Periodic Points Of Complex Quadratic Mappings

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 document explores the '' Periodic Point s'' of this Map ping - 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

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=rac{-B\pm\sqrt{B^2-4AC}}{2A}

In our case, we have

A=1, B=-1, C=c

, so we will write

:

\alpha_1 = rac{1-\sqrt{1-4c}}{2}

and

\alpha_2 = rac{1+\sqrt{1-4c}}{2}.

So we have two Finite fixed points

\alpha_1 \,

and

\alpha_2\,

.

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 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

One must remember that Infinity is fixed point of Polynomial .

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

PERIOD-2 CYCLES

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

\beta_1

and

\beta_2

such that

f_c(\beta_1) = \beta_2

, and

f_c(\beta_2) = \beta_1

.

Let us start by writing

f_c(f_c(\beta_n)) = \beta_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

; 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

\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 as

:

(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 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 = rac{1-\sqrt{1-4c}}{2} + rac{1+\sqrt{1-4c}}{2} = rac{1+1}{2} = 1

and

:

\alpha_1 \alpha_2 = rac{(1-\sqrt{1-4c})(1+\sqrt{1-4c})}{4} = rac{1^2 - (\sqrt{1-4c})^2}{4}

= rac{1 - 1 + 4c}{4} = rac{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 = rac{-1 - \sqrt{-3 -4c}}{2}

and

\beta_2 = rac{-1 + \sqrt{-3 -4c}}{2}.

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

f_c(\beta_1) = \beta_2

and

f_c(\beta_2) = \beta_1

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

Special cases

Again, let us look at

c=0

. Then

:

\beta_1 = rac{-1 - i\sqrt{3}}{2}

and

\beta_2 = rac{-1 + i\sqrt{3}}{2}

	CATEGORIES ABOUT PERIODIC POINTS OF COMPLEX QUADRATIC MAPPINGS

	complex analysis

	fractals

	limit sets

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Information About

Periodic Points Of Complex Quadratic Mappings