Let $\left( a _ { n } \right) _ { n \geqslant 0 }$ be a real sequence with $\sum a_n x^n$ having radius of convergence 1, with sum $f(x) = \sum_{n=0}^{+\infty} a_n x^n$ satisfying $f(x) \sim \frac{1}{1-x}$ as $x \to 1^-$, and with $a_n \geqslant 0$ for all $n \in \mathbb{N}$. We denote $A _ { n } = \sum _ { k = 0 } ^ { n } a _ { k }$, $\widetilde { a } _ { n } = \frac { A _ { n } } { n + 1 }$, and $\mu > 0$ is an upper bound of the sequence $\left( \widetilde { a } _ { n } \right) _ { n \geqslant 0 }$: $\forall n \in \mathbb { N } , \widetilde { a } _ { n } \leqslant \mu$.
a) For all $x \in ] - 1,1 [$, show that $( 1 - x ) \sum _ { k = 0 } ^ { + \infty } A _ { k } x ^ { k } = f ( x )$. b) Deduce that for all $x \in \left[ 0,1 \left[ \right. \right.$ and all $N \in \mathbb { N } ^ { * }$ $$\frac { f ( x ) } { 1 - x } \leqslant A _ { N - 1 } \frac { 1 - x ^ { N } } { 1 - x } + \mu \sum _ { k = N } ^ { + \infty } ( k + 1 ) x ^ { k }$$ c) Deduce that for all $x \in \left[ 0,1 \left[ \right. \right.$ and all $N \in \mathbb { N } ^ { * }$ $$f ( x ) \leqslant A _ { N - 1 } + \mu \left( ( N + 1 ) x ^ { N } + \frac { x ^ { N + 1 } } { 1 - x } \right) .$$

Let $\left( a _ { n } \right) _ { n \geqslant 0 }$ be a real sequence with $\sum a_n x^n$ having radius of convergence 1, with sum $f(x) = \sum_{n=0}^{+\infty} a_n x^n$ satisfying $f(x) \sim \frac{1}{1-x}$ as $x \to 1^-$, and with $a_n \geqslant 0$ for all $n \in \mathbb{N}$.
Let $P$ be a polynomial with real coefficients. Show that $$( 1 - x ) \sum _ { n = 0 } ^ { + \infty } a _ { n } x ^ { n } P \left( x ^ { n } \right) \underset { \substack { x \rightarrow 1 \\ x < 1 } } { \longrightarrow } \int _ { 0 } ^ { 1 } P ( t ) \mathrm { d } t$$ We will first consider the special case $P ( x ) = x ^ { k }$, where $k \in \mathbb { N }$.

Let $\theta \in \mathbb { R }$. Let $g$ be the function from $] - 1,1 [$ to $\mathbb { C }$ defined by $$g ( x ) = \sum _ { n = 1 } ^ { + \infty } \frac { \mathrm { e } ^ { \mathrm { i } n \theta } } { n } x ^ { n }$$
a) Show that $g$ is of class $C ^ { 1 }$ on $] - 1,1 [$ and that, for all $x \in ] - 1,1 [$, $$g ^ { \prime } ( x ) = \frac { \mathrm { e } ^ { \mathrm { i } \theta } - x } { x ^ { 2 } - 2 x \cos \theta + 1 }$$
b) Show that, if $x \in ] - 1,1 [$, $$h ( x ) = - \frac { 1 } { 2 } \ln \left( x ^ { 2 } - 2 x \cos \theta + 1 \right) + \mathrm { i } \arctan \left( \frac { x \sin \theta } { 1 - x \cos \theta } \right)$$ is well defined and that $h ( x ) = g ( x )$.

We are given a power series with complex coefficients $\sum_{n \geqslant 0} a_n z^n$ with radius of convergence $R$ strictly positive, possibly equal to $+\infty$, and $\varphi(A) = \sum_{n=0}^{+\infty} a_n A^n$ defined on $\mathcal{B} = \{A \in \mathcal{M}_d(\mathbb{R}), \|A\| < R\}$.
Find a necessary and sufficient condition on the power series $\sum_{n \geqslant 0} a_n z^n$ for there to exist $P \in \mathbb{R}[X]$ such that $$\forall A \in \mathcal{M}_d(\mathbb{R}), \quad \varphi(A) = P(A)$$

State the theorem allowing the product of two series of complex numbers. (We admit in the rest of Part III that the result valid for series of complex numbers still holds for series of matrices in $\mathcal{M}_d(\mathbb{C})$.)

