We denote $D_n = \left\{ \sum_{j=1}^{n} \frac{x_j}{2^j},\, (x_j)_{j \in \llbracket 1,n \rrbracket} \in \{0,1\}^n \right\}$ and $\pi_n(x) = \frac{\lfloor 2^n x \rfloor}{2^n}$.
Let $n \in \mathbb{N}^{\star}$. Justify $x \in D_n \Longleftrightarrow 2^n x \in \llbracket 0, 2^n - 1 \rrbracket$.

We denote $\pi_k(x) = \frac{\lfloor 2^k x \rfloor}{2^k}$.
Let $n \in \mathbb{N}^{\star}$ and $x = \sum_{j=1}^{n} \frac{x_j}{2^j}$ with $(x_j)_{j \in \llbracket 1,n \rrbracket} \in \{0,1\}^n$. Show $$\forall k \in \mathbb{N}, \quad \pi_k(x) = \sum_{j=1}^{\min(n,k)} \frac{x_j}{2^j}.$$

Let $f$ and $g$ be two arithmetic functions with finite abscissas of convergence. Show that if, for all $s > \max\left(A_c(f), A_c(g)\right), L_f(s) = L_g(s)$, then $f = g$.

We assume that $\lambda$ is a real distinct from 1 and we set $w = \frac{1}{\lambda - 1}$. We further set $\mathbf{f} = (1 + w)\delta - w\mathbf{1}$.
Using the notations of Dirichlet series given in subsection I.E, express, for values of the real $s$ to be specified, $L_{\mathbf{f}}(s)$ in terms of $w$ and $L_{\mathbf{1}}(s)$.

We assume that $\lambda$ is a real distinct from 1 and we set $w = \frac{1}{\lambda - 1}$. We denote $\log_2$ the logarithm function in base 2. For $n \geq 1$, we set $S_k(n) = \sum_{m=2}^{n} D_k(m)$. Deduce from the previous question that
$$\chi_n(\lambda) = (\lambda - 1)^n - \sum_{k=1}^{\lfloor \log_2 n \rfloor} (\lambda - 1)^{n-k-1} S_k(n).$$

Let $f$ and $g$ be two arithmetic functions with finite abscissas of convergence. Show that if, for all $s > \max\left(A_c(f), A_c(g)\right), L_f(s) = L_g(s)$, then $f = g$.

Using the notations of Dirichlet series given in subsection I.E, and with $w = \frac{1}{\lambda-1}$ and $\mathbf{f} = (1+w)\delta - w\mathbf{1}$, express, for values of the real $s$ to be specified, $L_{\mathbf{f}}(s)$ in terms of $w$ and $L_{\mathbf{1}}(s)$.

We denote $\log_2$ the logarithm function in base 2. For $n \geq 1$, we set $S_k(n) = \sum_{m=2}^{n} D_k(m)$ where $D_k(m)$ is the number of ways to decompose the integer $m$ into a product of $k$ factors greater than or equal to 2, the order of these factors being important. Deduce from the previous question that
$$\chi_n(\lambda) = (\lambda - 1)^n - \sum_{k=1}^{\left\lfloor \log_2 n \right\rfloor} (\lambda - 1)^{n-k-1} S_k(n).$$

We assume that $f \in L^1(\mathbb{R})$ and $g \in L^\infty(\mathbb{R})$. The convolution product of $f$ and $g$ is defined by $$\forall x \in \mathbb{R}, \quad (f * g)(x) = \int_{-\infty}^{+\infty} f(t) g(x-t) \,\mathrm{d}t$$
Show that $f * g$ is defined on $\mathbb{R}$ and that $$\forall x \in \mathbb{R}, \quad (f * g)(x) = \int_{-\infty}^{+\infty} f(x-t) g(t) \,\mathrm{d}t = (g * f)(x)$$

Assume that $f \in L^1(\mathbb{R})$ and $g \in L^\infty(\mathbb{R})$. Show that $f * g$ is defined on $\mathbb{R}$ and that $$\forall x \in \mathbb{R}, \quad (f*g)(x) = \int_{-\infty}^{+\infty} f(x-t)g(t)\,\mathrm{d}t = (g*f)(x)$$

The Pochhammer symbol is defined, for any real number $a$ and any natural integer $n$, by $$[a]_n = \begin{cases} 1 & \text{if } n = 0 \\ a(a+1)\cdots(a+n-1) = \prod_{k=0}^{n-1}(a+k) & \text{otherwise} \end{cases}$$ Let $n \in \mathbb{N}$. Give an expression of $[a]_n$

• using factorials when $a \in \mathbb{N}^*$;
• using two values of the function $\Gamma$, when $a \in D$.

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.

