is bounded when starting from the point z₀=0. A significant subset of the Mandelbrot 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*² +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_n² + c
z* = z*² + c
Subtraction gives
z_n+1-z* = z_n² - z*² = (z_n-z*)(z_n+z*)

Thus

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

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

z* = ½e^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*² = = ½e^iφ(1-½e^iφ)
c = ½cos(φ)−¼cos(2φ) + i[½sin(φ)−¼sin(2φ)]

This equation is a parametric equation for the set of c values. It shows how the points on the circle of radius 1/2 in the z* space map into the c space. For example, z*=½ maps into c=¼, z*=−½ maps into c=−3/4, z*=i½ maps into c=½+i¼ and z*=−i½ maps into c=−½+i¼.

The plot below shows the full set of c values.

And there is the familiar cardioid shape that bounds the main body of the mandelbrot set, as seen below.

The Limiting 2-Cycles

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*² + c
and
z₁* = z₂*² + c

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

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

The deviations from the limit values satisfy the equations

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

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

|(z_n+z₂*)(z_n+1+z₁*)| < 1

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

|(2z₂*)(2z₁*)| < 1
or, equivalently
|z₂*z₁*| < ¼

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

|z₂*z₁*| = ¼

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

|(1+z₁*)z₁*| = ¼

The allowable values of z₁ are solutions to the quadratic equation

z(1+z) = ¼e^iφ
or, equivalently
z² + z - ¼e^iφ = 0

There are two solutions to this quadratic equation; one corresponding to z₁* and the other to z₂*.

It is not the solutions for z₁* or z₂* that are of most interest; it is the set of values of c corresponding to these solutions.

The defining equation

z₂* = z₁*² + c
combined with the result that
z₂* = −(1 + z₁*)
gives
−(1 + z₁*) = z₁*² + c.
Thus
c = −(1 + z₁* + z₁*²)

It is known that

z₁* + z₁*² = ¼e^iφ
therefore
c = −(1 + ¼e^iφ)

This is the locus of a circle of radius ¼, the center of which is shifted one unit to the left on the real axis. This is shown below in a diagram with the values of c corresponding to limit points.

The Limiting 3-Cycles

The analysis for 3-cycles is more difficult but by same procedures as in the previous it can be established that the boundaries between the stable and unstable cycles are circles of radius 1/8. In general the boundaries between stable and unstable m-cycles are circles of radius 1/2^m. The diagram below shows the placement of circles having radii of 1/8, 1/16 and 1/32.

There are also easily discernible circles in the Mandelbrot set of radius 1/64.

The limit points z₁*, z₂* and z₃* satisfy the equations

z₃* = z₂*² + c
z₂* = z₁*² + c
z₁* = z₃*² + c

The substitution of the equation for z₁* into the equation for z₂*² and the substitution of that equation into the equation for z₃* produces a polynomial equation of degree 8.

z*₃ = ((z*₃²+c)² + c)² + c
= ((z*₃⁴+2z*₃²c + c² + c)² + c
= (z*₃⁴+2z*₃²c)² + 2(z*₃⁴+2z*₃²c)(c² + c) + (c² + c)² + c

There will be eight solutions to this equation and two pairs of triples which satisfy the conditions for limiting 3-cycles for each given value of c. The product of the eight solutions will be equal to the constant term divided by the coefficient of the highest power of z*. This means that the product of the solutions is equal to (c² + c)² + c.

In order to establish the conditions for stability of the 3-cycles consider the deviations between z_n+3 and z₃*, z_n+2 and z₂*, z_n+1 and z₁*, and z_n and z₃*. These deviations satisfy the equations

z_n+3 − z₃* = (z_n+2 − z₂*)(z_n+2 + z₂*)
z_n+2 − z₂* = (z_n+1 − z₁*)(z_n+1 + z₁*)
z_n+1 − z₁* = (z_n − z₃*)(z_n + z₃*)

Successive substitutions establish that if |z_n+3 − z₃*| is to be less than |z_n − z₃*| it is necessary that

|(z_n+2 + z₂*)(z_n+1 + z₁*)(z_n + z*)| < 1

For values very close to the limiting 3-cycles this reduces to

|(2z₂*)(2z₁*)(2z₃*)| < 1
or, equivalently
2³|z₁*z₂*z₃*| < 1

and for the boundary between the stable and unstable 3-cycles

|z₁*z₂*z₃*| = 1/8

From the defining equations for the limit points it follows that

z₃* − z₂* = (z₂* − z₁*)(z₂* + z₁*)
z₂* − z₁* = (z₁* − z₃*)(z₁* + z₃*)
z₃* − z₁* = (z₂* − z₃*)(z₂* + z₃*)

and therefore

z₃* − z₁* = (z₁* − z₃*)(z₃* + z₂*)(z₂* + z₁*)(z₁* + z₃*)
and with division by z₃* − z₁*
1 = −(z₃* + z₂*)(z₂* + z₁*)(z₁* + z₃*)

(To be continued.)