For iterations of a quadratic map zk+1 =
azk2 + bzk + c
by a linear transformation t = Az + B (which includes two translations by
Re(B), Im(B), scaling by |A| and rotation by Arg(A) )
one can exclude two parameters from a, b, c and reduce the map
to iterations of tk+1 = tk2 + d
(see the proof in Appendix below).
Quadratic maps f(z) = P2(z) have the only
critical point (that is a point where
P'2(z) = dP2(z)/dz = 0) and
it is in z = 0 for the map f(z) = z2 + c.
The critical points are important due to Fatou theorem: every attracting
cycle for a polynomial or rational function attracts at least one critical
point. Thus, quadratic maps may have the only attracting cycle and
testing the critical point shows if there is any attracting cycle.
For a cubic map f(z) = P3(z) from 4 parameters one can exclude 2
so it remains two complex parameters. Usually maps f(z) = z3 -
3a2z + b are used with two critical points at z = +-a.
Unfortunately there are two complex parameters (4 real numbers), therefore we
can plot only two dimentional cross-sections of the 4D space. E.g. we can
watch iterations of real cubic maps.
The simplest polynomial maps are f(z) = zN + c . They have
degenerate critical point at z = 0.
Here you see the z3, z4 Mandelbrot
sets and corresponding Douady rabbits.
Bifurcations map of the x4 Mandelbrot set
I've "measured by scale" a = 1.69 and
d = 7.34. Unfortunately it isn't the
d N ~ 2N (24 = 16) rule.
Appendix Let us prove equivalence of the maps z = az2 +
bz + c and z = z2 + c directly
az = a(az2 +bz + c) = (az)2 +b(az) + ac =
[(az)2 +2(az) b/2 + (b/2)2] -
(b/2)2 + ac = (az + b/2)2 -(b/2)2 + ac
(az + b/2) =
(az + b/2)2 -(b/2)2 + ac +b/2 or
t = t2+ d, t = az + b/2,
d = - (b/2)2 + ac +b/2.
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updated 23 Sept 2002