Let $\sum _ { k \geqslant 0 } c _ { k } x ^ { k }$ be a power series with radius of convergence $R > 0$ and $r \in ] 0 , R [$. Show that there exists $C \in \mathbb { R }$ such that $$\forall k \in \mathbb { N } , \quad \left| c _ { k } \right| \leqslant \frac { C } { r ^ { k } }.$$

We now study the linearization problem in the case $|\lambda| = 1$, with $\lambda$ not a root of unity. We set, for $m \geqslant 1$, $$\alpha_m := \min(1/5, \omega_{m+1}, \omega_{m+2}, \ldots, \omega_{2m}), \quad \gamma_m := \alpha_m^{2/m},$$ where $\omega_k := |\lambda^k - \lambda|, k \geqslant 2$, and $r_0 > 0$ is as given by question (24).
Still for $F \in O_{m+1}, m \geqslant 1$, we set $$P := \sum_{k=m+1}^{2m} \frac{(F)_k}{\lambda^k - \lambda} z^k \in O_{m+1} \quad , \quad R := (I + P)^\dagger - I.$$ Show that $P \circ (\lambda I) - \lambda P - F \in O_{2m+1}$ and that $R + P \in O_{2m+1}$. Show that $\hat{P}(r) \leqslant \alpha_m r$ for all $r \in [0, \gamma_m r_0]$, and that $$\hat{R}(r) \leqslant \frac{\alpha_m}{1 - \alpha_m} r$$ for all $r \in [0, (1-\alpha_m)\gamma_m r_0]$.

Let $\sum _ { k \geqslant 0 } c _ { k } x ^ { k }$ be a power series with radius of convergence $R > 0$, $r \in ]0, R[$, and $f(x) = \sum_{k=0}^{+\infty} c_k x^k$ for $x \in ]-R, R[$. Let $C \in \mathbb{R}$ be such that $|c_k| \leq C/r^k$ for all $k \in \mathbb{N}$. Deduce that for all $x \in ] - r , r [$ and for all $n \in \mathbb { N }$, $$\left| f ^ { ( n ) } ( x ) \right| \leqslant \frac { n ! r C } { ( r - | x | ) ^ { n + 1 } }.$$

We now study the linearization problem in the case $|\lambda| = 1$, with $\lambda$ not a root of unity. We set, for $m \geqslant 1$, $$\alpha_m := \min(1/5, \omega_{m+1}, \omega_{m+2}, \ldots, \omega_{2m}), \quad \gamma_m := \alpha_m^{2/m},$$ where $\omega_k := |\lambda^k - \lambda|, k \geqslant 2$, and $P, R$ as defined in question (25).
For $F \in O_{m+1}, m \geqslant 1$, show that $$G := (I + P)^\dagger \circ (\lambda I + F) \circ (I + P) - \lambda I = (I + R) \circ (\lambda I + F) \circ (I + P) - \lambda I$$ satisfies $G \in O_{2m+1}$.

Let $\sum _ { k \geqslant 0 } c _ { k } x ^ { k }$ be a power series with radius of convergence $R > 0$, $f(x) = \sum_{k=0}^{+\infty} c_k x^k$ for $x \in ]-R, R[$, and $a > 0$. Assume that $a < R / 3$. Show that the sequence of polynomials $\left( P _ { n } \right) _ { n \in \mathbb { N } ^ { * } } = \left( \Pi _ { n } ( f ) \right) _ { n \in \mathbb { N } ^ { * } }$ converges uniformly towards $f$ on $[ - a , a ]$.

Let $a > 0$, $I = [-a, a]$. For all $n \in \mathbb { N } ^ { * }$, the Chebyshev points of order $n$ in $I$ are $$a _ { k , n } ^ { * } = a \cos \left( \frac { ( 2 k - 1 ) \pi } { 2 n } \right) , \quad \text { for } k \in \llbracket 1 , n \rrbracket ,$$ and $W _ { n } ^ { * } ( X ) = \prod _ { k = 1 } ^ { n } \left( X - a _ { k , n } ^ { * } \right)$. For all $x \in [ - a , a ]$, show that $\left| W _ { n } ^ { * } ( x ) \right| \leqslant 2 \left( \frac { a } { 2 } \right) ^ { n }$.

