Fractals/Mathematics/periodic points of complex quadratic map

=Periodic points of complex quadratic map= 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.

Finding periodic points
solve these equations using numerical methods for solving polynomials - and even something simple such as Newton's method is going to converge a lot faster than finding the cycles just by iterating a single point (as is how bifurcations diagrams are usually made) under fc itself. Milo Brandt

Methods:
 * simple iterating and checking convergence
 * numerical methods
 * symbolic computations, algebraic
 * numerical methods for finding roots of polynomial equations

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


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

Distance between fixed points:


 * $$\frac{1- \Delta}{2} < \frac{1}{2} < \frac{1 + \Delta}{2}$$

is delta $$\Delta$$

where


 * $$\Delta = \sqrt{1 - 4c}$$ so


 * for $$c = \frac{1}{4}$$ distance is equal to zero:  $$\Delta(\frac{1}{4}) = \sqrt{0} = 0 $$ =the points coincide ( parabolic case)
 * for $$c = 0$$ distance is equal to 1:  $$\Delta(1) = \sqrt{1} = 1 $$ = the superattracting case ( alfa is a center and beta in on the unit circle)

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


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

Method by Claude Heiland-Allen
What I do to create an periodic components of Mandelbrot set, for $$f_c(z) = z^2 + c$$:
 * start iteration from $$z_0 := 0$$, with $$m := \infty$$
 * for each $$n = 1, 2, 3, ...$$ in order
 * calculate $$z_n := f_c(z_{n-1})$$
 * if $$|z_n| < m$$
 * set $$m := |z_n|$$
 * use Newton's method to solve $$w = f_c^{\circ n}(w)$$ with initial guess $$w^{(0)} := z_n$$ (this may fail to converge, in which case continue with the next $$n$$), the steps are $$w^{(i+1)} := w^{(i)} - \frac{f_c^{\circ n}(w^{(i)}) - w^{(i)}}{{f_c^{\circ n}}'(w^{(i)}) - 1}$$
 * calculate the derivative of the cycle $$\lambda := {f_c^{\circ n}}'(w)$$
 * if $$|\lambda| < 1$$, then the cycle is attractive and $$c$$ is within a hyperbolic component of period $$n$$, stop (success).

Where:
 * $$\lambda$$ may used as "interior coordinates" within the hyperbolic component.
 * $$w$$ and $$n$$ can be used for interior distance estimation.

The point of using Newton's method is to accelerate the computation of $$w$$, a point in the limit cycle attractor. Computing $$w$$ just by iterating $$f_c$$ could take many 1000s of iterations, especially when $$\lambda$$ is close to $$1$$.

I have no complete proof of correctness (but this doesn't mean I think it is incorrect; the images seem plausible). It relies on the "atom domains" surrounding each hyperbolic component of a given period.

It also relies on the cycle reached by Newton's method being the same cycle as the limit cycle approached by iteration: this is true for the quadratic Mandelbrot set because there is only one finite critical point, $$0$$ ($$\infty$$ is a fixed point) and each attracting or parabolic cycle has a critical point in its immediate basin (see ), which means there can be at most one attracting or parabolic cycle.

For an implementation in C99 you can see my blog post at 

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