Ytukyg

We have function $f: [0,1] rightarrow [0,1]$ such that $f(x) = 4x(1-x)$. Find the number of solutions of $f^{circ n}(x)=f(f(...f(x)))...)$ ($n$ times took $f()$). 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?

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.

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

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