San José State University

applet-magic.com
Thayer Watkins
Silicon Valley
USA

 The Components of Mandelbrot Sets Associated with Limit 2-Cycles

Consider Mandelbrot generated by the following iteration

#### zn+1 = znw + c

where c is a complex number and w is as well.

The value of z may approach a limit cycle of period 2 in which it value oscillates between Z1 and Z2. This means that

#### Z2 = Z1w + c Z1 = Z2w + c

Subtracting the second equation from the first gives

#### Z2 − Z1 = Z1w − Z2w which may be rearranged to Z2 + Z2w = Z1 + Z1w

Let the common value of each side of the equation be denoted as λ. Thus

#### Z2 + Z2w = λ Z1 + Z1w = λ

Thus Z1 and Z2 are solutions to the equation

#### Z + Zw = λ

For w≠2 there may be more than two solutions, in which case Z1 and Z2 would be any pair of the set of solutions.

## Conditions for a Stable Limit Cycle

The deviations from the limit values satisfy the following three equations

#### zn+2 − Z2 = zn+1w − Z1w zn+1 − Z1 = znw − Z2w zn − Z2 = zn-1w − Z1w

The ratio |zn+2 − Z2|/|zn − Z2| can be expressed in the form

#### [|zn+2 − Z2|/|(zn+1 − Z1)|][|(zn+1 − Z1)|//|zn − Z2|]]

Then zn+2 − Z2 can be replaced by its equal value of zn+1w − Z1w. In the second term of the product zn+1 − Z1 can be replaced by its equal value of znw − Z2w.

Thus in order for |zn+2 − Z2| to be less than |zn − Z2| it is necessary that

#### |[(zn+1w − Z1w)/(zn+1 − Z1)][(znw − Z2w)/(zn − Z2)]| < 1

At the boundary between the stable and unstable values of c the inequality becomes an equality so

#### |[(zn+1w − Z1w)/(zn+1 − Z1)][(znw − Z2w)/(zn − Z2)]| = 1

Now consider the limits as n→∞. Both terms go to 0/0 so l'Hospital's rule applies and hence

#### lim |[(zn+1w − Z1w)/(zn+1 − Z1)| = |wZ1w-1| and lim [(znw − Z2w)/(zn − Z2)]|= |wZ2w-1|

Thus the condition to be satisfied for a limit 2-cycle is

#### |w²(Z1Z2)w-1| = 1

The complex numbers whose absolute value is equal to 1 are those given by exp(iφ) for 0≤φ≤2π. Thus

## The Case of a Positive Integer Exponent

As stated previously, the values of Z1 and Z2 are solutions to the equation

#### Z + Zw − λ = 0

When w is a positive integer the above equation is a monic polynomial. The product of the roots of a monic polynomial is equal to the constant term in the equation. For w=2 then

#### Z1Z2 = −λ

Also for a monic polynomial the sum of the roots is equal to the coefficient of the term with the next to largest exponent. For w=2 this is 1 and for w>2 it is equal to 0. For w=2 then

For w=2

Thus

#### Z1(1-Z1) = −(1/4)exp(iφ) and therefore Z1² − Z1 − (1/4)exp(iφ) = 0

This equation can be solved for Z1 as a function of φ, say Z1(φ).

Since for this case

#### Z1² + c = Z2 = 1 − Z1and hence c = 1 − (Z1(φ) + Z1(φ)²)

(To be continued.)