Show that $h(x) = \sum_{n=0}^{\infty} \frac{(2n)!(3n)!}{(n!)^5} x^n$ has integer coefficients.

By admitting that the result of question 7 is valid for the functions $P_{t}(f)$ and $1 + \ln\left(P_{t}(f)\right)$, show that
$$\forall t \in \mathbf{R}_{+}^{*}, \quad -S^{\prime}(t) = \mathrm{e}^{-2t} \int_{-\infty}^{+\infty} \frac{P_{t}\left(f^{\prime}\right)(x)^{2}}{P_{t}(f)(x)} \varphi(x) \mathrm{d}x$$

Show that a power series $f(x) = \sum_{n=0}^{\infty} \frac{c_n}{n!} x^n \in \mathbf{Q}\llbracket x \rrbracket$ is a solution of a differential equation if and only if its Laplace transform $$\widehat{f}(x) \stackrel{\text{def}}{=} \sum_{n=0}^{\infty} c_n x^n$$ is a solution of a differential equation.

By using the Cauchy-Schwarz inequality, show that
$$\forall t \in \mathbf{R}_{+}^{*}, \quad -S^{\prime}(t) \leq \mathrm{e}^{-2t} \int_{-\infty}^{+\infty} P_{t}\left(\frac{f^{\prime 2}}{f}\right)(x) \varphi(x) \mathrm{d}x$$

Let $r \geq 2$ be an integer and $a_1, \ldots, a_r \in \mathbf{Q}$ be distinct rationals. Let $b_1, \ldots, b_r \in \mathbf{Q}^{\times}$ be nonzero rationals. Set $e^{a_i x} \stackrel{\text{def}}{=} \sum_{n=0}^{\infty} \frac{a_i^n}{n!} x^n$ and consider the power series $$f(x) = \sum_{n=0}^{\infty} \frac{u_n}{n!} x^n \stackrel{\text{def}}{=} b_1 e^{a_1 x} + \cdots + b_r e^{a_r x}.$$ Show that the Laplace transform $\widehat{f}(x) = \sum_{n=0}^{\infty} u_n x^n$ is the power series expansion of the rational function $$\sum_{i=1}^{r} \frac{b_i}{1 - a_i x}.$$ Deduce that $f$ is not the zero power series.

Deduce that we have:
$$\forall t \in \mathbf{R}_{+}^{*}, \quad -S^{\prime}(t) \leq \mathrm{e}^{-2t} \int_{-\infty}^{+\infty} \frac{f^{\prime 2}(x)}{f(x)} \varphi(x) \mathrm{d}x$$

Consider the sequence $(v_n)_{n \geq 0}$ defined in terms of the coefficients $u_n$ by the formula $$v_n = n! \sum_{i=0}^{n} \frac{u_i}{i!}$$ and the power series $$v(x) = \sum_{n=0}^{\infty} v_n x^n \in \mathbf{Q}\llbracket x \rrbracket.$$ Show the equality of power series $$\sum_{n=0}^{\infty} (v_n - n v_{n-1}) x^n = \sum_{n=0}^{\infty} u_n x^n.$$

Establish the following inequality
$$\operatorname{Ent}_{\varphi}(f) \leq \frac{1}{2} \int_{-\infty}^{+\infty} \frac{f^{\prime 2}(x)}{f(x)} \varphi(x) \mathrm{d}x$$

With the notation of question 18 and 19, show that the differential operator $L = -x^2 \left(\frac{d}{dx}\right) + (1-x)$ acts on $v(x)$ by $$(L \cdot v)(x) = \sum_{i=1}^{r} \frac{b_i}{1 - a_i x}.$$

With the notation of questions 18--20, deduce that if $v(x)$ is the power series expansion of a rational fraction $P/Q$, then every element of the non-empty set $\{1/a_i \mid a_i \neq 0\}$ is a pole of $P/Q$.

Show that $v(x)$ is not the power series expansion of a rational fraction.

Show that Theorem 1 is equivalent to the following statement:
Let $f(x) \in \mathbf{Q}\llbracket x \rrbracket$ be an exponential polynomial such that $f(1) = \sum_{i=1}^{s} P_i(1) e^{c_i}$ vanishes. Then $f(x)/(x-1)$ is still an exponential polynomial.
(An exponential polynomial is any power series with rational coefficients of the form $f(x) = \sum_{i=1}^{s} P_i(x) e^{c_i x}$, where $c_1, \ldots, c_s \in \mathbf{Q}$ are rationals and $P_1, \ldots, P_s \in \mathbf{Q}[x]$ are polynomials.)

For all integers $n, k \geq 0$, define the rational number $v_n(k)$ as the coefficient of degree $n$ in the power series $$\left(1 - s_1 x - \cdots - s_r x^r\right)^k v(x) = \sum_{n=0}^{\infty} v_n(k) x^n.$$ Show that $v(x)$ is the power series expansion of a rational fraction if and only if there exists an integer $k \geq 0$ such that $\sum_{n=0}^{\infty} v_n(k) x^n$ is a polynomial.

3. Show that on $] 0 , \infty [$ we have $f = g$. For this you may use a differential equation satisfied by $( f - g )$ and use the behavior of $f$ and $g$ at $+ \infty$.

