For $n \in \mathbb{N}^*$, $T_n$ denotes the polynomial function satisfying $T_n(x) = 2^{1-n} F_n(x)$ where $F_n(x) = \cos(n \arccos x)$, and $x_{n,j}$ denotes the $j$-th zero of $T_n$ in increasing order. We denote by $E_{n-1}$ the vector subspace of polynomial functions of degree at most $n-1$. Let $n \in \mathbb{N}^*$, $x \in \mathbb{R} \setminus \{x_{n,j},\, 1 \leqslant j \leqslant n\}$ and $P \in E_{n-1}$. a) Show that: $$P(x) = \sum_{j=1}^{n} \frac{P(x_{n,j})}{T_n'(x_{n,j})} \frac{T_n(x)}{x - x_{n,j}}$$ b) Deduce that: $$P(x) = \frac{2^{n-1}}{n} \sum_{j=1}^{n} (-1)^{n-j} \sqrt{1 - x_{n,j}^2}\, P(x_{n,j}) \frac{T_n(x)}{x - x_{n,j}}$$
For $n \in \mathbb{N}^*$, $T_n$ denotes the polynomial function satisfying $T_n(x) = 2^{1-n} F_n(x)$ where $F_n(x) = \cos(n \arccos x)$, and $x_{n,j}$ denotes the $j$-th zero of $T_n$ in increasing order. We denote by $E_{n-1}$ the vector subspace of polynomial functions of degree at most $n-1$.
Let $n \in \mathbb{N}^*$, $x \in \mathbb{R} \setminus \{x_{n,j},\, 1 \leqslant j \leqslant n\}$ and $P \in E_{n-1}$.
a) Show that:
$$P(x) = \sum_{j=1}^{n} \frac{P(x_{n,j})}{T_n'(x_{n,j})} \frac{T_n(x)}{x - x_{n,j}}$$
b) Deduce that:
$$P(x) = \frac{2^{n-1}}{n} \sum_{j=1}^{n} (-1)^{n-j} \sqrt{1 - x_{n,j}^2}\, P(x_{n,j}) \frac{T_n(x)}{x - x_{n,j}}$$