For every integer $n \in \mathbb{N}$, we set $F_n(x) = \cos(n \arccos x)$.
Deduce from the above that $F_n$ extends to $\mathbb{R}$ as a unique polynomial function, whose degree and leading coefficient should be specified.

Let $z \in \mathbb{C}$. Consider the function of the real variable $x$ $$G_z : x \mapsto \sum_{p=0}^{+\infty} \left(x^p(2z - x)^p\right)$$ Deduce (from II.C.5) that $G_z$ admits a Taylor expansion to any order at 0. We denote it $$G_z(x) = \sum_{k=0}^{n} a_k x^k + o\left(x^n\right) \quad x \to 0$$ Determine the coefficients $a_k$ for $k \in \mathbb{N}$.

Show that, for any real $\xi$, there exists a real sequence $\left(c_{p}(\xi)\right)_{p \in \mathbb{N}}$ such that $$\forall x \in \mathbb{R}, \quad \exp\left(-x^{2}\right) \cos(2\pi \xi x) = \sum_{p=0}^{+\infty} c_{p}(\xi) \exp\left(-x^{2}\right) x^{2p}$$

Let $f$ be a function that expands as a power series on $D(0,R)$, i.e., there exists a complex sequence $(a_n)$ such that $$\forall (x,y) \in D(0,R), \quad f(x,y) = \sum_{n=0}^{+\infty} a_n (x + \mathrm{i}y)^n$$ Show that for all $r \in [0, R[$, we have $f(0) = \frac{1}{2\pi} \int_0^{2\pi} f(r\cos(t), r\sin(t))\, \mathrm{d}t$.

Throughout this part, $f$ denotes a function that expands in a power series on $D(0,R)$, i.e., $$\forall (x,y) \in D(0,R), \quad f(x,y) = \sum_{n=0}^{+\infty} a_n (x + \mathrm{i} y)^n$$ Show that for all $r \in [0, R[$, we have $f(0) = \frac{1}{2\pi} \int_0^{2\pi} f(r\cos(t), r\sin(t)) \, \mathrm{d}t$.

We define the function $\varphi : \mathbb { R } \rightarrow \mathbb { R }$ by $$\begin{cases} \varphi ( x ) = \exp \left( \frac { - x } { \sqrt { 1 - x } } \right) & \text { if } x < 1 \\ \varphi ( x ) = 0 & \text { if } x \geqslant 1 \end{cases}$$
Demonstrate, for all $x \in ] - 1,1 [$, $$\varphi ( x ) = \sum _ { q = 0 } ^ { + \infty } \frac { ( - 1 ) ^ { q } } { q ! } x ^ { q } ( 1 - x ) ^ { - q / 2 }$$

Using the decomposition of $f(x) = g(x) = \frac{\sin x + 1}{\cos x}$ into even and odd parts, deduce $$\forall x \in I, \quad \tan(x) = \sum_{n=0}^{+\infty} \frac{\alpha_{2n+1}}{(2n+1)!} x^{2n+1} \quad \text{and} \quad \frac{1}{\cos x} = \sum_{n=0}^{+\infty} \frac{\alpha_{2n}}{(2n)!} x^{2n}.$$

Let $n \in \mathbb{N}$ be a non-zero natural integer. We define, for any real number $x$, $$\Phi_n(x) = \mathrm{e}^{-x} x^n \quad \text{and} \quad L_n(x) = \frac{\mathrm{e}^x}{n!} \Phi_n^{(n)}(x).$$ Using Leibniz's formula, prove that the function $L_n$ is polynomial of degree $n$. Determine the coefficients $c_{n,k}$ such that $L_n(x) = \sum_{k=0}^{n} c_{n,k} x^k$.

Using the result that for all $x \in ]0,1[$: $$\frac{\pi}{\sin(\pi x)} = \sum_{n=0}^{+\infty} \frac{(-1)^n}{n+x} + \sum_{n=0}^{+\infty} \frac{(-1)^n}{n+1-x},$$ deduce that, for $x \in ]-\frac{1}{2}, \frac{1}{2}[$: $$\frac{\pi}{\cos(\pi x)} = \sum_{k=0}^{+\infty} \left(\sum_{n=0}^{+\infty} \frac{(-1)^n}{(2n+1)^{2k+1}}\right) 2^{2k+2} x^{2k}.$$

We consider a power series $g \in O_1$ with non-negative real coefficients. We assume that there exist $a > 0, b > 0$ such that $$g \prec a\left(I + \frac{g^2}{b - g}\right)$$ Show that there exist $r > 0$ and a function $h: ]-r, r[ \longrightarrow \mathbb{R}$, expandable as a power series at 0, satisfying $h(0) = 0$ and such that $$h(x) = a\left(x + \frac{h(x)^2}{b - h(x)}\right)$$ for all $x \in ]-r, r[$. We also denote by $h$ the element of $O_1$ associated with the function $h$.

