Let $\left( c _ { k } \right) _ { k \in \mathbb{N} }$ be a sequence of elements of $\mathbb{R}^{+}$ such that the power series $\sum c _ { k } x ^ { k }$ has radius of convergence 1 and the series $\sum c _ { k }$ diverges. Show that $$\sum _ { k = 0 } ^ { + \infty } c _ { k } x ^ { k } \underset { x \rightarrow 1 ^ { - } } { \longrightarrow } + \infty$$ With the element $A$ of $\mathbb{R}^{+*}$ fixed, one will show that there exists $\alpha \in ]0,1[$ such that $$\forall x \in ]1 - \alpha , 1[ , \quad \sum _ { k = 0 } ^ { + \infty } c _ { k } x ^ { k } > A$$

Let $\left( a _ { n } \right) _ { n \in \mathbb{N} }$ and $\left( b _ { n } \right) _ { n \in \mathbb{N} }$ be two sequences of elements of $\mathbb{R}^{+*}$. We assume that $\left( a _ { n } \right) _ { n \in \mathbb{N} }$ is decreasing and that $$\forall n \in \mathbb{N} , \quad \sum _ { k = 0 } ^ { n } a _ { k } b _ { n - k } = 1 .$$ We set, for $n \in \mathbb{N}$: $$B _ { n } = \sum _ { k = 0 } ^ { n } b _ { k } .$$ Let $m$ and $n$ be two natural integers such that $m > n$. Show that $$a _ { n } \leq \frac { 1 } { B _ { n } } \quad \text{and} \quad 1 \leq a _ { n } B _ { m - n } + a _ { 0 } \left( B _ { m } - B _ { m - n } \right) .$$

Let $f$ be an arithmetic function. We define, for all real $s$ such that the series converges,
$$L_f(s) = \sum_{k=1}^{\infty} \frac{f(k)}{k^s}$$
We call abscissa of convergence of $L_f$
$$A_c(f) = \inf\left\{s \in \mathbb{R} \mid \text{the series } \sum \frac{f(k)}{k^s} \text{ converges absolutely}\right\}.$$
Show that if $s > A_c(f)$, then the series $\sum \frac{f(k)}{k^s}$ converges absolutely.

Let $\left( a _ { n } \right) _ { n \in \mathbb{N} }$ and $\left( b _ { n } \right) _ { n \in \mathbb{N} }$ be two sequences of elements of $\mathbb{R}^{+*}$. We assume that $\left( a _ { n } \right) _ { n \in \mathbb{N} }$ is decreasing and that $$\forall n \in \mathbb{N} , \quad \sum _ { k = 0 } ^ { n } a _ { k } b _ { n - k } = 1 .$$ We set, for $n \in \mathbb{N}$: $$B _ { n } = \sum _ { k = 0 } ^ { n } b _ { k } .$$ We assume in this question that there exists a sequence $\left( m _ { n } \right) _ { n \in \mathbb{N} }$ satisfying $m _ { n } > n$ for $n$ large enough and $$B _ { m _ { n } - n } \underset { n \rightarrow + \infty } { \sim } B _ { n } \quad \text{and} \quad B _ { m _ { n } } - B _ { m _ { n } - n } \underset { n \rightarrow + \infty } { \longrightarrow } 0 .$$ Show that $$a _ { n } \underset { n \rightarrow + \infty } { \sim } \frac { 1 } { B _ { n } } .$$

Let $f$ be an arithmetic function. We define, for all real $s$ such that the series converges,
$$L_f(s) = \sum_{k=1}^{\infty} \frac{f(k)}{k^s}$$
The abscissa of convergence of $L_f$ is defined as
$$A_c(f) = \inf\left\{s \in \mathbb{R} \mid \text{the series } \sum \frac{f(k)}{k^s} \text{ converges absolutely}\right\}.$$
Show that if $s > A_c(f)$, then the series $\sum \frac{f(k)}{k^s}$ converges absolutely.

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 $f = g$.

Let $\left( a _ { n } \right) _ { n \in \mathbb{N} }$ and $\left( b _ { n } \right) _ { n \in \mathbb{N} }$ be two sequences of elements of $\mathbb{R}^{+*}$. We assume that $\left( a _ { n } \right) _ { n \in \mathbb{N} }$ is decreasing and that $$\forall n \in \mathbb{N} , \quad \sum _ { k = 0 } ^ { n } a _ { k } b _ { n - k } = 1 .$$ We set, for $n \in \mathbb{N}$: $$B _ { n } = \sum _ { k = 0 } ^ { n } b _ { k } .$$ We assume in this question that there exists $C > 0$ such that $$b _ { n } \underset { n \rightarrow + \infty } { \sim } \frac { C } { n }$$ Using question 17 for a well-chosen sequence $\left( m _ { n } \right) _ { n \in \mathbb{N} }$, show that $$a _ { n } \underset { n \rightarrow + \infty } { \sim } \frac { 1 } { C \ln ( n ) }$$

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 $f = g$.

