Express the derivatives $f^{\prime}, f^{\prime\prime}$ and $f^{(3)}$ using usual functions, where $f$ is defined on $I = ]-\pi/2, \pi/2[$ by $$\forall x \in I, \quad f(x) = \frac{\sin x + 1}{\cos x}.$$
Let $f$ be defined on $I = ]-\pi/2, \pi/2[$ by $f(x) = \frac{\sin x + 1}{\cos x}$. Show that there exists a sequence of polynomials $\left(P_n\right)_{n \in \mathbb{N}}$ with real coefficients such that $$\forall n \in \mathbb{N}, \forall x \in I, \quad f^{(n)}(x) = \frac{P_n(\sin x)}{(\cos x)^{n+1}}$$ Make explicit the polynomials $P_0, P_1, P_2, P_3$ and, for every natural integer $n$, express $P_{n+1}$ as a function of $P_n$ and $P_n^{\prime}$.
Using the sequence of polynomials $(P_n)$ defined by $f^{(n)}(x) = \frac{P_n(\sin x)}{(\cos x)^{n+1}}$ for $f(x) = \frac{\sin x + 1}{\cos x}$ on $I = ]-\pi/2, \pi/2[$, justify that, for every integer $n \geqslant 1$, the polynomial $P_n$ is monic, of degree $n$ and that its coefficients are natural integers.
Let $f$ be defined on $I = ]-\pi/2, \pi/2[$ by $f(x) = \frac{\sin x + 1}{\cos x}$, and set $\alpha_n = f^{(n)}(0) = P_n(0)$ for every natural integer $n$. Using the identity $2f^{\prime}(x) = f(x)^2 + 1$, show $2\alpha_1 = \alpha_0^2 + 1$ and $$\forall n \in \mathbb{N}^{\star}, \quad 2\alpha_{n+1} = \sum_{k=0}^{n} \binom{n}{k} \alpha_k \alpha_{n-k}.$$
Let $\alpha_n = f^{(n)}(0)$ where $f(x) = \frac{\sin x + 1}{\cos x}$ on $I = ]-\pi/2, \pi/2[$. Let $R$ be the radius of convergence of the power series $\sum_{n \in \mathbb{N}} \frac{\alpha_n}{n!} x^n$ and $g$ its sum. Using Taylor's formula with integral remainder, show $$\forall N \in \mathbb{N}, \forall x \in \left[0, \pi/2\left[, \quad \sum_{n=0}^{N} \frac{\alpha_n}{n!} x^n \leqslant f(x)\right.\right.$$
Using the result of Q6, deduce the lower bound $R \geqslant \pi/2$ for the radius of convergence $R$ of the power series $\sum_{n \in \mathbb{N}} \frac{\alpha_n}{n!} x^n$.
Let $g$ be the sum of the power series $\sum_{n \in \mathbb{N}} \frac{\alpha_n}{n!} x^n$ with radius of convergence $R \geqslant \pi/2$. Show $$\forall x \in I, \quad 2g^{\prime}(x) = g(x)^2 + 1.$$
Let $f(x) = \frac{\sin x + 1}{\cos x}$ on $I = ]-\pi/2, \pi/2[$ and $g$ the sum of the power series $\sum_{n \in \mathbb{N}} \frac{\alpha_n}{n!} x^n$. Both satisfy $2h^{\prime}(x) = h(x)^2 + 1$. By considering the functions $\arctan f$ and $\arctan g$, show $$\forall x \in I, \quad f(x) = g(x).$$
Using the fact that $f(x) = g(x)$ on $I = ]-\pi/2, \pi/2[$ where $f(x) = \frac{\sin x + 1}{\cos x}$ and $g$ is the sum of the power series $\sum_{n \in \mathbb{N}} \frac{\alpha_n}{n!} x^n$, deduce that $R = \pi/2$.
Justify that every function $h : I \rightarrow \mathbb{R}$ can be written uniquely in the form $h = p + i$ with $p : I \rightarrow \mathbb{R}$ an even function and $i : I \rightarrow \mathbb{R}$ an odd function.
