I have several questions relating to Nth root fractals, to which I have not been able to find answers. Some background describing what I call Nth root fractals (my terminology, not a universally accepted one).

Nth root fractals are generated via use of the Newton–Raphson method for determining all the roots of a complex number. For those not familiar with such fractals, consider the followingPick a point in the complex plane & use it as the initial value in a Newton–Raphson method for determining the cube root of one.


If the root found is 1, color the initial point red.


If the root found is [-½ + ½ *SquareRoot(3)], color the initial point blue.


If the root found is [-½ - ½ *SquareRoot(3)], color the initial point yellow.


Pick another point as an initial value & repeat the above.If the above is done for every pixal in a monitor display, the result will be a cube root fractal. Similarly, 4th, 5th, et cetera root fractals can be generated.

Examples of such fractals can be found by searching the web.

The cube root fractal has a counter-intuitive property: All the boundaries between colored regions are triple point boundaries. Consider a circle cut into 3 equal pie-shaped sections. There are three linear boundaries & one triple point boundary. The center of the circle is the triple point boundary.

For the cube root fractal, there are no linear boundaries between colored regions. There are only triple point boundaries. Some questions for which I have not been able to find answers.If a triple boundary point happened to be chosen as an initial value, would the Newton-Raphson method fail to converge to a root? It seems possible, but not necessary that each iteration would find another triple point.


Would a 4th root fractal only have quadruple boundary points? Would a 5th root fractal only have quintuple boundary points?


Perhaps odd root factals would not have any linear boundaries, while even root fractals would have linear boundaries.


If there are linear boundaries for some value of N, could there also be multiple point boundaries? The answer to this seems to be yes, but I would not bet a lot of money on it.Does anybody have any thoughts on the above?

The easiest point on the boundries to find are z=zero (which is unfortunate) and z=- \frac{1}{\sqrt[3]{2}}. The behavior around the second of those shows the recursion takes a long time to settle down when near it.

Table[Length[FixedPointList[(# - (#^3-1)/(3#^2))&, N[-(2)^(-1/3) + 0.0000001 x], 200]],{x,-5,5}]

But the recursion formula maps points near - \frac{1}{\sqrt[3]{2}} to points near 0, which is the natural Grand Central Station of this map, mapping points to far distant points. Since the recursion: z - \frac{f(z)}{f'(z)} = z - \frac{x^3 - 1}{3 x^2} = \frac{1}{3}(2 x + \frac{1}{x^2}) maps z = r e^{i\theta} to w = \frac{1}{3}(2 r e^{i\theta} + \frac{e^{- 2 i\theta}}{r^2}) which strongly changes the argument of the complex number when z is small and not on the real axis.