We consider the case where $I = [-1,1]$ and $w(x) = 1$. Let $(p_n)_{n \in \mathbb{N}}$ be the sequence of orthogonal polynomials associated with the weight $w$.
Deduce explicitly a quadrature formula of order 5 (you will determine the points $x_j$ and the coefficients $\lambda_j$).

We assume that $f \in L^1(\mathbb{R})$ and $g \in L^\infty(\mathbb{R})$ and we further assume that $g \in L^1(\mathbb{R})$ and $f * g \in L^1(\mathbb{R})$. Admitting that, for all real $\xi$, $$\int_{-\infty}^{+\infty} \left(\int_{-\infty}^{+\infty} \mathrm{e}^{-\mathrm{i}x\xi} f(t) g(x-t) \,\mathrm{d}t\right) \mathrm{d}x \quad \text{and} \quad \int_{-\infty}^{+\infty} \left(\int_{-\infty}^{+\infty} \mathrm{e}^{-\mathrm{i}x\xi} f(t) g(x-t) \,\mathrm{d}x\right) \mathrm{d}t$$ exist and are equal, show that $\widehat{f * g} = \hat{f}\hat{g}$.

We assume that $f \in L^1(\mathbb{R})$ and $g \in L^\infty(\mathbb{R})$ and we further assume that $g \in L^1(\mathbb{R})$ and $f * g \in L^1(\mathbb{R})$. Admitting that, for all real $\xi$, $$\int_{-\infty}^{+\infty} \left(\int_{-\infty}^{+\infty} \mathrm{e}^{-\mathrm{i}x\xi} f(t) g(x-t)\,\mathrm{d}t\right)\mathrm{d}x \quad \text{and} \quad \int_{-\infty}^{+\infty} \left(\int_{-\infty}^{+\infty} \mathrm{e}^{-\mathrm{i}x\xi} f(t) g(x-t)\,\mathrm{d}x\right)\mathrm{d}t$$ exist and are equal, show that $\widehat{f*g} = \hat{f}\hat{g}$.

We classify cycles of length $k$ into three subsets:

• the set $\mathcal{A}_{k}$, consisting of cycles where at least one edge appears only once;
• the set $\mathcal{B}_{k}$, consisting of cycles where all edges appear exactly twice;
• the set $\mathcal{C}_{k}$, consisting of cycles where all edges appear at least twice and there exists at least one that appears at least three times.

What can be said about $\mathcal{B}_{k}$ if $k$ is odd? Deduce that $\lim_{n \rightarrow +\infty} \mathbb{E}\left(\frac{1}{n} \sum_{i=1}^{n} \Lambda_{i,n}^{k}\right) = 0$ in this case.

Cycles of length $k$ are classified into three subsets: the set $\mathcal{A}_{k}$, consisting of cycles where at least one edge appears only once; the set $\mathcal{B}_{k}$, consisting of cycles where all edges appear exactly twice; the set $\mathcal{C}_{k}$, consisting of cycles where all edges appear at least twice and there exists at least one that appears at least three times.
What can be said about $\mathcal{B}_{k}$ if $k$ is odd? Deduce that $\lim_{n \rightarrow +\infty} \mathbb{E}\left(\frac{1}{n} \sum_{i=1}^{n} \Lambda_{i,n}^{k}\right) = 0$ in this case.

We now assume that all coefficients $m _ { i , j } ( 1 \leqslant i , j \leqslant n )$ of the stochastic matrix $M$ are strictly positive. We set $\varepsilon = \min _ { 1 \leqslant i , j \leqslant n } m _ { i , j }$. The sequence $\left( M ^ { k } \right)$ converges to a stochastic matrix $B = \left( \begin{array} { l l l } b _ { 1 } & \cdots & b _ { n } \\ b _ { 1 } & \cdots & b _ { n } \\ b _ { 1 } & \cdots & b _ { n } \end{array} \right)$ all of whose rows are equal. We denote by $P ^ { \infty }$ the row $\left( b _ { 1 } , \ldots , b _ { n } \right)$.
Prove that, $\forall i \in \llbracket 1 , n \rrbracket$, $b _ { i } > 0$.

Assume that $k$ is even and that $\vec{\imath} \in \mathcal{B}_{k}$ is a cycle passing through $\frac{k}{2} + 1$ distinct vertices (i.e. $|\vec{\imath}| = \frac{k}{2} + 1$). We traverse the edges of $\vec{\imath}$ in order. To each edge of $\vec{\imath}$ we associate an opening parenthesis if this edge appears for the first time and a closing parenthesis if it appears for the second time.
Justify that we thus obtain a well-parenthesized word of length $k$.

Assume $k$ is even and $\vec{\imath} \in \mathcal{B}_{k}$ is a cycle passing through $\frac{k}{2}+1$ distinct vertices (i.e. $|\vec{\imath}| = \frac{k}{2}+1$). We traverse the edges of $\vec{\imath}$ in order. To each edge of $\vec{\imath}$ we associate an opening parenthesis if this edge appears for the first time and a closing parenthesis if it appears for the second time.
Justify that we thus obtain a well-parenthesized word of length $k$.

