We define a sequence $(a_n)_{n \geqslant 1}$ by setting $$\forall n \in \mathbb{N}^*, \quad a_n = \frac{(-n)^{n-1}}{n!}.$$ We define, when possible, $S(x) = \sum_{n=1}^{+\infty} a_n x^n$, with radius of convergence $R$. Prove that $$\forall x \in ]-R, R[, \quad x(1 + S(x))S'(x) = S(x).$$ One may use the result from Question 26.
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 the function $f : \left|]-\frac{1}{4}, \frac{1}{4}[ \begin{array}{cc} & \rightarrow \\ x & \mathbb{R} \\ & \mapsto 2xF(x) - 1 \end{array}\right.$ does not vanish.
We set, for all $t \in I$, $$f ( t ) = \sum _ { n = 0 } ^ { + \infty } C _ { n } t ^ { n } \quad \text { and } \quad g ( t ) = 2 t f ( t ) .$$ Show that $\forall t \in I , g ( t ) ^ { 2 } = 2 g ( t ) - 4 t$.
Let $z \in D$. Show the convergence of the series $\sum _ { n \geq 1 } \frac { z ^ { n } } { n }$. Specify the value of its sum when $z \in ] - 1,1 [$. We denote $$L ( z ) : = \sum _ { n = 1 } ^ { + \infty } \frac { z ^ { n } } { n }$$
Let $z \in D$. Show the convergence of the series $\sum _ { n \geq 1 } \frac { z ^ { n } } { n }$. Specify the value of its sum when $z \in ] - 1,1 [$. We denote $$L ( z ) : = \sum _ { n = 1 } ^ { + \infty } \frac { z ^ { n } } { n }$$
For $( n , N ) \in \mathbf { N } \times \mathbf { N } ^ { * }$, we denote by $P _ { n , N }$ the set of lists $\left( a _ { 1 } , \ldots , a _ { N } \right) \in \mathbf { N } ^ { N }$ such that $\sum _ { k = 1 } ^ { N } k a _ { k } = n$. If this set is finite, we denote by $p _ { n , N }$ its cardinality. We denote by $p _ { n }$ the final value of $\left( p _ { n , N } \right) _ { N \geq 1 }$, and $P ( z ) := \exp \left[ \sum _ { n = 1 } ^ { + \infty } L \left( z ^ { n } \right) \right]$ for all $z \in D$. We fix $\ell \in \mathbf { N }$ and $x \in [ 0,1 [$. Using the result of the previous question, establish the bound $\sum _ { n = 0 } ^ { \ell } p _ { n } x ^ { n } \leq P ( x )$. Deduce the radius of convergence of the power series $\sum _ { n } p _ { n } z ^ { n }$.
For $( n , N ) \in \mathbf { N } \times \mathbf { N } ^ { * }$, we denote by $P _ { n , N }$ the set of lists $\left( a _ { 1 } , \ldots , a _ { N } \right) \in \mathbf { N } ^ { N }$ such that $\sum _ { k = 1 } ^ { N } k a _ { k } = n$. If this set is finite, we denote by $p _ { n , N }$ its cardinality. We denote by $p _ { n }$ the final value of $\left( p _ { n , N } \right) _ { N \geq 1 }$, and $P ( z ) := \exp \left[ \sum _ { n = 1 } ^ { + \infty } L \left( z ^ { n } \right) \right]$ for all $z \in D$. Let $z \in D$. By examining the difference $\sum _ { n = 0 } ^ { + \infty } p _ { n } z ^ { n } - \sum _ { n = 0 } ^ { + \infty } p _ { n , N } z ^ { n }$, prove that $$P ( z ) = \sum _ { n = 0 } ^ { + \infty } p _ { n } z ^ { n }$$
Let $z \in D$. Show the convergence of the series $\sum_{n \geq 1} \frac{z^n}{n}$. Specify the value of its sum when $z \in ]-1,1[$. We denote $$L(z) := \sum_{n=1}^{+\infty} \frac{z^n}{n}$$
For $(n,N) \in \mathbf{N} \times \mathbf{N}^*$, we denote by $P_{n,N}$ the set of lists $(a_1, \ldots, a_N) \in \mathbf{N}^N$ such that $\sum_{k=1}^{N} k a_k = n$. If this set is finite, we denote by $p_{n,N}$ its cardinality. We denote by $p_n$ the final value of $(p_{n,N})_{N \geq 1}$. We fix $\ell \in \mathbf{N}$ and $x \in [0,1[$. Using the result of the previous question, establish the bound $\sum_{n=0}^{\ell} p_n x^n \leq P(x)$. Deduce the radius of convergence of the power series $\sum_n p_n z^n$.