For $(x,y) \in D(0,1)$ fixed, we define the complex number $z = x + iy$ and we set for $t$ real: $$\mathrm{N}(x,y,t) = \frac{1 - |z|^2}{|z - e^{it}|^2} = \frac{1 - (x^2 + y^2)}{(x - \cos t)^2 + (y - \sin t)^2}$$
Let $t \in [0, 2\pi]$ be fixed. Determine two complex numbers $\alpha$ and $\beta$, independent of $t$ and $z$, such that $$\mathrm{N}(x,y,t) = -1 + \frac{\alpha}{1 - ze^{-it}} + \frac{\beta}{1 - \bar{z}e^{it}}$$

We assume $m>1$. We study the Galton-Watson process starting with $k$ individuals in generation 0. We define $u_n$, $u_n^{(r)}$, $U(s)=\sum_{n=1}^{+\infty}u_n s^n$ and $U_r(s)=\sum_{n=1}^{+\infty}u_n^{(r)}s^n$ for $s\in[-1,1]$.
Deduce that, for every strictly positive integer $r$, $U_r=U^r$ ($U^r$ denotes $U\times U\times\cdots\times U$ $r$ times).

We define $\varphi$ the function defined on $] - 1 , + \infty [$ by $\varphi ( x ) = ( \psi ( 1 + x ) - \psi ( 1 ) ) ^ { 2 } + \left( \psi ^ { \prime } ( 1 ) - \psi ^ { \prime } ( 1 + x ) \right)$.
Show that $\varphi$ is $\mathcal { C } ^ { \infty }$ on its domain of definition and give for every natural integer $n \geqslant 2$ the value of $\varphi ^ { ( n ) } ( 0 )$ as a function of the successive derivatives of $\psi$ at the point 1.

We are given $f : \mathbb { R } ^ { 2 } \rightarrow \mathbb { C }$ and $g : \mathbb { R } ^ { 2 } \rightarrow \mathbb { R } ^ { 2 }$ two functions continuous and 1-periodic in each of their arguments, and $\omega \in \mathbb { R } ^ { 2 }$. We assume $x \neq 0$ (but close to 0). We consider a new unknown function $\tilde { \alpha } : \mathbb { R } \rightarrow \mathbb { R } ^ { 2 }$, of the form $$\tilde { \alpha } ( t ) = \alpha ( t ) + x h ( \alpha ( t ) ) ,$$ where $h : \mathbb { R } ^ { 2 } \rightarrow \mathbb { R } ^ { 2 }$ is an auxiliary function, 1-periodic in each of its arguments, of class $\mathscr { C } ^ { 1 }$ and of zero average, and which moreover, for some $\nu \in \mathbb { R } ^ { 2 }$, satisfies the equation $$\forall \theta \in \mathbb { R } ^ { 2 } , \quad d h ( \theta ) \cdot \omega + g ( \theta ) = \nu . \tag{4}$$
Determine $\nu$ as a function of $g$. In the case where the two components $g _ { 1 }$ and $g _ { 2 }$ of $g$ are trigonometric polynomials, deduce the existence of a solution $h$ of equation (4), which you will make explicit.

We set for every integer $n \geqslant 0$, $$B _ { n } = \sum _ { k = 0 } ^ { n } S ( n , k )$$ where $S(n,k)$ denotes the number of partitions of $\llbracket 1, n \rrbracket$ into $k$ parts. Let $R$ be the radius of convergence of the power series $\sum _ { n \geqslant 0 } \frac { B _ { n } } { n ! } z ^ { n }$, and for $x \in ] - R , R [$, set $f ( x ) = \sum _ { n = 0 } ^ { + \infty } \frac { B _ { n } } { n ! } x ^ { n }$.
Show that for all $x \in ] - R , R [ , f ^ { \prime } ( x ) = \mathrm { e } ^ { x } f ( x )$.

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 $k \in \mathbb { N } ^ { * }$, $$H _ { k } ( X ) = X ( X - 1 ) \cdots ( X - k + 1 )$$
Show that the family $( H _ { 0 } , \ldots , H _ { n } )$ is a basis of the space $\mathbb { R } _ { n } [ X ]$.