Let $a > 0$, $I = [-a,a]$, and $f(x) = \dfrac{1}{1+x^2}$ for $x \in \mathbb{R}$. For all $n \in \mathbb{N}^*$, the Chebyshev points of order $n$ in $I$ are $a_{k,n}^* = a\cos\left(\frac{(2k-1)\pi}{2n}\right)$ for $k \in \llbracket 1,n \rrbracket$, and $\Pi_n^*(f)$ denotes the interpolation polynomial of $f$ at these points. Show that, if $a < 2$, the sequence $\left( \Pi _ { n } ^ { * } ( f ) \right) _ { n \in \mathbb { N } ^ { * } }$ converges uniformly towards $f$ on $[ - a , a ]$.

Let $\sum_{k \geqslant 0} c_k x^k$ be a power series with radius of convergence $R > 0$, $f(x) = \sum_{k=0}^{+\infty} c_k x^k$ for $x \in ]-R,R[$, and $a > 0$. For all $n \in \mathbb{N}^*$, the Chebyshev points of order $n$ in $[-a,a]$ are $a_{k,n}^* = a\cos\left(\frac{(2k-1)\pi}{2n}\right)$ for $k \in \llbracket 1,n \rrbracket$, and $\Pi_n^*(f)$ denotes the interpolation polynomial of $f$ at these points. Show that, if $a < 2R/3$, the sequence $\left( \Pi _ { n } ^ { * } ( f ) \right) _ { n \in \mathbb { N } ^ { * } }$ converges uniformly towards $f$ on $[ - a , a ]$.

For $p \in \mathbb { N } ^ { * }$, consider a polynomial $P \in \mathbb { R } [ X ]$ such that the polynomial function $x \mapsto P ( x )$ is a solution of the equation $\left( E _ { p } \right) : x \left( y ^ { \prime \prime } - y ^ { \prime } \right) + p y = 0$. For all $x \in \mathbb { R }$, we denote $h ( x ) = \mathrm { e } ^ { - x } P ( x )$. Justify that the function $h$ is developable as a power series on $\mathbb { R }$.

For $p \in \mathbb { N } ^ { * }$, let $h(x) = \mathrm{e}^{-x} P(x)$ where $P$ is a polynomial solution of $(E_p)$, with power series coefficients satisfying $b _ { n } = \frac { ( - 1 ) ^ { n - 1 } ( n + p - 1 ) ! } { p ! n ! ( n - 1 ) ! } b _ { 1 }$ for all $n \in \mathbb{N}^*$. We set $g _ { p } ( x ) = x ^ { p - 1 } \mathrm { e } ^ { - x }$. Justify that $g _ { p } ^ { ( p ) }$ is developable as a power series and deduce from Question 34 that, for all $x \in \mathbb { R }$, $$P ( x ) = C x \mathrm { e } ^ { x } g _ { p } ^ { ( p ) } ( x )$$ where $C$ is a real constant whose expression in terms of $b _ { 1 }$ and $p$ we will specify.

Let $t \in \mathbf{R}$ and $(i,j) \in \llbracket 1;N \rrbracket^2$, justify that the series $\sum_{n \geq 0} \frac{t^n K^n[i,j]}{n!}$ converges. We denote by $H_t \in \mathscr{M}_N(\mathbf{R})$ the matrix defined by $$\forall (i,j) \in \llbracket 1;N \rrbracket^2, H_t[i,j] = e^{-t} \sum_{n=0}^{+\infty} \frac{t^n K^n[i,j]}{n!}$$

Show that $$\forall x \in ]-1,1[ \quad s(x)e^{x} = \frac{1}{1-x}$$ Deduce that $R = 1$.

