Roots of iterations of polynomials

Question

Let $f \in \Bbb Q[X]$ a polynomial, and let denote by $f^n$ the composition $\underbrace{f \circ \cdots \circ f}_{n \text{ times }}$.

Let $R(f^n) \subset \Bbb C$ the roots of $f^n$. I'm interested in knowing how $R(f^n)$ behaves as $n \to \infty$.

Here is an example :

• $g(x) = 0.5 x^5-0.5x-1$, with $n=1,2,3$

Here are some questions :

1. Is it true that $R(g^n)$ is "uniformly bounded", i.e. there is a ball of radius $r>0$ that contains all the $R(g^n)$ ? This is clearly false for the polynomials $X-a, (a≠0)$. But it seems that this could be true for the $g(x) = 0.5 x^5-0.5x-1$ (and also for many other examples I've tried).

   I think that the average of the roots of $f^n$ is constant if $\deg(f) ≥ 2$, but I don't know if this helps.

2. [soft question] Looking at the "fractal" pattern for $g^3$ (see above ; again this happens for other polynomials), I have somehow the intuition that $R(g^n)$ "converges" to some set $R \subset \Bbb C$.

   How could I formalize this idea, and then possibly prove that my intuition can be turned into a "theorem" ?

   In order to formalize this idea, my attempt was to consider a collection $(x_{m,k})_{k≥1}$ of Cauchy sequences such that $\bigcup_{n≥1} R(g^n) = \bigcup_{m≥1} \{x_{m,k} \mid k ≥ 1\}$ and $x_{m,k} \in R(g^k)$. There should be better (and more correct) ways to formalize my intuition…

Thank you for your comments !

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I did the pictures with Mathematica. I tried to see what happens for $g^4$, but it was completely wrong (because $g^4$ has degree $625$, which is pretty large, I presume). I hope that there is no error with the fact that $g^3$ has degree $125$.

Answer 1

Edit: After reading the comment of lhf to the original question, I figured I might point out that what's going on here is simply inverse iteration.

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How might we compute the roots of $f^n$, given its very high degree? Well, computing the roots of $f$ isn't hard, say we get $$z_{1},z_{2},z_{3},z_{4},z_{5}.$$ Now, each of those points has five more preimages. The collection of all of those will be the roots of $f^2$. Each of those has five more preimages, giving 125 roots of $f^3$. The process we're describing here is exactly inverse iteration - a well known method for generating an approximation to the Julia set of $f$. Thus, that's exactly the set your process is converging to.

Since you mention Mathematica, I might mention that we can check this with just a few lines of Mathematica code:

invImage[z_] := w /. NSolve[w^5/2 - w/2 - 1 == z, w];
invImage[zs_List] := invImage /@ zs;
roots = Flatten[Nest[invImage, 0, 4]];
JuliaSetPlot[z^5/2 - z/2 - 1, z, ColorFunction -> None,
 Epilog -> {Red, Opacity[0.5], Point[{Re[#], Im[#]} & /@ roots]}]

Here's another way to think about why this might work. The roots of a polynomial $f$ are exactly the fixed points of the polynomial $g(z)=f(z)+z$, since $$f(z_0)=0 \implies g(z_0) = f(z_0)+z_0=0+z_0=z_0.$$ Thus, the roots of your iterated function $F=f^n$ mostly lie on the Julia set for the function $F(z)+z$. There might be a few, isolated attractive points in the Fatou set.

Let's examine the implications for the fourth iterate - a polynomial of 620 terms whose coefficients have an absolute value of average size around $10^{23}$. The Julia set of a polynomial does not depend continuously on its coefficients, but it's actually close to continuous. We might hope that changing the coefficient of $z$ from $-4$ to $-3$ won't change the Julia set too much. We might even hope that, as we increase the number of iterations, the affect of adding 1 to the $z$ coefficient becomes less and less. The point being that we could examine the Julia set of $F$ itself, which is the same as the Julia set of the original $f$. Thus, we might hope that your roots are clustering on the Julia set of $f$.