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 $k \in \mathbb { N } ^ { * }$, $$H _ { k } ( X ) = X ( X - 1 ) \cdots ( X - k + 1 )$$ It has been established that for all $k \in \mathbb { N }$ and $x \in ] - 1,1 [$, $$\frac { \left( \mathrm { e } ^ { x } - 1 \right) ^ { k } } { k ! } = \sum _ { n = k } ^ { + \infty } S ( n , k ) \frac { x ^ { n } } { n ! }$$
III.D.1) For $x \in ] - 1,1 [$ and $\alpha \in \mathbb { R }$, simplify $\sum _ { k = 0 } ^ { + \infty } H _ { k } ( \alpha ) \frac { x ^ { k } } { k ! }$.
III.D.2) Show that for $u < \ln 2$ $$\mathrm { e } ^ { u \alpha } = \sum _ { k = 0 } ^ { + \infty } H _ { k } ( \alpha ) \frac { \left( \mathrm { e } ^ { u } - 1 \right) ^ { k } } { k ! }$$

We fix $n \in \mathbb { N }$. We define the linear map: $$\begin{aligned} \Delta : \mathbb { R } [ X ] & \rightarrow \mathbb { R } [ X ] \\ P ( X ) & \mapsto P ( X + 1 ) - P ( X ) \end{aligned}$$ We denote $F = \left\{ P \in \mathbb { R } _ { n } [ X ] \mid P ( 0 ) = 0 \right\}$, then $G = \operatorname { Vect } \left( X ^ { 2 k + 1 } ; 0 \leqslant k \leqslant n - 1 \right)$.
Let $Q ( X )$ be the polynomial such that $\forall p \in \mathbb { N } , Q ( p ) = \sum _ { k = 0 } ^ { p } k$.
V.D.1) Recall the explicit expression of the polynomial $Q ( X )$.
V.D.2) Show that the map: $$\begin{aligned} \Phi : F & \rightarrow G \\ P ( X ) & \mapsto \Delta ( P ( Q ( X - 1 ) ) ) \end{aligned}$$ is an isomorphism.
V.D.3) Deduce that for all $r \in \mathbb { N }$, there exists a unique polynomial $P _ { r } ( X )$ such that $$\forall p \in \mathbb { N } , \quad \sum _ { k = 1 } ^ { p } k ^ { 2 r + 1 } = P _ { r } \left( \frac { p ( p + 1 ) } { 2 } \right)$$

We fix $n \in \mathbb { N }$. For all $r \in \mathbb { N }$, there exists a unique polynomial $P _ { r } ( X )$ such that $$\forall p \in \mathbb { N } , \quad \sum _ { k = 1 } ^ { p } k ^ { 2 r + 1 } = P _ { r } \left( \frac { p ( p + 1 ) } { 2 } \right)$$
V.E.1) Determine the leading term in $P _ { r } ( X )$.
V.E.2) Show that for $r \geqslant 1$, $X ^ { 2 }$ divides $P _ { r } ( X )$.
V.E.3) Explicitly give the polynomials $P _ { 1 } ( X )$ and $P _ { 2 } ( X )$.

We assume $c(x) = 0$ for all $x \in [0,1]$. Let $u$ be the solution of problem (1) and $u_0, \ldots, u_n$ solutions of system (2) with $c = 0$. Define: $$\hat { B } _ { n + 1 } u ( X ) = \sum _ { k = 0 } ^ { n } u _ { k } \binom { n + 1 } { k } X ^ { k } ( 1 - X ) ^ { n + 1 - k }$$
Show that for all $n \in \mathbb { N } ^ { * }$ and all $x \in ]0,1[$ we have:
$$\left( \hat { B } _ { n + 1 } u \right) ^ { \prime \prime } ( x ) = - \frac { n } { n + 1 } \sum _ { \ell = 0 } ^ { n - 1 } f \left( \frac { \ell + 1 } { n + 1 } \right) \binom { n - 1 } { \ell } x ^ { \ell } ( 1 - x ) ^ { n - 1 - \ell }$$

We consider the Hilbert polynomials $$\left\{ \begin{array} { l } H _ { 0 } ( X ) = 1 \\ H _ { n } ( X ) = \frac { 1 } { n ! } \prod _ { k = 0 } ^ { n - 1 } ( X - k ) = \frac { X ( X - 1 ) \cdots ( X - n + 1 ) } { n ! } \quad \text { for all } n \in \mathbb { N } ^ { * } \end{array} \right.$$
Deduce $$\forall x \in ] - 1,1 \left[ , \quad \varphi ( x ) = 1 + \sum _ { i = 0 } ^ { + \infty } \left( \sum _ { j = 0 } ^ { + \infty } a _ { i , j } ( x ) \right) \right.$$ where we have set $$\forall ( i , j ) \in \mathbb { N } ^ { 2 } , \quad a _ { i , j } ( x ) = \frac { ( - 1 ) ^ { i + 1 } } { ( i + 1 ) ! } H _ { j } \left( \frac { i - 1 } { 2 } + j \right) x ^ { i + j + 1 }$$

