Let $n \in \mathbf{N}$. Show that for all real $t > 0$, $$p_n = \frac{e^{nt} P(e^{-t})}{2\pi} \int_{-\pi}^{\pi} e^{-in\theta} \frac{P(e^{-t}e^{i\theta})}{P(e^{-t})} \mathrm{d}\theta \tag{1}$$

We assume that $|\lambda| \notin \{0,1\}$ and that $\rho(f) > 0$. We consider the series $H \in O_2$ from part E. Show that the series $H$ satisfies $\hat{H} \prec \frac{1}{\omega} \hat{F} \circ (I + \hat{H})$.

We assume that $|\lambda| \notin \{0,1\}$ and that $\rho(f) > 0$. Using the result of question (22), conclude that the power series $h$ and $H$ of part E have strictly positive radius of convergence.

Taking $t = \frac { \pi } { \sqrt { 6 n } }$ in the formula
$$p _ { n } = \frac { e ^ { n t } P \left( e ^ { - t } \right) } { 2 \pi } \int _ { - \pi } ^ { \pi } e ^ { - i n \theta } \frac { P \left( e ^ { - t } e ^ { i \theta } \right) } { P \left( e ^ { - t } \right) } \mathrm { d } \theta$$
conclude that
$$p _ { n } = O \left( \frac { \exp \left( \pi \sqrt { \frac { 2 n } { 3 } } \right) } { n } \right) \quad \text { when } n \text { tends to } + \infty$$

By taking $t = \frac{\pi}{\sqrt{6n}}$ in formula $$p_n = \frac{e^{nt} P(e^{-t})}{2\pi} \int_{-\pi}^{\pi} e^{-in\theta} \frac{P(e^{-t}e^{i\theta})}{P(e^{-t})} \mathrm{d}\theta,$$ conclude that $$p_n = O\left(\frac{\exp\left(\pi\sqrt{\frac{2n}{3}}\right)}{n}\right) \quad \text{when } n \text{ tends to } +\infty.$$

We admit that $P \left( e ^ { - t } \right) \sim \sqrt { \frac { t } { 2 \pi } } \exp \left( \frac { \pi ^ { 2 } } { 6 t } \right)$ as $t$ tends to $0 ^ { + }$.
By applying formula (1) to $t = \frac { \pi } { \sqrt { 6 n } }$, prove that
$$p _ { n } \sim \frac { \exp \left( \pi \sqrt { \frac { 2 n } { 3 } } \right) } { 4 \sqrt { 3 } n } \quad \text { as } n \rightarrow + \infty$$

For $p \in \mathbb { N } ^ { * }$, let $h(x) = \mathrm{e}^{-x} P(x)$ where $P$ is a polynomial solution of $(E_p)$. We denote by $\left( b _ { n } \right) _ { n \in \mathbb { N } }$ the sequence of coefficients of the power series expansion of $h$, so that for all $x \in \mathbb { R }$, $h ( x ) = \sum _ { n = 0 } ^ { + \infty } b _ { n } x ^ { n }$. These coefficients satisfy $$\left\{ \begin{array} { l } b _ { 0 } = 0 \\ n ( n + 1 ) b _ { n + 1 } = - ( n + p ) b _ { n } , \quad \forall n \in \mathbb { N } ^ { * } . \end{array} \right.$$ Establish that, for all $n \in \mathbb { N } ^ { * } , b _ { n } = \frac { ( - 1 ) ^ { n - 1 } ( n + p - 1 ) ! } { p ! n ! ( n - 1 ) ! } b _ { 1 }$.

For all $n \in \mathbb { N } ^ { * }$, consider the points $a _ { k , n } = \frac { 2 k + 1 } { 2 n }$ for $k \in \llbracket 0 , n - 1 \rrbracket$. Let $\gamma > 0$ be such that $J_\alpha > 0$ for all $\alpha \in ]0, \gamma[$. Show that, for $\alpha \in ] 0 , \gamma [$, the sequence $\left( \left| \prod _ { k = 0 } ^ { n - 1 } \frac { 1 - a _ { k , n } ^ { 2 } } { \alpha ^ { 2 } + a _ { k , n } ^ { 2 } } \right| \right) _ { n \in \mathbb { N } ^ { * } }$ diverges to $+ \infty$.

