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 $M \in \mathcal{S}_{2n}(\mathbb{R}) \cap \mathrm{Sp}_{2n}(\mathbb{R})$. In this question $\lambda = 1$. Show that $E_{1}$ has even dimension and that there exists a basis of $E_{1}$ that is orthonormal and of the form $(X_{1}, \ldots, X_{p}, J_{n} X_{1}, \ldots, J_{n} X_{p})$ where $2p$ is the dimension of $E_{1}$.

Let $M \in \mathcal{S}_{2n}(\mathbb{R}) \cap \mathrm{Sp}_{2n}(\mathbb{R})$. What about $E_{-1}$? (i.e., does $E_{-1}$ have even dimension and does there exist an orthonormal basis of $E_{-1}$ of the form $(X_{1}, \ldots, X_{p}, J_{n} X_{1}, \ldots, J_{n} X_{p})$?)

Prove the following property: if $M \in \mathcal{S}_{2n}(\mathbb{R}) \cap \mathrm{Sp}_{2n}(\mathbb{R})$, there exists $P \in \mathcal{O}_{2n}(\mathbb{R}) \cap \mathrm{Sp}_{2n}(\mathbb{R})$ such that $P^{\top} M P$ is diagonal with diagonal coefficients $d_{1}, \ldots, d_{2n}$ satisfying for all $k \in \{1, \ldots, n\}$, $d_{k+n} = 1/d_{k}$.

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 $$A_0 = 1 \quad \text{and, for all } k \in \llbracket 1, n \rrbracket, \quad A_k = \frac{1}{k!} X(X - ka)^{k-1}.$$ Let $P$ be an element of $\mathbb{C}_n[X]$ and let $\alpha_0, \ldots, \alpha_n$ be complex numbers such that $$P = \sum_{k=0}^{n} \alpha_k A_k.$$ Prove that, for all $j \in \llbracket 0, n \rrbracket$, $\alpha_j = P^{(j)}(ja)$.

Let $$A = \frac{1}{8} \left(\begin{array}{llll} 9 & 1 & 3 & 3 \\ 1 & 9 & 3 & 3 \\ 3 & 3 & 9 & 1 \\ 3 & 3 & 1 & 9 \end{array}\right).$$ Show that $A \in \mathcal{S}_{4}(\mathbb{R}) \cap \mathrm{Sp}_{4}(\mathbb{R})$.

For all $n$ in $\mathbb{N}$, determine the degree of $T_n$, then show that $\left(T_k\right)_{0 \leqslant k \leqslant n}$ is a basis of $\mathbb{C}_n[X]$.
The sequence of polynomials $\left(T_n\right)_{n \in \mathbb{N}}$ is defined by $T_0 = 1, T_1 = X$ and $\forall n \in \mathbb{N}, T_{n+2} = 2X T_{n+1} - T_n$.

Deduce that, for all $n \in \mathbb{N}$ and $P \in \mathbb{C}_n[X]$, the function from $\mathbb{R}$ to $\mathbb{C}$, $\theta \mapsto P(\cos\theta)$ is in $\mathcal{S}_n$.
Recall that $\mathcal{S}_n$ is the $\mathbb{C}$-vector space of functions $f : \mathbb{R} \rightarrow \mathbb{C}$ satisfying $$\exists (a_0, \ldots, a_n) \in \mathbb{C}^{n+1}, \quad \exists (b_1, \ldots, b_n) \in \mathbb{C}^n, \quad \forall t \in \mathbb{R}, \quad f(t) = a_0 + \sum_{k=1}^{n}\left(a_k \cos(kt) + b_k \sin(kt)\right)$$

For $n \in \mathbb{N}$, calculate $\left\|T_n\right\|_{L^\infty([-1,1])}$.
The sequence of polynomials $\left(T_n\right)_{n \in \mathbb{N}}$ is defined by $T_0 = 1, T_1 = X$ and $\forall n \in \mathbb{N}, T_{n+2} = 2X T_{n+1} - T_n$.

