We consider the family of polynomials $$\left\{ \begin{array}{l} H_0 = 1 \\ H_k = \frac{1}{k!} \prod_{j=0}^{k-1} (X - j) \quad \text{for } k \in \llbracket 1, n \rrbracket \end{array} \right.$$
Determine the real sequences $\left(u_k\right)_{k \in \mathbb{N}}$ such that $$u_{k+2} - 2u_{k+1} + u_k = k^3 + 2k^2 + 5k + 7 \quad (k \in \mathbb{N})$$

We consider the family of polynomials $$\left\{ \begin{array}{l} H_0 = 1 \\ H_k = \frac{1}{k!} \prod_{j=0}^{k-1} (X - j) \quad \text{for } k \in \llbracket 1, n \rrbracket \end{array} \right.$$
Deduce that $H_n(\mathbb{Z}) \subset \mathbb{Z}$, that is, $H_n$ is integer-valued on the integers.

We consider the family of polynomials $$\left\{ \begin{array}{l} H_0 = 1 \\ H_k = \frac{1}{k!} \prod_{j=0}^{k-1} (X - j) \quad \text{for } k \in \llbracket 1, n \rrbracket \end{array} \right.$$
Let $P \in \mathbb{R}_n[X]$ be integer-valued on the integers. Show that $\delta(P)$ is also integer-valued on the integers.

We place ourselves in the particular case where $E = \mathbb{R}_{2m}[X]$, with $m \geq 2$ a fixed natural integer. This vector space is equipped with the scalar product $$\forall (P,Q) \in E^2, \quad (P \mid Q) = \int_{-1}^{1} P(t)Q(t)\,dt$$ The two endomorphisms $T$ and $M$ of $E$ are defined by $$\forall P \in \mathbb{R}_{2m}[X], \quad T(P) = P' \text{ and } M(P) = P^*$$ where $P^*(X) = P(-X)$.
Show that $T$ and $M$ satisfy hypotheses (H1), (H2), (H3) and (H4).

We place ourselves in the particular case where $E = \mathbb{R}_{2m}[X]$, with $m \geq 2$ a fixed natural integer. This vector space is equipped with the scalar product $$\forall (P,Q) \in E^2, \quad (P \mid Q) = \int_{-1}^{1} P(t)Q(t)\,dt$$ The two endomorphisms $T$ and $M$ of $E$ are defined by $$\forall P \in \mathbb{R}_{2m}[X], \quad T(P) = P' \text{ and } M(P) = P^*$$ where $P^*(X) = P(-X)$. We set $F^+ = \operatorname{ker}(M - \operatorname{Id}_E)$, $F^- = \operatorname{ker}(M + \operatorname{Id}_E)$.
What are the spaces $F^+$ and $F^-$ in this case?

We place ourselves in the particular case where $E = \mathbb{R}_{2m}[X]$, with $m \geq 2$ a fixed natural integer. This vector space is equipped with the scalar product $$\forall (P,Q) \in E^2, \quad (P \mid Q) = \int_{-1}^{1} P(t)Q(t)\,dt$$ The two endomorphisms $T$ and $M$ of $E$ are defined by $$\forall P \in \mathbb{R}_{2m}[X], \quad T(P) = P' \text{ and } M(P) = P^*$$ where $P^*(X) = P(-X)$. The map $S$ is defined by $S(v,w) = (v \mid T(w)) + (T(v) \mid w)$.
Show that $$\forall (P,Q) \in E^2, \quad S(P,Q) = P(1)Q(1) - P(-1)Q(-1)$$

We place ourselves in the particular case where $E = \mathbb{R}_{2m}[X]$, with $m \geq 2$ a fixed natural integer. This vector space is equipped with the scalar product $$\forall (P,Q) \in E^2, \quad (P \mid Q) = \int_{-1}^{1} P(t)Q(t)\,dt$$ The two endomorphisms $T$ and $M$ of $E$ are defined by $T(P) = P'$ and $M(P) = P^*$ where $P^*(X) = P(-X)$. We set $$\mathbb{R}_k^0[X] = \{P \in \mathbb{R}_k[X] \mid P(-1) = 0 \text{ and } P(1) = 0\}$$ The subspace $G$ consists of elements $u \in E$ satisfying (a) $u \in \operatorname{Im}(T)$ and (b) $\forall v \in E, S(u,v) = 0$, where $S(P,Q) = P(1)Q(1) - P(-1)Q(-1)$.
Determine the subspace $G$. Is hypothesis (H5) satisfied?

