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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
Such a limit point z* satisfies the equation
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.,
Thus
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
The values of c which give those limit points are simply
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:
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
From these equations defining z1* and z2* it follows that
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
Therefore if |zn+2-z2*| is to be less than |zn-z2*| it must be that
For values very close to a cycle pair this reduces to:
For the boundary of the stable set equality prevails; i.e.,
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
where r=1/(2³√2).
The values sought also satisfy the condition
If this equation is multiplied by z1*³ the result can be put into the form
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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