Using the expression of $t^{\prime}$ as a function of $t$ and the power series expansion $\tan(x) = \sum_{n=0}^{+\infty} \frac{\alpha_{2n+1}}{(2n+1)!} x^{2n+1}$, deduce $$\forall n \in \mathbb{N}^{\star}, \quad \alpha_{2n+1} = \sum_{k=1}^{n} \binom{2n}{2k-1} \alpha_{2k-1} \alpha_{2n-2k+1}.$$

For every natural integer $n$ and every real $x$, set $I_n(x) = \int_0^{\pi/2} \cos(2xt)(\cos t)^n \, \mathrm{d}t$. Show $$\forall n \in \mathbb{N}^{\star}, \forall x \in \mathbb{R}, \quad \sin(\pi x) = \pi x \frac{I_{2n}(x)}{I_{2n}(0)} \prod_{k=1}^{n} \left(1 - \frac{x^2}{k^2}\right)$$

For every natural integer $n$ and every real $x$, set $I_n(x) = \int_0^{\pi/2} \cos(2xt)(\cos t)^n \, \mathrm{d}t$. Using the result of Q20, deduce $$\forall n \in \mathbb{N}^{\star}, \forall x \in ]0,1[, \quad \cos(\pi x) = \frac{1}{2} \frac{I_{4n}(2x)}{I_{4n}(0)} \frac{I_{2n}(0)}{I_{2n}(x)} \prod_{p=1}^{n} \left(1 - \frac{4x^2}{(2p-1)^2}\right)$$

Using the power series expansion $\tan(x) = \sum_{n=0}^{+\infty} \frac{\alpha_{2n+1}}{(2n+1)!} x^{2n+1}$ and the formula $\pi \tan(\pi x) = \sum_{p=1}^{+\infty} 2(2^{2p}-1)\zeta(2p) x^{2p-1}$, show $$\forall n \in \mathbb{N}, \quad \alpha_{2n+1} = \frac{2\left(2^{2n+2} - 1\right)(2n+1)!}{\pi^{2n+2}} \zeta(2n+2).$$

2. a. Show that the radius of convergence of the power series $\sum \frac{E_n}{n!} x^n$ is $\geq 1$. b. For $|x| < 1$, we denote by $f(x)$ the sum of the preceding power series. Prove that $$2f'(x) = f^2(x) + 1, \quad \forall x \in ]-1, 1[$$ c. Deduce that $f(x) = \tan\left(\frac{x}{2} + \frac{\pi}{4}\right) = \frac{1}{\cos x} + \tan x, \quad \forall x \in ]-1, 1[$, then that $$\frac{1}{\cos x} = \sum_{n=0}^{\infty} \frac{E_{2n}}{(2n)!} x^{2n}, \quad \tan x = \sum_{n=0}^{\infty} \frac{E_{2n+1}}{(2n+1)!} x^{2n+1}, \quad \forall x \in ]-1, 1[$$

Show that for all $a \in \mathbb{R}$, $$\int_0^a \sin(x^2) \mathrm{d}x = \sum_{n=0}^{+\infty} (-1)^n \frac{a^{4n+3}}{(2n+1)!(4n+3)}$$

Show that the limits $$\lim_{a \rightarrow +\infty} \int_0^a \sin(x^2) \mathrm{d}x \text{ and } \lim_{a \rightarrow +\infty} \int_0^a \cos(x^2) \mathrm{d}x$$ exist and are finite.

We admit the identities: $$\lim_{a \rightarrow +\infty} \int_0^a \sin(x^2) \mathrm{d}x = \lim_{a \rightarrow +\infty} \int_0^a \cos(x^2) \mathrm{d}x = \frac{\sqrt{2\pi}}{4}$$
Show that there exist real numbers $c, c' \in \mathbb{R}$ such that, as $a \rightarrow +\infty$, $$\int_0^a \sin(x^2) \mathrm{d}x = \frac{\sqrt{2\pi}}{4} + \frac{c}{a} \cos(a^2) + \frac{c'}{a^3} \sin(a^2) + O\left(\frac{1}{a^5}\right)$$

