Consider the well-known iteration $f:z\to z^2 + c,$ and consider the values of $c$ for which $0$ is a periodic point. Experiment shows that most such values of $c$ (about $480$ out of $512$ for period $10$) lie on the real axis, and those that don't lie near the cusp part of the Mandelbrot set cardioid part, along the boundary (the cardioid itself). I assume all this is well-known, but not to me, so any enlightenment/references welcome.
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asked Apr 11, 2018 at 0:33
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3$\begingroup$ According to Wikipedia each bulb of the Mandelbrot set contains one such point, and they are precisely the roots of all the polynomials $Q_n(c)$, with $Q_{n+1}(c)=Q_n(c)^2+c$. The number of new such points produced at the $n$th step is [A000740](oeis.org/A000740). I believe this means that the closure of the set of periodic points is the whole regular boundary, so they can be found arbitrarily near any point of the boundary of the closure of the interior of $M$. $\endgroup$– მამუკა ჯიბლაძეApr 11, 2018 at 4:57
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$\begingroup$ All I know is what I see, and what I see is described in the question. Maybe it's a mathematica bug, but I am seeing an awful lot of real roots. $\endgroup$– Igor RivinApr 11, 2018 at 5:05
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$\begingroup$ It should be something like this; most of your reals must be between $-1.5$ and $-2$. $\endgroup$– მამუკა ჯიბლაძეApr 11, 2018 at 5:06
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$\begingroup$ In fact there are also unstable (repelling) periodic orbits, but I believe they should not be detectable... $\endgroup$– მამუკა ჯიბლაძეApr 11, 2018 at 5:17
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1$\begingroup$ @მამუკაჯიბლაძე $0$ cannot be an unstable periodic point, because the period $x_1$, $x_2$, ..., $x_N$ would contain $0$, making $\prod_k f'(x_k)$ equal to $0$, and thus making the orbit superattracting. It can be an unstable preperiodic point. $\endgroup$– David E SpeyerApr 11, 2018 at 15:24
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3 Answers
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The principal source is
MR0762431 Douady, A.; Hubbard, J. H. Étude dynamique des polynômes complexes. Partie I. Publications Mathématiques d'Orsay, 84-2. Université de Paris-Sud, Département de Mathématiques, Orsay, 1984. 75 pp.
MR0812271 Douady, A.; Hubbard, J. H. Étude dynamique des polynômes complexes. Partie II. Publications Mathématiques d'Orsay, 85-4. U Orsay, 1985. v+154 pp.
Available on Internet.
These values are called "centers of the hyperbolic components", and there is a combinatorial algorithm which determines for each $n$ how many of them are real. They did not study statistics but it is unlikely that a majority of them are real. Anyway, the algorithm ("Hubbard trees") permits in principle to reduce this question to combinatorics and number theory. The algorithm encodes all solutions $c$ by certain trees embedded in the plane, and the question is equivalent to counting trees with certain symmetry property.
EDIT. On my request, Gena Levin wrote the following:
There is a recursive formula for the number $H(n)$ of "real" hyperbolic components of the Mandelbrot corresponding to cycles of exact period $n$ via the number $P(n)$ of cycles of exact period $n$ for the Tchebyshev polynomial $x^2-2$: $H(n)=P(n)/2$, if $n$ is odd, and $H(n)=(P(n)+H(n/2))/2$ if $n$ even.
For example, if $n$ is a prime number then $H(n)=(2^{n-1}-1)/n.$
answered Apr 11, 2018 at 13:59
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The number of real centers of hyperbolic components of exact period $n$ is OEIS A000048. This isn't bad to work out from the combinatorics of kneading sequences which Rivin links to sources for. A number of references linked by OEIS imply this, although I oddly can't find one which explicitly states it and gives a proof. I know this because Sarah Koch, Dylan Thurston and I were computing these numbers earlier this year and I worked out the combinatorics, which I'm willing to write up if you care enough.
As OEIS says, this gives the formula $$\frac{1}{2n} \sum_{d|n,\ d \ \mbox{odd}} \mu(d) 2^{n/d} \approx \frac{2^n}{2n}.$$
By contrast, the number of $c$ of exact period $n$ is $$\sum_{d|n} \mu(d) 2^{n/d} \approx 2^n.$$
So the fraction of real roots is roughly $\tfrac{1}{2n}$.
answered Apr 11, 2018 at 15:33
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2$\begingroup$ I would be curious to see more details if you have the time. $\endgroup$– j.c.Apr 11, 2018 at 15:52
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$\begingroup$ I also obtain sometimes $1/(2n)$, sometimes $2/n$, sometimes $1/n$ :-) But how could Igor Rivin obtain 480 out of 512 by his computation?? $\endgroup$ Apr 13, 2018 at 2:16
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1$\begingroup$ I don't know where $480$ comes from. I get $\tfrac{1}{20} (2^{10} - 2^2)=51$ real roots. Mathematica confirms this. $\endgroup$ Apr 13, 2018 at 3:20
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$\begingroup$ What are cases where you get $2/n$ or $1/n$? $\endgroup$ Apr 13, 2018 at 3:23
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Upon investigation:
- Mathematica is screwing up. Using it's own function
CountRoots[], which counts real roots (exactly), the number of real roots, while quite far from zero (there are $30$ for $n=9,$ for example) is not the majority of the fixed points. - The parameter values corresponding to the periodic points are equidistributed on the boundary of $M$ with respect to harmonic measure. References are:
Ahlfors, Lars V., Conformal invariants. Topics in geometric function theory, McGraw-Hill Series in Higher Mathematics. New York etc.: McGraw-Hill Book Company. VII, 157 p. $ 10.95 (1973). ZBL0272.30012.
and
Brolin, H., Invariant sets under iteration of rational functions, Ark. Mat. 6, 103-144 (1965). ZBL0127.03401.
- Real fixed points correspond to the fixed points of the quadratic iteration on the intervals - a beautiful subject, one of the most beautiful papers on which is
Milnor, John; Thurston, William, On iterated maps of the interval, Dynamical systems, Proc. Spec. Year, College Park/Maryland, Lect. Notes Math. 1342, 465-563 (1988). ZBL0664.58015.
(many thanks to Curt McMullen for the refs).
answered Apr 11, 2018 at 15:16