Periodic points of complex quadratic mappings

This article describes periodic points of some complex quadratic maps. A map is a formula for computing a value of a variable based on its own previous value or values; a quadratic map is one that involves the previous value raised to the powers one and two; and a complex map is one in which the variable and the parameters are complex numbers. A periodic point of a map is a value of the variable that occurs repeatedly after intervals of a fixed length.

These periodic points play a role in the theories of Fatou and Julia sets.

Definitions
Let


 * $$f_c(z) = z^2+c\,$$

be the complex quadric mapping, where $$z$$ and $$c$$ are complex numbers.

Notationally, $$f^{(k)} _c (z)$$ is the $$k$$-fold composition of $$f_c$$ with itself (not to be confused with the $$k$$th derivative of $$f_c$$)—that is, the value after the k-th iteration of the function $$f _c.$$ Thus


 * $$f^{(k)} _c (z) = f_c(f^{(k-1)} _c (z)).$$

Periodic points of a complex quadratic mapping of period $$p$$ are points $$z$$ of the dynamical plane such that


 * $$f^{(p)} _c (z) = z,$$

where $$p$$ is the smallest positive integer for which the equation holds at that z.

We can introduce a new function:


 * $$F_p(z,f) = f^{(p)} _c (z) - z,$$

so periodic points are zeros of function $$F_p(z,f)$$: points z satisfying


 * $$F_p(z,f) = 0,$$

which is a polynomial of degree $$2^p.$$

Number of periodic points
The degree of the polynomial $$F_p(z,f)$$ describing periodic points is $$d = 2^p$$ so it has exactly $$d = 2^p$$ complex roots (= periodic points), counted with multiplicity.

Stability of periodic points (orbit) - multiplier


The multiplier (or eigenvalue, derivative) $$m(f^p,z_0)=\lambda$$ of a rational map $$f$$ iterated $$p$$ times at cyclic point $$z_0$$ is defined as:


 * $$m(f^p,z_0) = \lambda = \begin{cases}

f^{p \prime}(z_0), &\mbox{if }z_0 \ne \infty \\ \frac{1}{f^{p \prime} (z_0)}, & \mbox{if }z_0 = \infty \end{cases}$$

where $$f^{p\prime} (z_0)$$ is the first derivative of $$f^p$$ with respect to $$z$$ at $$z_0$$.

Because the multiplier is the same at all periodic points on a given orbit, it is called a multiplier of the periodic orbit.

The multiplier is:
 * a complex number;
 * invariant under conjugation of any rational map at its fixed point;
 * used to check stability of periodic (also fixed) points with stability index $$abs(\lambda). \,$$

A periodic point is
 * attracting when $$abs(\lambda) < 1;$$
 * super-attracting when $$abs(\lambda) = 0;$$
 * attracting but not super-attracting when $$0 < abs(\lambda) < 1;$$
 * indifferent when $$abs(\lambda) = 1;$$
 * rationally indifferent or parabolic if $$\lambda$$ is a root of unity;
 * irrationally indifferent if $$abs(\lambda)=1$$ but multiplier is not a root of unity;
 * repelling when $$abs(\lambda) > 1.$$

Periodic points
 * that are attracting are always in the Fatou set;
 * that are repelling are in the Julia set;
 * that are indifferent fixed points may be in one or the other. A parabolic periodic point is in the Julia set.

Finite fixed points
Let us begin by finding all finite points left unchanged by one application of $$f$$. These are the points that satisfy $$f_c(z)=z$$. That is, we wish to solve


 * $$z^2+c=z,\,$$

which can be rewritten as


 * $$\ z^2-z+c=0.$$

Since this is an ordinary quadratic equation in one unknown, we can apply the standard quadratic solution formula:


 * $$\alpha_1 = \frac{1-\sqrt{1-4c}}{2}$$ and $$\alpha_2 = \frac{1+\sqrt{1-4c}}{2}.$$

So for $$c \in \mathbb{C} \setminus \{1/4\}$$ we have two finite fixed points $$\alpha_1$$ and $$\alpha_2$$.

Since
 * $$\alpha_1 = \frac{1}{2}-m$$ and $$\alpha_2 = \frac{1}{2}+m$$ where $$m = \frac{\sqrt{1-4c}}{2},$$

we have $$\alpha_1 + \alpha_2 = 1$$.

Thus fixed points are symmetrical about $$z = 1/2$$.



