Let $f(x) = xe^x$, and let $V$ and $W$ denote the inverses of $f|_{]-\infty,-1]}$ and $f|_{[-1,+\infty[}$ respectively. For a real parameter $m$, we consider the equation with unknown $x \in \mathbb { R }$ $$x \mathrm { e } ^ { x } = m \tag{I.1}$$ Determine, as a function of $m$, the number of solutions of (I.1). Explicitly express the possible solutions using the functions $V$ and $W$.
Let $V$ and $W$ denote the inverses of $f|_{]-\infty,-1]}$ and $f|_{[-1,+\infty[}$ respectively, where $f(x) = xe^x$. For non-zero real parameters $a$ and $b$, we consider the equation with unknown $x \in \mathbb { R }$ $$\mathrm { e } ^ { a x } + b x = 0 \tag{I.3}$$ Determine, according to the values of $a$ and $b$, the number of solutions of (I.3). Explicitly express the possible solutions using the functions $V$ and $W$.
Let $n \in \mathbb{N}$ be a non-zero natural integer. We define, for any real number $x$, $$\Phi_n(x) = \mathrm{e}^{-x} x^n \quad \text{and} \quad L_n(x) = \frac{\mathrm{e}^x}{n!} \Phi_n^{(n)}(x).$$ Use the equality $\Phi_{n+1}^{(n+2)}(x) = \frac{\mathrm{d}^{n+1} \Phi_{n+1}^{(1)}}{\mathrm{d}x^{n+1}}(x)$ to prove the equality $$L_{n+1}'(x) = L_n'(x) - L_n(x)$$ valid for any real number $x$.
Let $n \in \mathbb{N}$ be a non-zero natural integer. We define, for any real number $x$, $$\Phi_n(x) = \mathrm{e}^{-x} x^n \quad \text{and} \quad L_n(x) = \frac{\mathrm{e}^x}{n!} \Phi_n^{(n)}(x).$$ Deduce that $L_n$ is a solution of the differential equation $$x L_n''(x) + (1-x) L_n'(x) + n L_n(x) = 0.$$
Let $n \in \mathbb{N}$ be a non-zero natural integer. We define, for any real number $x$, $$\Phi_n(x) = \mathrm{e}^{-x} x^n \quad \text{and} \quad L_n(x) = \frac{\mathrm{e}^x}{n!} \Phi_n^{(n)}(x).$$ The confluent hypergeometric function $M_{a,c}$ is the solution of $$x y''(x) + (c-x) y'(x) - a y(x) = 0$$ satisfying $M_{a,c}(0) = 1$. Show that $L_n$ is a confluent hypergeometric function.
If $\phi : \mathbb { R } \rightarrow \mathbb { R }$ is of class $\mathcal { C } ^ { 1 }$ and satisfies $$\sup \left\{ \left| \phi ^ { \prime } ( x ) \right| ; x \in \mathbb { R } \right\} < 1$$ show that $\phi$ has at least one fixed point (one may study the sign of $x - \phi ( x )$ for $| x |$ sufficiently large). Show that this fixed point is unique.
By means of the function $\psi ( x ) = \sqrt { 1 + x ^ { 2 } }$, show that in the previous question hypothesis (1) cannot be replaced by $$\forall x \in \mathbb { R } , \left| \phi ^ { \prime } ( x ) \right| < 1$$
Let $\ell$ be a strictly positive integer. Let $F$ be a closed subset of $\mathbb { R } ^ { \ell }$ and let $\phi : F \rightarrow F$ be a map. We assume that there exists $k \in [ 0,1 [$ such that $$\forall x \in F , \forall y \in F , \quad \| \phi ( y ) - \phi ( x ) \| \leqslant k \| y - x \| .$$ (a) We choose a point $x _ { 0 } \in F$. Show that the formula $x _ { n + 1 } = \phi \left( x _ { n } \right)$ defines a sequence $\left( x _ { n } \right) _ { n \geqslant 0 }$ of elements of $F$, and that this sequence is convergent in $F$. (b) Deduce that $\phi$ has a unique fixed point in $F$. (c) This fixed point being denoted $x ^ { * }$, bound $\left\| x _ { n } - x ^ { * } \right\|$ as a function of $\left\| x _ { 0 } - x ^ { * } \right\|$. (d) In what precedes, we assume that $$\phi = \underbrace { \theta \circ \cdots \circ \theta } _ { m \text { times } } ,$$ where $\theta : F \rightarrow F$ is a map and $m \geqslant 2$ is an integer. Show that $\theta$ has a fixed point, and a unique one, in $F$.
