In this part, $E$ denotes the vector space of functions $f : [0,1] \rightarrow \mathbb{R}$ continuous, equipped with the inner product defined by, $$\forall (f,g) \in E^2, \quad \langle f, g \rangle = \int_0^1 f(t) g(t) \, \mathrm{d}t$$ For all $s \in [0,1]$, we define the function $k_s$ by, $$\forall t \in [0,1], \quad k_s(t) = \begin{cases} t(1-s) & \text{if } t < s \\ s(1-t) & \text{if } t \geqslant s. \end{cases}$$ For all $f \in E$, we set, $$\forall s \in [0,1], \quad T(f)(s) = \int_0^1 k_s(t) f(t) \, \mathrm{d}t$$ For all $f \in E$, show that $T(f)$ is of class $\mathcal{C}^2$ then that $T(f)'' = -f$.

For all $f \in E$, we set, $$\forall s \in [0,1], \quad T(f)(s) = \int_0^1 k_s(t) f(t)\,\mathrm{d}t$$ Show that $T$ is injective.

In this part, $E$ denotes the vector space of functions $f : [0,1] \rightarrow \mathbb{R}$ continuous, equipped with the inner product defined by, $$\forall (f,g) \in E^2, \quad \langle f, g \rangle = \int_0^1 f(t) g(t) \, \mathrm{d}t$$ For all $s \in [0,1]$, we define the function $k_s$ by, $$\forall t \in [0,1], \quad k_s(t) = \begin{cases} t(1-s) & \text{if } t < s \\ s(1-t) & \text{if } t \geqslant s. \end{cases}$$ For all $f \in E$, we set, $$\forall s \in [0,1], \quad T(f)(s) = \int_0^1 k_s(t) f(t) \, \mathrm{d}t$$ Show that $T$ is injective.

For all $f \in E$, we set, $$\forall s \in [0,1], \quad T(f)(s) = \int_0^1 k_s(t) f(t)\,\mathrm{d}t$$ Determine the image of $T$.

In this part, $E$ denotes the vector space of functions $f : [0,1] \rightarrow \mathbb{R}$ continuous, equipped with the inner product defined by, $$\forall (f,g) \in E^2, \quad \langle f, g \rangle = \int_0^1 f(t) g(t) \, \mathrm{d}t$$ For all $s \in [0,1]$, we define the function $k_s$ by, $$\forall t \in [0,1], \quad k_s(t) = \begin{cases} t(1-s) & \text{if } t < s \\ s(1-t) & \text{if } t \geqslant s. \end{cases}$$ For all $f \in E$, we set, $$\forall s \in [0,1], \quad T(f)(s) = \int_0^1 k_s(t) f(t) \, \mathrm{d}t$$ Determine the image of $T$.

We define a sequence $(a_n)_{n \geqslant 1}$ by setting $$\forall n \in \mathbb{N}^*, \quad a_n = \frac{(-n)^{n-1}}{n!}.$$ We define, when possible, $S(x) = \sum_{n=1}^{+\infty} a_n x^n$, with radius of convergence $R$. We consider the function $$h : \begin{array}{ccc} ]-R,R[ & \rightarrow & \mathbb{R} \\ x & \mapsto & S(x)\mathrm{e}^{S(x)} \end{array}$$ Prove that $h$ is a solution on $]-R, R[$ of the differential equation $xy' - y = 0$.

Solve the differential equation $xy' - y = 0$ on each of the intervals $]0, R[$ and $]-R, 0[$ then on the interval $]-R, R[$.

We assume that $I = [a,b]$ with $a < b$, $\forall x \in I, w(x) = 1$ (general weight $w$ in the formula for $e(f)$), and that $f$ is of class $\mathcal{C}^{m+1}$ on $I$, where $m$ is the order of the quadrature formula $I_n(f) = \sum_{j=0}^n \lambda_j f(x_j)$.
For every natural number $m$, consider the function $\varphi_m : \mathbb{R}^2 \rightarrow \mathbb{R}$ defined by $$\forall (x,t) \in \mathbb{R}^2, \quad \varphi_m(x,t) = \begin{cases} (x-t)^m & \text{if } x \geqslant t, \\ 0 & \text{if } x < t. \end{cases}$$
Using the Taylor formula with integral remainder, show that $e(f) = e(R_m)$, where $R_m$ is defined by $$\forall x \in [a,b], \quad R_m(x) = \frac{1}{m!} \int_a^b \varphi_m(x,t) f^{(m+1)}(t)\,\mathrm{d}t.$$

