Formal power series manipulation (Cauchy product, algebraic identities)
The question involves algebraic operations on formal power series such as Cauchy products, deriving recurrence relations for coefficients, or proving identities like the Taylor shift formula for polynomials.
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 $$\forall x \in \mathbb{R}, \quad \exp\left(-x^{2}\right) \cos(2\pi \xi x) = \sum_{p=0}^{+\infty} c_{p}(\xi) \exp\left(-x^{2}\right) x^{2p}$$
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 $f(0) = \frac{1}{2\pi} \int_0^{2\pi} f(r\cos(t), r\sin(t)) \, \mathrm{d}t$.
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 $$\forall t \in ] - 1,1 \left[ , \quad \varphi ( t ) = \mathrm { e } \left( 1 - \sum _ { k = 1 } ^ { + \infty } b _ { k } t ^ { k } \right) \right.$$