We now assume that all coefficients $m _ { i , j } ( 1 \leqslant i , j \leqslant n )$ of the stochastic matrix $M$ are strictly positive. The sequence $\left( M ^ { k } \right)$ converges to a stochastic matrix $B$ all of whose rows are equal to $P ^ { \infty } = \left( b _ { 1 } , \ldots , b _ { n } \right)$.
Prove that the sequence $\left( P ^ { ( k ) } \right) _ { k \in \mathbb { N } } = \left( P ^ { ( 0 ) } M ^ { k } \right) _ { k \in \mathbb { N } }$ converges to $P ^ { \infty }$, regardless of the initial probability distribution $P ^ { ( 0 ) }$.

We set $$\forall ( P , Q ) \in \mathbb { R } [ X ] \times \mathbb { R } [ X ] , \quad ( P \mid Q ) = \frac { 2 } { \pi } \int _ { 0 } ^ { 1 } P ( 4 x ) Q ( 4 x ) \frac { \sqrt { 1 - x } } { \sqrt { x } } \mathrm { ~d} x .$$ Show that $( \cdot \mid \cdot )$ is an inner product on $\mathbb { R } [ X ]$.

We now assume that all coefficients $m _ { i , j } ( 1 \leqslant i , j \leqslant n )$ of the stochastic matrix $M$ are strictly positive. The sequence $\left( M ^ { k } \right)$ converges to a stochastic matrix $B$ all of whose rows are equal to $P ^ { \infty } = \left( b _ { 1 } , \ldots , b _ { n } \right)$.
Prove that $P ^ { \infty }$ is the unique probability distribution $P$ invariant by $M$, that is, satisfying $P M = P$.

Deduce from the above that $$\lim_{n \rightarrow +\infty} \mathbb{E}\left(\frac{1}{n} \sum_{i=1}^{n} \Lambda_{i,n}^{k}\right) = C_{k/2}$$

Deduce from the above that $$\lim_{n \rightarrow +\infty} \mathbb{E}\left(\frac{1}{n} \sum_{i=1}^{n} \Lambda_{i,n}^{k}\right) = C_{k/2}$$

We assume that the numpy module of Python has been imported using the instruction \texttt{import numpy as np}. This thus gives access to the following functions:

• \texttt{np.identity( n )} creates $I _ { n }$, the identity matrix of order n;
• \texttt{A.shape} gives, in the form of a tuple, the size of array A, for example \texttt{np.identity(5).shape} $\rightarrow ( 5,5 )$;
• \texttt{np.dot(A, B)} computes the matrix product of A and B if A and B are two-dimensional arrays compatible with matrix multiplication.

The naive algorithm for computing $B ^ { k }$ is based on its definition, namely: $$\left\{ \begin{array} { l } B ^ { 0 } = I _ { n } \\ B ^ { k } = B B ^ { k - 1 } \quad \forall k \geqslant 1 \end{array} \right.$$ Write a Python function \texttt{puissance1(B, k)} that takes as arguments a square matrix $B$ and a natural number k and returns the matrix $B ^ { k }$ computed using the naive algorithm.

We assume that the numpy module of Python has been imported using the instruction \texttt{import numpy as np}. This thus gives access to the following functions:

• \texttt{np.identity( n )} creates $I _ { n }$, the identity matrix of order n;
• \texttt{A.shape} gives, in the form of a tuple, the size of array A, for example \texttt{np.identity(5).shape} $\rightarrow ( 5,5 )$;
• \texttt{np.dot(A, B)} computes the matrix product of A and B if A and B are two-dimensional arrays compatible with matrix multiplication.

The fast exponentiation algorithm is based on the following principle: $$\begin{cases} B ^ { 0 } = I _ { n } & \\ B ^ { k } = \left( B ^ { 2 } \right) ^ { \frac { k } { 2 } } & \text { if } k \text { is even } \\ B ^ { k } = B \left( B ^ { 2 } \right) ^ { \frac { k - 1 } { 2 } } & \text { if } k \text { is odd.} \end{cases}$$ Write a Python function \texttt{puissance2(B, k)} that takes as arguments a square matrix B and a natural number $k$ and returns the matrix $B ^ { k }$ computed using the fast exponentiation algorithm.

We assume that the numpy module of Python has been imported using the instruction \texttt{import numpy as np}. This thus gives access to the following functions:

• \texttt{np.identity( n )} creates $I _ { n }$, the identity matrix of order n;
• \texttt{A.shape} gives, in the form of a tuple, the size of array A, for example \texttt{np.identity(5).shape} $\rightarrow ( 5,5 )$;
• \texttt{np.dot(A, B)} computes the matrix product of A and B if A and B are two-dimensional arrays compatible with matrix multiplication.