For $p \notin \mathbb{N}^*$, $p \neq 0$, let $f(x) = \sum_{n=0}^{+\infty} a_n x^n$ be a non-zero power series solution of $(E_p) : x(y'' - y') + py = 0$, with coefficients satisfying $n(n+1)a_{n+1} = (n-p)a_n$ for all $n \in \mathbb{N}^*$. Show that there exists a natural integer $q > p$ such that, for all integer $n \geqslant q$, $$\left| a _ { n + 1 } \right| \geqslant \frac { \left| a _ { n } \right| } { 2 ( n + 1 ) }.$$

For $p \notin \mathbb{N}^*$, $p \neq 0$, let $f(x) = \sum_{n=0}^{+\infty} a_n x^n$ be a non-zero power series solution of $(E_p)$, with $q$ a natural integer such that $q > p$ and $\left| a _ { n + 1 } \right| \geqslant \frac { \left| a _ { n } \right| } { 2 ( n + 1 ) }$ for all $n \geqslant q$. Deduce that, for all integer $n \geqslant q , \left| a _ { n } \right| \geqslant \frac { q ! \left| a _ { q } \right| } { 2 ^ { n - q } n ! }$.

For $p \notin \mathbb{N}^*$, $p \neq 0$, let $f(x) = \sum_{n=0}^{+\infty} a_n x^n$ be a non-zero power series solution of $(E_p)$, with $q$ a natural integer such that $q > p$ and $\left| a _ { n } \right| \geqslant \frac { q ! \left| a _ { q } \right| } { 2 ^ { n - q } n ! }$ for all $n \geqslant q$. Show that the function $\psi : \left\lvert\, \begin{array} { c c c } \mathbb { R } _ { + } ^ { * } & \rightarrow & \mathbb { R } \\ x & \mapsto & \sum _ { n = q } ^ { + \infty } \left| a _ { n } \right| x ^ { n } \end{array} \right.$ is not an element of $E$.

Let $I = [-1,1]$, $\alpha > 0$, $a_{k,n} = \frac{2k+1}{2n}$ for $k \in \llbracket 0, n-1 \rrbracket$, $R_n \in \mathbb{R}_{2n-1}[X]$ the interpolating polynomial of $f_\alpha(x) = \frac{1}{\alpha^2+x^2}$ at $\{\pm a_{k,n} \mid k \in \llbracket 0,n-1\rrbracket\}$, and $\gamma > 0$ such that $J_\alpha > 0$ for all $\alpha \in ]0,\gamma[$. Suppose that $\alpha < \gamma$. Show that $$\lim _ { n \rightarrow + \infty } \left| f _ { \alpha } ( 1 ) - R _ { n } ( 1 ) \right| = + \infty.$$

For $p \notin \mathbb{N}^*$, $p \neq 0$, let $f(x) = \sum_{n=0}^{+\infty} a_n x^n$ be a non-zero power series solution of $(E_p)$. Using the result of Question 39, deduce finally that the function $f$ is not an element of $E$.

To each function $f \in E$, we associate the endomorphism $U$ of $E$. Let $\lambda \in \mathbb{R}^*$ be an eigenvalue of $U$ with eigenvector $f$, which is a solution of $(E_{1/\lambda})$. We assume that $f$ is developable as a power series on $\mathbb { R } _ { + } ^ { * }$, that is, there exists a power series $\sum _ { n \geqslant 0 } a _ { n } x ^ { n }$ of infinite radius of convergence such that $$\forall x \in \mathbb { R } _ { + } ^ { * } , \quad f ( x ) = \sum _ { n = 0 } ^ { + \infty } a _ { n } x ^ { n } .$$ Using the results of Part IV, show that the only possible eigenvalues of $U$ are of the form $\lambda = 1 / p$ with $p \in \mathbb { N } ^ { * }$.

Determine the domain of definition of $\sigma$ and justify that $\sigma$ is continuous on it, where $\sigma(x) = \sum_{k=1}^{+\infty} \frac{x^k}{k^2}$.

