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Monday, 26 May 2008 23:47
Quadratic maps 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.

Polynomial maps

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
then
    (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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