9
$\begingroup$

Consider the function family given by $f_\lambda(z) = z - p_\lambda(z)/p_\lambda'(z)$ where $p_\lambda(z) = (z^2 - 1)(z - \lambda)$. Every attracting cycle and every rational neutral cycle of $f_\lambda$ attracts the one critical point of $f_\lambda$, which is $\lambda/3$ (see this other MathOverflow question and Alexandre Eremenko's answer for context).

If the sequence of iterates, $f^n_\lambda(\lambda/3)$, converges to a fixed point of $f_\lambda$ then the fixed point is one of the roots of $p_\lambda$, that is $1$, $-1$, or $\lambda$. If we mark the points $\lambda$ in $\mathbb{C}$ such that $f^n_\lambda(\lambda/3)$ converges to each of these fixed points then the result is an image that is reminiscent of a Newton basin fractal. Let $K$ be the set of values $\lambda\in\mathbb{C}$ such that $f^n_\lambda(\lambda/3)$ converges to a fixed point of $f_\lambda$. My question is:

Is the border, $\partial K$, a Julia set, and if so then for what function family $g$ does $J(g) = \partial K$?

I suspect that the answer is "yes" and that $g$ is a family of iterates of a rational function. In the attached images, the very bright green points are the ones nearest to the border $\partial K$.

A note about the little Mandelbrot sets visible in two of the attached images: In these images the white points indicate parameters $\lambda$ such that $f^n_\lambda(\lambda/3)$ converges to a rational neutral cycle of period greater than $1$. The white points indicate the bifurcation locus of $f_\lambda$ ($J(f_\lambda)$ is not continuously determined, in the sense of the Hausdorff metric, by the parameter at these points). The white points appear to be composed of many scaled and rotated copies of the boundary of the Mandelbrot set and I believe they are contained by $\partial K$. Perhaps that is a useful thing to know when searching for $g$.

Improve this question
$\endgroup$
7
  • $\begingroup$ It is very unlikely that the set defined in the parameter plane $\lambda$ is the Julia set of anything. Mandelbrot set is not a Julia set of anything. Why are you asking such a strange question? $\endgroup$
    – Alexandre Eremenko
    Commented Dec 3, 2012 at 2:56
  • $\begingroup$ @Alexandre Eremenko: I'm mostly interested in the bifurcation locus of $f_\lambda$ (and, in a more general way, in scenarios in which the Julia set of the family of Newton-Raphson iterates of a polynomial is not continuously determined by the roots of the polynomial). In this case the bifurcation locus (the boundaries of the little Mandelbrot sets) is contained by something that reminds me of a Julia set I recognize (a Newton basin fractal) so I thought maybe it would actually turn out to be the Julia set of some related family, which would give me some insights into the bifurcation locus. $\endgroup$
    – Aaron Golden
    Commented Dec 3, 2012 at 3:21
  • $\begingroup$ I agree with Eremenko; this is a Mandelbrot-style construction. Your pictures are quite interesting: they really DO locally look like Newton fractals, analogously to julia/mandelbrot correlation mentioned below. You can try to "fit" your data to a rational function, or at least estimate its possibly degree by counting possible fixed points in your image, which should be dark centers (dark centers also corresponds to attracting periodic points). Poles should be in regions which are black, as in one of the middle pictures. But your initial picture really do not look like a rational julia fractal $\endgroup$
    – Per Alexandersson
    Commented Dec 3, 2012 at 7:23
  • $\begingroup$ Parts of pictures in the parameter plane really look "somewhat like" the parts of the pictures in dynamical plane, and there are theorems of this sort. However, it is extremelly unlikely that that boundaries of some regions in parameter plane exactly coincide with any Julia sets. (This can happen, see my reply to Rodrigo's answer, but very rarely, and I think such coincidences can have no significance). $\endgroup$
    – Alexandre Eremenko
    Commented Dec 3, 2012 at 14:45
  • $\begingroup$ What do you mean by "a function family"? Certainly it is not true that your set is the Julia set of a rational function. (If you care enough, you should be able to fashion a formal proof, e.g. from the fact that the complement contains some domains bounded by analytic curves, in the small Mandelbrot copies.) On the other hand, a one-dimensional bifurcation locus can usually be described as the locus of non-normality of a suitable family of analytic functions, formed by taking the iterates of the free critical point. $\endgroup$
    – Lasse Rempe
    Commented Dec 3, 2012 at 22:47

2 Answers 2

Reset to default
10
$\begingroup$

Your $f_\lambda$ implements Newton's method for $p_\lambda$. As you said, the zeros $\{ 1,-1,\lambda \}$ of $p_\lambda$ are fixed points of $f_\lambda$, but even more, they are superattracting fixed points; that is, $f'_\lambda(1) = f'_\lambda(-1) = f'_\lambda(\lambda) = 0$.

If you choose a fixed $\lambda$ and iterate different points to see whether they are attracted to $1$, $-1$, or $\lambda$, coloring them accordingly, you will be plotting the Julia set of $f_\lambda$; this looks like the Newton fractals that you mention. Indeed, the first fractal in the Wikipedia page you linked to is the Julia set of the Newton method function associated to $z \mapsto z^3-1$.

Instead, what you are doing is plotting the results of iterating one particular value for many different parameters $\lambda$. This is not a dynamical procedure because you use a different $f_\lambda$ every time. This makes it very unlikely that your pictures are Julia sets. However, it can be shown that locally, near the critical value $f_{\lambda}(\lambda/3)$, the Julia set of $f_\lambda$ looks similar to this parametric picture, and this is why your pictures look like Newton fractals. For a proof of the similarity statement in the quadratic family, see Tan Lei's paper: Similarity between the Mandelbrot set and Julia sets Comm. Math. Phys. Volume 134, Number 3 (1990), 587-617.