Show that $$\int _ { 1 } ^ { + \infty } \frac { 1 } { \left( 1 + t ^ { 2 } \right) ^ { n } } \mathrm {~d} t = O \left( \frac { 1 } { n 2 ^ { n } } \right) .$$ One may lower bound $1 + t ^ { 2 }$ by a polynomial of degree 1.

Let $f \in \mathscr { D } _ { \rho } ( \mathbb { R } )$ such that $f ( 0 ) \neq 0$. The purpose of this question is to show that there exists $r \in \mathbb { R } _ { + } ^ { * } , r \leqslant \rho$ such that $\frac { 1 } { f } \in \mathscr { D } _ { r } ( \mathbb { R } )$.
6a. Show that we can assume without loss of generality that $f ( 0 ) = 1$.
We now write $f ( t ) = \sum _ { i = 0 } ^ { \infty } a _ { i } t ^ { i }$ and we assume that $a _ { 0 } = 1$.
6b. Only in this sub-question, we assume that there exists $r \in \mathbb { R } _ { + } ^ { * }$ such that $r \leqslant \rho$ and $g \in \mathscr { D } _ { r } ( \mathbb { R } )$ such that $f ( t ) g ( t ) = 1$ for all $t \in U _ { r }$. We write $g ( t ) = \sum _ { i = 0 } ^ { \infty } b _ { i } t ^ { i }$. Show that: $$\left\{ \begin{aligned} & b _ { 0 } = 1 \\ & \text { for } n \geqslant 1 , b _ { n } = - \left( b _ { 0 } a _ { n } + \ldots + b _ { n - 1 } a _ { 1 } \right) \end{aligned} \right.$$
We now define the sequence $\left( b _ { n } \right) _ { n \geqslant 0 }$ by the above recurrence formula.
6c. Show that there exists $c \in \mathbb { R } _ { + } ^ { * }$ such that $\left| a _ { n } \right| \leqslant c ^ { n }$ for all $n \in \mathbb { N }$.
6d. Show that $\left| b _ { n } \right| \leqslant ( 2 c ) ^ { n }$ for all $n \in \mathbb { N }$.
6e. Conclude.

Let $\left(a_k\right)_{k \in \mathbb{N}}$ be a sequence of elements of $\mathbb{K}$. For every polynomial $p \in \mathbb{K}[X]$, show that the expression $$\sum_{k=0}^{+\infty} a_k D^k p$$ makes sense and defines a polynomial of $\mathbb{K}[X]$.

Show that, for every sequence $\left(a_k\right)_{k \in \mathbb{N}}$ of elements of $\mathbb{K}$, $\sum_{k=0}^{+\infty} a_k D^k$ is a shift-invariant endomorphism.

Let $\left(a_k\right)_{k \in \mathbb{N}}$ and $\left(b_k\right)_{k \in \mathbb{N}}$ be sequences of elements of $\mathbb{K}$ such that $\sum_{k=0}^{+\infty} a_k D^k = \sum_{k=0}^{+\infty} b_k D^k$.
Show that, for all $k \in \mathbb{N}$, $a_k = b_k$.

For every $n \in \mathbb{N}$, define the polynomial $q_n = \frac{X^n}{n!}$. Let $T$ be an endomorphism of $\mathbb{K}[X]$.
Show that $T$ is a shift-invariant endomorphism if, and only if, $$T = \sum_{k=0}^{+\infty} \left(T q_k\right)(0) D^k$$