Let $a$ and $b$ be two real numbers. Show that the function $\left\lvert\, \begin{array} { r l l } \mathbb { R } _ { + } ^ { * } & \rightarrow & \mathbb { R } \\ t & \mapsto & a \mathrm { e } ^ { t } + b \end{array} \right.$ belongs to $E$ if and only if $a = b = 0$, where $E$ is the set of continuous functions $f$ from $\mathbb { R } _ { + } ^ { * }$ to $\mathbb { R }$ such that the integral $\int _ { 0 } ^ { + \infty } f ^ { 2 } ( t ) \frac { \mathrm { e } ^ { - t } } { t } \mathrm {~d} t$ converges.
For all $x \in \mathbb { R } _ { + } ^ { * }$ and all $t \in \mathbb { R } _ { + } ^ { * }$, we denote $k _ { x } ( t ) = \mathrm { e } ^ { \min ( x , t ) } - 1$ where $\min ( x , t )$ denotes the smaller of the real numbers $x$ and $t$. Draw a graph of the function $k _ { x }$. Show that $k _ { x }$ belongs to $E$, where $E$ is the set of continuous functions $f$ from $\mathbb { R } _ { + } ^ { * }$ to $\mathbb { R }$ such that the integral $\int _ { 0 } ^ { + \infty } f ^ { 2 } ( t ) \frac { \mathrm { e } ^ { - t } } { t } \mathrm {~d} t$ converges.
We assume that $f$ is a function from $\mathbb { R } _ { + } ^ { * }$ to $\mathbb { R }$ of class $\mathcal { C } ^ { 1 }$ satisfying $$\left\{ \begin{array} { l } \lim _ { x \rightarrow 0 } f ( x ) = 0 \\ \exists C > 0 ; \forall x > 0 , \quad \left| f ^ { \prime } ( x ) \right| \leqslant C \frac { \mathrm { e } ^ { x / 2 } } { \sqrt { x } } \end{array} \right.$$ For $x \in \mathbb { R } _ { + } ^ { * }$, we set $\Phi ( x ) = \frac { 4 \sqrt { x } \mathrm { e } ^ { x / 2 } } { 1 + x } - \int _ { 0 } ^ { x } \frac { \mathrm { e } ^ { t / 2 } } { \sqrt { t } } \mathrm {~d} t$. Show that $\Phi$ is of class $\mathcal { C } ^ { 1 }$ on $\mathbb { R } _ { + } ^ { * }$, that $\lim _ { x \rightarrow 0 } \Phi ( x ) = 0$ and that, for all $x > 0 , \Phi ^ { \prime } ( x ) \geqslant 0$. Deduce that $\Phi ( x ) \geqslant 0$ for all $x > 0$.
We assume that $f$ is a function from $\mathbb { R } _ { + } ^ { * }$ to $\mathbb { R }$ of class $\mathcal { C } ^ { 1 }$ satisfying $$\left\{ \begin{array} { l } \lim _ { x \rightarrow 0 } f ( x ) = 0 \\ \exists C > 0 ; \forall x > 0 , \quad \left| f ^ { \prime } ( x ) \right| \leqslant C \frac { \mathrm { e } ^ { x / 2 } } { \sqrt { x } } \end{array} \right.$$ Show that, for all $x > 0 , | f ( x ) | \leqslant 4 C \frac { \sqrt { x } \mathrm { e } ^ { x / 2 } } { 1 + x }$.
To each function $f \in E$, we associate the function $U ( f )$ with derivative $U(f)^\prime(x) = \mathrm { e } ^ { x } \int _ { x } ^ { + \infty } f ( t ) \frac { \mathrm { e } ^ { - t } } { t } \mathrm {~d} t$. Show that for all $f \in E$ and for all $x > 0$, $$\left| U ( f ) ^ { \prime } ( x ) \right| \leqslant \mathrm { e } ^ { x } \| f \| \left( \int _ { x } ^ { + \infty } \frac { \mathrm { e } ^ { - t } } { t } \mathrm {~d} t \right) ^ { 1 / 2 } \leqslant \| f \| \frac { \mathrm { e } ^ { x / 2 } } { \sqrt { x } }$$
To each function $f \in E$, we associate the function $U ( f )$ defined for all $x > 0$ by $U ( f ) ( x ) = \int _ { 0 } ^ { + \infty } \left( \mathrm { e } ^ { \min ( x , t ) } - 1 \right) f ( t ) \frac { \mathrm { e } ^ { - t } } { t } \mathrm {~d} t$. It has been shown that $\left| U ( f ) ^ { \prime } ( x ) \right| \leqslant \| f \| \frac { \mathrm { e } ^ { x / 2 } } { \sqrt { x } }$ and $\lim_{x\to 0} U(f)(x) = 0$. Deduce from the above that $U$ is an endomorphism of $E$ and that, for all $f \in E$ and all $x > 0$, $$| U ( f ) ( x ) | \leqslant 4 \| f \| \frac { \sqrt { x } \mathrm { e } ^ { x / 2 } } { 1 + x }$$
To each function $f \in E$, we associate the function $U ( f )$ defined for all $x > 0$ by $U ( f ) ( x ) = \int _ { 0 } ^ { + \infty } \left( \mathrm { e } ^ { \min ( x , t ) } - 1 \right) f ( t ) \frac { \mathrm { e } ^ { - t } } { t } \mathrm {~d} t$. It has been shown that $| U ( f ) ( x ) | \leqslant 4 \| f \| \frac { \sqrt { x } \mathrm { e } ^ { x / 2 } } { 1 + x }$ for all $x > 0$. Deduce that $$\| U ( f ) \| \leqslant 4 \| f \|.$$
To each function $f \in E$, we associate the function $U ( f )$ defined for all $x > 0$ by $U ( f ) ( x ) = \int _ { 0 } ^ { + \infty } \left( \mathrm { e } ^ { \min ( x , t ) } - 1 \right) f ( t ) \frac { \mathrm { e } ^ { - t } } { t } \mathrm {~d} t$. Show that $U$ is injective.
