$S(n)$ the set of all real roots of $p_n(x)$ where $p_{n+1}(x)=p(p_n(x))$, show that $S(n)\subseteq S(2n)$ I was trying this problem :
Let $p(x)=p_1(x)=4x^3-3x$ and $p_{n+1}(x)=p(p_n(x))$ for each positive integer $n$. Also, let $S(n)$ be the set of all the real roots of the equation $p_n(x)=x$.Prove that $S(n)\subseteq S(2n)$ and that the product of the elements of $S(n)$ is the average of the elements of $S(2n)$.
Can anyone help, with some hints probably? 
 A: Some hints: you should recognize $p(x) = 4x^3 - 3x$ as a Chebyshev polynomial. That is, it satisfies $p(\cos \theta) = \cos (3\theta)$. 
By induction, a similar identity holds for $p_n(x)$. This should let you characterize solutions to $p_n(x) = x$ as $\cos \theta$ for some set of angles $\theta$, from which at least proving that $S(n) \subseteq S(2n)$ shouldn't be too hard.
I guess you can also prove $S(n) \subseteq S(2n)$ from first principles without even knowing anything about $p(x)$, but that's no fun.
The second part of the question is stated unhelpfully; it becomes easier once you realize what the product of the elements of $S(n)$ is.
A: The key observation is that, $p(x) = 4x^3 - 3x = T_3(x)$ is nothing but a Chebyshev's polynomial. One way to introduce them is as $T_n(x) = \cos(n\arccos x) $ or $T_n(\cos\theta) = \cos n\theta$ (putting $x=\cos \theta$), and it is obvious that $T_n \circ T_m = T_{nm}$, thus $p_n = T_3^n = T_{3^n}$.
Now, $S(n)=\{x|~p_n(x)=x\}$ is equivalent to $ \cos 3^n\theta = \cos \theta$(put, $x=\cos\theta$). Now this is just an trigonometry problem, can you continue from here? 
