Periodic points of complex quadratic mappings

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This article on periodic points of complex quadratic mappings describes periodic points of some quadratic polynomial mappings on the complex numbers. This theory is applied in relation with the theories of Julia sets, and the Mandelbrot set.

Note: This is unfinished.

1 Definitions
2 Period-1 points (fixed points)
- 2.1 Special Cases
- 2.2 Only 1 fixed point
3 Period-2 Cycles
- 3.1 Special Cases

[edit] Definitions

Let

f c (z) = z 2 + c,

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

[edit] Period-1 points (fixed points)

Let us begin by finding all 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 the solutions of $A x 2 + B x + C = 0$ are given by

$x=\frac{-B\pm\sqrt{B^2-4AC}}{2A}$

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

$\alpha_1 = \frac{1-\sqrt{1-4a}}{2}$ and $\alpha_1 = \frac{1+\sqrt{1-4a}}{2}$

[edit] Special Cases

An important case of the quadratic mapping is $c = 0$ . In this case, we get $α 1 = 0$ and $α 2 = 1$ . In this case, 0 is a superattractive fixed-point, and 1 belongs to the Julia set.

[edit] Only 1 fixed point

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

[edit] Period-2 Cycles

Suppose next that we wish to look at period-2 cycles. That is, we want to find two points $β 1$ and $β 2$ such that $f c (β 1) = β 2$ , and $f c (β 2) = β 1$ .

Let us start by writing $f c (f c (β n)) = β n$ , and see where trying to solve this leads.

f c (f c (z)) = (z 2 + c) 2 + c = z 4 + 2 z 2 c + c 2 + c

Thus, the equation we wish to solve is actually $z 4 + 2 c z 2 - z + c 2 + c = 0$ ; a formidable equation indeed!

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 $α 1$ and $α 2$ , 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 as

(z - α 1)(z - α 2)(z - β 1)(z - β 2) = 0

This expands directly as $x 4 - A x 3 + B x 2 - C x + D = 0$ (note the alternating signs), where

$D = α 1 α 2 β 1 β 2$
$C = α 1 α 2 β 1 + α 1 α 2 β 2 + α 1 β 1 β 2 + α 2 β 1 β 2$
$B = α 1 α 2 + α 1 β 1 + α 1 β 2 + α 2 β 1 + α 2 β 2 + β 1 β 2$
$A = α 1 + α 2 + β 1 + β 2$

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 $D = c β 1 β 2$ and $A = 1 + β 1 + β 2$ . Matching these against the coefficients from expanding $f$ , we get

D = c β 1 β 2 = c 2 + c

and

A = 1 + β 1 + β 2 = 0

From this, we easily get $β 1 β 2 = c + 1$ and $b e t a 1 + β 2 = - 1$ . 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 (the formulas are a tad messy) that $f c (β 1) = β 2$ and $f c (β 2) = β 1$ , meaning these two points are the two halves of a single period-2 cycle.

[edit] 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. By doing a little algebra, we find $| β 1 | = | β 2 | = 1$ . Thus, both these points are "hiding" in the Julia set.

Another special case is $c = - 1$ , which gives $β 1 = 0$ and $β 2 = - 1$ . This gives the well-known superattractive cycle found in the largest period-2 lobe of the quadratic Mandelbrot set.

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Category: Complex analysis