Periodic points of complex quadratic mappings

Definitions

Let

be the complex quadric mapping, where z and c are complex-valued.

Notationally,   f c ( k ) ( z ) is the   k -fold composition of f c with itself—that is, the value after the k-th iteration of function f c . Thus

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

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

We can introduce a new function:

so periodic points are zeros of function   F p ( z , f ) : points z satisfying

which is a polynomial of degree 2 p .

Stability of periodic points (orbit) - multiplier

The multiplier (or eigenvalue, derivative) m ( f p , z 0 ) = λ of a rational map f iterated p times at cyclic point z 0 is defined as:

where f p ′ ( 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 periodic point is

Periodic points

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

which can be rewritten as

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

So for c ∈ C ∖ [ 1 / 4 , + inf ] we have two finite fixed points α 1 and α 2 .

Since

then α 1 + α 2 = 1. .

Thus fixed points are symmetrical around z = 1 / 2.

Complex dynamics

Here different notation is commonly used:

and

Using Viète's formulas one can show that:

Since the derivative with respect to z is

then

This implies that P c can have at most one attractive fixed point.

These points are distinguished by the facts that:

Special cases

An important case of the quadratic mapping is c = 0 . In this case, we get α 1 = 0 and α 2 = 1 . In this case, 0 is a superattractive fixed point, and 1 belongs to the Julia set.

Only one fixed point

We have α 1 = α 2 exactly when 1 − 4 c = 0. This equation has one solution, c = 1 / 4 , in which case α 1 = α 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 C to the Riemann sphere (extended complex plane) C ^ by adding infinity :

and extending polynomial f c such that f c ( ∞ ) = ∞ .

Then infinity is :

Period-2 cycles

Period-2 cycles are two distinct points β 1 and β 2 such that f c ( β 1 ) = β 2 and f c ( β 2 ) = β 1 .

We write f c ( f c ( β n ) ) = β n :

Equating this to z, we obtain

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 α 1 and α 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

This expands directly as x 4 − A x 3 + B x 2 − C x + D = 0 (note the alternating signs), where

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

and

Adding these to the above, we get D = c β 1 β 2 and A = 1 + β 1 + β 2 . Matching these against the coefficients from expanding f , we get

From this, we easily get

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

Closer examination shows that :

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 − α 1 ) and ( z − α 2 ) , which account for the two fixed points α 1 and α 2 (whose values were given earlier and which still remain at the fixed point after two iterations):

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

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

both of which are complex numbers. We have | β 1 | = | β 2 | = 1 . Thus, both these points are "hiding" in the Julia set. Another special case is c = − 1 , which gives β 1 = 0 and β 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 = 4 x n ( 1 − x n ) . Here the equivalence is given by z = 2 − 4 x . One of the k-cycles of the logistic variable x (all of which cycles are repelling) is