The original Mandelbrot set is constructed by iterating 0 with different parameters $c$ in the family $z \mapsto z^2+c$; a parametric construction just as above. The appearance of little Mandelbrots sets in your pictures signals a region of parameters $\lambda$ where an iterate of $f_\lambda$ has some (local) behavior that is conjugate to a quadratic polynomial. If you pick $\lambda$ in one of those small Mandelbrots and draw the Julia set of $f_\lambda$, you will find small regions that looks like the Julia set of that quadratic polynomial!

Added: Eremenko is right. Just because a picture is not generated by a dynamical process, it doesn't necessarily follow that it can't be a Julia set. A heuristic argument to support that conclusion is as follows: Even though Julia sets of rational functions have structure at all scales, this structure can be described by a sequence of finite combinatorial objects (for quadratic maps, we have kneading sequences and external angles). Once a picture contains something as complicated as a Mandelbrot set, it is possible to find infinitely many different such sequences of combinatorial data. It is in this sense that I say it is very unlikely that a parametric fractal like yours can be generated as a Julia set.

Improve this answer
$\endgroup$
4
  • 1
    $\begingroup$ Your explanation is correct, except the sentence which begins with "Therefore...". How do you really know that these pictures (or parts of them) are not the Julia sets of some other function? For example, the boundary of the Mandelbrot set of $\lambda z(1-z)$ contains a circle. And this surcle is the Julia set of $z^2$. Of course this is very unlikely, and probably can be proved in this example, if anyone cares about such...strange, to say the least problem. $\endgroup$
    – Alexandre Eremenko
    Commented Dec 3, 2012 at 14:39
  • 1
    $\begingroup$ Thank you for the explanation, Rodrigo, but I am confused by the statement, "locally, near $\lambda$, the Julia set of $f_\lambda$ looks similar to this parametric picture." Since $\lambda$ is a (super)attracting fixed point of $f_\lambda$, the Julia set of $f_\lambda$ is bound away from $\lambda$, right? What does it mean for the $J(f_\lambda)$ to look like anything near $\lambda$? Did you mean near a different point? If you can point me to a reference that explains this visual correspondence between the Julia sets and the parametric picture I would greatly appreciate it. $\endgroup$
    – Aaron Golden
    Commented Dec 3, 2012 at 18:54
  • $\begingroup$ @Aaron: I corrected the answer. The similarity between the dynamical and parameter pictures exists near the critical point (the one iterated in the parametric process). My mistake stems from thinking in terms of the quadratic family, where the normalization $f_c:z \mapsto z^2+c$ is chosen so that the critical point of $f_c$ is precisely $c$ itself. $\endgroup$
    – Rodrigo A. Pérez
    Commented Dec 4, 2012 at 0:01
  • $\begingroup$ @Rodrigo: I think you mean that the similarity will occur near the critical value, not the critical point? $\endgroup$
    – Lasse Rempe
    Commented Dec 4, 2012 at 9:10
5
$\begingroup$

The set in question is the bifurcation locus of the family $f_{\lambda}$. It is hence the set of non-normality of the family $$\bigl(\lambda\mapsto f_{\lambda}^n(\lambda/3)\bigr)_{n\in\mathbb{N}};$$

see Theorem 4.2 in McMullen's book "Complex Dynamics and Renormalization".

As has been pointed out, you cannot expect the set to exactly coincide with the Julia set of a rational function. It should be possible to prove this formally. Indeed, your set has complementary components bounded by analytic curves, or regions bounded by curves analytic except at a single cusp. Such curves cannot bound Fatou components of a rational map, unless the map is a Blaschke product and the Julia set is itself a circle.

(I cannot seem to find a reference for this fact right now. However, the boundary of a Siegel disk may be smooth, but cannot be analytic anywhere; otherwise, the conjugacy to a rotation would extend beyond the boundary. On the other hand, boundaries of attracting basins are even known to have Hausdorff dimension strictly greater than one; see Przytycki, "On hyperbolic Hausdorff dimension of the boundary of a basin of attraction for a holomorphic map and of quasirepellers". Showing that the boundary cannot be analytic is much easier, both for attracting and parabolic basins.)

Improve this answer
$\endgroup$
3
  • $\begingroup$ Thank you for the reference, Lasse. This may be a silly question, but how do you know that $\partial K$ is the bifurcation locus of the family $f_\lambda$? Is it sufficient to know that the limit of the iterates $f^n_\lambda(\lambda/3)$ changes as $\lambda$ crosses between connected components of $K$? I ask because I started out looking for $\lambda$ such that $f_\lambda$ had neutral cycles of period greater than one, thinking that the bifurcation locus could include only those parameters. In the images, those parameters are indicated in white, the boundaries of the little Mandelbrot sets. $\endgroup$
    – Aaron Golden
    Commented Dec 7, 2012 at 20:43
  • $\begingroup$ @Aaron, the fact that $\partial K$ is contained in the bifurcation locus is clear (since the family in question cannot be normal near a point of the boundary). The other direction follows from Montel's theorem. I am not sure what you mean about the neutral cycles. There are many parameters in the bifurcation locus without neutral cycles (e.g. parameters for which the free critical point is eventually mapped to a repelling periodic cycle). $\endgroup$
    – Lasse Rempe
    Commented Dec 9, 2012 at 23:12
  • 1
    $\begingroup$ @Aaron - do you have any further questions on the topic, or might you consider accepting an answer? $\endgroup$
    – Lasse Rempe
    Commented Dec 26, 2012 at 17:47

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .