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

z_n+1 = z_n.

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.,

z_n+1 = z_n³ + c
z* = z*³ + c
Subtraction gives
z_n+1-z* = z_n³ - z*³ = (z_n-z*)(z_n² + z_nz* + z*²)

Thus

|z_n+1-z*| will be less than |z_n-z*| if |z_n² + z_nz* + z*²|<1

For values of z_n 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^iφ
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
and
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)
or
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 z_n-z* satisfy the equations

z_n+2-z₂* = (z_n+1-z₁*)(z_n+1²+z_n+1z₁*+z₁*²)
and
z_n+1-z₁* = (z_n-z₂*)(z_n*²+z_n*z₂*+z₂*²)
hence
z_n+2-z₂* = (z_n-z₂*)(z_n+1²+z_n+1z₁*+z₁*²)(z_n*²+z_n*z₂*+z₂*²)

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

|(z_n+1²+z_n+1z₁*+z₁*²)(z_n*²+z_n*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^iφ

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^iφ

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

The expansion of the square on the left gives

(z₁*²)² + 2z₁*²(1/3)e^iφ + (1/9)e^i2φ = z₁*²(−1 + (1/3)e^iφ)

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

(z₁*²)² + 2z₁*²(1/3)e^iφ + (1/9)e^i2φ = z₁*²(−1 + (1/3)e^iφ)
(z₁*²)² + z₁*²(1+(1/3)e^iφ) + (1/9)e^i2φ = 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*³
and
c = z₂* − z₁*³

Multiplying these expressions gives

c² = z₁*z₂* −(z₁*⁴+z₂*⁴) + (z₁*z₂*)³
and hence
c² = (1/3)e^iφ −(z₁*⁴+z₂*⁴) + (1/27)e^i3φ.

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^iφ)² − (1/9)e^i2φ
= (1+(2/3)e^iφ)

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

c² = −1 − (1/3)e^iφ + (1/27)e^i3φ

The curve for c² is shown below:

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

(To be continued.)