Periodic points of complex quadratic mappings

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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-valued.

Notationally,  \ f^{(k)} _c (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

 \ 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.

Stability of periodic points (orbit) - multiplier

File:Multiplier4 f.png
Stability index of periodic points along horizontal axis
File:Mandelbrot set Components.jpg
boundaries of regions of parameter plane with attracting orbit of periods 1-6
Critical orbit of discrete dynamical system based on complex quadratic polynomial. It tends to weakly attracting fixed point with abs(multiplier)=0.99993612384259

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;[1]
  • used to check stability of periodic (also fixed) points with stability index abs(\lambda). \,

A periodic point is[2]

  • 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; \,
  • 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.[3] A parabolic periodic point is in the Julia set.

Period-1 points (fixed 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

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 C \setminus [1/4,+\inf ] 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}

then \alpha_1 + \alpha_2 = 1 .\,.

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

File:JuliaRay3.png
This image shows fixed points (both repelling)

Complex dynamics

File:Fixed.png
Fixed points for c along horizontal axis
File:Julia0bb.jpg
Fatou set for F(z)=z*z with marked fixed point

Here different notation is commonly used:[4]

\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}.\,

Using Viète's formulas one can show that:

 \alpha_c + \beta_c = 1 .

Since the derivative with respect to z is

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

then

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 external ray for angle=0 for c \in M \setminus  \left \{ \frac{1}{4} \right \}
    • the most repelling fixed point, belongs to 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).[5]
  • \alpha_c \, is:
    • the landing point of several rays
    • attracting when c is in main cardioid of Mandelbrot set, then it is in interior of Filled-in Julia set, it means belongs to Fatou set ( strictly to basin of attraction of finite fixed point )
    • parabolic at the root point of the limb of Mandelbrot set
    • repelling for other c values

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 extending polynomial f_c\, such that f_c(\infty)=\infty. \,

Then infinity is :

f_c(\infty)=\infty=f^{-1}_c(\infty).\,

Period-2 cycles

File:Bifurcation1-2.png
Bifurcation from period 1 to 2 for complex quadratic map

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.

We write f_c(f_c(\beta_n)) = \beta_n:

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 2 applications (or more).

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}{2} \pm (-\frac{3}{4} - c)^\frac{1}{2}. \,

These two roots, which are the same as those found by the first method, form the period-2 orbit.[7]

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.[8]

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).

References

Further reading

External links