From now on, $f$ denotes an infinitely differentiable function from $[0,1]$ to $\mathbb{R}$. We assume that there exists a unique point $x_0 \in [0,1]$ where $f'$ vanishes. We also assume that $f''(x_0) > 0$. We are also given an infinitely differentiable function $g : [0,1] \rightarrow \mathbb{R}$.
We admit that there exist real numbers $d, d' \in \mathbb{R}$ such that, as $a \rightarrow +\infty$, $$\int_0^a \cos(x^2) \mathrm{d}x = \frac{\sqrt{2\pi}}{4} + \frac{d}{a} \sin(a^2) + \frac{d'}{a^3} \cos(a^2) + O\left(\frac{1}{a^5}\right)$$
Show that, as $t \rightarrow +\infty$, $$\int_{x_0}^1 g(x) \sin(tf(x)) \mathrm{d}x = g(x_0) \int_{x_0}^1 \sin(tf(x)) \mathrm{d}x + O\left(\frac{1}{t}\right)$$

From now on, $f$ denotes an infinitely differentiable function from $[0,1]$ to $\mathbb{R}$. We assume that there exists a unique point $x_0 \in [0,1]$ where $f'$ vanishes. We also assume that $f''(x_0) > 0$. We are also given an infinitely differentiable function $g : [0,1] \rightarrow \mathbb{R}$.
For all $x \in [x_0, 1]$, we define $$h(x) = \sqrt{|f(x) - f(x_0)|}$$ We admit that the bijection $h : [x_0, 1] \rightarrow [0, h(1)]$ admits an inverse application $h^{-1} : [0, h(1)] \rightarrow [x_0, 1]$ that is infinitely differentiable.
We assume that $x_0 \in ]0,1[$. Show that, as $t \rightarrow +\infty$, $$\int_0^1 g(x) \sin(tf(x)) \mathrm{d}x = g(x_0) \sin\left(tf(x_0) + \frac{\pi}{4}\right) \sqrt{\frac{2\pi}{tf''(x_0)}} + O\left(\frac{1}{t}\right)$$

We define a sequence $(a_n)_{n \geqslant 1}$ by setting $$\forall n \in \mathbb{N}^*, \quad a_n = \frac{(-n)^{n-1}}{n!}.$$ We define, when possible, $S(x) = \sum_{n=1}^{+\infty} a_n x^n$, with radius of convergence $R$. Prove that the function $S$ is defined and continuous on $[-R, R]$.

We define a sequence $(a_n)_{n \geqslant 1}$ by setting $$\forall n \in \mathbb{N}^*, \quad a_n = \frac{(-n)^{n-1}}{n!}.$$ We define, when possible, $S(x) = \sum_{n=1}^{+\infty} a_n x^n$, with radius of convergence $R$. Prove that $$\forall x \in ]-R, R[, \quad x(1 + S(x))S'(x) = S(x).$$ One may use the result from Question 26.

We define a sequence $(a_n)_{n \geqslant 1}$ by setting $$\forall n \in \mathbb{N}^*, \quad a_n = \frac{(-n)^{n-1}}{n!}.$$ We define, when possible, $S(x) = \sum_{n=1}^{+\infty} a_n x^n$, with radius of convergence $R$. Let $W$ be the Lambert function defined in Part I (inverse of $f|_{[-1,+\infty[}$ where $f(x)=xe^x$). Using the results of Questions 31 and 32, deduce that $$\forall x \in ]-R, R[, \quad S(x) = W(x).$$

We define a sequence $(a_n)_{n \geqslant 1}$ by setting $$\forall n \in \mathbb{N}^*, \quad a_n = \frac{(-n)^{n-1}}{n!}.$$ We define, when possible, $S(x) = \sum_{n=1}^{+\infty} a_n x^n$, with radius of convergence $R$. It has been shown that $S(x) = W(x)$ for all $x \in ]-R,R[$. Does this result remain true on $[-R, R]$?

Given three real numbers $a, b$ and $c$ with $c \in D$, the Gauss hypergeometric function is defined by $$F_{a,b,c}(x) = \sum_{n=0}^{+\infty} \frac{[a]_n [b]_n}{[c]_n} \frac{x^n}{n!}$$ Determine the radius of convergence of the power series $\sum \frac{[a]_n [b]_n}{[c]_n} \frac{x^n}{n!}$.

Given three real numbers $a, b$ and $c$ with $c \in D$, the Gauss hypergeometric function is defined by $$F_{a,b,c}(x) = \sum_{n=0}^{+\infty} \frac{[a]_n [b]_n}{[c]_n} \frac{x^n}{n!}$$ Express the function $x \mapsto F_{\frac{1}{2}, 1, \frac{3}{2}}\left(-x^2\right)$ using usual functions.