Complex dynamics


Here different notation is commonly used:


 * $$\alpha_c = \frac{1-\sqrt{1-4c}}{2}$$ with multiplier $$\lambda_{\alpha_c} = 1-\sqrt{1-4c}$$

and


 * $$\beta_c = \frac{1+\sqrt{1-4c}}{2}$$ with multiplier $$\lambda_{\beta_c} = 1+\sqrt{1-4c}.$$

Again we have


 * $$\alpha_c + \beta_c = 1 .$$

Since the derivative with respect to z is


 * $$P_c'(z) = \frac{d}{dz}P_c(z) = 2z ,$$

we have


 * $$P_c'(\alpha_c) + P_c'(\beta_c)= 2 \alpha_c + 2 \beta_c = 2 (\alpha_c + \beta_c) = 2 .$$

This implies that $$P_c$$ can have at most one attractive fixed point.

These points are distinguished by the facts that:
 * $$\beta_c$$ is:
 * the landing point of the external ray for angle=0 for $$c \in M \setminus \left\{ 1/4 \right\}$$
 * the most repelling fixed point of the Julia set
 * the one on the right (whenever fixed point are not symmetrical around the real axis), it is the extreme right point for connected Julia sets (except for cauliflower).
 * $$\alpha_c$$ is:
 * the landing point of several rays
 * attracting when $$c$$ is in the main cardioid of the Mandelbrot set, in which case it is in the interior of a filled-in Julia set, and therefore belongs to the Fatou set (strictly to the basin of attraction of finite fixed point)
 * parabolic at the root point of the limb of the Mandelbrot set
 * repelling for other values of $$c$$

Special cases
An important case of the quadratic mapping is $$c=0$$. In this case, we get $$\alpha_1 = 0$$ and $$\alpha_2=1$$. In this case, 0 is a superattractive fixed point, and 1 belongs to the Julia set.

Only one fixed point
We have $$\alpha_1=\alpha_2$$ exactly when $$1-4c=0.$$ This equation has one solution, $$c=1/4,$$ in which case $$\alpha_1=\alpha_2=1/2$$. In fact $$c=1/4$$ is the largest positive, purely real value for which a finite attractor exists.

Infinite fixed point
We can extend the complex plane $$\mathbb{C}$$ to the Riemann sphere (extended complex plane) $$\mathbb{\hat{C}}$$ by adding infinity:


 * $$\mathbb{\hat{C}} = \mathbb{C} \cup \{ \infty \}$$

and extend $$f_c$$ such that $$f_c(\infty)=\infty.$$

Then infinity is:
 * superattracting
 * a fixed point of $$f_c$$: $$f_c(\infty)=\infty=f^{-1}_c(\infty).$$

Period-2 cycles
Period-2 cycles are two distinct points $$\beta_1$$ and $$\beta_2$$ such that $$f_c(\beta_1) = \beta_2$$ and $$f_c(\beta_2) = \beta_1$$, and hence


 * $$f_c(f_c(\beta_n)) = \beta_n$$

for $$n \in \{1, 2\}$$:


 * $$f_c(f_c(z)) = (z^2+c)^2+c = z^4 + 2cz^2 + c^2 + c.$$

Equating this to z, we obtain


 * $$z^4 + 2cz^2 - z + c^2 + c = 0.$$

This equation is a polynomial of degree 4, and so has four (possibly non-distinct) solutions. However, we already know two of the solutions. They are $$\alpha_1$$ and $$\alpha_2$$, computed above, since if these points are left unchanged by one application of $$f$$, then clearly they will be unchanged by more than one application of $$f$$.

Our 4th-order polynomial can therefore be factored in 2 ways:

First method of factorization

 * $$(z-\alpha_1)(z-\alpha_2)(z-\beta_1)(z-\beta_2) = 0.\,$$

This expands directly as $$x^4 - Ax^3 + Bx^2 - Cx + D = 0$$ (note the alternating signs), where


 * $$D = \alpha_1 \alpha_2 \beta_1 \beta_2, \,$$


 * $$C = \alpha_1 \alpha_2 \beta_1 + \alpha_1 \alpha_2 \beta_2 + \alpha_1 \beta_1 \beta_2 + \alpha_2 \beta_1 \beta_2, \,$$


 * $$B = \alpha_1 \alpha_2 + \alpha_1 \beta_1 + \alpha_1 \beta_2 + \alpha_2 \beta_1 + \alpha_2 \beta_2 + \beta_1 \beta_2, \,$$