We are given $f \in \mathcal{C}^1(\mathbb{R})$ such that $f'$ is $L$-Lipschitzian, with $L > 0$, and we fix $\tau$ such that $0 < \tau \leq 2/L$. We further suppose that $f$ is $\alpha$-convex, with $\alpha > 0$, that is $$g(x) := f(x) - \frac{1}{2}\alpha x^2 \quad \text{is a convex function on } \mathbb{R}$$ Justify that $f'(x) - \alpha x$ is an increasing function of $x \in \mathbb{R}$. Deduce that $\alpha \leq L$.

4. Using the expression of $g$ obtained in question 2.b., show that $g$ is continuous at 0.

We are given $f \in \mathcal{C}^1(\mathbb{R})$ such that $f'$ is $L$-Lipschitzian, with $L > 0$, and $f$ is $\alpha$-convex with $\alpha > 0$, that is $g(x) := f(x) - \frac{1}{2}\alpha x^2$ is a convex function on $\mathbb{R}$. Show that $f(x) \geq f(0) + f'(0)x + \alpha x^2/2$ for all $x \in \mathbb{R}$. Deduce that $f$ admits a minimizer on $\mathbb{R}$.

We are given $f \in \mathcal{C}^1(\mathbb{R})$ such that $f'$ is $L$-Lipschitzian, with $L > 0$, $f$ is $\alpha$-convex with $\alpha > 0$, and $x_*$ denotes a minimizer of $f$. Show that for all $x, y \in \mathbb{R}$ $$\alpha|x-y|^2 \leq \left(f'(x) - f'(y)\right)(x-y)$$

We are given $f \in \mathcal{C}^1(\mathbb{R})$ such that $f'$ is $L$-Lipschitzian, with $L > 0$, $f$ is $\alpha$-convex with $\alpha > 0$, and $x_*$ denotes a minimizer of $f$. Deduce that for all $x, y \in \mathbb{R}$, denoting $\tilde{x} := x - \tau f'(x)$ and $\tilde{y} := y - \tau f'(y)$, we have $$|\tilde{x} - \tilde{y}|^2 \leq |x-y|^2(1 - \alpha\tau(2 - L\tau)).$$

We are given $f \in \mathcal{C}^1(\mathbb{R})$ such that $f'$ is $L$-Lipschitzian, with $L > 0$, $f$ is $\alpha$-convex with $\alpha > 0$, and $x_*$ denotes a minimizer of $f$. We suppose $0 < \tau < 2/L$. Show that $\left|x_n - x_*\right| \leq \rho^n \left|x_0 - x_*\right|$, where $\rho$ is a constant that we will specify, and such that $0 \leq \rho < 1$.

Consider the function $$f(x) := \frac{1}{3}x^3 \quad \text{if } x \geq 0, \quad f(x) := 0 \quad \text{if } x < 0$$ and the sequence $(x_n)_{n \in \mathbb{N}}$ defined by $x_{n+1} := x_n - \tau f'(x_n)$. We suppose only $\tau > 0$. Show that for all $x_0 \in \mathbb{R}$, the sequence $\left(x_n\right)_{n \in \mathbb{N}}$ converges to a minimizer of $f$.

Deduce, using question 6, that if $p$ has a stable root then $J(p)$ is not invertible.

Suppose that $S _ { 0 } = 0$. Give the expression of the solution triplet $( S , I , R )$ of system $( F )$.
The system $(F)$ is: $$( F ) : \left\{ \begin{array} { l } S ^ { \prime } ( x ) = - I ( x ) S ( x ) \\ I ^ { \prime } ( x ) = I ( x ) S ( x ) - I ( x ) \\ R ^ { \prime } ( x ) = I ( x ) \\ S ( 0 ) = S _ { 0 } , \quad I ( 0 ) = I _ { 0 } , \quad R ( 0 ) = R _ { 0 } \end{array} \right.$$ with $S_0 + I_0 + R_0 = 1$ and $S_0, I_0, R_0 \in [0,1]$.

Show that if $S _ { 0 } > 0$ then the function $S$ of the solution triplet $( S , I , R )$ of $( F )$ never vanishes, and deduce that $S$ is strictly positive.
The system $(F)$ is: $$( F ) : \left\{ \begin{array} { l } S ^ { \prime } ( x ) = - I ( x ) S ( x ) \\ I ^ { \prime } ( x ) = I ( x ) S ( x ) - I ( x ) \\ R ^ { \prime } ( x ) = I ( x ) \\ S ( 0 ) = S _ { 0 } , \quad I ( 0 ) = I _ { 0 } , \quad R ( 0 ) = R _ { 0 } \end{array} \right.$$

We are given $f \in \mathcal{C}^1(\mathbb{R})$, convex, admitting a minimizer $x_* \in \mathbb{R}$, with $f'$ being $L$-Lipschitzian, and $0 < \tau < 2/L$. The sequence $(x_n)_{n \in \mathbb{N}}$ is defined by $x_{n+1} := x_n - \tau f'(x_n)$. Show that $\frac{\tau}{2}(2 - \tau L)\sum_{0 \leq i < n}\left|f'(x_i)\right|^2 \leq \left(f(x_0) - f(x_n)\right)$ for all $n \in \mathbb{N}^*$. Deduce that $f'(x_n) \rightarrow 0$ when $n \rightarrow \infty$.