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

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

We admit that $(g_k)_{k \in \mathbb{N}^*}$ is a total sequence, where $g_k(x) = \sqrt{2}\sin(k\pi x)$. For all $f \in E$, we set, $$\forall x \in [0,1], \quad \Phi(x) = \sum_{k=1}^{+\infty} \frac{1}{k^2\pi^2} \langle f, g_k \rangle g_k(x)$$ Show that $\Phi$ is continuous.

In this part, $E$ denotes the vector space of functions $f : [0,1] \rightarrow \mathbb{R}$ continuous, equipped with the inner product defined by, $$\forall (f,g) \in E^2, \quad \langle f, g \rangle = \int_0^1 f(t) g(t) \, \mathrm{d}t$$ For all $k \in \mathbb{N}^*$, we set $g_k(x) = \sqrt{2} \sin(k\pi x)$. We admit that $(g_k)_{k \in \mathbb{N}^*}$ is a total sequence. For all $f \in E$, we set, $$\forall x \in [0,1], \quad \Phi(x) = \sum_{k=1}^{+\infty} \frac{1}{k^2 \pi^2} \langle f, g_k \rangle g_k(x)$$ Show that $\Phi$ is continuous.

We admit that $(g_k)_{k \in \mathbb{N}^*}$ is a total sequence, where $g_k(x) = \sqrt{2}\sin(k\pi x)$. For all $f \in E$, we set, $$\forall x \in [0,1], \quad \Phi(x) = \sum_{k=1}^{+\infty} \frac{1}{k^2\pi^2} \langle f, g_k \rangle g_k(x)$$ For all $N \in \mathbb{N}$, we set $f_N = \sum_{k=1}^N \langle f, g_k \rangle g_k$. Show that $$\lim_{N \rightarrow +\infty} \left\| T(f_N) - \Phi \right\| = 0$$

In this part, $E$ denotes the vector space of functions $f : [0,1] \rightarrow \mathbb{R}$ continuous, equipped with the inner product defined by, $$\forall (f,g) \in E^2, \quad \langle f, g \rangle = \int_0^1 f(t) g(t) \, \mathrm{d}t$$ For all $k \in \mathbb{N}^*$, we set $g_k(x) = \sqrt{2} \sin(k\pi x)$. We admit that $(g_k)_{k \in \mathbb{N}^*}$ is a total sequence. For all $f \in E$, we set, $$\forall x \in [0,1], \quad \Phi(x) = \sum_{k=1}^{+\infty} \frac{1}{k^2 \pi^2} \langle f, g_k \rangle g_k(x)$$ For all $N \in \mathbb{N}$, we set $f_N = \sum_{k=1}^N \langle f, g_k \rangle g_k$. Show that $$\lim_{N \rightarrow +\infty} \left\| T(f_N) - \Phi \right\| = 0$$

We admit that $(g_k)_{k \in \mathbb{N}^*}$ is a total sequence, where $g_k(x) = \sqrt{2}\sin(k\pi x)$. For all $f \in E$, we set, $$\forall x \in [0,1], \quad \Phi(x) = \sum_{k=1}^{+\infty} \frac{1}{k^2\pi^2} \langle f, g_k \rangle g_k(x)$$ For all $N \in \mathbb{N}$, we set $f_N = \sum_{k=1}^N \langle f, g_k \rangle g_k$. Deduce $T(f) = \Phi$.

In this part, $E$ denotes the vector space of functions $f : [0,1] \rightarrow \mathbb{R}$ continuous, equipped with the inner product defined by, $$\forall (f,g) \in E^2, \quad \langle f, g \rangle = \int_0^1 f(t) g(t) \, \mathrm{d}t$$ For all $k \in \mathbb{N}^*$, we set $g_k(x) = \sqrt{2} \sin(k\pi x)$. We admit that $(g_k)_{k \in \mathbb{N}^*}$ is a total sequence. For all $f \in E$, we set, $$\forall x \in [0,1], \quad \Phi(x) = \sum_{k=1}^{+\infty} \frac{1}{k^2 \pi^2} \langle f, g_k \rangle g_k(x)$$ For all $N \in \mathbb{N}$, we set $f_N = \sum_{k=1}^N \langle f, g_k \rangle g_k$. Deduce $T(f) = \Phi$.