To each function $f \in E$, we associate the function $U ( f )$ defined for all $x > 0$ by $U ( f ) ( x ) = \int _ { 0 } ^ { + \infty } \left( \mathrm { e } ^ { \min ( x , t ) } - 1 \right) f ( t ) \frac { \mathrm { e } ^ { - t } } { t } \mathrm {~d} t$. Is the endomorphism $U$ surjective?
We consider the Cauchy problem associated with $F_0$ defined by: $$\forall y \in ]0, +\infty[, \quad F_0(y) = ay\ln\left(\frac{\theta}{y}\right)$$ with $a, \theta > 0$ and $0 < y_{\text{init}} < \theta$. We denote $\phi_0$ the solution of this problem on $[0, +\infty[$. (a) Show that there exists $\varepsilon > 0$ such that for all $t \in ]0, \varepsilon]$ we have $y_{\text{init}} < \phi_0(t) < \theta$. (b) By considering the function $z_0(t) = \ln\left(\phi_0(t)/\theta\right)$ find the expression of $\phi_0$. (c) Deduce that $\phi_0$ satisfies $y_{\text{init}} < \phi_0(t) < \theta$ for all $t \in ]0, +\infty[$ and that moreover $\phi_0$ is strictly increasing.
Show that a power series $f(x) = \sum_{n=0}^{\infty} c_n x^n \in \mathbf{Q}\llbracket x \rrbracket$ is a solution of a differential equation if and only if there exist an integer $d \geq 0$ and polynomials not all zero $S_0, \ldots, S_d \in \mathbf{Z}[x]$ such that: for all $n \geq 0$, $$S_0(n) c_n + \cdots + S_d(n) c_{n+d} = 0.$$
Show that $S$ is continuous on $\mathbf{R}_{+}$, where $S(t) = \operatorname{Ent}_{\varphi}\left(P_{t}(f)\right)$. Hint: You may first show that, if $x \in \mathbf{R}$, then $t \mapsto P_{t}(f)(x)$ is continuous on $\mathbf{R}_{+}$.
Give a new proof, based on questions 12 and 13 above, of the fact that the power series $\sum_{m=0}^{\infty} x^{m^2}$ is not the expansion of a rational function.
Verify that we have $S(0) = \operatorname{Ent}_{\varphi}(f)$ and $\lim_{t \rightarrow +\infty} S(t) = 0$, where $S(t) = \operatorname{Ent}_{\varphi}\left(P_{t}(f)\right)$.
Show that the power series $$h(x) = \sum_{n=0}^{\infty} \frac{(2n)!(3n)!}{(n!)^5} x^n$$ is a solution of a differential equation, then make one explicit.
We admit that $S$ is of class $C^{1}$ on $\mathbf{R}_{+}^{*}$ and that $$\forall t \in \mathbf{R}_{+}^{*}, \quad S^{\prime}(t) = \int_{-\infty}^{+\infty} \frac{\partial P_{t}(f)(x)}{\partial t} \left(1 + \ln\left(P_{t}(f)(x)\right)\right) \varphi(x) \mathrm{d}x$$ Show that $$\forall t \in \mathbf{R}_{+}^{*}, \quad S^{\prime}(t) = \int_{-\infty}^{+\infty} L\left(P_{t}(f)\right)(x) \left(1 + \ln\left(P_{t}(f)(x)\right)\right) \varphi(x) \mathrm{d}x$$