We denote by $\phi_0$ and $\phi_1$ the maps from $\mathbf{C}$ to $\mathbf{C}$ defined by: $\phi_0(z) = \frac{1+\mathrm{i}}{2}\bar{z} + \frac{-1+\mathrm{i}}{2}$ and $\phi_1(z) = \frac{1-\mathrm{i}}{2}\bar{z} + \frac{1+\mathrm{i}}{2}$. The map $f\in\mathcal{E}$ satisfies $Tf = f$, i.e. $f(x) = \phi_0(f(2x))$ for $x\in[0,\frac{1}{2}]$ and $f(x) = \phi_1(f(2x-1))$ for $x\in]\frac{1}{2},1]$. Let $(r_n)_{n\geq 1}\in\{0,1\}^{\mathbf{N}^*}$. a) Show that the series with general term $\frac{r_n}{2^n}$ converges and that its sum $x$ belongs to $[0,1]$. b) By setting for every natural number $p$, $x_p = \sum_{n=1}^{\infty}\frac{r_{n+p}}{2^n}$, prove the relation: $$f(x) = \phi_{r_1}\circ\phi_{r_2}\circ\ldots\phi_{r_p}\left(f\left(x_p\right)\right)$$ for every non-zero natural number $p$.
We denote by $\phi_0$ and $\phi_1$ the maps from $\mathbf{C}$ to $\mathbf{C}$ defined by: $\phi_0(z) = \frac{1+\mathrm{i}}{2}\bar{z} + \frac{-1+\mathrm{i}}{2}$ and $\phi_1(z) = \frac{1-\mathrm{i}}{2}\bar{z} + \frac{1+\mathrm{i}}{2}$. The map $f\in\mathcal{E}$ satisfies $Tf = f$, i.e. $f(x) = \phi_0(f(2x))$ for $x\in[0,\frac{1}{2}]$ and $f(x) = \phi_1(f(2x-1))$ for $x\in]\frac{1}{2},1]$.
Let $(r_n)_{n\geq 1}\in\{0,1\}^{\mathbf{N}^*}$.
a) Show that the series with general term $\frac{r_n}{2^n}$ converges and that its sum $x$ belongs to $[0,1]$.
b) By setting for every natural number $p$, $x_p = \sum_{n=1}^{\infty}\frac{r_{n+p}}{2^n}$, prove the relation:
$$f(x) = \phi_{r_1}\circ\phi_{r_2}\circ\ldots\phi_{r_p}\left(f\left(x_p\right)\right)$$
for every non-zero natural number $p$.