We now study the linearization problem in the case $|\lambda| = 1$, with $\lambda$ not a root of unity. We set, for $m \geqslant 1$, $$\alpha_m := \min(1/5, \omega_{m+1}, \omega_{m+2}, \ldots, \omega_{2m}), \quad \gamma_m := \alpha_m^{2/m},$$ where $\omega_k := |\lambda^k - \lambda|, k \geqslant 2$, and $r_0 > 0$ is as given by question (24).
Still for $F \in O_{m+1}, m \geqslant 1$, we set $$P := \sum_{k=m+1}^{2m} \frac{(F)_k}{\lambda^k - \lambda} z^k \in O_{m+1} \quad , \quad R := (I + P)^\dagger - I.$$ Show that $P \circ (\lambda I) - \lambda P - F \in O_{2m+1}$ and that $R + P \in O_{2m+1}$. Show that $\hat{P}(r) \leqslant \alpha_m r$ for all $r \in [0, \gamma_m r_0]$, and that $$\hat{R}(r) \leqslant \frac{\alpha_m}{1 - \alpha_m} r$$ for all $r \in [0, (1-\alpha_m)\gamma_m r_0]$.

We now study the linearization problem in the case $|\lambda| = 1$, with $\lambda$ not a root of unity. We set, for $m \geqslant 1$, $$\alpha_m := \min(1/5, \omega_{m+1}, \omega_{m+2}, \ldots, \omega_{2m}), \quad \gamma_m := \alpha_m^{2/m},$$ where $\omega_k := |\lambda^k - \lambda|, k \geqslant 2$, and $P, R$ as defined in question (25).
For $F \in O_{m+1}, m \geqslant 1$, show that $$G := (I + P)^\dagger \circ (\lambda I + F) \circ (I + P) - \lambda I = (I + R) \circ (\lambda I + F) \circ (I + P) - \lambda I$$ satisfies $G \in O_{2m+1}$.

For every integer $k \geqslant 2$, we set $\zeta(k) = \sum_{n=1}^{+\infty} n^{-k}$.
Let $h$ be the function from $\mathbb{R}$ to $\mathbb{R}$ defined by $$\forall x \in \mathbb{R}, \quad h(x) = \begin{cases} \frac{x}{e^x - 1} & \text{if } x \neq 0 \\ 1 & \text{if } x = 0 \end{cases}$$
Show that for all $z \in \mathbb{C}$ such that $|z| < 2\pi$, we have $$z = \left(e^z - 1\right)\left(1 - \frac{z}{2} + \sum_{k=1}^{+\infty} \frac{(-1)^{k-1} \zeta(2k)}{2^{2k-1}\pi^{2k}} z^{2k}\right).$$

For every integer $k \geqslant 2$, we set $\zeta(k) = \sum_{n=1}^{+\infty} n^{-k}$. Let $h$ be the function from $\mathbb{R}$ to $\mathbb{R}$ defined by $$\forall x \in \mathbb{R}, \quad h(x) = \begin{cases} \frac{x}{e^x - 1} & \text{if } x \neq 0 \\ 1 & \text{if } x = 0 \end{cases}$$ Show that for all $z \in \mathbb{C}$ such that $|z| < 2\pi$, we have $$z = \left(e^z - 1\right)\left(1 - \frac{z}{2} + \sum_{k=1}^{+\infty} \frac{(-1)^{k-1}\zeta(2k)}{2^{2k-1}\pi^{2k}} z^{2k}\right)$$

For all $p \in \mathbb{K}[X]$ non-zero and $a \in \mathbb{K}$, show, using question 11, that $$p(X+a) = \sum_{k=0}^{\deg(p)} \frac{a^k}{k!} p^{(k)}$$ where $p^{(k)}$ denotes the $k$-th derivative of the polynomial $p$. Recognize this formula.

For all $n \in \mathbb{N}$ and all $z \in \mathbb{C}$, the Bernoulli polynomial is defined by $$B_{n}(z) = n! \int_{0}^{1} \frac{\mathrm{e}^{z\omega(t)}}{(\mathrm{e}^{\omega(t)} - 1)\omega(t)^{n-1}} \,\mathrm{d}t$$ where $\omega(t) = e^{2i\pi t}$. Show that, for all $n \in \mathbb{N}^{*}$ and all $z \in \mathbb{C}$, $$B_{n}(z+1) - B_{n}(z) = n z^{n-1}.$$

Let $\varphi$ be the function defined by
$$\forall t \in ] - 1,1 \left[ \backslash \{ 0 \} , \quad \varphi ( t ) = ( 1 - t ) ^ { 1 - 1 / t } \right.$$
We define the sequence $\left( b _ { n } \right) _ { n \in \mathbb { N } }$ by
$$\left\{ \begin{array} { l } b _ { 0 } = - 1 \\ \forall n \in \mathbb { N } ^ { * } , \quad b _ { n } = - \frac { 1 } { n } \sum _ { k = 1 } ^ { n } \frac { 1 } { k + 1 } b _ { n - k } \end{array} \right.$$
Conclude that
$$\forall t \in ] - 1,1 \left[ , \quad \varphi ( t ) = \mathrm { e } \left( 1 - \sum _ { k = 1 } ^ { + \infty } b _ { k } t ^ { k } \right) \right.$$