Number of zeros of compositions of $f(x) = 4x(1-x)$

Question

We have the function $f: [0,1] \rightarrow [0,1]$ such that $f(x) = 4x(1-x)$. Find the number of zeros of $$f^{\circ n}(x)=f(f(...f(x)))...)$$ ($n$ times composition of $f()$). I think that the answer is $2^n$ and I have tried to prove it by induction, but I did not succeed. Do you have an idea how to solve it?

Answer 1

It should be fairly simple to prove informally using an induction-like argument. If you have a function g(x) that has the following properties:

• continuous
• n minimal values of zero at x1, x2, x3...xn
• n-1 maximal values of 1 at y1, y2, y3,...yn-1, where xk < yk < xk+1
• monotonic in each of the intervals xk < x < yk, and yk < xk+1 < yk+1

then

applying your f to that function results in another function that has the following properties:

• continuous
• 2n-1 minimal values of zero at new x1, x2, x3...x2n-1
• 2n maximal values of 1 at y1, y2, y3,...y2n, where xk < yk < xk+1
• monotonic in each of the intervals xk < x < yk, and yk < xk+1 < yk+1

P{roof of ther first result is elementary. Proof of the second result is simply because your f(x) produces zeros at x=0 and x=1 (so your new x1..x2n-1 will consist your old x1..xn plus your old y1..yn-1. Proof of the third result comes from the fact that f(x)=1 iff x=1/2 and the continuity and monotonicity of g(x) in each sub-interval. Proof of the fourth result should also be fairly elementary.

Answer 2

From this question, for $\sin^2{\frac{\alpha}{2}} \in [0,1]$

$$f^{\circ n}\left(\color{red}{\sin^2{\frac{\alpha}{2}}}\right)=\color{blue}{\sin^2{\left(2^{n-1}\alpha\right)}}$$ and from $$\color{blue}{\sin^2{\left(2^{n-1}\alpha\right)}}=0 \Rightarrow 2^{n-1}\alpha=k\pi \Rightarrow \alpha=\frac{k\pi}{2^{n-1}},k\in\mathbb{Z}$$ But $$\color{red}{\sin^2{\frac{\alpha}{2}}}=\sin^2{\frac{k\pi}{2^n}} \tag{1}$$ and $$\sin^2{\frac{(2^n-k)\pi}{2^n}}= \sin^2{\left(\pi-\frac{k\pi}{2^n}\right)}= \sin^2{\frac{k\pi}{2^n}}$$ Thus, for $k\in \left\{0,1,...,2^{n-1}\right\}$ we have $2^{n-1}+1$ different values for $(1)$ and the same number of zero's for $f^{\circ n}\left(x\right)$.