Starting from the relation $(1-x)s(x) = e^{-x}$ for $x \in ]-1,1[$, express $\frac{d_n}{n!}$ for natural number $n$, in the form of a sum.

Let $\varphi_{0}$ be the function defined on $\mathbb{R}$ by
$$\left\{ \begin{array}{l} \varphi_{0}(x) = e^{-1/x^{2}} \text{ if } x \neq 0 \\ \varphi_{0}(0) = 0 \end{array} \right.$$
a. Show that for all $n \in \mathbb{N}$ there exists a polynomial $P_{n}$ such that for $x \neq 0$ we have
$$\varphi_{0}^{(n)}(x) = P_{n}\left(\frac{1}{x}\right) e^{-1/x^{2}}$$
b. Show that $\varphi_{0}$ is of class $C^{\infty}$ on $\mathbb{R}$.

Throughout this problem, $I = ]-1, +\infty[$, and $f(x) = \int_0^{\pi/2} (\sin(t))^x \mathrm{~d}t$.
Give a simple asymptotic equivalent of $f(x)$ as $x$ tends to $-1$.

Let $f : \mathbb{R} \rightarrow \mathbb{R}$ be a function of class $C^{\infty}$ with compact support. For all $n \in \mathbb{N}$, we set $$M_{n} = \sup_{x \in \mathbb{R}} \left|f^{(n)}(x)\right| = \left\|f^{(n)}\right\|_{\infty}$$ In this part we assume that $f$ is not identically zero, and $x_{0} \in \operatorname{Supp}(f)$ is such that for all integers $n \geqslant 0$, $f^{(n)}(x_{0}) = 0$.
Show that if there exist constants $A > 0$ and $B > 0$, and a subsequence $(n_{j})_{j \geqslant 1}$ such that $M_{n_{j}} \leqslant A B^{n_{j}} (n_{j})!$, then $f$ is identically zero on the interval $]x_{0} - 1/B,\, x_{0} + 1/B[$.

Let $f : \mathbb{R} \rightarrow \mathbb{R}$ be a function of class $C^{\infty}$ with compact support. For all $n \in \mathbb{N}$, we set $$M_{n} = \sup_{x \in \mathbb{R}} \left|f^{(n)}(x)\right| = \left\|f^{(n)}\right\|_{\infty}$$ In this part we assume that $f$ is not identically zero.
Deduce that for all $B > 0$ we have
$$\frac{M_{n}}{B^{n} n!} \underset{n \rightarrow \infty}{\longrightarrow} +\infty$$

Throughout this problem, $I = ]-1, +\infty[$, and $f(x) = \int_0^{\pi/2} (\sin(t))^x \mathrm{~d}t$.
Prove that $f$ is expandable as a power series on $]-1, 1[$.

For all $p \in \mathbb{K}[X]$ non-zero and $a \in \mathbb{K}$, show, using question 11, that $$p(X+a) = \sum_{k=0}^{\deg(p)} \frac{a^k}{k!} p^{(k)}$$ where $p^{(k)}$ denotes the $k$-th derivative of the polynomial $p$. Recognize this formula.

For all $n \in \mathbb { N } ^ { * }$ and all $k \in \llbracket 0 , n \rrbracket$, we set $x _ { n , k } = - \sqrt { n } + \frac { 2 k } { \sqrt { n } }$. The function $B _ { n }$ is defined as in Q19. We assume $k \in I _ { n } = \left\{ k \in \llbracket 0 , n \rrbracket \mid x _ { n , k } \in [ 0 , \ell + 1 ] \right\}$.
Deduce that $$B _ { n } \left( x _ { n , k } \right) = \frac { 1 } { \sqrt { 2 \pi } } \frac { 1 + O \left( \frac { 1 } { n } \right) } { \left( 1 - \frac { x _ { n , k } ^ { 2 } } { n } \right) ^ { \frac { n + 1 } { 2 } } \left( 1 + \frac { x _ { n , k } } { \sqrt { n } } \right) ^ { \frac { x _ { n , k } } { 2 } \sqrt { n } } \left( 1 - \frac { x _ { n , k } } { \sqrt { n } } \right) ^ { - \frac { x _ { n , k } } { 2 } \sqrt { n } } }$$ then that $$B _ { n } \left( x _ { n , k } \right) = \frac { 1 } { \sqrt { 2 \pi } } \exp \left( - \frac { x _ { n , k } ^ { 2 } } { 2 } \right) \left( 1 + O \left( \frac { 1 } { \sqrt { n } } \right) \right)$$