We assume that $I = [a,b]$ with $a < b$, and that $f$ is of class $\mathcal{C}^{m+1}$ on $I$, where $m \geqslant 1$ is the order of the quadrature formula $I_n(f) = \sum_{j=0}^n \lambda_j f(x_j)$.
For every natural number $m$, consider the function $\varphi_m : \mathbb{R}^2 \rightarrow \mathbb{R}$ defined by $$\forall (x,t) \in \mathbb{R}^2, \quad \varphi_m(x,t) = \begin{cases} (x-t)^m & \text{if } x \geqslant t, \\ 0 & \text{if } x < t. \end{cases}$$
Deduce that, if $m \geqslant 1$, $$e(f) = \frac{1}{m!} \int_a^b K_m(t) f^{(m+1)}(t)\,\mathrm{d}t$$ where the function $K_m : [a,b] \rightarrow \mathbb{R}$ is defined by $$\forall t \in [a,b], \quad K_m(t) = e\left(x \mapsto \varphi_m(x,t)\right) = \int_a^b \varphi_m(x,t) w(x)\,\mathrm{d}x - \sum_{j=0}^n \lambda_j \varphi_m(x_j, t).$$ You may use the following admitted result: for every continuous function $g : [a,b]^2 \rightarrow \mathbb{R}$, we have $$\int_a^b \left(\int_a^b g(x,t)\,\mathrm{d}t\right)\mathrm{d}x = \int_a^b \left(\int_a^b g(x,t)\,\mathrm{d}x\right)\mathrm{d}t.$$

We assume that $I = [0,1]$, $\forall x \in I, w(x) = 1$, and we consider the quadrature formula $$I_1(g) = \frac{g(0) + g(1)}{2},$$ which is of order $m = 1$.
Calculate the associated Peano kernel $t \mapsto K_1(t)$ and show that, for every function $g$ of class $\mathcal{C}^2$ from $[0,1]$ to $\mathbb{R}$, we have the following bound on the associated quadrature error: $$|e(g)| \leqslant \frac{1}{12} \sup_{x \in [0,1]} |g''(x)|.$$

Let $n \in \mathbb{N}$ be a non-zero natural integer. We define, for any real number $x$, $$\Phi_n(x) = \mathrm{e}^{-x} x^n \quad \text{and} \quad L_n(x) = \frac{\mathrm{e}^x}{n!} \Phi_n^{(n)}(x).$$ For any real number $x$, express $\Phi_n^{(n)}(x)$ and $\Phi_n^{(n+1)}(x)$ in terms of $L_n(x)$ and $L_n'(x)$.

Let $n \in \mathbb{N}$ be a non-zero natural integer. We define, for any real number $x$, $$\Phi_n(x) = \mathrm{e}^{-x} x^n \quad \text{and} \quad L_n(x) = \frac{\mathrm{e}^x}{n!} \Phi_n^{(n)}(x).$$ Use the equality $\Phi_{n+1}^{(n+1)}(x) = \frac{\mathrm{d}^{n+1} x\Phi_n(x)}{\mathrm{d}x^{n+1}}$, which we will justify, to prove the equality $$L_{n+1}(x) = \left(1 - \frac{x}{n+1}\right) L_n(x) + \frac{x}{n+1} L_n'(x)$$ valid for any real number $x$.

Let $n \in \mathbb{N}$ be a non-zero natural integer. We define, for any real number $x$, $$\Phi_n(x) = \mathrm{e}^{-x} x^n \quad \text{and} \quad L_n(x) = \frac{\mathrm{e}^x}{n!} \Phi_n^{(n)}(x).$$ Use the equality $\Phi_{n+1}^{(n+2)}(x) = \frac{\mathrm{d}^{n+1} \Phi_{n+1}^{(1)}}{\mathrm{d}x^{n+1}}(x)$ to prove the equality $$L_{n+1}'(x) = L_n'(x) - L_n(x)$$ valid for any real number $x$.

Let $n \in \mathbb{N}$ be a non-zero natural integer. We define, for any real number $x$, $$\Phi_n(x) = \mathrm{e}^{-x} x^n \quad \text{and} \quad L_n(x) = \frac{\mathrm{e}^x}{n!} \Phi_n^{(n)}(x).$$ Deduce that $L_n$ is a solution of the differential equation $$x L_n''(x) + (1-x) L_n'(x) + n L_n(x) = 0.$$

Construct a function $\rho \in \mathcal{C}^\infty(\mathbb{R})$, constant equal to 1 on $[-1,1]$ and constant equal to 0 on $\mathbb{R} \setminus [-2,2]$.

Construct a function $\rho \in \mathcal{C}^\infty(\mathbb{R})$, constant equal to 1 on $[-1,1]$ and constant equal to 0 on $\mathbb{R} \setminus [-2,2]$.

Let $n \in \mathbb{N}$ be a non-zero natural integer. We define, for any real number $x$, $$\Phi_n(x) = \mathrm{e}^{-x} x^n \quad \text{and} \quad L_n(x) = \frac{\mathrm{e}^x}{n!} \Phi_n^{(n)}(x).$$ The confluent hypergeometric function $M_{a,c}$ is the solution of $$x y''(x) + (c-x) y'(x) - a y(x) = 0$$ satisfying $M_{a,c}(0) = 1$. Show that $L_n$ is a confluent hypergeometric function.

