Prove that $$\frac { 2 } { 0 ! + 1 ! + 2 ! } + \frac { 3 } { 1 ! + 2 ! + 3 ! } + \cdots + \frac { n } { ( n - 2 ) ! + ( n - 1 ) ! + n ! } = 1 - \frac { 1 } { n ! }$$

If $f ( x ) = \frac { x ^ { n } } { n ! } + \frac { x ^ { n - 1 } } { ( n - 1 ) ! } + \cdots + x + 1$, then show that $f ( x ) = 0$ has no repeated roots.

Let $ABCD$ be a parallelogram. Let $O$ be a point in its interior such that $\angle AOB + \angle DOC = 180^{\circ}$. Show that $\angle ODC = \angle OBC$.

Let $a_{1}, \ldots, a_{n}$ be distinct complex numbers. Show that the functions $e^{a_{1} z}, \ldots, e^{a_{n} z}$ are linearly independent over $\mathbb{C}$.

The Frattini subgroup of a finite group $G$ is the intersection of all its proper maximal subgroups. Let $p$ be a prime number. Show that the Frattini subgroup of $\mathbb{Z} / p^{n}$, $n \geq 2$, is generated by $p$.

21. (Total Score: 16 points) Subquestion 1 is worth 4 points, Subquestion 2 is worth 6 points, Subquestion 3 is worth 6 points As shown in the figure, $P - ABC$ is a regular triangular pyramid with base edge length 1. Points $D$, $E$, $F$ are on edges $PA$, $PB$, $PC$ respectively. The cross-section $DEF$ is parallel to the base $ABC$, and the sum of edge lengths of the frustum $DEF - ABC$ equals the sum of edge lengths of the pyramid $P - ABC$. (The sum of edge lengths is the sum of the lengths of all edges of a polyhedron)
(1) Prove: $P - ABC$ is a regular tetrahedron;
(2) If $PD = \frac { 1 } { 2 } PA$, find the dihedral angle $D - BC - A$. (Express the result using inverse trigonometric functions)
(3) Let the volume of the frustum $DEF - ABC$ be $V$. Does there exist a right parallelepiped with all edges equal and volume $V$ such that it has the same sum of edge lengths as the frustum $DEF - ABC$? If it exists, construct such a parallelepiped explicitly and provide a proof; if it does not exist, explain why. [Figure]

Justify that the triangles $B C M$ and $A B M$ have the same area without calculating it.

Prove that the quadrilateral ABFE is a trapezoid with two sides of equal length.

Establish that $\tau = \tau _ { 0 } \cup \tau _ { 1 }$.

Let $q \in Q(E)$.
Show that there exists a unique symmetric bilinear form on $E$, denoted $\varphi$, such that $q = q_\varphi$.

For the rest of this problem, we assume that $\varphi$ is a non-degenerate symmetric bilinear form on $E$, and we denote by $q$ its quadratic form.
Let $e = (e_1, \ldots, e_n)$ be a basis of $E$. We still denote by $e^* = (e_1^*, \ldots, e_n^*)$ the dual basis of $e$. Let $p \in \{1, \ldots, n\}$. We denote by $F$ the space spanned by $e_1, \ldots, e_p$.
a) Show that $F^\perp$ is the preimage under $h$ of $\operatorname{Vect}(e_{p+1}^*, \ldots, e_n^*)$, where $h$ is defined in I.A.1.
b) Show that $\operatorname{dim}(F) + \operatorname{dim}(F^\perp) = n$.
c) Show that $(F^\perp)^\perp = F$.

For the rest of this problem, we assume that $\varphi$ is a non-degenerate symmetric bilinear form on $E$, and we denote by $q$ its quadratic form.
Let $F$ and $G$ be two vector subspaces of $E$.
a) Show that $(F+G)^\perp = F^\perp \cap G^\perp$.
b) Show that $(F \cap G)^\perp = F^\perp + G^\perp$.

We wish to prove the Cartan-Dieudonné theorem, whose statement is: ``if $f \in O(E,q)$, $f$ is the composition of at most $n$ reflections, where $n = \operatorname{dim}(E)$, with the convention that $\operatorname{Id}_E$ is the composition of 0 reflections.''
Prove the Cartan-Dieudonné theorem when $n = 1$.