We place ourselves in the particular case where $E = \mathbb{R}_{2m}[X]$, with $m \geq 2$ a fixed natural integer. This vector space is equipped with the scalar product $$\forall (P,Q) \in E^2, \quad (P \mid Q) = \int_{-1}^{1} P(t)Q(t)\,dt$$ We define for any natural integer $n$ the polynomial $R_n$ as follows $$R_n(X) = (X^2 - 1)^n$$ and we now set $$L_n(X) = \frac{1}{2^n n!} R_n^{(n)}(X)$$
Let $n \in \mathbb{N}$.
(a) What is the degree of the polynomial $L_n$? Express $M(L_n)$ in terms of $L_n$.
(b) Show that if $n \geq 1$ then $$\forall P \in \mathbb{R}_{n-1}[X], \quad (L_n \mid P) = 0$$ (c) Show that for every integer $k$ such that $0 \leq k \leq n$, we have $$L_n^{(k)}(1) = \frac{(n+k)!}{(n-k)!} \frac{1}{k! \, 2^k}$$ (d) Show that for every natural integer $k$, we have $$S\left(L_n, L_n^{(2k+1)}\right) = 2 L_n^{(2k+1)}(1)$$

We place ourselves in the particular case where $E = \mathbb{R}_{2m}[X]$, with $m \geq 2$ a fixed natural integer. This vector space is equipped with the scalar product $$\forall (P,Q) \in E^2, \quad (P \mid Q) = \int_{-1}^{1} P(t)Q(t)\,dt$$ The endomorphisms $T(P) = P'$ and $M(P) = P^*$ are defined as before. The polynomials $L_n$ are defined by $L_n(X) = \frac{1}{2^n n!} R_n^{(n)}(X)$ where $R_n(X) = (X^2-1)^n$.
Show that the pair $(L_{2m}, L_{2m-1})$ is a characterizing pair of $G$.

We place ourselves in the particular case where $E = \mathbb{R}_{2m}[X]$, with $m \geq 2$ a fixed natural integer. This vector space is equipped with the scalar product $$\forall (P,Q) \in E^2, \quad (P \mid Q) = \int_{-1}^{1} P(t)Q(t)\,dt$$ The endomorphisms $T(P) = P'$ and $M(P) = P^*$ are defined as before. We set $$\mathbb{R}_{2m-1}^0[X] = \{P \in \mathbb{R}_{2m-1}[X] \mid P(-1) = 0 \text{ and } P(1) = 0\}$$ The polynomials $L_n$ are defined by $L_n(X) = \frac{1}{2^n n!} R_n^{(n)}(X)$ where $R_n(X) = (X^2-1)^n$.
Let $\lambda \in \mathbb{R}$. We consider the problem: find $P \in \mathbb{R}_{2m-1}^0[X]$ such that $$\forall Q \in \mathbb{R}_{2m-1}^0[X], \int_{-1}^{1} P(t)Q(t)\,dt - \lambda \int_{-1}^{1} P'(t)Q'(t)\,dt = 0$$ Show that this problem admits a non-identically zero solution $P$ if and only if $\lambda$ is a root of the polynomial $$K(X) = \left(\sum_{k=0}^{m-1} (-1)^k L_{2m-1}^{(2k+1)}(1) X^k\right) \cdot \left(\sum_{k=0}^{m-1} (-1)^k L_{2m}^{(2k+1)}(1) X^k\right)$$

We place ourselves in the particular case where $E = \mathbb{R}_{2m}[X]$, with $m \geq 2$ a fixed natural integer. This vector space is equipped with the scalar product $$\forall (P,Q) \in E^2, \quad (P \mid Q) = \int_{-1}^{1} P(t)Q(t)\,dt$$ We set $$\mathbb{R}_{2m-1}^0[X] = \{P \in \mathbb{R}_{2m-1}[X] \mid P(-1) = 0 \text{ and } P(1) = 0\}$$
Show that $$\forall P \in \mathbb{R}_{2m-1}^0[X], \quad (P \mid P) \leq 4 (P' \mid P')$$ with strict inequality if $P$ is non-zero.