Let $r$ be the function from $\mathbb{R}$ to $\mathbb{C}$ such that, for all real $x$, $$r(x) = \frac{1}{2\pi} \int_{-\infty}^{+\infty} \mathrm{e}^{\mathrm{i}x\xi} \rho(\xi) \,\mathrm{d}\xi$$ where $\rho \in \mathcal{C}^\infty(\mathbb{R})$ is constant equal to 1 on $[-1,1]$ and constant equal to 0 on $\mathbb{R} \setminus [-2,2]$.
Show that $r$ is differentiable on $\mathbb{R}$ and give an expression for its derivative function (possibly involving an integral).

Let $\rho$ be the function constructed in question 30. Let $r$ be the function from $\mathbb{R}$ to $\mathbb{C}$ such that, for all real $x$, $$r(x) = \frac{1}{2\pi} \int_{-\infty}^{+\infty} \mathrm{e}^{\mathrm{i}x\xi} \rho(\xi)\,\mathrm{d}\xi$$ Show that $r$ is differentiable on $\mathbb{R}$ and give an expression for its derivative function (possibly involving an integral).

Let $r$ be the function from $\mathbb{R}$ to $\mathbb{C}$ such that, for all real $x$, $$r(x) = \frac{1}{2\pi} \int_{-\infty}^{+\infty} \mathrm{e}^{\mathrm{i}x\xi} \rho(\xi) \,\mathrm{d}\xi$$ where $\rho \in \mathcal{C}^\infty(\mathbb{R})$ is constant equal to 1 on $[-1,1]$ and constant equal to 0 on $\mathbb{R} \setminus [-2,2]$.
Show that $x \mapsto x^2 r(x)$ is bounded on $\mathbb{R}$ and deduce that $r$ is integrable and bounded on $\mathbb{R}$.

Let $\rho$ be the function constructed in question 30. Let $r$ be the function from $\mathbb{R}$ to $\mathbb{C}$ such that, for all real $x$, $$r(x) = \frac{1}{2\pi} \int_{-\infty}^{+\infty} \mathrm{e}^{\mathrm{i}x\xi} \rho(\xi)\,\mathrm{d}\xi$$ Show that $x \mapsto x^2 r(x)$ is bounded on $\mathbb{R}$ and deduce that $r$ is integrable and bounded on $\mathbb{R}$.

Let $\lambda > 0$ and let $f \in L^1(\mathbb{R}) \cap \mathcal{C}^1(\mathbb{R})$ such that $\hat{f} \in L^1(\mathbb{R})$ and such that $\hat{f}$ is zero outside the segment $[-\lambda, \lambda]$. We denote by $r_\lambda$ the function from $\mathbb{R}$ to $\mathbb{C}$ such that $r_\lambda(x) = r(\lambda x)$ for all real $x$, where $r(x) = \frac{1}{2\pi} \int_{-\infty}^{+\infty} \mathrm{e}^{\mathrm{i}x\xi} \rho(\xi) \,\mathrm{d}\xi$.
We admit that $f * r_\lambda$ is integrable. Show that $f = \lambda f * r_\lambda$.

Let $\lambda > 0$ and let $f \in L^1(\mathbb{R}) \cap \mathcal{C}^1(\mathbb{R})$ such that $\hat{f} \in L^1(\mathbb{R})$ and such that $\hat{f}$ is zero outside the segment $[-\lambda, \lambda]$. We denote by $r_\lambda$ the function from $\mathbb{R}$ to $\mathbb{C}$ such that $r_\lambda(x) = r(\lambda x)$ for all real $x$, where $r(x) = \frac{1}{2\pi}\int_{-\infty}^{+\infty} \mathrm{e}^{\mathrm{i}x\xi}\rho(\xi)\,\mathrm{d}\xi$ and $\rho$ is the function from question 30. We admit that $f * r_\lambda$ is integrable. Show that $f = \lambda f * r_\lambda$.

We assume that $f$ is a function from $\mathbb { R } _ { + } ^ { * }$ to $\mathbb { R }$ of class $\mathcal { C } ^ { 1 }$ satisfying $$\left\{ \begin{array} { l } \lim _ { x \rightarrow 0 } f ( x ) = 0 \\ \exists C > 0 ; \forall x > 0 , \quad \left| f ^ { \prime } ( x ) \right| \leqslant C \frac { \mathrm { e } ^ { x / 2 } } { \sqrt { x } } \end{array} \right.$$ Show that, for all $x > 0 , | f ( x ) | \leqslant 4 C \frac { \sqrt { x } \mathrm { e } ^ { x / 2 } } { 1 + x }$.

For all $n \in \mathbb{N}^{\star}$, justify that there exists a unique function $f_n \in \mathcal{C}^{2}(]0, +\infty[)$ satisfying $f_n(1) = 0$, $f_n(2) = 0$ and $f_n^{\prime\prime}(x) = (-1)^n 2^{-nx^2}$ for all $x > 0$.