Deduce that, for every polynomial $P \in \mathbb{R}[X]$, $$\lim_{n \rightarrow +\infty} \mathbb{E}\left(\frac{1}{n} \sum_{i=1}^{n} P\left(\Lambda_{i,n}\right)\right) = \frac{1}{2\pi} \int_{-2}^{2} P(x) \sqrt{4 - x^{2}} \, \mathrm{d}x$$

Deduce that, for any polynomial $P \in \mathbb{R}[X]$, $$\lim_{n \rightarrow +\infty} \mathbb{E}\left(\frac{1}{n} \sum_{i=1}^{n} P\left(\Lambda_{i,n}\right)\right) = \frac{1}{2\pi} \int_{-2}^{2} P(x) \sqrt{4 - x^{2}} \, \mathrm{d}x$$

Give without proof the value of $C _ { 3 }$ and represent all Dyck paths of length 6.

For all $n \in \mathbb { N }$, set $\mu _ { n } = \left( X ^ { n } \mid 1 \right)$ with the inner product $$( P \mid Q ) = \frac { 2 } { \pi } \int _ { 0 } ^ { 1 } P ( 4 x ) Q ( 4 x ) \frac { \sqrt { 1 - x } } { \sqrt { x } } \mathrm { ~d} x .$$ Deduce that $\forall n \in \mathbb { N } , \mu _ { n } = C _ { n }$.

Let $n \in \mathbb { N }$. Deduce from the previous parts the value of the determinant $$H _ { n } = \operatorname { det } \left( C _ { i + j - 2 } \right) _ { 1 \leqslant i , j \leqslant n + 1 } = \left| \begin{array} { c c c c c c } C _ { 0 } & C _ { 1 } & C _ { 2 } & \ldots & C _ { n - 1 } & C _ { n } \\ C _ { 1 } & \therefore & & \therefore & \therefore & C _ { n + 1 } \\ C _ { 2 } & & \ddots & \ddots & \therefore & \vdots \\ \vdots & \therefore & \ddots & \ddots & & C _ { 2 n - 2 } \\ C _ { n - 1 } & \therefore & \ddots & & \therefore & C _ { 2 n - 1 } \\ C _ { n } & C _ { n + 1 } & \cdots & C _ { 2 n - 2 } & C _ { 2 n - 1 } & C _ { 2 n } \end{array} \right|.$$

Using the relation $b_0 = 1$ and $\sum_{p=0}^{n-1} \binom{n}{p} b_p = 0$ for all integer $n \geqslant 2$, deduce the value of $b_1, b_2, b_3$ and $b_4$.

The polynomials $B_m$ are defined by $$\forall m \in \mathbb{N}, \quad B_m(x) = \sum_{k=0}^m \binom{m}{k} b_k x^{m-k}.$$
Determine $B_0, B_1, B_2$ and $B_3$.

Let $s > 1$ be a real number and let $X$ be a random variable taking values in $\mathbb{N}^*$ following the zeta distribution with parameter $s$.
Deduce that $$\zeta(s)^{-1} = \lim_{n \rightarrow +\infty} \prod_{k=1}^{n}\left(1 - p_k^{-s}\right).$$

Let $s > 1$ be a real number and let $X$ be a random variable taking values in $\mathbb{N}^*$ following the zeta distribution with parameter $s$.
We denote $r(n)$ the number of divisors $d \geqslant 1$ of $n$. Show that the series $\sum_{n=1}^{+\infty} r(n) n^{-s}$ converges and that its sum equals $\zeta(s)^2$.

Let $s > 1$ be a real number and let $X$ be a random variable taking values in $\mathbb{N}^*$ following the zeta distribution with parameter $s$. We recall that $\chi_4(2n) = 0$ and $\chi_4(2n-1) = (-1)^{n-1}$ for $n \in \mathbb{N}^*$. We set $g(n) = r_1(n) - r_3(n)$ where $r_i(n) = \operatorname{Card}\{d \in \mathbb{N} : d \equiv i [4] \text{ and } d \mid n\}$.
Deduce that the series $$\sum_{n=0}^{+\infty} \frac{(-1)^n}{(2n+1)^s}$$ is convergent and that its sum equals $E(g(X))$.

Deduce that, for all $x \in \mathbb{R}$, $$\sin(\pi x) = \pi x \lim_{n \rightarrow +\infty} \prod_{k=1}^{n}\left(1 - \frac{x^2}{k^2}\right).$$

Show that for all $x \in ]0,1[$: $$\frac{\pi}{\sin(\pi x)} = \sum_{n=0}^{+\infty} \frac{(-1)^n}{n+x} + \sum_{n=0}^{+\infty} \frac{(-1)^n}{n+1-x}.$$

We set $C = \exp\left(\frac{I}{2\pi}\right)$ with: $$I = \int_0^{2\pi} \ln\left(\max\left\{\left|e^{i\theta} - 1\right|, \left|e^{i\theta} + 1\right|\right\}\right) d\theta$$ Show that: $$I = 4\sum_{k=0}^{\infty} \frac{(-1)^k}{(2k+1)^2}$$ You may use the result from question 2.13.