We define the sequence of polynomials $\left( H _ { k } \right) _ { k \in \mathbb { N } }$ in $\mathbb { R } [ X ]$ by $H _ { 0 } ( X ) = 1$ and, for all $k \in \mathbb { N } ^ { * }$, $$H _ { k } ( X ) = X ( X - 1 ) \cdots ( X - k + 1 )$$ and $S(n,k)$ denotes the number of partitions of $\llbracket 1, n \rrbracket$ into $k$ parts.
III.B.1) For all $k \in \mathbb { N }$, establish a simplified expression for $H _ { k + 1 } ( X ) + k H _ { k } ( X )$.
III.B.2) Deduce that, for every natural integer $n$ $$X ^ { n } = \sum _ { k = 0 } ^ { n } S ( n , k ) H _ { k } ( X )$$

We introduce a uniformly distributed random variable $Z : \Omega \rightarrow \{-1,1\}^{n}$.
(a) Show that for $m \in \{0, \ldots, n-1\}$, we have $$\sum_{k=0}^{m} (n - 2k) \binom{n}{k} = n \binom{n-1}{m}.$$
(b) Deduce that for all $A \in \mathcal{M}_{n}(\{-1,1\})$, $$\mathbb{E}\left[g_{A}(Z)\right] = \frac{n^{2}}{2^{n-1}} \binom{n-1}{\left\lfloor \frac{n}{2} \right\rfloor}$$ where $\left\lfloor \frac{n}{2} \right\rfloor$ denotes the floor of $\frac{n}{2}$.

For every natural integer $n$ and every real $x$, set $I_n(x) = \int_0^{\pi/2} \cos(2xt)(\cos t)^n \, \mathrm{d}t$. By differentiating $x \mapsto \ln(\cos(\pi x))$, show $$\forall x \in J, \quad \pi \tan(\pi x) = -\frac{2I_{4n}^{\prime}(2x)}{I_{4n}(2x)} + \frac{I_{2n}^{\prime}(x)}{I_{2n}(x)} + \sum_{k=1}^{n} \frac{8x}{(2k-1)^2} \frac{1}{1 - \frac{4x^2}{(2k-1)^2}}.$$

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 $$\forall n \in \mathbb{N}, \quad \alpha_{2n+1} = \frac{2\left(2^{2n+2} - 1\right)(2n+1)!}{\pi^{2n+2}} \zeta(2n+2).$$

For $j \in \mathbb{N}$, we denote by $Y_{n,j}$ the set of partitions whose first term $\alpha_1$ is less than or equal to $j$ and by $y_{n,j}$ the cardinality of $Y_{n,j}$; we set $y_{0,0} = 1$.
We propose to show that, if $2 \leqslant j \leqslant n$, then $y_{n,j} = y_{n,j-1} + y_{n-j,\min(j,n-j)}$.
Prove that this equality is true for $j = n$.

For $j \in \mathbb{N}$, we denote by $Y_{n,j}$ the set of partitions whose first term $\alpha_1$ is less than or equal to $j$ and by $y_{n,j}$ the cardinality of $Y_{n,j}$; we set $y_{0,0} = 1$.
We propose to show that, if $2 \leqslant j \leqslant n$, then $y_{n,j} = y_{n,j-1} + y_{n-j,\min(j,n-j)}$.
For $j < n$, verify that $y_{n,j} = y_{n,j-1} + y_{n-j,j}$. Conclude.

Using the Cauchy product of power series, deduce that, for all integers $n$ and all real numbers $\alpha$ and $\beta$, $$L_{n}(\alpha + \beta) = \sum_{k=0}^{n} \binom{n}{k} L_{k}(\alpha) L_{n-k}(\beta).$$

In Friedman's urn model ($a_{0} = 1, b_{0} = 0, a = d = 0, b = c = 1$), we have $\operatorname{card}(\Omega_{n}) = n!$. For $0 < u < v$ and $|x|$ sufficiently small, $$H(x,u,v) = \sum_{n=0}^{+\infty} \frac{x^{n}}{n!} \left( \sum_{p=1}^{+\infty} p^{n} (v-u)^{n+1} \left(\frac{u}{v}\right)^{p} \right)$$
Using question 6, justify that, for all integers $n$ and all $u$ and $v$ such that $0 < u < v$, the sum $$\sum_{p=1}^{+\infty} p^{n} (v-u)^{n+1} \left(\frac{u}{v}\right)^{p}$$ is a polynomial function of $u$ and $v$.

