Let $a > 0$, $I = [-a, a]$. For all $n \in \mathbb { N } ^ { * }$, the Chebyshev points of order $n$ in $I$ are $$a _ { k , n } ^ { * } = a \cos \left( \frac { ( 2 k - 1 ) \pi } { 2 n } \right) , \quad \text { for } k \in \llbracket 1 , n \rrbracket ,$$ and $W _ { n } ^ { * } ( X ) = \prod _ { k = 1 } ^ { n } \left( X - a _ { k , n } ^ { * } \right)$. For all $x \in [ - a , a ]$, show that $\left| W _ { n } ^ { * } ( x ) \right| \leqslant 2 \left( \frac { a } { 2 } \right) ^ { n }$.

We consider a power series $f = \lambda I + F$ with $F \in O_2, \rho(F) > 0$. We still assume that $\lambda$ has modulus 1 and is not a root of unity. We consider the real $r_0 > 0$ given by question (24) (applied for $m = 1$) and the sequence $r_k$ defined by recursion from $r_0$ by the relation $$r_{k+1} = (1 - \alpha_{2^k})(1 + \alpha_{2^k}^2)^{-1}(1 + \alpha_{2^k})^{-1} \gamma_{2^k} r_k.$$ Show that there exist sequences $F_k$ and $P_k$ of elements of $O_2$, defined for $k \geqslant 0$, such that $F_0 = F$ and, for all $k \geqslant 0$, $$\begin{aligned} & \lambda I + F_{k+1} = (I + P_k)^\dagger \circ (\lambda I + F_k) \circ (I + P_k), \\ & F_k \in O_{1+2^k}, \quad P_k \in O_{1+2^k}, \\ & \widehat{F_k}(r_k) \leqslant r_k, \quad \widehat{P_k}(r_{k+1}) \leqslant r_k - r_{k+1}. \end{aligned}$$

Let $a > 0$, $I = [-a,a]$, and $f(x) = \dfrac{1}{1+x^2}$ for $x \in \mathbb{R}$. For all $n \in \mathbb{N}^*$, the Chebyshev points of order $n$ in $I$ are $a_{k,n}^* = a\cos\left(\frac{(2k-1)\pi}{2n}\right)$ for $k \in \llbracket 1,n \rrbracket$, and $\Pi_n^*(f)$ denotes the interpolation polynomial of $f$ at these points. Show that, if $a < 2$, the sequence $\left( \Pi _ { n } ^ { * } ( f ) \right) _ { n \in \mathbb { N } ^ { * } }$ converges uniformly towards $f$ on $[ - a , a ]$.

We set $r_\infty := \lim r_k$ and $$h_k := (I + P_0) \circ (I + P_1) \circ \cdots \circ (I + P_{k-1}).$$ Explain why $r_\infty$ is well defined, and show that $\hat{h}_k(r_k) \leqslant r_0$ for all $k \geqslant 1$. Deduce that the series $h$ of question E satisfies $\hat{h}(r_\infty) \leqslant r_0$, thus that $\rho(h) \geqslant r_\infty$.

Let $\sum_{k \geqslant 0} c_k x^k$ be a power series with radius of convergence $R > 0$, $f(x) = \sum_{k=0}^{+\infty} c_k x^k$ for $x \in ]-R,R[$, and $a > 0$. For all $n \in \mathbb{N}^*$, the Chebyshev points of order $n$ in $[-a,a]$ are $a_{k,n}^* = a\cos\left(\frac{(2k-1)\pi}{2n}\right)$ for $k \in \llbracket 1,n \rrbracket$, and $\Pi_n^*(f)$ denotes the interpolation polynomial of $f$ at these points. Show that, if $a < 2R/3$, the sequence $\left( \Pi _ { n } ^ { * } ( f ) \right) _ { n \in \mathbb { N } ^ { * } }$ converges uniformly towards $f$ on $[ - a , a ]$.

If $n \in \mathbf{N}$, we denote by $D_n$ the improper integral $\int_0^{\pi/2} (\ln(\sin(t)))^n \mathrm{~d}t$.
Verify that if $n \in \mathbf{N}^*$, then $$(-1)^n D_n = \int_0^{+\infty} \frac{u^n}{\sqrt{\mathrm{e}^{2u} - 1}} \mathrm{~d}u$$ then that $$D_n \underset{n \to +\infty}{\sim} (-1)^n n!$$

Throughout this problem, $I = ]-1, +\infty[$, and $f(x) = \int_0^{\pi/2} (\sin(t))^x \mathrm{~d}t$.
Prove that $f$ is expandable as a power series on $]-1, 1[$.

Throughout this problem, $I = ]-1, +\infty[$, and $f(x) = \int_0^{\pi/2} (\sin(t))^x \mathrm{~d}t$. We call $\tilde{f}$ the application from $\mathbf{R}_+$ to $\mathbf{R}$, defined by: $$\forall x \in \mathbf{R}^+, \tilde{f}(x) = \ln(f(2x))$$
Show that $$\forall p \in \mathbf{N}^*, \forall x \in \mathbf{R}_+, \tilde{f}(x+p) = \tilde{f}(x) + \sum_{k=0}^{p-1} \ln\left(\frac{2x+2k+1}{2x+2k+2}\right)$$

Give an example of a power series expansion of a rational function whose antiderivative is not the expansion of a rational function.
(The antiderivative of a power series $f(x) = \sum_{n=0}^{\infty} c_n x^n$ is defined as $\int_0^x f(t)\,dt \stackrel{\text{def}}{=} \sum_{n=0}^{\infty} \frac{c_n}{n+1} x^{n+1}$.)