We consider the Hilbert polynomials $$\left\{ \begin{array} { l } H _ { 0 } ( X ) = 1 \\ H _ { n } ( X ) = \frac { 1 } { n ! } \prod _ { k = 0 } ^ { n - 1 } ( X - k ) = \frac { X ( X - 1 ) \cdots ( X - n + 1 ) } { n ! } \quad \text { for all } n \in \mathbb { N } ^ { * } \end{array} \right.$$
Use the results admitted in the preamble to establish the equality $$\forall x \in ] - 1,1 \left[ , \quad \varphi ( x ) = \sum _ { n = 0 } ^ { + \infty } a _ { n } x ^ { n } \right.$$ where $$\left\{ \begin{array} { l } a _ { 0 } = 1 \\ a _ { n } = \sum _ { k = 0 } ^ { n - 1 } \frac { ( - 1 ) ^ { n - k } } { ( n - k ) ! } H _ { k } \left( \frac { n + k } { 2 } - 1 \right) \quad \text { for all } n \in \mathbb { N } ^ { * } \end{array} \right.$$

Recall the definition of the Cauchy product of two power series and state the theorem relating to it.

We consider a general balanced urn with parameters $a_{0}, b_{0}, a, b, c, d \in \mathbb{N}$ satisfying $a + b = c + d = s$. For all real $u$ and $v$, we set $P_{0}(u,v) = u^{a_{0}} v^{b_{0}}$ and $P_{n}(u,v) = \sum_{\omega \in \Omega_{n}} u^{b(\omega)} v^{n(\omega)}$.
Prove that, for all integers $n$, $$P_{n+1}(u,v) = u^{a+1} v^{b} \frac{\partial P_{n}}{\partial u}(u,v) + u^{c} v^{d+1} \frac{\partial P_{n}}{\partial v}(u,v)$$

In the general model of a Pólya urn ($b = c = 0$, $a = d$), the function $G$ is defined on $U$ by $$G(x,u,v) = u^{a_{0}} v^{b_{0}} (1 - axu^{a})^{-a_{0}/a} (1 - axv^{a})^{-b_{0}/a}$$ and admits the expansion $G(x,u,v) = \sum_{n=0}^{+\infty} Q_{n}(u,v) \frac{x^{n}}{n!}$ on $D_{\rho}$. The function $H(x,u,v) = \sum_{n=0}^{+\infty} P_{n}(u,v) \frac{x^{n}}{n!}$ was defined in part III.
Deduce that, for all integers $n$, $P_{n} = Q_{n}$, and then that $H$ and $G$ coincide on $D_{\rho}$.

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 assume that $\lambda$ is a real distinct from 1 and we set $w = \frac{1}{\lambda - 1}$. We further set $\mathbf{f} = (1 + w)\delta - w\mathbf{1}$. We denote $\log_2$ the logarithm function in base 2, defined by $\log_2(x) = \frac{\ln(x)}{\ln(2)}$ for all real $x > 0$.
Show that, for $s$ real sufficiently large,
$$\frac{1}{L_{\mathbf{f}}(s)} = 1 + \sum_{m=2}^{\infty} m^{-s} \sum_{k=1}^{\lfloor \log_2 m \rfloor} w^k D_k(m)$$
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.

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 denote $\log_2$ the logarithm function in base 2, defined by $\log_2(x) = \frac{\ln(x)}{\ln(2)}$ for all real $x > 0$. With $w = \frac{1}{\lambda-1}$ and $\mathbf{f} = (1+w)\delta - w\mathbf{1}$, show that, for $s$ real sufficiently large,
$$\frac{1}{L_{\mathbf{f}}(s)} = 1 + \sum_{m=2}^{\infty} m^{-s} \sum_{k=1}^{\left\lfloor \log_2 m \right\rfloor} w^k D_k(m)$$
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.

Let $A > 2$. Let $f$ be a continuous and bounded function from $\mathbb{R}$ to $\mathbb{R}$ and $P$ a polynomial of degree $p$. Justify that there exists a constant $K$ such that $$\forall x \in \mathbb{R} \setminus ]-A, A[, \quad |f(x) - P(x)| \leqslant K|x|^{p}.$$