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 $t$ be the function defined on $I = ]-\pi/2, \pi/2[$ by $t(x) = \tan(x)$. For every natural integer $n$, express $t^{(n)}(0)$ as a function of the reals $(\alpha_i)_{i \in \mathbb{N}}$.
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 $s > 1$, let $\zeta(s) = \sum_{n=1}^{+\infty} \frac{1}{n^s}$. Bound $\sum_{n=2}^{+\infty} \frac{1}{n^s}$ by two integrals and deduce $\lim_{s \rightarrow +\infty} \zeta(s) = 1$.
For every $s > 1$, let $\zeta(s) = \sum_{n=1}^{+\infty} \frac{1}{n^s}$. Determine $C(s)$ such that $$\forall s \in ]1, +\infty[, \quad \sum_{k=1}^{+\infty} \frac{1}{(2k-1)^s} = C(s) \zeta(s).$$
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 $$\forall n \in \llbracket 2, +\infty\llbracket, \forall x \in \mathbb{R}, \quad \left(1 - \frac{4x^2}{n^2}\right) I_n(x) = \frac{n-1}{n} I_{n-2}(x) \quad \text{and} \quad \left(1 - \frac{4x^2}{n^2}\right) \frac{I_n(x)}{I_n(0)} = \frac{I_{n-2}(x)}{I_{n-2}(0)}.$$
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 $$\forall n \in \mathbb{N}^{\star}, \forall x \in ]0,1[, \quad \cos(\pi x) = \frac{1}{2} \frac{I_{4n}(2x)}{I_{4n}(0)} \frac{I_{2n}(0)}{I_{2n}(x)} \prod_{p=1}^{n} \left(1 - \frac{4x^2}{(2p-1)^2}\right)$$
For every natural integer $n$ and every real $x$ in $J = [0, 1/2[$, set $$S_n(x) = \sum_{p=1}^{+\infty} \left(\sum_{k=n+1}^{+\infty} \frac{2^{2p+1} x^{2p-1}}{(2k-1)^{2p}}\right).$$ Justify that, for every natural integer $n$, the function $S_n$ is defined on $J$.
For every natural integer $n$ and every real $x$ in $J = [0, 1/2[$, set $$S_n(x) = \sum_{p=1}^{+\infty} \left(\sum_{k=n+1}^{+\infty} \frac{2^{2p+1} x^{2p-1}}{(2k-1)^{2p}}\right).$$ Show that the sequence $(S_n)$ converges pointwise on $J$ to the zero function.
For every natural integer $n$ and every real $x$, set $I_n(x) = \int_0^{\pi/2} \cos(2xt)(\cos t)^n \, \mathrm{d}t$. By differentiating $x \mapsto \ln(\cos(\pi x))$, show $$\forall x \in J, \quad \pi \tan(\pi x) = -\frac{2I_{4n}^{\prime}(2x)}{I_{4n}(2x)} + \frac{I_{2n}^{\prime}(x)}{I_{2n}(x)} + \sum_{k=1}^{n} \frac{8x}{(2k-1)^2} \frac{1}{1 - \frac{4x^2}{(2k-1)^2}}.$$
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 inequality $t\cos(t) \leqslant \sin(t)$ for $t \in [0, \pi/2]$, deduce $$\forall n \in \mathbb{N}^{\star}, \forall x \in [0,1], \quad 0 \leqslant -I_n^{\prime}(x) \leqslant \frac{4x}{n} I_n(x)$$ then, for $x \in [0,1]$, the limit $\lim_{n \rightarrow +\infty} \frac{I_n^{\prime}(x)}{I_n(x)}$.
Using the results of Q26 and Q28, deduce the equality $$\forall x \in J, \quad \pi \tan(\pi x) = \sum_{p=1}^{+\infty} 2\left(2^{2p} - 1\right) \zeta(2p) x^{2p-1}$$
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 $$\forall n \in \mathbb{N}, \quad \alpha_{2n+1} = \frac{2\left(2^{2n+2} - 1\right)(2n+1)!}{\pi^{2n+2}} \zeta(2n+2).$$
Using the result $\alpha_{2n+1} = \frac{2(2^{2n+2}-1)(2n+1)!}{\pi^{2n+2}} \zeta(2n+2)$ and the fact that $\lim_{s \to +\infty} \zeta(s) = 1$, deduce an equivalent of $\alpha_{2n+1}$ as $n$ tends to infinity.
