In a previous question on MO I mentioned that I had convinced myself of the following:

When $f_p(z) = z - p(z)/p'(z)$ and $p$ is a complex polynomial, the Julia set, $J(f_p)$ is not continuously determined (in the sense of the Hausdorff metric) by the roots of $p$ if any of those roots is not simple.

I can think of a few different ways for a proof to go, but the main idea is the same in each one: A multiple root can be "separated" by an arbitrarily small perturbation to form a set of simple roots, and that causes a change in the dynamics.

I have worked out the details (I think) for one proof along those lines, but per the advice I received on meta I will not ask for anyone to read my work in order to answer my question. My question is:

Does this result appear in the literature (or is it a trivial application of something that appears in the literature)?

Alternatively, if the result is actually wrong then I would of course accept that as an answer.

Edit: I feel like the result should be in Paul Blanchard's The Dynamics of Newton's Method. Maybe for experts it is there and I'm just not knowledgeable enough to connect the dots. For example, Blanchard notes that the simple roots of the polynomial are superattracting fixed points while the multiple roots are merely attracting fixed points. He notes that the critical points are the simple roots and inflection points of the polynomial, and that the degree of the map $z\to z-p(z)/p'(z)$ is equal to the degree of $p$ only when the roots of $p$ are all simple.

The nature of fixed points changing, the set of critical points changing, and the degree of the map itself changing all hint that "something drastic" will happen to the global dynamics of the map when a multiple root is separated (or alternatively, when a set of simple roots collide to form a multiple root), but is it obvious that the Julia set could never remain stable in this case?


1 Answer 1


I think for multiple roots this is clear. At a multiple root $z_0$, you have an attracting fixed point. Take a small disc $D$ around this root; then (as you note) for a suitable perturbation $\tilde{p}$ of your polynomial, there will be at least two different roots well inside the disc $D$. These are (super-)attracting fixed points of the corresponding Newton's method, and hence the boundary of their basins (which is in the Julia set) must also have some points near $z_0$. This shows the Julia set does not depend continuously.

I do not know the literature on Newton's method sufficiently well to know whether this has been explicitly observed anywhere, but it seems essentially self-evident (assuming enough background knowledge in complex dynamics, of course).

Also note that, at a multiple root, the degree of the corresponding Newton's method drops. So these are really places where you approach the boundary of the "natural" parameter space (of rational functions of the expected degree). There has been quite a lot of work on possible limits of the dynamics as one approaches this boundary in general, although I can't say whether any of it is directly relevant.

  • $\begingroup$ Thank you for the answer! Your comments make my edit (which I posted before I realized you had already answered) mostly irrelevant. I suspected this might be the outcome, as I mentioned in the early question: Whether or not the result appears explicitly in the literature it seems that it's an obvious observation for experts. Your last paragraph is interesting, and I would love to read more about the dynamics near these boundaries of the parameter space, if you have a moment to point me to a good starting point for the "quite a lot of work" you mentioned. $\endgroup$ Commented Nov 16, 2015 at 7:28
  • $\begingroup$ Laura deMarco's and Jan Kiwi's work might be relevant, if memory serves. $\endgroup$ Commented Nov 19, 2015 at 1:04

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.