We consider a natural integer $n$ and a complex number $a$. We define a family of polynomials $(A_0, A_1, \ldots, A_n)$ by setting $$A_0 = 1 \quad \text{and, for all } k \in \llbracket 1, n \rrbracket, \quad A_k = \frac{1}{k!} X(X - ka)^{k-1}.$$ Let $P$ be an element of $\mathbb{C}_n[X]$ and let $\alpha_0, \ldots, \alpha_n$ be complex numbers such that $$P = \sum_{k=0}^{n} \alpha_k A_k.$$ Prove that, for all $j \in \llbracket 0, n \rrbracket$, $\alpha_j = P^{(j)}(ja)$.

We consider a natural integer $n$ and a complex number $a$. We define a family of polynomials $(A_0, A_1, \ldots, A_n)$ by setting $$A_0 = 1 \quad \text{and, for all } k \in \llbracket 1, n \rrbracket, \quad A_k = \frac{1}{k!} X(X - ka)^{k-1}.$$ Using the result of Question 24, deduce Abel's binomial identity: $$\forall (a, x, y) \in \mathbb{C}^3, \quad (x+y)^n = y^n + \sum_{k=1}^{n} \binom{n}{k} x(x - ka)^{k-1}(y + ka)^{n-k}.$$

We consider a natural integer $n$ and a complex number $a$. Using Abel's binomial identity $$\forall (a, x, y) \in \mathbb{C}^3, \quad (x+y)^n = y^n + \sum_{k=1}^{n} \binom{n}{k} x(x - ka)^{k-1}(y + ka)^{n-k},$$ establish the relation $$\forall (a, y) \in \mathbb{C}^2, \quad ny^{n-1} = \sum_{k=1}^{n} \binom{n}{k} (-ka)^{k-1}(y + ka)^{n-k}.$$

1. We denote $E_n = \operatorname{Card} \operatorname{MD}(n)$ and $\mathscr{I}_n$ the set of odd numbers in $\Delta_n$. a. Prove that for $n \geq 1$: $$E_{n+1} = \sum_{i \in \mathscr{I}_{n+1}} \binom{n}{i-1} E_{i-1} E_{n+1-i}$$ b. Deduce that for $n \geq 1$: $$2E_{n+1} = \sum_{i=0}^{n} \binom{n}{i} E_i E_{n-i}$$

For every integer $n \geqslant 1$, we denote $C_{n}$ the number of well-parenthesized words of length $2n$. We set by convention $C_{0} = 1$. For every $x \in ]-\frac{1}{4}, \frac{1}{4}[$, we set $F(x) = \sum_{k=0}^{+\infty} C_{k} x^{k}$.
Show that, for every $x \in ]-\frac{1}{4}, \frac{1}{4}[$, $F(x) = 1 + x(F(x))^{2}$.

We classify cycles of length $k$ into three subsets:

• the set $\mathcal{A}_{k}$, consisting of cycles where at least one edge appears only once;
• the set $\mathcal{B}_{k}$, consisting of cycles where all edges appear exactly twice;
• the set $\mathcal{C}_{k}$, consisting of cycles where all edges appear at least twice and there exists at least one that appears at least three times.