We place ourselves in the particular case where $E = \mathbb{R}_{2m}[X]$, with $m \geq 2$ a fixed natural integer. This vector space is equipped with the scalar product $$\forall (P,Q) \in E^2, \quad (P \mid Q) = \int_{-1}^{1} P(t)Q(t)\,dt$$ The polynomial $K(X)$ is defined as in question 23.
Deduce that the roots of $K$ are all real and belong to the interval $]0, 4[$.

Let $n \in \mathbb{N}^*$, $h = \frac{1}{n+1}$, and $x_i = ih$ for all $i \in \{0, \ldots, n+1\}$. Show that there exists a unique family of real numbers $\left( u _ { i } \right) _ { 0 \leq i \leq n + 1 }$ satisfying
$$\left\{ \begin{array} { l } - \frac { 1 } { h ^ { 2 } } \left( u _ { i + 1 } + u _ { i - 1 } - 2 u _ { i } \right) + c \left( x _ { i } \right) u _ { i } = f \left( x _ { i } \right) , \text { for } 1 \leq i \leq n \\ u _ { 0 } = u _ { n + 1 } = 0 \end{array} \right.$$

We assume (in this question only) that $c ( x ) = 0$ and $f ( x ) = 1$ for all $x \in [ 0,1 ]$. Let $n \in \mathbb{N}^*$, $h = \frac{1}{n+1}$, and $x_i = ih$. We denote by $u$ the exact solution of problem (1): $$\left\{ \begin{array} { l } - u ^ { \prime \prime } ( x ) + c ( x ) u ( x ) = f ( x ) , x \in [ 0,1 ] \\ u ( 0 ) = u ( 1 ) = 0 \end{array} \right.$$ and $(u_i)_{0 \leq i \leq n+1}$ the unique family satisfying $$\left\{ \begin{array} { l } - \frac { 1 } { h ^ { 2 } } \left( u _ { i + 1 } + u _ { i - 1 } - 2 u _ { i } \right) + c \left( x _ { i } \right) u _ { i } = f \left( x _ { i } \right) , \text { for } 1 \leq i \leq n \\ u _ { 0 } = u _ { n + 1 } = 0 \end{array} \right.$$ Show that for all $i \in \{ 0 , \ldots , n + 1 \}$, we have
$$u _ { i } = u \left( x _ { i } \right) = \frac { 1 } { 2 } x _ { i } \left( 1 - x _ { i } \right)$$

Show that if $f$ is positive, then $u _ { i } \geq 0$ for all $i \in \{ 0 , \ldots , n + 1 \}$.

Let $u$ be the unique solution of problem (1) and $\left( u _ { i } \right) _ { 0 \leq i \leq n + 1 }$ the family defined by relation (2) for $n \in \mathbb { N } ^ { * }$. Show that there exists a constant $\tilde { C } > 0$, independent of $n$, such that
$$\max _ { 0 \leq i \leq n + 1 } \left| u \left( x _ { i } \right) - u _ { i } \right| \leq \frac { \tilde { C } } { n ^ { 2 } }$$
Hint: one may introduce the vector $X = {}^{ t } \left( \varepsilon _ { 1 } , \ldots , \varepsilon _ { n } \right)$ where we set $\varepsilon _ { i } = u \left( x _ { i } \right) - u _ { i }$ and compute $A _ { n } X$.

Let $\alpha$ be a real number. We denote $f_{\alpha} : x \longmapsto (1-x)^{-\alpha}$.
Specify the domain of definition $D$ of $f_{\alpha}$. Justify that $f_{\alpha}$ is of class $C^{1}$ on $D$ and give a first-order linear differential equation satisfied by $f_{\alpha}$ on $D$.

We consider a general balanced urn with parameters $a_{0}, b_{0}, a, b, c, d \in \mathbb{N}$ satisfying $a + b = c + d = s$. For all real $u$ and $v$, we set $P_{0}(u,v) = u^{a_{0}} v^{b_{0}}$ and $P_{n}(u,v) = \sum_{\omega \in \Omega_{n}} u^{b(\omega)} v^{n(\omega)}$.
Prove that, for all integers $n$, $$P_{n+1}(u,v) = u^{a+1} v^{b} \frac{\partial P_{n}}{\partial u}(u,v) + u^{c} v^{d+1} \frac{\partial P_{n}}{\partial v}(u,v)$$

We consider a general balanced urn. For all real $x, u$ and $v$, we set, provided it exists, $$H(x, u, v) = \sum_{n=0}^{+\infty} P_{n}(u,v) \frac{x^{n}}{n!}$$ Let $\rho > 0$. We set $D_{\rho} = ]-\rho, \rho[ \times ]0,2[^{2} = \{(x,u,v) \in \mathbb{R}^{3} ; |x| < \rho, 0 < u < 2, 0 < v < 2\}$.
Justify that, for $\rho$ sufficiently small, the function $H$ is well defined on $D_{\rho}$.

