We denote $\pi_k(x) = \frac{\lfloor 2^k x \rfloor}{2^k}$.
Let $n \in \mathbb{N}^{\star}$ and $x = \sum_{j=1}^{n} \frac{x_j}{2^j}$ with $(x_j)_{j \in \llbracket 1,n \rrbracket} \in \{0,1\}^n$. Show
$$\forall k \in \mathbb{N}, \quad \pi_k(x) = \sum_{j=1}^{\min(n,k)} \frac{x_j}{2^j}.$$