Let $\psi(x) = \begin{cases} \frac{x}{e^x-1} & x\neq 0 \\ 1 & x=0 \end{cases}$, $u(x,t) = \psi(x)e^{tx}$, and for all $n \in \mathbb{N}$, let $A_{n}$ be the function from $\mathbb{R}$ to $\mathbb{R}$ such that, for all $t \in \mathbb{R}$, $$A_{n}(t) = \frac{\partial^{n} u}{\partial x^{n}}(0,t).$$ Show that, for all $n \in \mathbb{N}$, $A_{n} = B_{n}$, where $(B_n)_{n\in\mathbb{N}}$ are the Bernoulli polynomials.

We denote by $\mathcal{E}$ the set of functions $f : \mathbb{C} \rightarrow \mathbb{C}$ expandable as a power series with radius of convergence infinity. For all $n \in \mathbb{N}$, we define $$\gamma_{n} : \begin{cases} [0,1] \rightarrow \mathbb{C} \\ t \mapsto (2n+1)\pi\, \mathrm{e}^{2\mathrm{i}\pi t} \end{cases}$$ and for all $n \in \mathbb{N}$ and all $z \in \mathbb{C}$, $$Q_{n}(z) = n! \int_{0}^{1} \frac{\mathrm{e}^{z\gamma_{n}(t)}}{(\mathrm{e}^{\gamma_{n}(t)} - 1)\gamma_{n}(t)^{n-1}} \,\mathrm{d}t.$$ Show that, for all $n \in \mathbb{N}$, $Q_{n} \in \mathcal{E}$.

For all $n \in \mathbb{N}$ and all $z \in \mathbb{C}$, let $$Q_{n}(z) = n! \int_{0}^{1} \frac{\mathrm{e}^{z\gamma_{n}(t)}}{(\mathrm{e}^{\gamma_{n}(t)} - 1)\gamma_{n}(t)^{n-1}} \,\mathrm{d}t$$ where $\gamma_{n}(t) = (2n+1)\pi\,\mathrm{e}^{2\mathrm{i}\pi t}$. Show that $$\forall n \in \mathbb{N}^{*},\, \forall z \in \mathbb{C}, \quad Q_{n}(z+1) - Q_{n}(z) = n z^{n-1}.$$

For all $n \in \mathbb{N}$ and all $z \in \mathbb{C}$, let $$Q_{n}(z) = n! \int_{0}^{1} \frac{\mathrm{e}^{z\gamma_{n}(t)}}{(\mathrm{e}^{\gamma_{n}(t)} - 1)\gamma_{n}(t)^{n-1}} \,\mathrm{d}t$$ where $\gamma_{n}(t) = (2n+1)\pi\,\mathrm{e}^{2\mathrm{i}\pi t}$. Using the property $\mathcal{P}$: $$\exists c > 0,\, \forall n \in \mathbb{N},\, \forall z \in \mathbb{C},\, \left(|z| = (2n+1)\pi \Rightarrow |\mathrm{e}^{z} - 1| \geqslant c\right),$$ show that there exist two constants $a, b \in \mathbb{R}_{+}^{*}$ such that, for all $n \in \mathbb{N}^{*}$ and all $z \in \mathbb{C}$, $$|Q_{n}(z)| \leqslant a\,\mathrm{e}^{bn|z|}.$$

15. Show that for $x \in ] - 1,1 \left[ \right.$, the function $r \mapsto F _ { \lambda } ( x , r )$ is expandable as a power series in a neighborhood of $\mathbf { 0 }$. For $x \in ] - 1,1$, we denote $a _ { n } ^ { ( \lambda ) } ( x )$ the $n$-th coefficient of the expansion of the function $r \mapsto F _ { \lambda } ( x , r )$ so that, for $r$ in a neighborhood of 0 ,
$$F _ { \lambda } ( x , r ) = \sum _ { n \geqslant 0 } a _ { n } ^ { ( \lambda ) } ( x ) r ^ { n } .$$
16a. For $x \in ] - 1,1 \left[ \right.$, show that $a _ { 1 } ^ { ( \lambda ) } ( x ) = 2 \lambda x a _ { 0 } ^ { ( \lambda ) } ( x )$ and that, for every integer $n \geqslant 1$,
$$( n + 1 ) a _ { n + 1 } ^ { ( \lambda ) } ( x ) = 2 ( n + \lambda ) x a _ { n } ^ { ( \lambda ) } ( x ) - ( n + 2 \lambda - 1 ) a _ { n - 1 } ^ { ( \lambda ) } ( x ) .$$
Hint: one may begin by computing $\left( 1 - 2 x r + r ^ { 2 } \right) \frac { \partial F _ { \lambda } } { \partial r } ( x , r )$. 16b. Deduce that, for all $n \geqslant 0$, the function $a _ { n } ^ { ( \lambda ) }$ is a polynomial of degree $n$ whose leading coefficient and parity will be determined. We now assume that $\lambda > \frac { 1 } { 2 }$. For $P , Q \in \mathbb { R } [ X ]$, we set

Let $\alpha, \beta \in \mathbb{R}$ be such that $\lim_{x \rightarrow 0} \frac{x^2 \sin(\beta x)}{\alpha x - \sin x} = 1$. Then $6(\alpha + \beta)$ equals