I was looking at the Newton Fractal in my fractal viewer (an earlier version is given here on the website—the most recent is fancier will soon go up). I was struck by its symmetry and thought to investigate it a little bit. I have collected together here some fascinating little facts about it. First we must give a brief definition of the fractal. We consider the function on $z$ such that:

$z^{\nu }=1;z\in ℂ;\nu \in \mathbb{N};\nu >2$

An illustration of the cubic Newton fractal near the origin, with each colour representing one of the three roots. The three main attractors at the edge of the figure are the three first-order iterates.

We would normally think of the Newton fractal as the boundary of the basins of attraction under the standard Newton iterations, but it is more useful to actually look at reverse iterations of Newton’s method, i.e. take the set to be the union of all the iterates, where the $n+1$^th iterates are the points which map to the $n$^th iterates under this transformation:

$(\nu -1)z^{\nu }-\nu z_0{z}^{\nu -1}+1=0;z,z_0\in ℂ$

(i.e. given a starting value, this equation determines that point’s parents). By inspection, we confirm our intuition that the fractal must have order $\nu$ rotational symmetry, and reflection under complex conjugation. Now, on looking at the fractal, I was certainly expecting to find that each ‘loop’ is a scaled-up version of the previous one. Actually, the relationship is not linear. As we let the number of (reverse) iterations tend to infinity,

$z_0=\frac{\nu -1}{\nu }z+O\left(z^{1-\nu }\right).$

There is indeed a tendancy towards a linear relationship for the regions on the complex plane far from the origin, but it is non-linear locally (i.e., on the large scale, the fractal is invariant under $z\mapsto \frac{\nu }{\nu -1}z$).

Now, the result I was really after when after looking at the image was to find which regions are fractal and which are not (i.e. what is supremum the arguments of the members of the fractal in each branch). We can solve this by looking at the initial equation, this time considering $n^0$ tending to infinity. We obtain that

$z=\frac{1}{\sqrt[\nu -1]{\nu z_0}}+O\left({z_0}^{-\frac{2\nu -1}{{(\nu -1)}^2}}\right)$

This is very useful, since we know that the scaling is linear for large numbers of iterations, so the ‘loops’ are self-similar. Each starting value here has $\nu$-fold rotational symmetry locally, so we can consider regular axes about point. Solving this and taking the argument, we find that all the points approaching the origin must do so on lines with argument $\frac{\pi (1+2n)}{\nu (\nu -1)}$.

In our figure above of the symmetric cubic, the points on the boundary are all contained within segments of angle $\frac{\pi }{3}$. Interesting enough, (and no justification is given here), the rotational symmetry on a local scale is always an integer multiple of the global asymptotic symmetry, even for the case of arbitrary polynomials, which have very strangely shaped attraction basins.

References

1. Drexler, M., Sobey, I. J. and Bracher, C. Fracal characteristics of Newton’s method on polynomials. Oxford. 1996