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-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

Stability index of periodic points along horizontal axis

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 $abs(\lambda) = 1; \,$
  • 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.^[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 $z = 1/2.\,$

This image shows fixed points (both repelling)

Complex dynamics

Fixed points for c along horizontal axis

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 :

• superattracting
• a fixed point of polynomial $f_c: \,$^[6]

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

Period-2 cycles

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

1. Alan F. Beardon, Iteration of Rational Functions, Springer 1991, ISBN 0-387-95151-2, p. 41
2. Alan F. Beardon, Iteration of Rational Functions, Springer 1991, ISBN 0-387-95151-2, page 99
3. Some Julia sets by Michael Becker
4. On the regular leaf space of the cauliflower by Tomoki Kawahira Source: Kodai Math. J. Volume 26, Number 2 (2003), 167-178.
5. Periodic attractor by Evgeny Demidov
6. R L Devaney, L Keen (Editor): Chaos and Fractals: The Mathematics Behind the Computer Graphics. Publisher: Amer Mathematical Society July 1989, ISBN 0-8218-0137-6 , ISBN 978-0-8218-0137-6
7. Period 2 orbit by Evgeny Demidov
8. Gvozden Rukavina : Quadratic recurrence equations - exact explicit solution of period four fixed points functions in bifurcation diagram

Further reading

• Alan F. Beardon, Iteration of Rational Functions, Springer 1991, ISBN 0-387-95151-2
• Michael F. Barnsley (Author), Stephen G. Demko (Editor), Chaotic Dynamics and Fractals (Notes and Reports in Mathematics in Science and Engineering Series) Academic Pr (April 1986), ISBN 0-12-079060-2
• Wolf Jung : Homeomorphisms on Edges of the Mandelbrot Set. Ph.D. thesis of 2002
• The permutations of periodic points in quadratic polynominials by J Leahy

External links

• Algebraic solution of Mandelbrot orbital boundaries by Donald D. Cross
• Brown Method by Robert P. Munafo
• arXiv:hep-th/0501235v2 V.Dolotin, A.Morozov: Algebraic Geometry of Discrete Dynamics. The case of one variable.