# Periodic points of complex quadratic mappings

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Periodic points of complex quadratic mappings

This article describes periodic points of some complex quadratic map. This theory is applied in relation with the theories of Fatou and Julia sets.

Definitions

Let

:$f_c\left(z\right)=z^2+c,$

where $z$ and $c$ are complex-valued. (This $f$ is the "complex quadratic mapping" mentioned in the title.) This article explores the "periodic points" of this mapping - that is, the points that form a periodic cycle when $f$ is repeatedly applied to them.

$f^\left\{\left(k\right)\right\} _c \left(z\right)$ is the $k$ -fold compositions of $f _c,$ with itself = iteration of function $f _c,$

$f^\left\{\left(k\right)\right\} _c \left(z\right) = f_c\left(f^\left\{\left(k-1\right)\right\} _c \left(z\right)\right)$

then periodic points of complex quadratic mapping of period $p$ are points $z$ of dynamical plane such that :

$z : f^\left\{\left(p\right)\right\} _c \left(z\right) = z$

where $p$ is the smallest positive integer.

We can introduce new function:

$F_p\left(z,f\right) = f^\left\{\left(p\right)\right\} _c \left(z\right) - z$

so periodic points are zeros of function $F_p\left(z,f\right)$ :

$z : F_p\left(z,f\right) = 0$

which is polynomial of degree $= 2^p$

Stability of periodic points ( orbit )

The multiplier $m\left(f,z_0\right)=lambda ,$ of rational map $f,$ at fixed point $z_0,$ is defined as :

where $f_c\text{'}\left(z_0\right),$ is first derivative of $f_c$ with respect to $z,$ at $z_0,$.

Because multiplier is the same at all points of peiodic one can name it multiplier of periodic orbit.

Multiplier is:
*complex number,
*invariant under conjugation of any rational map at its fixed point [Alan F. Beardon, Iteration of Rational Functions, Springer 1991, ISBN 0-387-95151-2, p. 41]
*used to check stability of periodic (also fixed) points.

=Period-1 points (fixed points)=

Finite fixed points

Let us begin by finding all finite points left unchanged by 1 application of $f$. These are the points that satisfy $f_c\left(z\right)=z$. That is, we wish to solve

: $z^2+c=z,$

which can be rewritten

: $z^2-z+c=0.$

Since this is an ordinary quadratic equation in 1 unknown, we can apply the standard quadratic solution formula. Look in any standard mathematics textbook, and you will find that there are two solutions of $Ax^2+Bx+C=0$ are given by

: $x=frac\left\{-Bpmsqrt\left\{B^2-4AC\left\{2A\right\}$

In our case, we have $A=1, B=-1, C=c$, so we will write

: $alpha_1 = frac\left\{1-sqrt\left\{1-4c\left\{2\right\}$ and $alpha_2 = frac\left\{1+sqrt\left\{1-4c\left\{2\right\}.$So for $c in C setminus \left[1/4,+inf \right]$ we have two finite fixed points $alpha_1 ,$ and $alpha_2,$.

Since : $alpha_1 = frac\left\{1\right\}\left\{2\right\}-m$ and $alpha_2 = frac\left\{1\right\}\left\{2\right\}+ m$ where $m = frac\left\{sqrt\left\{1-4c\left\{2\right\}$

then $alpha_1 + alpha_2 = 1 ,$.

It means that fixed points are symmetrical around $z = 1/2,$.

Complex dynamics

Here different notation is commonly used:

: $alpha_c = frac\left\{1-sqrt\left\{1-4c\left\{2\right\}$ and

Using Viète's formulas one can show that:

:

Since derivative with respect to z is :

:$P_c\text{'}\left(z\right) = frac\left\{d\right\}\left\{dz\right\}P_c\left(z\right) = 2z$

then

:

It implies that $P_c ,$ can have at most one attractive fixed point.

This points are distinguished by the facts that:
* is :
**the landing point of external ray for angle=0 for $c in M setminus left \left\{ frac\left\{1\right\}\left\{4\right\} ight \right\}$
**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) [ [http://www.ibiblio.org/e-notes/MSet/Attractor.htm Periodic attractor by Evgeny Demidov] ] .
* $alpha_c ,$ is:
** landing point of several rays
** is :
***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 Mandelbtot set
***repelling for other c values

pecial 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 might wonder what value $c$ should have to cause $alpha_1=alpha_2$. The answer is that this will happen exactly when $1-4c=0$. This equation has 1 solution: $c=1/4$ (in which case, $alpha_1=alpha_2=1/2$). This is interesting, since $c=1/4$ is the largest positive, purely-real value for which a finite attractor exists.