For $(n,N) \in \mathbf{N} \times \mathbf{N}^*$, we denote by $P_{n,N}$ the set of lists $(a_1, \ldots, a_N) \in \mathbf{N}^N$ such that $\sum_{k=1}^{N} k a_k = n$. If this set is finite, we denote by $p_{n,N}$ its cardinality. We denote by $p_n$ the final value of $(p_{n,N})_{N \geq 1}$. Let $z \in D$. By examining the difference $\sum_{n=0}^{+\infty} p_n z^n - \sum_{n=0}^{+\infty} p_{n,N} z^n$, prove that $$P(z) = \sum_{n=0}^{+\infty} p_n z^n$$
Let $f$ be a power series and $z$ a complex number such that $\hat{f}(|z|) < \infty$. Show then that the series $f$ converges at $z$ and that $|f(z)| \leqslant \hat{f}(|z|)$. Give an example where this inequality is strict.
We consider a power series $f = \lambda z + F, F \in O_2, \lambda = (f)_1 \neq 0$, with $g = f^\dagger$ the reciprocal series. Show that $\hat{g} \prec (1/\lambda)(I + \hat{F} \circ \hat{g})$, conclude using part C that $\rho(g) > 0$ if $\rho(f) > 0$.
We set $\lambda = (f)_1$ and denote $f = \lambda z + F$, with $F \in O_2$. We assume that $\lambda \neq 0$ and that $\lambda$ is not a complex root of unity, that is, $\lambda^n \neq 1$ for all integer $n \geqslant 1$. We propose to show that there exists a unique power series of the form $h = I + H, H \in O_2$ satisfying $h^\dagger \circ f \circ h = \lambda z$. Show that there exists a unique series $H \in O_2$ such that $H \circ (\lambda I) - \lambda H = F \circ (I + H)$.
We set $\lambda = (f)_1$ and denote $f = \lambda z + F$, with $F \in O_2$. We assume that $\lambda \neq 0$ and that $\lambda$ is not a complex root of unity. We have shown in question (19) that there exists a unique series $H \in O_2$ such that $H \circ (\lambda I) - \lambda H = F \circ (I + H)$. Conclude that there exists a unique power series of the form $h = I + H, H \in O_2$ satisfying $h^\dagger \circ f \circ h = \lambda z$.
We consider a power series $f = \lambda I + F$ with $F \in O_2, \rho(F) > 0$. We still assume that $\lambda$ has modulus 1 and is not a root of unity. We consider the real $r_0 > 0$ given by question (24) (applied for $m = 1$) and the sequence $r_k$ defined by recursion from $r_0$ by the relation $$r_{k+1} = (1 - \alpha_{2^k})(1 + \alpha_{2^k}^2)^{-1}(1 + \alpha_{2^k})^{-1} \gamma_{2^k} r_k.$$ Show that there exist sequences $F_k$ and $P_k$ of elements of $O_2$, defined for $k \geqslant 0$, such that $F_0 = F$ and, for all $k \geqslant 0$, $$\begin{aligned}
& \lambda I + F_{k+1} = (I + P_k)^\dagger \circ (\lambda I + F_k) \circ (I + P_k), \\
& F_k \in O_{1+2^k}, \quad P_k \in O_{1+2^k}, \\
& \widehat{F_k}(r_k) \leqslant r_k, \quad \widehat{P_k}(r_{k+1}) \leqslant r_k - r_{k+1}.
\end{aligned}$$
Let $(a_n)_{n \in \mathbb{N}}$ be a real sequence such that the series $\sum a_n^2$ converges. Justify the existence of a strictly increasing sequence of natural integers $(\phi(j))_{j \in \mathbb{N}}$ satisfying $$\forall j \in \mathbb{N}, \quad \sum_{n > \phi(j)}^{+\infty} a_n^2 \leqslant \frac{1}{8^j}.$$
Let $f \in \mathscr { D } _ { \rho } ( \mathbb { R } )$ such that $f ( 0 ) > 0$. Show that there exists $\rho _ { f } \in \mathbb { R } _ { + } ^ { * }$ such that $\rho _ { f } \leqslant \rho$ and such that $f > 0$ on $U _ { \rho _ { f } }$ and $\sqrt { f } \in \mathscr { D } _ { \rho _ { f } } ( \mathbb { R } )$.
