San José State University
Thayer Watkins
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The Structure of the Cubic Mandelbrot Set

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

zn+1 = zn3 + 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*³ +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 = zn³ + c
z* = z*³ + c
Subtraction gives
zn+1-z* = zn³ - z*³ = (zn-z*)(zn² + znz* + z*²)


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

For values of zn close to z* this reduces to 3|z*²|<1. The boundary between the stable and unstable limit points is given by |z*|=1/√3. Such limit points are given by the equation

z* = (1/√3)e
for 0≤φ≤2π

The question is what are the values of c which give those limit points. Those values of c are simply

c = z* - z*³ = z*(1 − z*²)
c = (1/√3)cos(φ)−(1/3√3)cos(3φ)
+ i[(1/√3)sin(φ)−(1/3√3)sin(3φ)]

This equation is a parametric equation for the set of c values. It shows how the points on the circle of radius (1/√3) in the z* space map into the c space. For example, z*=1/√3 maps into c=2/(3√3), and z*=−1/√3 maps into c=−2/(3√3). On the other hand z*=±i/√3 map into ±i4/(3√3), respectively.

The plot below shows the full set of c values.

This is the double lobed shape that bounds the main body of the cubic mandelbrot set.

The 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*³ + c
z1* = z2*³ + c

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

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

When z1*z2* is added to both sides of the above equation the result can be put into the form

(z1*+z2*)² = z1*z2* - 1

The deviations zn-z* satisfy the equations

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

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

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

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

|(3z2*²)(3z1*²)| < 1
or, equivalently
|z2*z1*|² < (1/9)
and hence
|z2*z1*| < (1/3)

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

|z2*z1*| = (1/3)

This means that z2*z1* is on a circle in the complex plane of radius (1/3). This means that

z2*z1* = (1/3)e

It was previously determined that (z1*+z2*)² = z1*z2* - 1 therefore the condition for the boundary of stability reduces to:

(z2*+z1*)² = −1 + (1/3)e

This says that (z2*+z1*)² is on a circle of radius (1/3) centered at (−1,0).

It the above equation is multiplied by z1*² the result reduces as follows:

z1*²(z2*+z1*)² = z1*²(−1+ (1/3)e)
(z1*² +z1*z2*)² = z1*²(−1+ (1/3)e)
(z1*² + (1/3)e)² = z1*²(−1 + (1/3)e)

The expansion of the square on the left gives

(z1*²)² + 2z1*²(1/3)e + (1/9)ei2φ = z1*²(−1 + (1/3)e)

This reduces to a quadratic equation in z1*²; i.e.,

(z1*²)² + 2z1*²(1/3)e + (1/9)ei2φ = z1*²(−1 + (1/3)e)
(z1*²)² + z1*²(1+(1/3)e) + (1/9)ei2φ = 0

This quadratic can be solved for z1* (and z2*), but their values are not of interest. It is the values of c which they correspond to that are of interest.

From the original equations

c = z1* − z2
c = z2* − z1

Multiplying these expressions gives

c² = z1*z2* −(z1*4+z2*4) + (z1*z2*)³
and hence
c² = (1/3)e −(z1*4+z2*4) + (1/27)ei3φ.

The term (z1*4+z2*4) is the sum of the squares of the roots of the quadratic equation. It can be the shown that the sum of the squares of the roots of the quadratic equation ax²+bx+c=0 is equal to (b/a)²−(c/a). This means that

(z1*4+z2*4) = (1+(1/3)e)² − (1/9)ei2φ
= (1+(2/3)e)

When this expression is substituted into the equation for c² the result is:

c² = −1 − (1/3)e + (1/27)ei3φ

The curve for c² is shown below:

But it is the c curve which is important. That is shown below:

(To be continued.)

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