Infinite fixed point

We can extend complex plane $mathbb\left\{C\right\}$ to Riemann sphere (extended complex plane) $mathbb\left\{hat\left\{C$ by adding infinity

$mathbb\left\{hat\left\{C = mathbb\left\{C\right\} cup \left\{ infty \right\}$

and extend polynomial $f_c,$ such that $f_c\left(infty\right)=infty,$

Then infinity is :
*superattracting
*fixed point $f_c\left(infty\right)=infty=f^\left\{-1\right\}_c\left(infty\right),$of polynomial $f_c,$ [R L Devaney, L Keen (Editor}: Chaos and Fractals: The Mathematics Behind the Computer Graphics. Publisher: Amer Mathematical Society July 1989, ISBN-10: 0821801376 , ISBN-13: 9780821801376] .

Period-2 cycles

Suppose next that we wish to look at "period-2 cycles". That is, we want to find two points and such that , and .

Let us start by writing , and see where trying to solve this leads.

: $f_c\left(f_c\left(z\right)\right) = \left(z^2+c\right)^2+c = z^4 + 2z^2c + c^2 + c.,$

Thus, the equation we wish to solve is actually $z^4 + 2cz^2 - z + c^2 + c = 0$.

This equation is a polynomial of degree 4, and so has 4 (possibly non-distinct) solutions. "However", actually, we already know 2 of the solutions. They are $alpha_1$ and $alpha_2$, computed above. It is simple to see why this is; if these points are left unchanged by 1 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

:

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

:

:

:

:

We already have 2 solutions, and only need the other 2. This is as difficult as solving a quadratic polynomial. In particular, note that

: $alpha_1 + alpha_2 = frac\left\{1-sqrt\left\{1-4c\left\{2\right\} + frac\left\{1+sqrt\left\{1-4c\left\{2\right\} = frac\left\{1+1\right\}\left\{2\right\} = 1$

and

: $alpha_1 alpha_2 = frac\left\{\left(1-sqrt\left\{1-4c\right\}\right)\left(1+sqrt\left\{1-4c\right\}\right)\right\}\left\{4\right\} = frac\left\{1^2 - \left(sqrt\left\{1-4c\right\}\right)^2\right\}\left\{4\right\}= frac\left\{1 - 1 + 4c\right\}\left\{4\right\} = frac\left\{4c\right\}\left\{4\right\} = c.$

Adding these to the above, we get and . Matching these against the coefficients from expanding $f$, we get

: and

From this, we easily get : and .

From here, we construct a quadratic equation with $A\text{'} = 1, B = 1, C = c+1$ and apply the standard solution formula to get

: and

Closer examination shows (the formulas are a tad messy) that :

and

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

econd method of factorization

$\left(z^2+c\right)^2 + c -z = \left(z^2 + c - z\right)\left(z^2 + z + c +1 \right) ,$

The roots of the first factor are the two fixed points $z_\left\{1,2\right\},$ . They are repelling outside the main cardioid.

The second factor has two roots

$z_\left\{3,4\right\} = -frac\left\{1\right\}\left\{2\right\} pm \left(-frac\left\{3\right\}\left\{4\right\} - c\right)^frac\left\{1\right\}\left\{2\right\} ,$

These two roots form period-2 orbit. [ [http://www.ibiblio.org/e-notes/MSet/Attractor.htm Period 2 orbit by Evgeny Demidov] ]

pecial cases

Again, let us look at $c=0$. Then

: and

both of which are complex numbers. By doing a little algebra, we find . Thus, both these points are "hiding" in the Julia set.

Another special case is $c=-1$, which gives and . This gives the well-known superattractive cycle found in the largest period-2 lobe of the quadratic Mandelbrot set.

Cycles for period>2

There is no general solution in radicals to polynomial equations of degree five or higher, so it must be computed using numerical methods.

References

*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-10: 0120790602
* [http://www.math.sunysb.edu/cgi-bin/thesis.pl?thesis02-3 Wolf Jung : Homeomorphisms on Edges of the Mandelbrot Set. Ph.D. thesis of 2002]
* [http://hdl.handle.net/10090/3895| The permutations of periodic points in quadratic polynominials by J Leahy]

* [http://cosinekitty.com/mandel_orbits_analysis.html "Algebraic solution of Mandelbrot orbital boundaries" by Donald D. Cross ]
* [http://www.mrob.com/pub/muency/brownmethod.html "Brown Method" by Robert P. Munafo]
* [http://arxiv.org/abs/hep-th/0501235 arXiv:hep-th/0501235v2] V.Dolotin, A.Morozov: "Algebraic Geometry of Discrete Dynamics". The case of one variable.
* [http://arxiv.org/abs/0802.2565 Gvozden Rukavina : Quadratic recurrence equations - exact explicit solution of period four fixed points functions in bifurcation diagram]

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