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*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.,

z_n+1 = z_n⁴ + c
z* = z*⁴ + c
Subtraction gives
z_n+1-z* = z_n⁴ - z*⁴
= (z_n² - z*²)(z_n² + z*²)
= (z_n-z*)(z_n + z*)(z_n² + z*²)

Thus

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

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

z_n+2-z₂* = (z_n+1-z₁*)(z_n+1+z₁*)(z_n+1²+z₁*²)
and
z_n+1-z₁* = (z_n-z₂*)(z_n+z₂*)(z_n²+z₂*²)
hence
z_n+2-z₂* = (z_n-z₂*)(z_n+z₂*)(z_n²+z₂*²)(z_n+1+z₁*)(z_n+1²+z₁*²)

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

|(z_n+z₂*)(z_n²+z₂*²)(z_n+1+z₁*)(z_n+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
and
|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^iφ

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^iφ + z₁*²)(r²e^i2φ + 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.)