Let $n \in \mathbb{N}$. We consider $n+1$ distinct points in $I$, denoted $x_0 < x_1 < \cdots < x_n$, and a continuous function $f$ from $I$ to $\mathbb{R}$.
Show that the linear map $\varphi : \left|\,\begin{array}{ccl} \mathbb{R}_n[X] & \rightarrow & \mathbb{R}^{n+1} \\ P & \mapsto & \left(P(x_0), P(x_1), \ldots, P(x_n)\right) \end{array}\right.$ is an isomorphism.

We choose $I = [-1,1]$ and write $C_{n,m}$ instead of $C_{n,m}^I$. We fix an extremal pair $(Q_0, R_0)$, i.e., $Q_0$ and $R_0$ are monic and $\frac{\|Q_0\|_I \|R_0\|_I}{\|Q_0 R_0\|_I} = C_{n,m}$.
Let $J$ be a segment contained in $I$ such that $\left\|Q_0\right\|_J = \left\|Q_0\right\|_I$ and $\left\|R_0\right\|_J = \left\|R_0\right\|_I$. Show that: $$\left\|Q_0 R_0\right\|_J = \left\|Q_0 R_0\right\|_I$$

We choose $I = [-1,1]$ and write $C_{n,m}$ instead of $C_{n,m}^I$. We fix an extremal pair $(Q_0, R_0)$. Deduce from questions 4.27 and 4.29 that there exists an extremal pair $(Q_1, R_1)$ such that: $$\left\|Q_1\right\|_I = \left|Q_1(-1)\right| \quad \text{and} \quad \left\|R_1\right\|_I = \left|R_1(1)\right|$$

We choose $I = [-1,1]$ and write $C_{n,m}$ instead of $C_{n,m}^I$. Let $(Q_1, R_1)$ be an extremal pair such that $\left\|Q_1\right\|_I = \left|Q_1(-1)\right|$ and $\left\|R_1\right\|_I = \left|R_1(1)\right|$. Let $n_1$ and $m_1$ be the degrees of $Q_1$ and $R_1$ respectively. We set $Q_2 = X^{n-n_1} Q_1$ and $R_2 = X^{m-m_1} R_1$.
Show that $(Q_2, R_2)$ is a good extremal pair, i.e., $Q_2$ and $R_2$ are monic of degrees $n$ and $m$ respectively, $\frac{\|Q_2\|_I \|R_2\|_I}{\|Q_2 R_2\|_I} = C_{n,m}$, and $\|Q_2\|_I = |Q_2(-1)|$, $\|R_2\|_I = |R_2(1)|$.

We choose $I = [-1,1]$ and write $C_{n,m}$ instead of $C_{n,m}^I$. Let $(Q_2, R_2)$ be a good extremal pair. Let $w$ be a root of $Q_2$ and let $S \in \mathbb{C}[X]$ be such that: $$Q_2(X) = (X - w)S(X)$$ By setting: $$S_2(X) = (X + 1 - |w+1|)S(X)$$ show that $(S_2, R_2)$ is a good extremal pair.

We choose $I = [-1,1]$ and write $C_{n,m}$ instead of $C_{n,m}^I$. Deduce from question 4.32 that there exists a polynomial $Q_3$ whose roots are all in $[-1, +\infty[$ and such that the pair $(Q_3, R_2)$ forms a good extremal pair.

We choose $I = [-1,1]$ and write $C_{n,m}$ instead of $C_{n,m}^I$. Show that there exists a polynomial $Q_4$ whose roots are all in $I$ and such that the pair $(Q_4, R_2)$ forms a good extremal pair.
To do this, given a root $w$ of $Q_3$ that is not in $I$, one may introduce the polynomial: $$S_3(X) = \frac{X-1}{X-w} Q_3(X)$$ then one may follow the method used in the two previous questions.