We denote by $\mathcal{E}$ the set of functions $f : \mathbb{C} \rightarrow \mathbb{C}$ expandable as a power series with radius of convergence infinity. Using the functions $Q_n \in \mathcal{E}$ satisfying $Q_n(z+1) - Q_n(z) = nz^{n-1}$ for all $n \in \mathbb{N}^*$ and $z \in \mathbb{C}$, and the bound $|Q_n(z)| \leqslant a\,\mathrm{e}^{bn|z|}$ for constants $a,b \in \mathbb{R}_+^*$, deduce the existence of a solution in $\mathcal{E}$ to the equation $(E_h)$: $$\forall z \in \mathbb{C},\, f(z+1) - f(z) = h(z)$$ when $h \in \mathcal{E}$.
Show that if $f \in \mathbf{Q}\llbracket x \rrbracket$ is the power series expansion of a rational function with rational coefficients, then $f$ is globally bounded. (A power series $f(x) = \sum_{n=0}^{\infty} c_n x^n \in \mathbf{Q}\llbracket x \rrbracket$ is globally bounded if there exist integers $A, B \geq 1$ such that $A f(Bx) \stackrel{\text{def}}{=} \sum_{n=0}^{\infty} (B^n A c_n) x^n$ is a power series with integer coefficients.)
For all integers $n, k \geq 0$, define the rational number $v_n(k)$ as the coefficient of degree $n$ in the power series $$\left(1 - s_1 x - \cdots - s_r x^r\right)^k v(x) = \sum_{n=0}^{\infty} v_n(k) x^n.$$ Show that $v(x)$ is the power series expansion of a rational fraction if and only if there exists an integer $k \geq 0$ such that $\sum_{n=0}^{\infty} v_n(k) x^n$ is a polynomial.
For all integers $n, k \geq 0$, define the rational number $v_n(k)$ as the coefficient of degree $n$ in the power series $$\left(1 - s_1 x - \cdots - s_r x^r\right)^k v(x) = \sum_{n=0}^{\infty} v_n(k) x^n.$$ Observe the equality: for all $n \geq r$ and $k \geq 0$, $$v_n(k+1) = v_n(k) - s_1 v_{n-1}(k) - \cdots - s_r v_{n-r}(k).$$
A function $E$ (with rational coefficients) is a power series $f(x) = \sum_{n=0}^{\infty} \frac{b_n}{n!} x^n \in \mathbf{Q}\llbracket x \rrbracket$ satisfying: (a) $f$ is a solution of a differential equation; (b) there exists a real number $C > 0$ such that $|b_n| \leq C^n$ and $\operatorname{denom}(b_0, \ldots, b_n) \leq C^n$ for all $n \geq 1$. Show that if $f$ is a function $E$, then the numerical series $\sum_{n=0}^{\infty} \frac{b_n}{n!} \alpha^n$ converges for every real number $\alpha$. We denote its value by $f(\alpha)$.
A function $E$ (with rational coefficients) is a power series $f(x) = \sum_{n=0}^{\infty} \frac{b_n}{n!} x^n \in \mathbf{Q}\llbracket x \rrbracket$ satisfying: (a) $f$ is a solution of a differential equation; (b) there exists a real number $C > 0$ such that $|b_n| \leq C^n$ and $\operatorname{denom}(b_0, \ldots, b_n) \leq C^n$ for all $n \geq 1$. Let $f$ be a function $E$ that is not a polynomial. Show that there exists $R > 0$ such that the numerical series $\sum_{n=0}^{\infty} b_n \alpha^n$ diverges for every real number $\alpha$ with $|\alpha| > R$.
A function $E$ (with rational coefficients) is a power series $f(x) = \sum_{n=0}^{\infty} \frac{b_n}{n!} x^n \in \mathbf{Q}\llbracket x \rrbracket$ satisfying: (a) $f$ is a solution of a differential equation; (b) there exists a real number $C > 0$ such that $|b_n| \leq C^n$ and $\operatorname{denom}(b_0, \ldots, b_n) \leq C^n$ for all $n \geq 1$. Recall that $\widehat{f}(x)$ denotes the Laplace transform $\widehat{f}(x) \stackrel{\text{def}}{=} \sum_{n=0}^{\infty} b_n x^n$. Which functions $E$ are such that $\widehat{f}$ is also a function $E$?