- Periodic points of complex quadratic mappings
This article describes

periodic point s of some complex quadratic map. This theory is applied in relation with the theories of Fatou andJulia set s.**Definitions**Let

:$f\_c(z)=z^2+c,$

where $z$ and $c$ are complex-valued. (This $f$ is the "complex quadratic mapping" mentioned in the title.) This article explores the "

periodic point s" of this mapping - that is, the points that form a periodic cycle when $f$ is repeatedly applied to them.$f^\{(k)\}\; \_c\; (z)$ is the $k$ -fold compositions of $f\; \_c,$ with itself = iteration of function $f\; \_c,$

$f^\{(k)\}\; \_c\; (z)\; =\; f\_c(f^\{(k-1)\}\; \_c\; (z))$

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

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

where $p$ is the smallest positive integer.

We can introduce new function:

$F\_p(z,f)\; =\; f^\{(p)\}\; \_c\; (z)\; -\; z$

so periodic points are zeros of function $F\_p(z,f)$ :

$z\; :\; F\_p(z,f)\; =\; 0$

which is polynomial of degree $=\; 2^p$

**Stability of periodic points ( orbit )**The

**multiplier**$m(f,z\_0)=lambda\; ,$ of rational map $f,$ at fixed point $z\_0,$ is defined as :$m(f,z\_0)=lambda\; =\; egin\{cases\}\; f\_c\text{'}(z\_0),\; mbox\{if\; \}z\_0\; e\; infty\; \backslash \; frac\{1\}\{f\_c\text{'}(z\_0)\},\; mbox\{if\; \}z\_0\; =\; infty\; end\{cases\}$

where $f\_c\text{'}(z\_0),$ 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 $f\_c(z)=z$. 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

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

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

: $alpha\_1\; =\; frac\{1-sqrt\{1-4c\{2\}$ 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 $alpha\_1\; +\; alpha\_2\; =\; 1\; ,$.

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

Complex dynamics Here different notation is commonly used:

: $alpha\_c\; =\; frac\{1-sqrt\{1-4c\{2\}$ and $eta\_c\; =\; frac\{1+sqrt\{1-4c\{2\}.$

Using

Viète's formulas one can show that::$alpha\_c\; +\; eta\_c\; =\; -frac\{B\}\{A\}\; =\; 1$

Since derivative with respect to z is :

:$P\_c\text{'}(z)\; =\; frac\{d\}\{dz\}P\_c(z)\; =\; 2z$

then

:$P\_c\text{'}(alpha\_c)\; +\; P\_c\text{'}(eta\_c)=\; 2\; alpha\_c\; +\; 2\; eta\_c\; =\; 2\; (alpha\_c\; +\; eta\_c)\; =\; 2\; ,$

It implies that $P\_c\; ,$ can have at most one attractive fixed point.

This points are distinguished by the facts that:

* $eta\_c\; ,$ is :

**the landing point ofexternal ray for angle=0 for $c\; in\; M\; setminus\; left\; \{\; frac\{1\}\{4\}\; ight\; \}$

**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 $mathbb\{C\}$ toRiemann sphere (extended complex plane) $mathbb\{hat\{C$ by adding infinity$mathbb\{hat\{C\; =\; mathbb\{C\}\; cup\; \{\; infty\; \}$

and extend polynomial $f\_c,$ such that $f\_c(infty)=infty,$

Then infinity is :

*superattracting

*fixed point $f\_c(infty)=infty=f^\{-1\}\_c(infty),$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 $eta\_1$ and $eta\_2$ such that $f\_c(eta\_1)\; =\; eta\_2$, and $f\_c(eta\_2)\; =\; eta\_1$.

Let us start by writing $f\_c(f\_c(eta\_n))\; =\; eta\_n$, and see where trying to solve this leads.

: $f\_c(f\_c(z))\; =\; (z^2+c)^2+c\; =\; z^4\; +\; 2z^2c\; +\; c^2\; +\; c.,$

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**: $(z-alpha\_1)(z-alpha\_2)(z-eta\_1)(z-eta\_2)\; =\; 0.,$

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

: $D\; =\; alpha\_1\; alpha\_2\; eta\_1\; eta\_2,$

: $C\; =\; alpha\_1\; alpha\_2\; eta\_1\; +\; alpha\_1\; alpha\_2\; eta\_2\; +\; alpha\_1\; eta\_1\; eta\_2\; +\; alpha\_2\; eta\_1\; eta\_2,$

: $B\; =\; alpha\_1\; alpha\_2\; +\; alpha\_1\; eta\_1\; +\; alpha\_1\; eta\_2\; +\; alpha\_2\; eta\_1\; +\; alpha\_2\; eta\_2\; +\; eta\_1\; eta\_2,$

: $A\; =\; alpha\_1\; +\; alpha\_2\; +\; eta\_1\; +\; eta\_2.,$

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

: $alpha\_1\; +\; alpha\_2\; =\; frac\{1-sqrt\{1-4c\{2\}\; +\; frac\{1+sqrt\{1-4c\{2\}\; =\; frac\{1+1\}\{2\}\; =\; 1$

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 $D\; =\; c\; eta\_1\; eta\_2$ and $A\; =\; 1\; +\; eta\_1\; +\; eta\_2$. Matching these against the coefficients from expanding $f$, we get

: $D\; =\; c\; eta\_1\; eta\_2\; =\; c^2\; +\; c$ and $A\; =\; 1\; +\; eta\_1\; +\; eta\_2\; =\; 0.$

From this, we easily get :$eta\_1\; eta\_2\; =\; c\; +\; 1$ and $eta\_1\; +\; eta\_2\; =\; -1$.

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

: $eta\_1\; =\; frac\{-1\; -\; sqrt\{-3\; -4c\{2\}$ and $eta\_2\; =\; frac\{-1\; +\; sqrt\{-3\; -4c\{2\}.$

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

$f\_c(eta\_1)\; =\; eta\_2$ and $f\_c(eta\_2)\; =\; eta\_1$

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

**econd method of factorization**$(z^2+c)^2\; +\; c\; -z\; =\; (z^2\; +\; c\; -\; z)(z^2\; +\; z\; +\; c\; +1\; )\; ,$

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

The second factor has two roots

$z\_\{3,4\}\; =\; -frac\{1\}\{2\}\; pm\; (-frac\{3\}\{4\}\; -\; c)^frac\{1\}\{2\}\; ,$

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

: $eta\_1\; =\; frac\{-1\; -\; isqrt\{3\{2\}$ and $eta\_2\; =\; frac\{-1\; +\; isqrt\{3\{2\}$

both of which are complex numbers. By doing a little algebra, we find $|\; eta\_1\; |\; =\; |\; eta\_2\; |\; =\; 1$. Thus, both these points are "hiding" in the Julia set.

Another special case is $c=-1$, which gives $eta\_1\; =\; 0$ and $eta\_2\; =\; -1$. 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****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-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*]**External links*** [

*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*]

*Wikimedia Foundation.
2010.*

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