We choose $I = [-1,1]$ and write $C_{n,m}$ instead of $C_{n,m}^I$. Briefly explain why there exists a polynomial $R_4$ such that the pair $(Q_4, R_4)$ forms a very good extremal pair, i.e., a good extremal pair in which all complex roots of $Q_4$ and $R_4$ are contained in $I$.

We choose $I = [-1,1]$ and fix any very good extremal pair $(Q, R)$. We set $P = QR$ and denote by $x_1 \leq \ldots \leq x_{n+m}$ the roots of $P$ counted with multiplicity. Show that: $$Q = \prod_{k=m+1}^{n+m}\left(X - x_k\right) \quad \text{and} \quad R = \prod_{k=1}^{m}\left(X - x_k\right)$$

We choose $I = [-1,1]$ and fix any very good extremal pair $(Q, R)$. We set $P = QR$ and denote by $x_1 \leq \ldots \leq x_{n+m}$ the roots of $P$ counted with multiplicity, with $Q = \prod_{k=m+1}^{n+m}(X-x_k)$ and $R = \prod_{k=1}^{m}(X-x_k)$.
Verify that for all $x \in ]-\infty, -1[$, we have $|Q(x)| > |Q(-1)|$.

We choose $I = [-1,1]$ and fix any very good extremal pair $(Q, R)$. We set $P = QR$. By contradiction, show that $|P(-1)| = \|P\|_I$.
To do this, one may choose a real number $\epsilon > 0$, introduce the segment $I_\epsilon = [-1-\epsilon, 1]$ and bound the quantity: $$\frac{\|Q\|_{I_\epsilon} \|R\|_{I_\epsilon}}{\|P\|_{I_\epsilon}}$$ using question 4.37.

We choose $I = [-1,1]$ and fix any very good extremal pair $(Q, R)$. We set $P = QR$. By considering the function: $$\begin{aligned} f : \mathbb{R} &\rightarrow \mathbb{R} \\ y &\mapsto P(\cos y), \end{aligned}$$ verify that for all $x \in [-1,1]$, $$P(x) = \|P\|_I \cos\left((n+m)\operatorname{Arccos} x\right).$$

We choose $I = [-1,1]$ and fix any very good extremal pair $(Q, R)$. We set $P = QR$. Deduce from question 4.43 that: $$C_{n,m} = 2^{n+m-1} \cdot \left[\prod_{k=1}^n \left(1 + \cos\left(\frac{2k-1}{2(n+m)}\pi\right)\right)\right] \cdot \left[\prod_{k=1}^m \left(1 + \cos\left(\frac{2k-1}{2(n+m)}\pi\right)\right)\right].$$

Show that, for all $n \in \mathbb{N}$, $\left\|T_n'\right\|_{L^\infty([-1,1])} = n^2$.
One may begin by establishing that, for all $n \in \mathbb{N}$ and $\theta \in \mathbb{R}$, $|\sin(n\theta)| \leqslant n|\sin\theta|$.
The sequence of polynomials $\left(T_n\right)_{n \in \mathbb{N}}$ is defined by $T_0 = 1, T_1 = X$ and $\forall n \in \mathbb{N}, T_{n+2} = 2X T_{n+1} - T_n$.

Let $n \in \mathbb{N}$. We consider $n+1$ distinct points in $I$, denoted $x_0 < x_1 < \cdots < x_n$.
Show that, for all $i \in \llbracket 0, n \rrbracket$, there exists a unique polynomial $L_i \in \mathbb{R}_n[X]$ such that $$\forall j \in \llbracket 0, n \rrbracket, \quad L_i(x_j) = \begin{cases} 0 & \text{if } j \neq i, \\ 1 & \text{if } j = i. \end{cases}$$

Let $n \in \mathbb{N}$. We consider $n+1$ distinct points in $I$, denoted $x_0 < x_1 < \cdots < x_n$, and the polynomials $L_0, \ldots, L_n$ defined in Q5.
Show that $(L_0, \ldots, L_n)$ is a basis of $\mathbb{R}_n[X]$.