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 $x \in ]-\frac{1}{2}, \frac{1}{2}[$: $$\frac{\pi}{\cos(\pi x)} = \sum_{k=0}^{+\infty} \left(\sum_{n=0}^{+\infty} \frac{(-1)^n}{(2n+1)^{2k+1}}\right) 2^{2k+2} x^{2k}.$$

Using the result that for $x \in ]-\frac{1}{2}, \frac{1}{2}[$: $$\frac{\pi}{\cos(\pi x)} = \sum_{k=0}^{+\infty} \left(\sum_{n=0}^{+\infty} \frac{(-1)^n}{(2n+1)^{2k+1}}\right) 2^{2k+2} x^{2k},$$ deduce that the function $$v : \begin{array}{ccc} ]-\frac{\pi}{2}, \frac{\pi}{2}[ & \longrightarrow & \mathbb{R} \\ x & \longmapsto & \frac{1}{\cos(x)} \end{array}$$ is expandable as a power series and that, for all $k \in \mathbb{N}$, $$\sum_{n=0}^{+\infty} \frac{(-1)^n}{(2n+1)^{2k+1}} = \frac{\pi^{2k+1}}{2^{2k+2}(2k)!} E_{2k}$$ where, for all $k \in \mathbb{N}$, $E_{2k} = v^{(2k)}(0)$.

Let $v(x) = \frac{1}{\cos(x)}$ on $]-\frac{\pi}{2}, \frac{\pi}{2}[$ and $E_{2k} = v^{(2k)}(0)$ for $k \in \mathbb{N}$.
Show that, for $n \in \mathbb{N}^*$, $$\sum_{k=0}^{n} (-1)^k \binom{2n}{2k} E_{2k} = 0$$ and deduce the values of $E_0$, $E_2$ and $E_4$.

We consider a power series $\sum_{n \geqslant 0} \alpha_n z^n$, with radius of convergence $R \neq 0$ and with $\alpha_0 = 1$. We denote by $S$ the sum of this power series on its disk of convergence.
Show that there exists a real number $q > 0$ such that $\forall n \in \mathbb{N}, |\alpha_n| \leqslant q^n$.

We consider a power series $\sum_{n \geqslant 0} \alpha_n z^n$, with radius of convergence $R \neq 0$ and with $\alpha_0 = 1$, and sum $S$. We assume that $\frac{1}{S}$ is expandable as a power series in a neighbourhood of 0 and we denote by $\sum_{n \geqslant 0} \beta_n z^n$ its expansion.
Calculate $\beta_0$ and, for all $n \in \mathbb{N}^*$, express $\beta_n$ in terms of $\alpha_1, \ldots, \alpha_n, \beta_1, \ldots, \beta_{n-1}$. Deduce that $$\forall n \in \mathbb{N}, \quad |\beta_n| \leqslant (2q)^n.$$

We consider a power series $\sum_{n \geqslant 0} \alpha_n z^n$, with radius of convergence $R \neq 0$, $\alpha_0 = 1$, and sum $S$.
Show that $\frac{1}{S}$ is expandable as a power series in a neighbourhood of 0.

Using the results of questions 28--30, show that there exists a unique complex sequence $(b_n)_{n \in \mathbb{N}}$ and a real number $r > 0$ such that, for all $z \in \mathbb{C}$, $$0 < |z| < r \Rightarrow \frac{z}{\mathrm{e}^z - 1} = \sum_{n=0}^{+\infty} \frac{b_n}{n!} z^n.$$

We fix a choice of $\lambda$ such that $P_a(x) = x - \lambda x(x-a)(x-1)$ satisfies $P([0,1])=[0,1]$ and $P$ is increasing on $[0,1]$. Let $\left(P_a^{\circ n}\right)_{n\geq 0}$ be the sequence of polynomials defined recursively by $P_a^{\circ 0}(x) = x$ and $P_a^{\circ n+1}(x) = P_a\left(P_a^{\circ n}(x)\right)$.
Show that $P_a^{\circ n}$ converges uniformly to 1 on every compact subset of $]a,1]$ and uniformly to 0 on every compact subset of $[0,a[$.

Let $b \in \mathbb{R}$ such that $\cos(b) \in ]0,1[$. Show that the sequence of functions $(f_{b,n})_{n\in\mathbb{N}}$ defined by $$f_{b,n}(t) = P_{\cos(b)}^{\circ n}\left(\cos^2\left(\frac{\pi}{2}t\right)\right)$$ converges uniformly to 1 on every compact subset of $]-\cos(b), \cos(b)[$ and converges uniformly to 0 on every compact subset of $[-1,-\cos(b)[\cup]\cos(b),1]$.