# For $p(x)= x (1/2-x)$ is the set of 0's of $p^n(x)$ dense in the Julia set?

I plotted the filled Julia set for $$p(x) = -x^2+ x/2$$ and the zero's of $$p^n(x)$$ for $$n=10$$, and the $$1024$$ $$0$$'s are mainly clustered around the boundary, as shown, where the filled Julia set, the basin of the fixed point at 0 is green, and the $$0$$'s of $$p^n(x)$$ are red. Topologically the zero's are all in the interior of the filled Julia set, but they appear to cover the boundary too. This surprised me and I can't figure out how to prove it.

• Yes, this makes sense. Essentially, you've discovered the inverse iteration technique to generate the Julia set. – Mark McClure Jun 5 '19 at 10:20

Yes, there is a theorem that states exactly that: if you take a polynomial $$p$$ and any point $$z_0$$, (you took $$z_0=0$$), then $$\{ z : p^n(z)=z_0 \text{ for some }n \}$$ accumulates on the Julia set. There is only one exception (up to change of variable): if you take $$f(z)=z^d$$ and $$z_0=0$$. This is a theorem first proved by a mathematician called Hans Brolin.
Just of point of clarification: as $$n$$ increases, (most) red dots will get closer and closer to the Julia set, but none of them are actually on the boundary. This is because preimages of points in the Fatou set stay in the Fatou set.