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: $$P(x) = \frac{2^{n-1}}{n} \sum_{j=1}^{n} (-1)^{n-j} \sqrt{1 - x_{n,j}^2}\, P(x_{n,j}) \frac{T_n(x)}{x - x_{n,j}}$$

Determine $T_0, T_1, T_2$ and $T_3$, where the Chebyshev polynomials of the first kind $(T_n)_{n \in \mathbb{N}}$ are defined by $$\forall n \in \mathbb{N}, \quad \forall \theta \in \mathbb{R}, \quad T_n(\cos\theta) = \cos(n\theta)$$

Let $\widehat { \mu } = \left( \mu _ { 1 } > \cdots > \mu _ { n } \right) \in \mathbb { R } ^ { n }$. We define the polynomials $$Q _ { 0 } = \prod _ { k = 1 } ^ { n } \left( X - \mu _ { k } \right) \quad \text { and } \quad \forall j \in \{ 1 , \ldots , n \} , \quad P _ { j } = \frac { Q _ { 0 } } { \left( X - \mu _ { j } \right) } .$$ (a) Show that the family $\left( Q _ { 0 } , P _ { 1 } , P _ { 2 } , \ldots , P _ { n } \right)$ is a basis of $\mathbb { R } _ { n } [ X ]$.
(b) Let $j \in \{ 1 , \ldots , n \}$. Verify that $( - 1 ) ^ { j - 1 } P _ { j } \left( \mu _ { j } \right) > 0$.

We consider a natural integer $n$ and a complex number $a$. We define a family of polynomials $(A_0, A_1, \ldots, A_n)$ by setting $$A_0 = 1 \quad \text{and, for all } k \in \llbracket 1, n \rrbracket, \quad A_k = \frac{1}{k!} X(X - ka)^{k-1}.$$ We denote by $\mathbb{C}_n[X]$ the $\mathbb{C}$-vector space of polynomials with complex coefficients and degree at most $n$. Prove that the family $(A_0, \ldots, A_n)$ is a basis of $\mathbb{C}_n[X]$.

Let $n$ be a non-zero natural number. Let $P$ be in $\mathbb{C}_{2n}[X]$, and for all $\lambda \in \mathbb{C}$, $P_\lambda(X) = P(\lambda X) - P(\lambda)$. For all $\lambda$ in $\mathbb{C}$, $Q_\lambda(X) = \frac{P(\lambda X) - P(\lambda)}{X - 1} \in \mathbb{C}_{2n-1}[X]$. We consider the polynomial $R(X) = X^{2n} + 1$. For $k$ in $\llbracket 1, 2n \rrbracket$, we denote $\varphi_k = \frac{\pi}{2n} + \frac{k\pi}{n}$ and $\omega_k = \mathrm{e}^{\mathrm{i}\varphi_k}$.
Using formula (I.1), show that $$\forall \lambda \in \mathbb{C}, \quad Q_\lambda(X) = -\frac{1}{2n} \sum_{k=1}^{2n} \frac{P(\lambda\omega_k) - P(\lambda)}{\omega_k - 1} \frac{X^{2n} + 1}{X - \omega_k} \omega_k$$ then deduce that $$\forall \lambda \in \mathbb{C}, \quad \lambda P'(\lambda) = \frac{1}{2n} \sum_{k=1}^{2n} P(\lambda\omega_k) \frac{2\omega_k}{(1 - \omega_k)^2} - \frac{P(\lambda)}{2n} \sum_{k=1}^{2n} \frac{2\omega_k}{(1 - \omega_k)^2}.$$

Let $n$ be a non-zero natural number. Let $P$ be in $\mathbb{C}_{2n}[X]$, and for all $\lambda$ in $\mathbb{C}$, let $Q_\lambda(X) = \frac{P(\lambda X) - P(\lambda)}{X-1} \in \mathbb{C}_{2n-1}[X]$. We consider the polynomial $R(X) = X^{2n} + 1$. For $k$ in $\llbracket 1, 2n \rrbracket$, we denote $\varphi_k = \frac{\pi}{2n} + \frac{k\pi}{n}$ and $\omega_k = \mathrm{e}^{\mathrm{i}\varphi_k}$.
Using formula $$\forall B \in \mathbb{C}_{2n-1}[X], \quad B(X) = \sum_{k=1}^{2n} B(\alpha_k) \frac{A(X)}{(X - \alpha_k) A'(\alpha_k)} \tag{I.1}$$ show that $$\forall \lambda \in \mathbb{C}, \quad Q_\lambda(X) = -\frac{1}{2n} \sum_{k=1}^{2n} \frac{P(\lambda\omega_k) - P(\lambda)}{\omega_k - 1} \frac{X^{2n}+1}{X - \omega_k} \omega_k$$ then deduce that $$\forall \lambda \in \mathbb{C}, \quad \lambda P'(\lambda) = \frac{1}{2n} \sum_{k=1}^{2n} P(\lambda\omega_k) \frac{2\omega_k}{(1-\omega_k)^2} - \frac{P(\lambda)}{2n} \sum_{k=1}^{2n} \frac{2\omega_k}{(1-\omega_k)^2}.$$

Let $a _ { 0 } , a _ { 1 } , \cdots , a _ { 19 } \in \mathbb { R }$ and
$$P ( x ) = x ^ { 20 } + \sum _ { i = 0 } ^ { 19 } a _ { i } x ^ { i } , \quad x \in \mathbb { R }$$
If $P ( x ) = P ( - x )$ for all $x \in \mathbb { R }$, and
$$P ( k ) = k ^ { 2 } , \text{ for } k = 0,1,2 \cdots , 9$$
then find
$$\lim _ { x \rightarrow 0 } \frac { P ( x ) } { \sin ^ { 2 } x }$$