We define a sequence of polynomials $\left( F _ { i } \right) _ { i \geqslant 0 }$ by the following recurrence formula: $$\begin{aligned} F _ { 0 } & = f _ { 0 } + f _ { 1 } X + \cdots + f _ { d } X ^ { d } \\ \text { for } i \geqslant 0 , \quad F _ { i + 1 } & = F _ { i } + R _ { i } \end{aligned}$$ where $R _ { i }$ denotes the remainder of the Euclidean division of $P$ by $F _ { i }$. We denote by $Q _ { i }$ the quotient of the Euclidean division of $P$ by $F _ { i }$. We are furthermore given $r , s \in \mathbb { R } _ { + } ^ { * }$ with $r < \rho$ and we set, for $i \in \mathbb { N }$: $$\alpha _ { i } = s ^ { - d } \cdot \left\| F _ { i } - X ^ { d } \right\| _ { r , s } ; \quad \beta _ { i } = \left\| 1 - Q _ { i } \right\| _ { r , s } ; \quad \varepsilon _ { i } = s ^ { - d } \cdot \left\| R _ { i } \right\| _ { r , s } .$$
Show that we can choose $r$ and $s$ such that $\alpha _ { 0 } + 2 \varepsilon _ { 0 } \leqslant \frac { 1 } { 3 }$ and $\beta _ { 0 } + \varepsilon _ { 0 } \leqslant \frac { 1 } { 3 }$.

We define a sequence of polynomials $\left( F _ { i } \right) _ { i \geqslant 0 }$ by the following recurrence formula: $$\begin{aligned} F _ { 0 } & = f _ { 0 } + f _ { 1 } X + \cdots + f _ { d } X ^ { d } \\ \text { for } i \geqslant 0 , \quad F _ { i + 1 } & = F _ { i } + R _ { i } \end{aligned}$$ where $R _ { i }$ denotes the remainder of the Euclidean division of $P$ by $F _ { i }$. We denote by $Q _ { i }$ the quotient of the Euclidean division of $P$ by $F _ { i }$. We set, for $i \in \mathbb { N }$: $$\alpha _ { i } = s ^ { - d } \cdot \left\| F _ { i } - X ^ { d } \right\| _ { r , s } ; \quad \beta _ { i } = \left\| 1 - Q _ { i } \right\| _ { r , s } ; \quad \varepsilon _ { i } = s ^ { - d } \cdot \left\| R _ { i } \right\| _ { r , s } .$$
Verify that, for all $i \in \mathbb { N }$, we have the relation: $$\left( 1 - Q _ { i } \right) \cdot R _ { i } = \left( Q _ { i + 1 } - Q _ { i } \right) \cdot F _ { i + 1 } + R _ { i + 1 }$$

We define a sequence of polynomials $\left( F _ { i } \right) _ { i \geqslant 0 }$ by the following recurrence formula: $$\begin{aligned} F _ { 0 } & = f _ { 0 } + f _ { 1 } X + \cdots + f _ { d } X ^ { d } \\ \text { for } i \geqslant 0 , \quad F _ { i + 1 } & = F _ { i } + R _ { i } \end{aligned}$$ where $R _ { i }$ denotes the remainder of the Euclidean division of $P$ by $F _ { i }$. We denote by $Q _ { i }$ the quotient of the Euclidean division of $P$ by $F _ { i }$. We set, for $i \in \mathbb { N }$: $$\alpha _ { i } = s ^ { - d } \cdot \left\| F _ { i } - X ^ { d } \right\| _ { r , s } ; \quad \beta _ { i } = \left\| 1 - Q _ { i } \right\| _ { r , s } ; \quad \varepsilon _ { i } = s ^ { - d } \cdot \left\| R _ { i } \right\| _ { r , s } .$$
Show that, for all $i \in \mathbb { N }$, we have $\alpha _ { i + 1 } \leqslant \alpha _ { i } + \varepsilon _ { i }$ and if $\alpha _ { i + 1 } < 1$ then: $$\beta _ { i + 1 } \leqslant \beta _ { i } + \frac { \beta _ { i } \varepsilon _ { i } } { 1 - \alpha _ { i + 1 } } \quad \text { and } \quad \varepsilon _ { i + 1 } \leqslant \frac { \beta _ { i } \varepsilon _ { i } } { 1 - \alpha _ { i + 1 } } .$$

For natural number $n$, we set $r_n = \sum_{k=n+1}^{+\infty} \frac{1}{k!}$. Prove the bound $$r_n \leq \frac{1}{(n+1)!} \sum_{k=0}^{+\infty} \frac{1}{(n+2)^k}$$ Deduce a simple equivalent of $r_n$ as $n$ tends to $+\infty$.