I was messing around with polynomials and their real roots when I, as recreational mathematicians do, asked myself the following random question:
Suppose I am given a polynomial $P(x)$. How can I find the number of real roots of the polynomial $P^{\circ n}(x)$, representing the n-fold composition of $P$ with itself (counting multiplicity)?
I started with the simple polynomial $P_1(x)=x^2-1$. This was an easy example, as it turned out that $P_1^{\circ n}(x)$ has $n+1$ real roots, which was simple to prove.
My next example was the polynomial $P_2(x)=x^2-2$. This one was more difficult, but I eventually determined that $P_2^{\circ n}(x)$ has $2F_{n+1}-2$ real roots, where $F_n$ represents the sequence of Fibonacci numbers with $F_0=F_1=1$.
In general, I am considering polynomials of the form $P_c(x)=x^2-c$. For $c$ less than $-1$, the iterates of this polynomial have no real roots, and for $c$ greater than $3$, the $n$th iterate of this polynomial seems to have $2^n$ real roots.
The polynomial with which I am stumped is $$P_{3/2}(x)=x^2-\frac{3}{2}$$ While I have been unable to derive a formula for the number of real roots of $P_{3/2}^{\circ n}(x)$, by observing the number of real roots for the first couple of iterations, I have come up with the remarkable conjecture that the number of real roots of the $n$th iterate is given by $2p(n)$, where $p(n)$ represents the number of partitions of $n$.
This conjecture, if true, would be truly amazing. How can I prove it?
NOTE: To prove the formulae I obtained for $P_1$ and $P_2$, I divided the real line up into intervals that the polynomials in question sent to one another, and from this I obtained a recursive formula for each. However, I cannot figure out how to nearly divide $\mathbb R$ into intervals in the same way.
