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.

DefinitionsEdit

Let

<math>f_c(z) = z^2+c\,</math>

be the complex quadratic mapping, where <math>z</math> and <math>c</math> are complex numbers.

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

<math>f^{(k)} _c (z) = f_c(f^{(k-1)} _c (z)).</math>

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

<math>f^{(p)} _c (z) = z,</math>

where <math>p</math> is the smallest positive integer for which the equation holds at that z.

We can introduce a new function:

<math>F_p(z,f) = f^{(p)} _c (z) - z,</math>

so periodic points are zeros of function <math>F_p(z,f)</math>: points z satisfying

<math>F_p(z,f) = 0,</math>

which is a polynomial of degree <math>2^p.</math>

Number of periodic pointsEdit

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

Stability of periodic points (orbit) - multiplierEdit

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
File:Critical orbit 3d.png
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) <math>m(f^p,z_0)=\lambda</math> of a rational map <math>f</math> iterated <math>p</math> times at cyclic point <math>z_0</math> is defined as:

<math>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}</math>

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

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;<ref>Alan F. Beardon, Iteration of Rational Functions, Springer 1991, Template:ISBN, p. 41</ref>
  • used to check stability of periodic (also fixed) points with stability index <math>abs(\lambda). \,</math>

A periodic point is<ref>Alan F. Beardon, Iteration of Rational Functions, Springer 1991, Template:ISBN, page 99</ref>

  • attracting when <math>abs(\lambda) < 1;</math>
    • super-attracting when <math>abs(\lambda) = 0;</math>
    • attracting but not super-attracting when <math>0 < abs(\lambda) < 1;</math>
  • indifferent when <math>abs(\lambda) = 1;</math>
  • repelling when <math>abs(\lambda) > 1.</math>

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.<ref>Some Julia sets by Michael Becker</ref> A parabolic periodic point is in the Julia set.

Period-1 points (fixed points)Edit

Finite fixed pointsEdit

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

<math>z^2+c=z,\,</math>

which can be rewritten as

<math>\ z^2-z+c=0.</math>

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

<math>\alpha_1 = \frac{1-\sqrt{1-4c}}{2}</math> and <math>\alpha_2 = \frac{1+\sqrt{1-4c}}{2}.</math>

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

Since

<math>\alpha_1 = \frac{1}{2}-m</math> and <math>\alpha_2 = \frac{1}{2}+m</math> where <math>m = \frac{\sqrt{1-4c}}{2},</math>

we have <math>\alpha_1 + \alpha_2 = 1</math>.

Thus fixed points are symmetrical about <math>z = 1/2</math>.

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

Complex dynamicsEdit

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:<ref>On the regular leaf space of the cauliflower by Tomoki Kawahira Source: Kodai Math. J. Volume 26, Number 2 (2003), 167-178. Template:Webarchive</ref>

<math>\alpha_c = \frac{1-\sqrt{1-4c}}{2}</math> with multiplier <math>\lambda_{\alpha_c} = 1-\sqrt{1-4c}</math>

and

<math>\beta_c = \frac{1+\sqrt{1-4c}}{2}</math> with multiplier <math>\lambda_{\beta_c} = 1+\sqrt{1-4c}.</math>

Again we have

<math>\alpha_c + \beta_c = 1 .</math>

Since the derivative with respect to z is

<math>P_c'(z) = \frac{d}{dz}P_c(z) = 2z ,</math>

we have

<math>P_c'(\alpha_c) + P_c'(\beta_c)= 2 \alpha_c + 2 \beta_c = 2 (\alpha_c + \beta_c) = 2 .</math>

This implies that <math>P_c</math> can have at most one attractive fixed point.