We denote by $C _ { 2 \pi , 2 \pi } \left( \mathbb { R } ^ { 2 } ; \mathbb { C } \right)$ the set of continuous functions $f : \mathbb { R } ^ { 2 } \rightarrow \mathbb { C }$ such that: $$\forall \left( \theta _ { 1 } , \theta _ { 2 } \right) \in \mathbb { R } ^ { 2 } , f \left( \theta _ { 1 } + 2 \pi , \theta _ { 2 } \right) = f \left( \theta _ { 1 } , \theta _ { 2 } \right) = f \left( \theta _ { 1 } , \theta _ { 2 } + 2 \pi \right)$$
Let $f \in C _ { 2 \pi , 2 \pi } \left( \mathbb { R } ^ { 2 } ; \mathbb { C } \right)$. Prove that $$\sup _ { \left( \theta _ { 1 } , \theta _ { 2 } \right) \in \mathbb { R } ^ { 2 } } \left| f \left( \theta _ { 1 } , \theta _ { 2 } \right) \right| = \sup _ { \left( \theta _ { 1 } , \theta _ { 2 } \right) \in [ 0,2 \pi ] ^ { 2 } } \left| f \left( \theta _ { 1 } , \theta _ { 2 } \right) \right|$$ Deduce that $\left( \theta _ { 1 } , \theta _ { 2 } \right) \mapsto \left| f \left( \theta _ { 1 } , \theta _ { 2 } \right) \right|$ is bounded on $\mathbb { R } ^ { 2 }$ and attains its supremum.

We denote $\mathcal{I} = \{(j, k) \in \mathbf{N}^{2} \mid j \in \mathbf{N} \text{ and } 0 \leq k < 2^{j}\}$; for $j \in \mathbf{N}$, $\mathcal{T}_{j} = \{k \in \mathbf{N} \mid 0 \leq k < 2^{j}\}$. Let $a_{j,k}$ be a family of real numbers indexed by $(j, k) \in \mathcal{I}$. We denote $b_{j} = \max_{k \in \mathcal{T}_{j}} |a_{j,k}|$, and we suppose that the series $\sum b_{j}$ is convergent.
For all $j \in \mathbf{N}$, let $f_{j}^{a}$ be the function defined by $$f_{j}^{a}(x) = \sum_{k \in \mathcal{T}_{j}} a_{j,k} \theta_{j,k}(x)$$ Show that the series $\sum f_{j}^{a}$ is uniformly convergent on $[0,1]$ towards a function denoted $f^{a}$, which belongs to $\mathcal{C}_{0}$ and which satisfies, for all $(j, k) \in \mathcal{I}$, $c_{j,k}(f^{a}) = a_{j,k}$.

For all $n \in \mathbf{N}$, let $S_{n} f$ be the function of $\mathcal{C}_{0}$ defined by $$S_{n} f = \sum_{j=0}^{n} \sum_{k \in \mathcal{T}_{j}} c_{j,k}(f) \theta_{j,k}$$ Show that for all $n \in \mathbf{N}$ and all $\ell \in \mathcal{T}_{n+1}$, the function $S_{n} f$ is affine on the interval $[\ell 2^{-n-1}, (\ell+1) 2^{-n-1}]$.

For all $n \in \mathbf{N}$, let $S_{n} f$ be the function of $\mathcal{C}_{0}$ defined by $$S_{n} f = \sum_{j=0}^{n} \sum_{k \in \mathcal{T}_{j}} c_{j,k}(f) \theta_{j,k}$$ where, for all $(j, k) \in \mathcal{I}$, $$c_{j,k}(f) = f\left(\left(k + \frac{1}{2}\right) 2^{-j}\right) - \frac{f(k 2^{-j}) + f((k+1) 2^{-j})}{2} .$$ Let $n \in \mathbf{N}$. Suppose that for all $\ell \in \mathcal{T}_{n}$, $(S_{n-1} f)(\ell 2^{-n}) = f(\ell 2^{-n})$. Show that we also have that for all $\ell \in \mathcal{T}_{n+1}$, $(S_{n} f)(\ell 2^{-n-1}) = f(\ell 2^{-n-1})$.
One may distinguish cases according to the parity of $\ell$.