Show that, if the cycle $(i_{1}, i_{2}, \ldots, i_{k}, i_{1})$ belongs to $\mathcal{A}_{k}$, then $$\mathbb{E}\left(X_{i_{1}i_{2}} X_{i_{2}i_{3}} \cdots X_{i_{k-1}i_{k}} X_{i_{k}i_{1}}\right) = 0.$$

Cycles of length $k$ are classified into three subsets: the set $\mathcal{A}_{k}$, consisting of cycles where at least one edge appears only once; the set $\mathcal{B}_{k}$, consisting of cycles where all edges appear exactly twice; the set $\mathcal{C}_{k}$, consisting of cycles where all edges appear at least twice and there exists at least one that appears at least three times.
Show that, if the cycle $(i_{1}, i_{2}, \ldots, i_{k}, i_{1})$ belongs to $\mathcal{A}_{k}$, then $$\mathbb{E}\left(X_{i_{1}i_{2}} X_{i_{2}i_{3}} \cdots X_{i_{k-1}i_{k}} X_{i_{k}i_{1}}\right) = 0.$$

We set, for all $n \in \mathbb { N }$, $$D _ { n } ( X ) = \left| \begin{array} { c c c c c c } C _ { 0 } & C _ { 1 } & C _ { 2 } & \ldots & C _ { n - 1 } & C _ { n } \\ C _ { 1 } & \therefore & & \therefore & \therefore & C _ { n + 1 } \\ C _ { 2 } & & \ddots & \therefore & \ddots & \vdots \\ \vdots & \ddots & \ddots & \ddots & & C _ { 2 n - 2 } \\ C _ { n - 1 } & C _ { n } & C _ { n + 1 } & \ldots & C _ { 2 n - 2 } & C _ { 2 n - 1 } \\ 1 & X & \cdots & X ^ { n - 2 } & X ^ { n - 1 } & X ^ { n } \end{array} \right|$$ with the inner product $( P \mid Q ) = \frac { 2 } { \pi } \int _ { 0 } ^ { 1 } P ( 4 x ) Q ( 4 x ) \frac { \sqrt { 1 - x } } { \sqrt { x } } \mathrm { ~d} x$. Let $( n , k ) \in \mathbb { N } ^ { 2 }$ such that $k < n$. Show $\left( D _ { n } \mid X ^ { k } \right) = 0$.

We set, for all $n \in \mathbb { N }$, $$D _ { n } ( X ) = \left| \begin{array} { c c c c c c } C _ { 0 } & C _ { 1 } & C _ { 2 } & \ldots & C _ { n - 1 } & C _ { n } \\ C _ { 1 } & \therefore & & \therefore & \therefore & C _ { n + 1 } \\ C _ { 2 } & & \ddots & \therefore & \ddots & \vdots \\ \vdots & \ddots & \ddots & \ddots & & C _ { 2 n - 2 } \\ C _ { n - 1 } & C _ { n } & C _ { n + 1 } & \ldots & C _ { 2 n - 2 } & C _ { 2 n - 1 } \\ 1 & X & \cdots & X ^ { n - 2 } & X ^ { n - 1 } & X ^ { n } \end{array} \right|$$ and $$H _ { n } ^ { \prime } = \left| \begin{array} { c c c c c c } C _ { 1 } & C _ { 2 } & C _ { 3 } & \ldots & C _ { n - 1 } & C _ { n } \\ C _ { 2 } & \therefore & & \ddots & \ddots & C _ { n + 1 } \\ C _ { 3 } & & \ddots & \ddots & \ddots & \vdots \\ \vdots & \ddots & \ddots & \ddots & & C _ { 2 n - 3 } \\ C _ { n - 1 } & \therefore & \ddots & & \ddots & C _ { 2 n - 2 } \\ C _ { n } & C _ { n + 1 } & \cdots & C _ { 2 n - 3 } & C _ { 2 n - 2 } & C _ { 2 n - 1 } \end{array} \right|.$$ Deduce that $\forall n \in \mathbb { N } , D _ { n } = U _ { n }$, then determine, for all $n \in \mathbb { N } ^ { * }$, the value of the determinant $H _ { n } ^ { \prime }$.

