The integer part of the real number $x$ is denoted $[x]$. For every real $x$ and every non-zero natural number $n$: $r_n(x) = [2^n x] - 2[2^{n-1}x]$. We denote by $\mathbf{Z}\left[\frac{1}{2}\right]$ the set of rationals of the form $\frac{k}{2^n}$ where $k\in\mathbf{Z}$ and $n\in\mathbf{N}$. The map $f\in\mathcal{E}$ satisfies $Tf = f$.
Conversely, let $x\in[0,1[$.
a) Establish that, for every non-zero natural number $n$, $r_n(x)\in\{0,1\}$.
b) Show that, for every non-zero natural number $N$ and every real $x\in[0,1[$: $$\frac{[2^N x]}{2^N} = \sum_{n=1}^{N}\frac{r_n(x)}{2^n} \quad \text{then} \quad x = \sum_{n=1}^{\infty}\frac{r_n(x)}{2^n}.$$
c) Show that if, moreover, $x\in\mathbf{Z}\left[\frac{1}{2}\right]$ then there exists $N\in\mathbf{N}$ such that $r_n(x) = 0$ for every natural number $n > N$.
d) Calculate $f\left(\frac{1}{2}\right)$ and $f\left(\frac{1}{4}\right)$. Recognize $\phi_0\circ\phi_0$ and deduce $f\left(\frac{1}{2^k}\right)$ for all $k\in\mathbf{N}$.