[Math] Quadratic Julia sets and periodic cycles

complex-dynamicsdynamical systemsfractals

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?

Best Answer

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."

Related Solutions

[Math] Fun with Newton’s Method – Infinitely many cycles

A general strategy to find functions with desired properties is to study a parametrized family of functions. This is exactly the path to the bifurcation diagram in real dynamics and the Mandelbrot set in complex dynamics.

A super-attractive orbit of period 3

After a little experimentation, I decided to look at the family of cubic polynomials $f_{\lambda}(x)=\lambda+x-x^3$, which has the associated Newton's method iteration function $$n_{\lambda}(x)=\frac{2 x^3 + \lambda}{3 x^2-1}.$$ Note that $$n_{\lambda}'(x)=\frac{6 x \left(x^3-x-\lambda \right)}{\left(1-3 x^2\right)^2}$$ so that $x=0$ is a critical point. Thus, to find a value of $\lambda$ so that $n_{\lambda}$ has a super-attractive orbit of period 3, it suffices to find a value of $\lambda$ such that $n_{\lambda}(n_{\lambda}(n_{\lambda}(0))) = 0$. Grabbing just the numerator of the left hand side, this equation is equivalent to $$-16 \lambda ^9+51 \lambda ^7-39 \lambda ^5+11 \lambda^3-\lambda = 0.$$ Using your favorite numerical solver, it turns out that $\lambda\approx -1.49175$ does the trick.

A repulsive orbit of period 3

If you stick to quadratic polynomials, you're not going to find any attractive orbits of period 3, but you can find repelling orbits of period 3. Indeed, consider the polynomial $p(x)=x^2+1$. This has no real roots so perhaps it's not surprising that the corresponding Newton's method iteration function $$n(x)=\frac{x}{2}-\frac{1}{2 x}$$ displays chaotic behavior on the real line. Thus, we might expect a point of period 3. To find it, simply solve the equation $n(n(n(x))) = x$. This is equivalent to $$\frac{x^8-28 x^6+70 x^4-28 x^2+1}{8 x \left(x^6-7 x^4+7x^2-1\right)} = x.$$ Or using your favorite numerical solver again, you can find the roots. There are six real roots forming two orbits of period 3 namely $$2.07652 \rightarrow 0.797473 \rightarrow -0.228243 \rightarrow 2.07652$$ and $$-2.07652 \rightarrow -0.797473 \rightarrow 0.228243 \rightarrow -2.07652.$$

Finding Periodic/Fixed Points in the Julia Sets close to the Period-3 Cardioid

For the promenade around the main cardioid, the point in the Julia set you want is a fixed point $z = f_c(z)$. By tuning (renormalization), the point you want in the "little Julia set" at the center of the Julia set is a periodic point $z = f_c^p(z)$, in this particular case $p = 3$. You can use Newton's method to find the root numerically, the degree is too high for symbolic solution via radicals. You need a good initial guess for Newton's method to converge to the "correct" tuned fixed point (periodic point), as there is another one at the "tip" of the little Julia set (just as there two fixed points for $z = f_c(z)$), and moreover the fixed points $z = f_c(z)$ are also (unwanted) solutions for $z = f_c^p(z)$.

Here is a small C99 program using my mandelbrot-numerics library:

/*
gcc julia-spiral-center.c `PKG_CONFIG_PATH=${HOME}/opt/lib/pkgconfig pkg-config --cflags --libs mandelbrot-numerics` -lm -fopenmp -O3 -std=c99 -Wall -Wextra -pedantic
LD_LIBRARY_PATH=${HOME}/opt/lib/ ./a.out |
ffmpeg -framerate 25 -i - -pix_fmt yuv420p -profile:v high -level:v 4.1 -crf:v 20 -movflags +faststart julia-spiral-center.mp4 -y
*/

#include <complex.h>
#include <math.h>
#include <stdio.h>
#include <stdlib.h>

#ifndef M_PI
#define M_PI 3.141592653589793
#endif

// git clone https://code.mathr.co.uk/mandelbrot-numerics.git
// make -C mandelbrot-numerics/c/lib prefix=${HOME}/opt install
#include <mandelbrot-numerics.h>

double cnorm(double _Complex z)
{
  double x = creal(z);
  double y = cimag(z);
  return x * x + y * y;
}

int main(int argc, char **argv)
{
  // silence -Wunused-parameter
  (void) argc;
  (void) argv;
  // number of animation frames
  const int frames = 25 * 30 * 1;
  // parameters for Julia set graphics rendering
  const int iterations = 1024;
  const double escape_radius_squared = 1 << 24;
  const int width = 640;
  const int height = 360;
  unsigned char *pgm = malloc(width * height);
  // how far from the target component to place the moving c
  // 1.0 is the boundary of the component
  const double radius = 1.25;
  // period of the target component
  const int period = 3;
  // these don't need to be exact, they will be changed by m_d_interior()
  double _Complex z = 0;
  double _Complex c = -1.754;
  // maximum number of steps of Newton's method
  const int steps = 64;
  // needs a good initial guess to converge to the correct attractor
  double _Complex w = -0.1 - 0.01 * I;
  for (int frame = 0; frame < frames; ++frame)
  {
    // path around the exterior of the component
    double t = (frame + 0.5) / frames;
    double _Complex q = radius * cexp(2 * M_PI * I * t);
    // find c,z such that f_c^p(z) = z and d/dz f_c^p(z) = q
    m_d_interior(&z, &c, z, c, q, period, steps);
    // find w such that f_c^p(w) = w
    // use previous frame's w as initial guess for continuity
    m_d_attractor(&w, w, c, period, steps);
    // scaling for view
    double r = cabs(w);
    // render Julia set
    #pragma omp parallel for
    for (int j = 0; j < height; ++j)
    {
      for (int i = 0; i < width; ++i)
      {
        // view centered on w
        double _Complex z0 =
          ( ((i + 0.5) / width - 0.5) * 2 * width / height
            - I * ((j + 0.5) / height - 0.5) * 2 ) * r + w;
        // initialize derivative with pixel spacing
        double _Complex dz = 2 * r / height;
        // unescaped pixels will be grey
        pgm[j * width + i] = 128;
        for (int k = 0; k < iterations; ++k)
        {
          dz = 2 * z0 * dz;
          z0 = z0 * z0 + c;
          if (cnorm(z0) > escape_radius_squared)
          {
            // compute exterior distance estimate
            double de = 2 * cabs(z0) * log(cabs(z0)) / cabs(dz);
            // colour boundary based on distance estimate
            pgm[j * width + i] = 255 * tanh(de);
            break;
          }
        }
      }
    }
    // output PGM stream to stdout
    fprintf(stdout, "P5\n%d %d\n255\n", width, height);
    fwrite(pgm, width * height, 1, stdout);
    fflush(stdout);
  }
  // cleanup
  free(pgm);
  return 0;
}

The source code for the two functions I used from the library can be browsed at:

They are both applications of Newton's root finding method, in 2 complex variables and 1 complex variables respectively.

You can view the video rendered by this program.

Best Answer

Related Solutions

[Math] Fun with Newton’s Method – Infinitely many cycles

Finding Periodic/Fixed Points in the Julia Sets close to the Period-3 Cardioid

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