Consider the function $f_c(z) = z^2 + c$. Applying this function repeatedly, we get the familiar quadratic Julia sets that fractal enthusiasts burn compute cycles plotting.

Infinity is always one attractor of the system. Depending on the choice of $c$, a finite attractor may also exist. Sometimes this is a fixed-point. Sometimes it is a repeating cycle of some finite number of points.

Consider the case of a fixed-point. The actual numerical value of this fixed point depends on $c$. So I set out to investigate a way to compute this number directly.

A fixed point of $f_c$ is simply any $z$ for which $f_c(z) = z$. In other words, we wish to solve $z^2 + c = z$. Rearranging as $z^2 - z + c = 0$, I was easily able to find

$$z_1 = \frac{1 \pm \sqrt{1 - 4c}}2$$

At this point, something struck me: First, there are obviously two such fixed-points, only one of which is the finite attractor. But, more conspicuously, these two fixed-points always exist. Even when there is no fixed-point attractor, there definitely are two fixed points.

What about a period-2 cycle? That is, we want to solve $f_c(f_c(z)) = z$. Solving $(z^2 + c)^2 + c = z$ is a little more tricky than the last equation - but the formula for $z_1$ gives us two of the solutions, and it's then fortunately easy to discover the other two:

$$z_2 = \frac{1 \pm \sqrt{-3-4c}}2$$

Again, this cycle always exists.

At this point, I tried to find a period-3 cycle. Clearly $((z^2 + c)^2 + c)^2 + c = z$ has 8 solutions, two of which are $z_1$, which leaves 6 remaining. At this point, I was unable to work out how to solve the equation. The mighty Mathematica™ also refused to give me a closed form. (I suppose it's plausible that none exists.)

It seems clear though that these solutions exist, even if I can't easily compute them. And if there's 6 of them, that's presumably a pair of period-3 cycles. More generally, it seems there is no reason why cycles of any finite length wouldn't exist all the time. So, my actual question is this: Where do all these periodic cycles "live" when they aren't the attractor of the system?


There is an awful lot of interesting stuff in this question, but I'll do my best to keep this answer organized.

(1) Since you are working over the complex numbers, which are algebraically closed, the polynomial equation $f^{\circ n}_c(z) = z$ always has $2^n$ solutions when counted with multiplicity, even if you can't explicitly solve for them.

(2) It is unfortunately not the case that $f_c$ has periodic cycles of every possible (exact) period. For example, you can check that $f(z) = z^2 - \frac{3}{4}$ has no period $2$ points which are not fixed points. However, this behavior has been studied, and we know exactly when a rational map can be missing points of a given period:

Suppose that $f\in \mathbb{C}(z)$ is a rational map of degree $d\geq 2$, and suppose that $f$ has no cycle of exact period $n$. Then the tuple $(n,d)$ is either $(2,2)$, $(2,3)$, $(3,2)$, or $(4,2)$. Moreover, if $f$ is a polynomial, only $(2,2)$ can occur.

This is a thorem of I.N. Baker proved in the paper "Fixpoints of polynomials and rational fucntions." (1964) In particular, for the maps $f_c$ you are considering, you definitely have points of exact period $n$ for all $n\geq 3$. For an excellent discussion of such topics, I recommend section 4.1 of Joe Silverman's book "The Arithmetic of Dynamical Systems."

(3) If I understand correctly, you are assuming that the $c$ you've chosen is such that $f_c$ has a finite attracting cycle. The Fatou-Shishikura theorem says that there are at most $2$ non-repelling cycles. You've identified these as $\infty$ and the finite attracting cycle. It follows that all other cycles are repelling, and hence live in the Julia set. I hope that answers your actual question. For a simple proof of Fatou-Shishikura in the case of polynomial maps, see Theorem VI.1.2 of Carleson and Gamelin's book "Complex Dynamics."

  • $\begingroup$ It seems the case of $z^2 - \frac34$ only "lacks" a period-2 cycle because both points of the cycle just happen to coincide in this case. I find it interesting that this choice of $c$ is also the exact place where the Julia set goes from connected to disconnected - that doesn't sound like a coincidence! ;-) $\endgroup$ May 8 '12 at 9:36
  • $\begingroup$ Indeed that is not a coincidence. $\endgroup$
    – mick
    Feb 11 '13 at 22:29

For a very simple case, pick $c=0$.

If $|z| > 1$, then the sequence of iterates generated by $z$ diverges to infinity.
If $|z| < 1$, then the sequence generated by $z$ converges to $0$.
The interesting part is what happens along the circle $|z|=1$.

In this circle, you get periodic points of any periods as well as points whose sequence is not periodic, ans sometimes dense is that circle. To see this, remark that $\arg(f(z)) = 2z$, so you may understand $f$ better if you look at what it does to $arg(z)/2\pi$ : it is the multiplication by $2$ map from $\mathbb{R}/\mathbb{Z}$ to itself.

If you write numbers of $\mathbb{R}/\mathbb{Z}$ in their binary expansion, points whose orbit is $k$-periodic are exactly the points with a $k$-periodic binary expansion. For example, the two $3$-periodic orbits correspond to the binary expansions : $(.001001 \ldots \to .010010 \ldots \to .100100 \ldots \to .001001 \ldots)$ and $(.011011 \ldots \to .110110 \ldots \to .101101 \ldots \to .011011 \ldots)$. Indeed, if a binary expansion is $3$-periodic without being $1$-periodic, then it has to be one of those $6$ numbers.

You can do the same for orbits of any length : you can easily find all the points on the circle who generates a periodic sequence for any period you want.
Additionnally to those points, you have all the points who correspond to ultimately periodic binary sequences, they are the points on the circle that will at some point land on one of those $k$-cycles.
All of those points corresponds to rational numbers of $\mathbb{R}/\mathbb{Z}$.

Then there are uncountably many points of the circle, so there are uncountably many points we've missed so far. Some of them will have an orbit that's dense in the circle. Some of them will have strange behaviour, for example if you start from $.01001000100001\ldots$, this one will get arbitrarily close to $0$ only to get farther and farther from it, get close to $.1$ and then jump back even closer to $0$.

So in this simple case, $\mathbb{C}$ is split into two open sets, the basin of attraction of $\infty$, the one for $0$, and between them is a closed subset of $\mathbb{C}$ where all the other cycles are hidden and chaotic behaviour happens.

In general, you will have the same kind of picture : $\mathbb{C}$ is split into one or more (actually I don't know if there is always an attractive cycle in $\mathbb{C}$) open subsets that are basins of attraction, while the frontier between them is where all the other cycles are hidden (but still there).

  • $\begingroup$ So every cycle that isn't an attractor is always a repellor? (And, thus, hides in the Julia set of the system.) $\endgroup$ May 8 '12 at 9:38
  • 1
    $\begingroup$ This is true for the maps $f_c$ you are considering if you know there is a finite attracting cycle. There are value of $c$ for which there are no attracting cycles, but there are indifferent cycles. When $c$ is the golden mean, for instance, you get a Siegel disc. Since $f_c$ has degree $2$, it can only have one finite non-repelling cycle, so almost every cycle is repelling. $\endgroup$
    – froggie
    May 8 '12 at 10:30

A computer can find solutions to this, but the solutions can't be put into a form which can map to any other form - the most you can do is find a periodic point of period 3 in the 2 bulbs at the top and bottom of the Mandelbrot set, and the one on the negative real axis by expanding the LHS, collecting on the LHS, and setting $z=0$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.