We consider a general balanced urn. For all real $x, u$ and $v$, we set $$H(x, u, v) = \sum_{n=0}^{+\infty} P_{n}(u,v) \frac{x^{n}}{n!}$$ defined on $D_{\rho} = ]-\rho, \rho[ \times ]0,2[^{2}$ for $\rho$ sufficiently small.
Justify that $H$ admits a first-order partial derivative with respect to $x$ on the domain $D_{\rho}$, obtained by term-by-term differentiation with respect to $x$ of the expression for $H$.

We consider a general balanced urn. For all real $x, u$ and $v$, we set $$H(x, u, v) = \sum_{n=0}^{+\infty} P_{n}(u,v) \frac{x^{n}}{n!}$$ defined on $D_{\rho} = ]-\rho, \rho[ \times ]0,2[^{2}$ for $\rho$ sufficiently small.
Prove that $H$ admits a first-order partial derivative with respect to $u$ on the domain $D_{\rho}$, obtained by term-by-term differentiation with respect to $u$ of the expression for $H$.

We consider a general balanced urn. For all real $x, u$ and $v$, we set $$H(x, u, v) = \sum_{n=0}^{+\infty} P_{n}(u,v) \frac{x^{n}}{n!}$$ defined on $D_{\rho} = ]-\rho, \rho[ \times ]0,2[^{2}$ for $\rho$ sufficiently small.
Verify that $H(0, u, v) = u^{a_{0}} v^{b_{0}}$ and then that $H$ is a solution on $D_{\rho}$ of the partial differential equation $$\frac{\partial H}{\partial x}(x,u,v) = u^{a+1} v^{b} \frac{\partial H}{\partial u}(x,u,v) + u^{c} v^{d+1} \frac{\partial H}{\partial v}(x,u,v).$$

In the general model of a Pólya urn, we consider the balanced urn model for which $b = c = 0$, so $a = d$. Each time we draw a ball, we add $a$ balls of its color to the urn. The initial composition is $a_{0}$ white balls and $b_{0}$ black balls. The function $G$ is defined on $U = \{(x,u,v) \in \mathbb{R} \times \mathbb{R}_{+}^{*} \times \mathbb{R}_{+}^{*} ; axu^{a} < 1, axv^{a} < 1\}$ by $$G(x,u,v) = u^{a_{0}} v^{b_{0}} (1 - axu^{a})^{-a_{0}/a} (1 - axv^{a})^{-b_{0}/a}$$ We use the notation $D_{\rho} = ]-\rho, \rho[ \times ]0,2[^{2}$.
Using the preliminary results, prove that there exists $\rho > 0$ such that $D_{\rho} \subset U$ and, for all $(x,u,v) \in D_{\rho}$, $$G(x,u,v) = \sum_{n=0}^{+\infty} Q_{n}(u,v) \frac{x^{n}}{n!}$$ where $Q_{n}$ is a polynomial function of two variables to be specified.

In the general model of a Pólya urn ($b = c = 0$, $a = d$), the function $G$ is defined on $U$ by $$G(x,u,v) = u^{a_{0}} v^{b_{0}} (1 - axu^{a})^{-a_{0}/a} (1 - axv^{a})^{-b_{0}/a}$$ and admits the expansion $G(x,u,v) = \sum_{n=0}^{+\infty} Q_{n}(u,v) \frac{x^{n}}{n!}$ on $D_{\rho}$.
Justify that $G$ admits a first-order partial derivative with respect to $x$ on the domain $D_{\rho}$, obtained by term-by-term differentiation with respect to $x$ of the expression for $G$.

Let $f(x) = xe^x$. Prove that the mapping $f$ establishes a bijection from the interval $] - \infty , - 1 ]$ onto the interval $\left[ - \mathrm { e } ^ { - 1 } , 0 [ \right.$. In the rest of the problem, the inverse of this bijection is denoted $V$.