 * $$A = \alpha_1 + \alpha_2 + \beta_1 + \beta_2.\,$$

We already have two solutions, and only need the other two. Hence the problem is equivalent to solving a quadratic polynomial. In particular, note that


 * $$\alpha_1 + \alpha_2 = \frac{1-\sqrt{1-4c}}{2} + \frac{1+\sqrt{1-4c}}{2} = \frac{1+1}{2} = 1$$

and


 * $$\alpha_1 \alpha_2 = \frac{(1-\sqrt{1-4c})(1+\sqrt{1-4c})}{4} = \frac{1^2 - (\sqrt{1-4c})^2}{4}= \frac{1 - 1 + 4c}{4} = \frac{4c}{4} = c.$$

Adding these to the above, we get $$D = c \beta_1 \beta_2$$ and $$A = 1 + \beta_1 + \beta_2$$. Matching these against the coefficients from expanding $$f$$, we get


 * $$D = c \beta_1 \beta_2 = c^2 + c$$ and $$A = 1 + \beta_1 + \beta_2 = 0.$$

From this, we easily get


 * $$\beta_1 \beta_2 = c + 1$$ and $$\beta_1 + \beta_2 = -1$$.

From here, we construct a quadratic equation with $$A' = 1, B = 1, C = c+1$$ and apply the standard solution formula to get


 * $$\beta_1 = \frac{-1 - \sqrt{-3 -4c}}{2}$$ and $$\beta_2 = \frac{-1 + \sqrt{-3 -4c}}{2}.$$

Closer examination shows that:


 * $$f_c(\beta_1) = \beta_2$$ and $$f_c(\beta_2) = \beta_1,$$

meaning these two points are the two points on a single period-2 cycle.

Second method of factorization
We can factor the quartic by using polynomial long division to divide out the factors $$(z-\alpha_1)$$ and $$(z-\alpha_2), $$ which account for the two fixed points $$\alpha_1$$ and $$\alpha_2$$ (whose values were given earlier and which still remain at the fixed point after two iterations):


 * $$(z^2+c)^2 + c -z = (z^2 + c - z)(z^2 + z + c +1 ). \,$$

The roots of the first factor are the two fixed points. They are repelling outside the main cardioid.

The second factor has the two roots


 * $$\frac{-1 \pm \sqrt{-3 -4c}}{2}. \,$$

These two roots, which are the same as those found by the first method, form the period-2 orbit.

Special cases
Again, let us look at $$c=0$$. Then


 * $$\beta_1 = \frac{-1 - i\sqrt{3}}{2}$$ and $$\beta_2 = \frac{-1 + i\sqrt{3}}{2},$$

both of which are complex numbers. We have $$| \beta_1 | = | \beta_2 | = 1$$. Thus, both these points are "hiding" in the Julia set. Another special case is $$c=-1$$, which gives $$\beta_1 = 0$$ and $$\beta_2 = -1$$. This gives the well-known superattractive cycle found in the largest period-2 lobe of the quadratic Mandelbrot set.

Cycles for period greater than 2
The degree of the equation $$f^{(n)}(z)=z$$ is 2n; thus for example, to find the points on a 3-cycle we would need to solve an equation of degree 8. After factoring out the factors giving the two fixed points, we would have a sixth degree equation.

There is no general solution in radicals to polynomial equations of degree five or higher, so the points on a cycle of period greater than 2 must in general be computed using numerical methods. However, in the specific case of period 4 the cyclical points have lengthy expressions in radicals.

In the case c = –2, trigonometric solutions exist for the periodic points of all periods. The case $$z_{n+1}=z_n^2-2$$ is equivalent to the logistic map case r = 4: $$x_{n+1}=4x_n(1-x_n).$$ Here the equivalence is given by $$z=2-4x.$$ One of the k-cycles of the logistic variable x (all of which cycles are repelling) is


 * $$\sin^2\left(\frac{2\pi}{2^k-1}\right), \, \sin^2\left(2\cdot\frac{2\pi}{2^k-1}\right), \, \sin^2\left(2^2\cdot\frac{2\pi}{2^k-1}\right), \, \sin^2\left(2^3\cdot\frac{2\pi}{2^k-1}\right), \dots, \sin^2\left(2^{k-1}\frac{2\pi}{2^k-1}\right).$$