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Let $f : \mathbb{C} \to \mathbb{C}$ be a polynomial map of degree $q > 1$. Consider $E_n \subset \mathbb{C}$ the set of periodic points with period (dividing) $n$; generally, $|E_n| = q^n$. Since all the peridic points are limited to a bounded subset of $\mathbb{C}$, we know that there are always two among them within a distance of $O(q^{-n/2})$.

Question. (i) How close can two distinct elements of $E_n$ get? In the asymptotic where $n \to \infty$ along a sequence, can it happen that $E_n$ contains a pair of points within a distance of $o(1/q^n)$?

(ii) Given any point $z_0 \in \mathbb{C}$ and $C$, does the disk $|z-z_0| < C/q^n$ contain only $O_{C,z_0}(1)$ points from $E_n$?

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Consider $f(x) = x^2 - 2$. This has the property that $f(2 \cos (x)) = 2 \cos(2x)$, so that $f_n(2 \cos (x)) = 2 \cos(2^n x)$ (where $f_n$ is $f$ iterated $n$ times). Thus $E_n = \{2 \cos (x): \cos(x) = \cos(2^n x)\}$, and the two greatest members of $E_n$ are both greater than $2 \cos(\pi 2^{1-n})\approx 2 - \pi^2 2^{2-2n}$

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    $\begingroup$ Thank you! This example clarifies a lot. I think it would be interesting to know also if $O(q^{-2n})$ is as small as the distance can get. $\endgroup$ – Vesselin Dimitrov Oct 26 '14 at 4:41
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The answer to both of your questions is negative, even if you replace $q^{-n}$ by any other sequence $(a_n)$ of natural numbers.

The reason is, broadly speaking, that you may have of course have maps having multiple periodic points, and under small perturbations these will yield maps having a cycle of potentially large period nearby.

A little bit more precisely, let us consider the family of quadratic polynomials, $f_c(z) = z^2+c$. The following is well-known:

Proposition. Suppose that $f_c$ has a periodic point $z_0$ of period $n$, with corresponding multiplier $\mu = (f_c^n)'(z_0) = e^{2\pi i p/q}$, for some $p,q\in\mathbb{N}$.

Then, for any neighbourhood $U$ of $z_0$, there is a parameter $c'$ (arbitrarily close to $c$) such that $f_{c'}$ has a periodic point $z'$ of period $nq$ such that $f_{c'}^{kn}(z')\in U$ for $k\geq 0$, and such that $z'$ itself has multiplier $\mu' = (f_{c'}^{nq})'(z') = e^{2\pi i /q'}$ for some (large) integer $q'$.

(In addition, the map $f_{c'}$ also has a (repelling) periodic point of period $n$ in $U$.)

(What happens is that the parameter $c$ is on the boundary of two components of the interior of the Mandelbrot set, one consisting of points having an attracting orbit of period $n$ and one having an attracting orbit of period $nq$. The perturbation is obtained by passing a little bit along the boundary of the latter component.)

Now start e.g. with the map $f_{1/4}$ and $z=1/2$, and apply this proposition inductively. Take a limit of the resulting parameters (taking care to make each perturbation small enough as to not destroy the features already constructed). Clearly in this way we can find a quadratic polynomial having periodic points $(z_j)_{j\geq 0}$ of periods $(n_j)$, where $$ n_j = \prod_{k=1}^j q_j$$. (Here $q_j$ is some rapidly increasing sequence of positive integers obtained in the construction.)

Given any sequence $(\varepsilon_n)$, we can inductively carry out the construction so that $q_j$ points of the orbit of $z_j$ are within distance $\varepsilon_{n_j}$ of each other. Letting $\varepsilon_n$ tend to zero faster than $a_n$, the claim follows.

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  • $\begingroup$ Thank you for explaining this! On the other hand, if $f$ is defined over $\bar{\mathbb{Q}}$, we have a diophantine bound and we may not take $a_n$ to decay too fast. Thus the limit value of the parameter $c$ must be transcendental. $\endgroup$ – Vesselin Dimitrov Oct 26 '14 at 22:00
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    $\begingroup$ @VesselinDimitrov that's an interesting point. The construction is a generalization of the Feigenbaum point (where we make a sequence of period 2 bifurcations, and take the limit). In this case, I believe that it is not known whether the limit parameter is transcendental. (More generally, one could ask whether every infinitely renormalizable parameter is transcendental.) $\endgroup$ – Lasse Rempe-Gillen Oct 26 '14 at 22:38
  • $\begingroup$ Take, say, $a_n := \exp(-5^{n})$. If $c \in \bar{\mathbb{Q}}$, then the points of period $n$ are algebraic numbers of degree $O(2^n)$ and bounded (absolute, multiplicative Weil) height. Then their pairwise differences are algebraic numbers of degrees $O(4^n)$ and still of bounded height. But a non-zero algebraic number of degree $d$ and absolute multiplicative height $H$ is at least $H^{-d}$ in absolute value: this is the Liouville diophantine bound. $\endgroup$ – Vesselin Dimitrov Oct 26 '14 at 23:04
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    $\begingroup$ @VesselinDimitrov Yes, that's what I took your first comment to mean. Interestingly, using recent work on 'near-parabolic renormalization', it could be possible to quantify the above construction, and hence - combined with your observation - give estimates on the combinatorics (i.e., the numbers $q_j$) that ensure that the resulting parameter is transcendental. That could be a publishable result, although depending on the technical work needed it may or may not be worthwhile to work it out. Davoud Cheraghi (Imperial College) is a world expert on this type of argument. $\endgroup$ – Lasse Rempe-Gillen Oct 27 '14 at 9:22
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While Robert Israel's answer is correct, $x^2-2$ is a very exceptional case.

In a way, how close do they get is not the interesting question: rather, what gets strange is how far apart do the closest ones remain? We know the periodic points are dense in the Julia set, but in the case of weird ones (like the ones with Cremer points, or even some with Siegel disks where the disk itself is very 'deep' within the Julia set, as measured by the external rays), the periodic points tend to avoid certain parts of the Julia set as long as possible. This is what causes the 'inverse method' of rendering images of Julia sets to be so bad for those cases.

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