Using the expansion $\frac{z}{\mathrm{e}^z - 1} = \sum_{n=0}^{+\infty} \frac{b_n}{n!} z^n$ (valid for $0 < |z| < r$), by performing a Cauchy product, show that $b_0 = 1$ and, for all integer $n \geqslant 2$, $$\sum_{p=0}^{n-1} \binom{n}{p} b_p = 0.$$

Using the expansion $\frac{z}{\mathrm{e}^z - 1} = \sum_{n=0}^{+\infty} \frac{b_n}{n!} z^n$ and a parity argument, show that $b_{2p+1} = 0$ for all integer $p \geqslant 1$.

The polynomials $B_m$ are defined by $$\forall m \in \mathbb{N}, \quad B_m(x) = \sum_{k=0}^m \binom{m}{k} b_k x^{m-k}.$$
Show that, for all integer $m \geqslant 2$, $B_m(1) = b_m$, then that, for all integer $m \geqslant 1$, $B_m' = m B_{m-1}$.

We fix an integer $n \in \mathbb{N}^*$ and consider a function $g : [0,n] \rightarrow \mathbb{R}$ of class $\mathcal{C}^\infty$. The polynomial $B_1$ is as defined in the sequence $(B_m)$.
Show that $$\int_0^n g(x)\,\mathrm{d}x = \sum_{k=0}^{n-1} \frac{g(k) + g(k+1)}{2} - \int_0^n B_1(x - \lfloor x \rfloor) g'(x)\,\mathrm{d}x.$$

We fix an integer $n \in \mathbb{N}^*$ and consider a function $g : [0,n] \rightarrow \mathbb{R}$ of class $\mathcal{C}^\infty$. The polynomials $B_m$ are defined by $B_m(x) = \sum_{k=0}^m \binom{m}{k} b_k x^{m-k}$.
Deduce that for all integer $m \geqslant 2$, $$\int_0^n g(x)\,\mathrm{d}x = \sum_{k=0}^{n-1} \frac{g(k)+g(k+1)}{2} + \sum_{p=2}^m \frac{(-1)^{p-1} b_p}{p!}\left(g^{(p-1)}(n) - g^{(p-1)}(0)\right) + \frac{(-1)^m}{m!} \int_0^n B_m(x - \lfloor x \rfloor) g^{(m)}(x)\,\mathrm{d}x.$$

Let the functions $f$, $g$ and $D$ be defined on $\mathbb{R} \backslash \mathbb{Z}$ by: $$f(x) = \pi \operatorname{cotan}(\pi x) = \pi \frac{\cos(\pi x)}{\sin(\pi x)}, \quad g(x) = \frac{1}{x} + \sum_{n=1}^{+\infty} \left(\frac{1}{x+n} + \frac{1}{x-n}\right).$$ We set $D = f - g$.
Show that for all $x \in \mathbb{R} \backslash \mathbb{Z}$, we have $$g\left(\frac{x}{2}\right) + g\left(\frac{1+x}{2}\right) = 2g(x).$$

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 $$\forall x \in \mathbb{R}, \quad h(x) = \begin{cases} \frac{x}{e^x - 1} & \text{if } x \neq 0 \\ 1 & \text{if } x = 0 \end{cases}$$
We define a sequence of real numbers $(b_n)_{n \in \mathbb{N}}$ by setting $b_0 = 1$, $b_1 = -\frac{1}{2}$, then $$\forall n \in \mathbb{N}^*, \quad b_{2n+1} = 0 \quad \text{and} \quad b_{2n} = \frac{(-1)^{n-1}(2n)! \zeta(2n)}{2^{2n-1}\pi^{2n}}.$$
Show that for all $n \in \mathbb{N}$: $$\sum_{k=0}^{n} \frac{b_k}{k!(n+1-k)!} = \begin{cases} 1 & \text{if } n = 0 \\ 0 & \text{if } n \geqslant 1 \end{cases}.$$

Let $g(x) = \frac{1}{x} + \sum_{n=1}^{+\infty}\left(\frac{1}{x+n} + \frac{1}{x-n}\right)$. Show that for all $x \in \mathbb{R} \backslash \mathbb{Z}$, we have $$g\left(\frac{x}{2}\right) + g\left(\frac{1+x}{2}\right) = 2g(x)$$