A permutation $\sigma$ of $\llbracket 1, n \rrbracket$ is called alternating up if the list $(\sigma(1), \ldots, \sigma(n))$ satisfies $x_1 < x_2 > x_3 < x_4 > \cdots$. Determine the alternating up permutations of $\llbracket 1, n \rrbracket$ for $n = 2$, $n = 3$, $n = 4$.
A permutation $\sigma$ of $\llbracket 1, n \rrbracket$ is called alternating up if $(\sigma(1), \ldots, \sigma(n))$ satisfies $x_1 < x_2 > x_3 < x_4 > \cdots$, and alternating down if it satisfies the reverse inequalities. Show, for every $n \geqslant 2$, that the number of alternating up permutations equals the number of alternating down permutations.
Let $\beta_n$ denote the number of alternating up permutations of $\llbracket 1, n \rrbracket$ (with $\beta_0 = \beta_1 = 1$). Let $k$ and $n$ be two integers such that $2 \leqslant k \leqslant n$ and $A$ a subset with $k$ elements of $\llbracket 1, n \rrbracket$. We consider the lists $(x_1, \ldots, x_k)$ consisting of $k$ pairwise distinct elements of $A$. Show that the number of these lists that are alternating up equals $\beta_k$.
Let $\beta_n$ denote the number of alternating up permutations of $\llbracket 1, n \rrbracket$ (with $\beta_0 = \beta_1 = 1$). For $k \in \llbracket 0, n \rrbracket$, enumerate the permutations $\sigma$ alternating (up or down) of $\llbracket 1, n+1 \rrbracket$ such that $\sigma(k+1) = n+1$. Show, for every integer $n \geqslant 1$, $$2\beta_{n+1} = \sum_{k=0}^{n} \binom{n}{k} \beta_k \beta_{n-k}.$$
Using the recurrence relation $2\beta_{n+1} = \sum_{k=0}^{n} \binom{n}{k} \beta_k \beta_{n-k}$ and the analogous relation $2\alpha_{n+1} = \sum_{k=0}^{n} \binom{n}{k} \alpha_k \alpha_{n-k}$ with $\alpha_0 = \beta_0 = 1$, deduce that $\beta_n = \alpha_n$ for every $n \in \mathbb{N}$.
For every integer $n \geqslant 2$, let $p_n$ be the probability that a uniformly random permutation of $\llbracket 1, n \rrbracket$ is alternating up (with $p_0 = p_1 = 1$). Show that the sequence $(p_n)$ tends to 0. Give an equivalent of $p_{2n+1}$ as $n$ tends to infinity.
For every integer $n \geqslant 2$, equip the set $\Omega_n$ of permutations of $\llbracket 1, n \rrbracket$ with the uniform probability. Let $p_i$ denote the probability that a permutation is alternating up (with $p_0 = p_1 = 1$). Define the random variable $M_n$ on $\Omega_n$ by: $M_n(\sigma) = k+1$ where $k$ is the largest integer such that $(\sigma(1), \ldots, \sigma(k))$ is alternating up. For every $i \in \llbracket 0, n \rrbracket$, show $\mathbb{P}(M_n > i) = p_i$.
For every integer $n \geqslant 2$, equip the set $\Omega_n$ of permutations of $\llbracket 1, n \rrbracket$ with the uniform probability. Let $p_i$ denote the probability that a permutation is alternating up (with $p_0 = p_1 = 1$). Define the random variable $M_n$ on $\Omega_n$ by: $M_n(\sigma) = k+1$ where $k$ is the largest integer such that $(\sigma(1), \ldots, \sigma(k))$ is alternating up. Express $\mathbb{E}(M_n)$ as a function of $p_0, p_1, \ldots, p_n$. Deduce $\lim_{n \rightarrow \infty} \mathbb{E}(M_n) = \frac{\sin(1) + 1}{\cos(1)}$.