Thayer Watkins
Silicon Valley
& Tornado Alley

The Structure of the Quartic Mandelbrot Set

The Quartic Mandelbrot set is the set of complex numbers c such that the iteration scheme

zn+1 = zn4 + c

is bounded when starting from the point z0=0. A significant subset of this set consists of those values of c such that the iteration scheme approaches limits for which

zn+1 = zn.

Such a limit point z* satisfies the equation

z* = z*4 +c

For any c there is a limit point z*; i.e., such that if z0=z* the iteration will remain at z* forever.

The crucial question is what are the limit points that are stable so that the iteration starting from z0=0 will approach them.

Consider the deviations of the iteration values from the corresponding limit point; i.e.,

zn+1 = zn4 + c
z* = z*4 + c
Subtraction gives
zn+1-z* = zn4 - z*4
= (zn2 - z*2)(zn2 + z*2)
= (zn-z*)(zn + z*)(zn2 + z*2)


|zn+1-z*| will be less than |zn-z*| if |(zn + z*)(zn2 + z*2)|<1

For values of zn close to z* this reduces to 4|z*3|<1. The boundary between the stable and unstable limit points is given by |z*|=(1/4)1/3. Such limit points are given by the equation

z* = (1/4)1/3*e
for 0≤φ≤2π

The values of c which give those limit points are simply

c = z* - z*4
c = (1/4)1/3cos(φ)−(1/4)4/3cos(4φ)
+ i[(1/4)1/3sin(φ)−(1/4)4/3sin(4φ)]

This equation is a parametric equation for the set of c values. It shows how the points on the circle of radius (1/4)1/3 in the z* space map into the c space.

The plot below shows the full set of c values:

Limiting Two-Cycles

An iteration may approach a limit cycle rather than a limit point. For a two-period cycle of z1* and z2* the values would have to satisfy the equations

z2* = z1*4 + c
z1* = z2*4 + c

From these equations defining z1* and z2* it follows that

z2*−z1* = (z2*−z1*)(z2*+z1*)(z2*2+z1*2)
and with division by (z1*−z2*)
which is valid if z2*≠z1*
−1 = (z2*+z1*)(z2*2+z1*2)

This is a condition satisfied by all two-cycle values. It will be used later.

First a condition for stability must be derived. The deviations zn-z* satisfy the equations

zn+2-z2* = (zn+1-z1*)(zn+1+z1*)(zn+12+z1*2)
zn+1-z1* = (zn-z2*)(zn+z2*)(zn2+z2*2)
zn+2-z2* = (zn-z2*)(zn+z2*)(zn2+z2*2)(zn+1+z1*)(zn+12+z1*2)

Therefore if |zn+2-z2*| is to be less than |zn-z2*| it must be that

|(zn+z2*)(zn2+z2*2)(zn+1+z1*)(zn+12+z1*2)| < 1

For values very close to a cycle pair this reduces to:

|(2z2*)(2z2*2)(2z1*)(2z1*2)| < 1
or, equivalently
16|z2*z1*|³ < 1
|z2*z1*| < 1/(2³√2)>;

For the boundary of the stable set equality prevails; i.e.,

|z2*z1*| = 1/(2³√2)

This means the values of z1* and z2* being sought must lie on circle in the complex plane whose center is the origin and whose radius is 1/(2³√2). This is alternatively expressed as

z2*z1* = re

where r=1/(2³√2).

The values sought also satisfy the condition

−1 = (z2*+z1*)(z2*2+z1*2)

If this equation is multiplied by z1*³ the result can be put into the form

z1*3 = (z1*z2* + z1*3)((z1*z2*)2 + z1*4)
and after substitution of the expression for z1*z2*
z1*3 = (re + z1*2)(r2ei2φ + z1*4)

where r=1/(2³√2). This is a sixth order polynomial that determines the values of z1 that correspond to a stable two-cycle. The polynomial for z2 is the same as the one for z1. The sixth order polynomial will have six roots, three sets of pairs.

(To be continued.)

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