The naive algorithm for computing $B^k$ uses: $$\left\{ \begin{array} { l } B ^ { 0 } = I _ { n } \\ B ^ { k } = B B ^ { k - 1 } \quad \forall k \geqslant 1 \end{array} \right.$$ The fast exponentiation algorithm uses: $$\begin{cases} B ^ { 0 } = I _ { n } & \\ B ^ { k } = \left( B ^ { 2 } \right) ^ { \frac { k } { 2 } } & \text { if } k \text { is even } \\ B ^ { k } = B \left( B ^ { 2 } \right) ^ { \frac { k - 1 } { 2 } } & \text { if } k \text { is odd.} \end{cases}$$ We assume that $k \geqslant 2$ and we denote by $p$ the unique integer such that $2 ^ { p } \leqslant k < 2 ^ { p + 1 }$. For each of the calls \texttt{puissance1(B, k)} and \texttt{puissance2(B, k)}, compute the number of calls to the function \texttt{np.dot}, in the worst case and in the best case.

Let $(E_n)_{n\in\mathbb{N}}$ be a sequence of finite subsets of $[-1,1]^2$ such that, for all $(u,v)\neq(0,0)$, $$\frac{1}{|E_n|}\sum_{(s,t)\in E_n} e_{u,v}(s,t) \underset{n\rightarrow+\infty}{\longrightarrow} 0.$$ Show that for all $f \in \mathcal{T}$, $$\frac{1}{|E_n|}\sum_{(s,t)\in E_n} f(s,t) \underset{n\rightarrow+\infty}{\longrightarrow} \frac{1}{4}\int_{-1}^{1}\int_{-1}^{1} f(s,t)\,\mathrm{d}s\,\mathrm{d}t.$$

Let $(E_n)_{n\in\mathbb{N}}$ be a sequence of finite subsets of $[-1,1]^2$ such that, for all $(u,v)\neq(0,0)$, $$\frac{1}{|E_n|}\sum_{(s,t)\in E_n} e_{u,v}(s,t) \underset{n\rightarrow+\infty}{\longrightarrow} 0.$$ Show that for all $a,b,c,d \in [-1,1]$ such that $a < b$ and $c < d$, $$\frac{|E_n \cap ([a,b]\times[c,d])|}{|E_n|} \underset{n\rightarrow+\infty}{\longrightarrow} \frac{|b-a||d-c|}{4}.$$

Let $z \in D$. Show that the function $t \in [ 0,1 ] \mapsto L ( t z )$ is differentiable and give a simple expression for its derivative. Deduce that $t \mapsto ( 1 - t z ) e ^ { L ( t z ) }$ is constant on $[ 0,1 ]$ and conclude that
$$\exp ( L ( z ) ) = \frac { 1 } { 1 - z }$$

Let $z \in D$. Show that the function $\Phi : t \mapsto L ( t z )$ is differentiable on an open interval including $[ - 1,1 ]$ and give a simple expression for its derivative on $[ - 1,1 ]$.

Let $z \in D$. Show that the function $\Phi : t \mapsto L(tz)$ is differentiable on an open interval including $[-1,1]$ and give a simple expression for its derivative on $[-1,1]$.

We are given two matrices $A$ and $B$ in $\mathcal{M}_n(\mathbf{K})$. We assume that $A$ and $B$ commute.
We define an application $$\begin{aligned} g : \mathbf{R} & \rightarrow \mathcal{M}_n(\mathbf{K}) \\ t & \longmapsto g(t) = e^{t(A+B)} e^{-tB} \end{aligned}$$
$\mathbf{2}$ ▷ Show that the application $g$, and the application $f_A$ defined in the preamble, are solutions to the same Cauchy problem. Deduce from this a proof of the relation $$\forall t \in \mathbf{R} \quad e^{t(A+B)} = e^{tA} e^{tB}.$$

Let $V$ and $V^{\prime}$ be two subspaces of $E$ of dimension $p$. Let $u = (u_1, \ldots, u_p)$ and $u^{\prime} = (u_1^{\prime}, \ldots, u_p^{\prime})$ be the orthonormal families constructed in question (1).
Show that if $\operatorname{dim}(V \cap V^{\prime}) \geqslant 1$, then $u_k = u_k^{\prime}$ for all $1 \leqslant k \leqslant \operatorname{dim}(V \cap V^{\prime})$.

If $f$ and $g$ are two power series such that $f \prec g$, show that $\rho(f) \geqslant \rho(g)$.

We fix two orthonormal families $u = (u_1, \ldots, u_p)$ of vectors of $V$ and $u^{\prime} = (u_1^{\prime}, \ldots, u_p^{\prime})$ of vectors of $V^{\prime}$ satisfying the conditions of question (1).
Show that if $\operatorname{dim}(V \cap V^{\prime}) \geqslant 1$, then $u_k = u_k^{\prime}$ for all $1 \leqslant k \leqslant \operatorname{dim}(V \cap V^{\prime})$.