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the Fully General Mandelbrot Set |
Fully general means that the Mandelbrot sets are generated from iteration schemes of the form
where c, z and w are complex numbers. (The integers and the real numbers are included within the set of complex numbers.) A symmetry is such that when a value c is transformed, such as by rotation, the transformed value has the same characteristics as does c with respect to inclusion in a set,
A rotation in the complex plane is equivalent to multiplication by a complex number, say t. Thus c→tc, z→tz etc. Let {z_{n}} be the iteration sequence for c. Consider the transformation of c and z_{n} for some n. Because multiplication of complex numbers is associative and commutative
If t^{w-1}=1 then
Thus if t^{w-1}=1 then the sequence associated with tc is just {tz_{n}}. If |t|=1 then c and tc would have the same sequence of absolute values and thus the same classification as c with respect to inclusion or exclusion in a set or other aspects such as escape time which is used in coloring the Mandelbrot set images.
If w is just a positive integer m then t would have to be an (m-1) root of unity and the Mandelbrot set has the symmetries involving rotations by integral multiples of an angle equal to 2π/(m-1).
Consider c and its associated iteration sequence {z_{n}}. Now consider the complex conjugates of c and the some member of the sequence z_{n}. For typographic convenience these will be denoted as c and z_{n}. Let z_{n}=re^{iφ}. Then the conjugate of z_{n} is z_{n}=re^{−iφ}. Therefore
This means that c generates the sequence made up of the conjugates of {z_{n}}. Since |z_{n}=||z_{n}| the conjugate sequence will have the same characteristics as {z_{n}} and c will be classified in the category as c. Thus any Mandelbrot set is symmetric under conjugation.
The condition that the iteration sequence for c be mapped into the transformed sequence when c is transformed to tc then z is mapped into tz. This means that
Without considering the matter of the absolute values of the transformed sequence let us find the values of t corresponding to the exponent in the iteration scheme. Let w=u+iv and express t in polar form as re^{iφ}. Then
where k is an integer. This requires that
This is simply two linear equations in two unknowns and the solution is
For k=0 the only solution is φ=0 and ln(r)=1 and hence r=1. This is the identity transformation, a trivial form of symmetry.
For k=1 and u=m, an integer, and v=0 the solution is ln(r)=0 (r=1) and φ=2π/(m-1). The Mandelbrot set for an exponent equal to an integer m is symmetric under rotations of an angle equal to a multiple of 2π/(m-1).
Without the condition that |t|=1 the symmetries are not strictly rotational, but there is a symmetry which is in the nature of a spiral. When c and {z_{n}} are transformed into tc and {tz_{n}} the convergence nature of the sequence is preserved. That is to say, if {z_{n}} converges to a fixed point z* then {tz_{n}} converges to a fixed point tz*. If {z_{n}} converges to a two-cycle then {tz_{n}} converges to the corresponding two-cycle. Thus tc will have the same classification as c. The only difference is that tc is located at a different distance from the origin than is c.
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