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, and $\omega(t) = e^{2i\pi t}$ for $t \in [0,1]$. Let $f \in \mathcal{E}$ whose power series expansion we denote $\sum a_{n} z^{n}$. Show that, for all $k \in \mathbb{Z}$: $$\int_{0}^{1} f(\omega(t)) \omega(t)^{-k} \,\mathrm{d}t = \begin{cases} a_{k} & \text{if } k \in \mathbb{N} \\ 0 & \text{otherwise} \end{cases}$$

Let $f \in C^{2}(\mathbf{R}) \cap CL(\mathbf{R})$ such that $f^{\prime}$ and $f^{\prime\prime}$ have slow growth and $t \in \mathbf{R}_{+}$.
Show that $x \in \mathbb{R} \mapsto P_{t}(f)(x)$ is of class $C^{2}$ on $\mathbf{R}$. Also show that
$$\forall x \in \mathbf{R}, \quad P_{t}(f)^{\prime}(x) = \mathrm{e}^{-t} \int_{-\infty}^{+\infty} f^{\prime}\left(\mathrm{e}^{-t} x + \sqrt{1 - \mathrm{e}^{-2t}} y\right) \varphi(y) \mathrm{d}y$$
and
$$\forall x \in \mathbf{R}, \quad P_{t}(f)^{\prime\prime}(x) = \mathrm{e}^{-2t} \int_{-\infty}^{+\infty} f^{\prime\prime}\left(\mathrm{e}^{-t} x + \sqrt{1 - \mathrm{e}^{-2t}} y\right) \varphi(y) \mathrm{d}y.$$

Recall that $x$ is a fixed element of $]0;1[$. Show that:
$$\int _ { 0 } ^ { 1 } \frac { t ^ { x - 1 } } { 1 + t } \mathrm {~d} t = \sum _ { k = 0 } ^ { + \infty } \frac { ( - 1 ) ^ { k } } { k + x }$$

Let $f \in C^2(\mathbf{R}) \cap CL(\mathbf{R})$ such that $f'$ and $f''$ have slow growth and $t \in \mathbf{R}_+$. Show that $x \in \mathbf{R} \mapsto P_t(f)(x)$ is of class $C^2$ on $\mathbf{R}$. Also show that $$\forall x \in \mathbf{R}, \quad P_t(f)'(x) = \mathrm{e}^{-t}\int_{-\infty}^{+\infty} f'\!\left(\mathrm{e}^{-t}x + \sqrt{1-\mathrm{e}^{-2t}}\,y\right)\varphi(y)\,\mathrm{d}y$$ and $$\forall x \in \mathbf{R}, \quad P_t(f)''(x) = \mathrm{e}^{-2t}\int_{-\infty}^{+\infty} f''\!\left(\mathrm{e}^{-t}x + \sqrt{1-\mathrm{e}^{-2t}}\,y\right)\varphi(y)\,\mathrm{d}y.$$

Give an example of a power series expansion of a rational function whose antiderivative is not the expansion of a rational function.
(The antiderivative of a power series $f(x) = \sum_{n=0}^{\infty} c_n x^n$ is defined as $\int_0^x f(t)\,dt \stackrel{\text{def}}{=} \sum_{n=0}^{\infty} \frac{c_n}{n+1} x^{n+1}$.)