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 $A_k'(X) = A_{k-1}(X-a)$, deduce, for $j$ and $k$ elements of $\llbracket 0, n \rrbracket$, the value of $A_k^{(j)}(ja)$. Distinguish according to whether $j < k$, $j = k$ or $j > k$.

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 $$\forall (a, y) \in \mathbb{C}^2, \quad ny^{n-1} = \sum_{k=1}^{n} \binom{n}{k} (-ka)^{k-1}(y + ka)^{n-k}.$$

We consider the set $E_2$ of functions from $]-1,1[$ to $\mathbb{R}$ of the form $$t \mapsto \sum_{n=0}^{+\infty} a_n t^n$$ where $(a_n)_n \in \mathbb{R}^{\mathbb{N}}$ and $\sum (a_n)^2$ is convergent. For $f, g \in E_2$, we set $$\langle f, g \rangle = \sum_{n=0}^{+\infty} a_n b_n \quad \text{where } f: t \mapsto \sum_{n=0}^{+\infty} a_n t^n \text{ and } g: t \mapsto \sum_{n=0}^{+\infty} b_n t^n.$$ Show that $E_2$ equipped with $\langle \cdot, \cdot \rangle$ is a real pre-Hilbert space.

We consider the set $E_2$ of functions from $]-1,1[$ to $\mathbb{R}$ of the form $$t \mapsto \sum_{n=0}^{+\infty} a_n t^n$$ where $(a_n)_n \in \mathbb{R}^{\mathbb{N}}$ and $\sum (a_n)^2$ is convergent. For $f, g \in E_2$, we set $$\langle f, g \rangle = \sum_{n=0}^{+\infty} a_n b_n \quad \text{where } f : t \mapsto \sum_{n=0}^{+\infty} a_n t^n \text{ and } g : t \mapsto \sum_{n=0}^{+\infty} b_n t^n.$$ Show that $E_2$ equipped with $\langle \cdot, \cdot \rangle$ is a real pre-Hilbert space.

We consider the set $E_2$ of functions from $]-1,1[$ to $\mathbb{R}$ of the form $$t \mapsto \sum_{n=0}^{+\infty} a_n t^n$$ where $(a_n)_n \in \mathbb{R}^{\mathbb{N}}$ and $\sum (a_n)^2$ is convergent. For $f, g \in E_2$, we set $$\langle f, g \rangle = \sum_{n=0}^{+\infty} a_n b_n \quad \text{where } f: t \mapsto \sum_{n=0}^{+\infty} a_n t^n \text{ and } g: t \mapsto \sum_{n=0}^{+\infty} b_n t^n.$$ Let $x \in ]-1,1[$. Determine $g_x \in E_2$ such that, for all $f \in E_2$, $$f(x) = \langle g_x, f \rangle$$

We consider the set $E_2$ of functions from $]-1,1[$ to $\mathbb{R}$ of the form $$t \mapsto \sum_{n=0}^{+\infty} a_n t^n$$ where $(a_n)_n \in \mathbb{R}^{\mathbb{N}}$ and $\sum (a_n)^2$ is convergent. For $f, g \in E_2$, we set $$\langle f, g \rangle = \sum_{n=0}^{+\infty} a_n b_n \quad \text{where } f : t \mapsto \sum_{n=0}^{+\infty} a_n t^n \text{ and } g : t \mapsto \sum_{n=0}^{+\infty} b_n t^n.$$ Let $x \in ]-1,1[$. Determine $g_x \in E_2$ such that, for all $f \in E_2$, $$f(x) = \langle g_x, f \rangle$$

We consider the set $E_2$ of functions from $]-1,1[$ to $\mathbb{R}$ of the form $$t \mapsto \sum_{n=0}^{+\infty} a_n t^n$$ where $(a_n)_n \in \mathbb{R}^{\mathbb{N}}$ and $\sum (a_n)^2$ is convergent. For $f, g \in E_2$, we set $$\langle f, g \rangle = \sum_{n=0}^{+\infty} a_n b_n \quad \text{where } f: t \mapsto \sum_{n=0}^{+\infty} a_n t^n \text{ and } g: t \mapsto \sum_{n=0}^{+\infty} b_n t^n.$$ Deduce that $E_2$ is a reproducing kernel Hilbert space and specify its kernel.