These points are distinguished by the facts that:

  • <math>\beta_c</math> is:
    • the landing point of the external ray for angle=0 for <math>c \in M \setminus \left\{ 1/4 \right\}</math>
    • 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).<ref>Periodic attractor by Evgeny Demidov Template:Webarchive</ref>
  • <math>\alpha_c</math> is:
    • the landing point of several rays
    • attracting when <math>c</math> 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 <math>c</math>

Special casesEdit

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

Only one fixed pointEdit

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

Infinite fixed pointEdit

We can extend the complex plane <math>\mathbb{C}</math> to the Riemann sphere (extended complex plane) <math>\mathbb{\hat{C}}</math> by adding infinity:

<math>\mathbb{\hat{C}} = \mathbb{C} \cup \{ \infty \}</math>

and extend <math>f_c</math> such that <math>f_c(\infty)=\infty.</math>

Then infinity is:

  • superattracting
  • a fixed point of <math>f_c</math>:<ref>R L Devaney, L Keen (Editor): Chaos and Fractals: The Mathematics Behind the Computer Graphics. Publisher: Amer Mathematical Society July 1989, Template:ISBN , Template:ISBN</ref><math display="block">f_c(\infty)=\infty=f^{-1}_c(\infty).</math>

Period-2 cyclesEdit

File:Bifurcation1-2.png
Bifurcation from period 1 to 2 for complex quadratic map
File:Bifurcation of periodic points from period 1 to 2 for fc(z)=z*z +c.gif
Bifurcation of periodic points from period 1 to 2 for fc(z)=z*z +c

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

<math>f_c(f_c(\beta_n)) = \beta_n</math>

for <math>n \in \{1, 2\}</math>:

<math>f_c(f_c(z)) = (z^2+c)^2+c = z^4 + 2cz^2 + c^2 + c.</math>

Equating this to z, we obtain

<math>z^4 + 2cz^2 - z + c^2 + c = 0.</math>

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 <math>\alpha_1</math> and <math>\alpha_2</math>, computed above, since if these points are left unchanged by one application of <math>f</math>, then clearly they will be unchanged by more than one application of <math>f</math>.

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

First method of factorizationEdit

<math>(z-\alpha_1)(z-\alpha_2)(z-\beta_1)(z-\beta_2) = 0.\,</math>

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

<math>D = \alpha_1 \alpha_2 \beta_1 \beta_2, \,</math>
<math>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, \,</math>
<math>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, \,</math>
<math>A = \alpha_1 + \alpha_2 + \beta_1 + \beta_2.\,</math>

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

<math>\alpha_1 + \alpha_2 = \frac{1-\sqrt{1-4c}}{2} + \frac{1+\sqrt{1-4c}}{2} = \frac{1+1}{2} = 1</math>

and

<math>\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.</math>

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

<math>D = c \beta_1 \beta_2 = c^2 + c</math> and <math>A = 1 + \beta_1 + \beta_2 = 0.</math>

From this, we easily get

<math>\beta_1 \beta_2 = c + 1</math> and <math>\beta_1 + \beta_2 = -1</math>.

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

<math>\beta_1 = \frac{-1 - \sqrt{-3 -4c}}{2}</math> and <math>\beta_2 = \frac{-1 + \sqrt{-3 -4c}}{2}.</math>

Closer examination shows that:

<math>f_c(\beta_1) = \beta_2</math> and <math>f_c(\beta_2) = \beta_1,</math>

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

Second method of factorizationEdit

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

<math>(z^2+c)^2 + c -z = (z^2 + c - z)(z^2 + z + c +1 ). \,</math>

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

<math>\frac{-1 \pm \sqrt{-3 -4c}}{2}. \,</math>

These two roots, which are the same as those found by the first method, form the period-2 orbit.<ref>Period 2 orbit by Evgeny Demidov Template:Webarchive</ref>

Special casesEdit

Again, let us look at <math>c=0</math>. Then

<math>\beta_1 = \frac{-1 - i\sqrt{3}}{2}</math> and <math>\beta_2 = \frac{-1 + i\sqrt{3}}{2},</math>

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

Cycles for period greater than 2Edit

File:Periodic points of f(z) = z*z-0.75 for period =6 as intersections of 2 implicit curves.svg
Periodic points of f(z) = z*z−0.75 for period =6 as intersections of 2 implicit curves

The degree of the equation <math>f^{(n)}(z)=z</math> 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.<ref>Gvozden Rukavina : Quadratic recurrence equations - exact explicit solution of period four fixed points functions in bifurcation diagram</ref>

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

<math>\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).</math>

ReferencesEdit

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Further readingEdit

External linksEdit

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