We denote $\alpha > -1/2$, $E$ the $\mathbb{R}$-vector space of functions of class $\mathcal{C}^\infty$ on $[-1,1]$ with real values, $$\varphi_\alpha(y) : t \mapsto \left(1-t^2\right)y''(t) - (2\alpha+1)t\,y'(t)$$ and $$S_\alpha(f,g) = \int_{-1}^{1} f(t)g(t)\left(1-t^2\right)^{\alpha - \frac{1}{2}} \mathrm{~d}t$$ Show that $$\forall (f,g) \in E^2, \quad S_\alpha\left(\varphi_\alpha(f), g\right) = S_\alpha\left(f, \varphi_\alpha(g)\right)$$ One may calculate the derivative of $t \mapsto \left(1-t^2\right)^{\alpha+\frac{1}{2}} f'(t)$.

Let $P \in \mathcal{P}$. Show that $P$ decomposes uniquely in the form: $$P(x,y) = H(x,y) + (1 - x^2 - y^2) Q(x,y)$$ where $H$ is a harmonic polynomial and $Q \in \mathcal{P}$.

In this part, we consider two real symmetric matrices $A, B \in \mathcal{S}_{n}(\mathbb{R})$ and their sum $C = A + B$. We denote by $a = s^{\downarrow}(A)$, $b = s^{\downarrow}(B)$ and $c = s^{\downarrow}(C)$.
Show that $$\sum_{i=1}^{n} c_{i} = \sum_{i=1}^{n} a_{i} + \sum_{i=1}^{n} b_{i}.$$

Let $M \in \mathscr{M}_{N,d}(\mathbb{R})$ defined by $M_{i,j} = g_{j}(i)$, $q \in \Sigma_{N}$, and for all $\theta \in \mathbb{R}^{d}$, $f(\theta) = M\theta$, $Z(\theta) = \sum_{i=1}^{N} e^{f_{i}(\theta)}$, $$p(\theta) = \left(\frac{e^{f_{1}(\theta)}}{Z(\theta)}, \ldots, \frac{e^{f_{N}(\theta)}}{Z(\theta)}\right) \in \Sigma_{N},$$ and $L(\theta) = \ln(Z(\theta)) - q^{T} M\theta$.
Show that $L$ is of class $\mathscr{C}^{2}$ and that for all integers $1 \leqslant l, k \leqslant d$ we have $$\frac{\partial^{2} L}{\partial \theta_{l} \partial \theta_{k}}(\theta) = \sum_{i=1}^{N} p_{i}(\theta)(M_{il} - m_{l}(\theta))(M_{ik} - m_{k}(\theta))$$ where $m(\theta) = M^{T} p(\theta)$.

The difference operator is the endomorphism $\delta$ of $\mathbb{R}_n[X]$ such that $\delta = \tau - \operatorname{Id}_{\mathbb{R}_n[X]}$: $$\delta : \left\{ \begin{array}{l} \mathbb{R}_n[X] \rightarrow \mathbb{R}_n[X] \\ P(X) \mapsto P(X+1) - P(X) \end{array} \right.$$ Let $P \in \mathbb{R}_{n-1}[X]$. Show that $$\sum_{j=0}^{n} (-1)^{n-j} \binom{n}{j} P(j) = 0$$

We consider the family of polynomials $$\left\{ \begin{array}{l} H_0 = 1 \\ H_k = \frac{1}{k!} \prod_{j=0}^{k-1} (X - j) \quad \text{for } k \in \llbracket 1, n \rrbracket \end{array} \right.$$
Show that, for $k, l \in \llbracket 0, n \rrbracket$, $$\delta^k\left(H_l\right)(0) = \begin{cases} 1 & \text{if } k = l \\ 0 & \text{if } k \neq l \end{cases}$$

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

Let $\Lambda$ be a non-empty subset of $\mathbb{R}_*^+$ such that $$\forall (x, y) \in \Lambda^2, \quad x + y \in \Lambda$$ We say that $\Lambda$ is closed under addition. Show that if $(x, y) \in \Lambda^2$, $(k, n) \in \mathbb{N} \times \mathbb{N}^*$ and $k \leqslant n$, then $nx + k(y-x) \in \Lambda$.