in the Mandelbrot set satisfying $|c-c'|<f(k)$?

Such fact would be very useful for rendering the Mandelbrot set, as it would give the exact count of iterations needed for given resolution.

  • $\begingroup$ What is the motivation for the condition $|z_k| < 2$? Also, do you mean that for any $k$ with $|z_k| < 2$, $|c - c'| < f(k)$, or do you mean that if $|z_k| < 2$ for all $k$< $|c - c'| < f(k)$ for all $k$? $\endgroup$ – 6005 Jun 12 '13 at 14:48
  • $\begingroup$ @Goos: I think neither? What the OP wants is probably "If the Mandelbrot iteration starting from $c$ stays within the critical circle $|z| < 2$ for $k$ steps, then there exists $c'$ in the Mandelbrot set such that $|c - c'| < f(k)$." $\endgroup$ – Willie Wong Jun 12 '13 at 14:59
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    $\begingroup$ On the contrary, you have Koebe $\frac14$ disks completely disjoint to $M$ for every point that does escape. $\endgroup$ – Hagen von Eitzen Jun 12 '13 at 15:00
  • $\begingroup$ @Goos: It's well known that if $|z_k|>2$, then $(z_n)$ diverges, that's why. I mean that for any k with $|z_k|<2$ you can find $c'$ (depending on that k) such that $|c-c'|<f(k)$. I don't get the second part of this question (from "or do you mean"), why is "for all $k$" twice and what's with "$k<|c-c'|$"? $\endgroup$ – janek37 Jun 12 '13 at 15:00
  • $\begingroup$ @HagenvonEitzen: can you elaborate? Is the radius of the Koebe disk independent of $k$? $\endgroup$ – janek37 Jun 12 '13 at 15:41

In principle, yes. The set of points whose orbits stay within $|z|< 4$ for the first $n$ iterations is a bounded open subset $M_n\subset\mathbb C$. We have $M_{n+1}\Subset M_n$ and $M=\bigcap _{n\in\mathbb N}M_n$. Let $f(n)=\max\{d(z,M)\mid z\in \overline{M_n}\}$. Then $z\in M_n$ implies there is a point $c\in M$ with $d(c,z)\le f(n)$. The sequence $(f(n))$ is nonincreasing, hence $a:=\lim_{n\to\infty} f(n)$ exists. Then for all $n$, we get a point $z_n\in M_n$ with $d(z_n,M)=a$. A subsequence of $(z_n)$ converges to some $\zeta$ with $d(\zeta,M)=a$, but we also have $\zeta\in\bigcap \overline{M_n}=M$. However, this mere existence of $f$ won't help you in practice.

  • $\begingroup$ You're right. I hoped that there are some upper bounds for the $f$ you defined. Hopefully in the form $f(n)<Cq^n$ for some $C>0$, $0<q<1$. $\endgroup$ – janek37 Jun 12 '13 at 17:15
  • $\begingroup$ I've found en.wikipedia.org/wiki/… which helps a little, but still doesn't really hint on the number of iterations. $\endgroup$ – janek37 Jun 12 '13 at 17:29

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