Let $n$ be in $\mathbb { N } ^ { * }$. We denote by $U _ { n }$ the open set $\left( \mathbb { R } _ { + } ^ { * } \right) ^ { n }$. Its closure, denoted $\overline { U _ { n } }$, is $\left( \mathbb { R } _ { + } \right) ^ { n }$. We consider the map $F _ { n }$ from $\overline { U _ { n } }$ to $\mathbb { R }$, defined by
$$\forall \left( x _ { 1 } , \ldots , x _ { n } \right) \in \overline { U _ { n } } , \quad F _ { n } \left( x _ { 1 } , \ldots , x _ { n } \right) = x _ { 1 } + \left( x _ { 1 } x _ { 2 } \right) ^ { 1 / 2 } + \left( x _ { 1 } x _ { 2 } x _ { 3 } \right) ^ { 1 / 3 } + \cdots + \left( x _ { 1 } \cdots x _ { n } \right) ^ { 1 / n } .$$
We denote by $h _ { n }$ the map from $\overline { U _ { n } }$ to $\mathbb { R }$, defined by
$$\forall \left( x _ { 1 } , \ldots , x _ { n } \right) \in \overline { U _ { n } } , \quad h _ { n } \left( x _ { 1 } , \ldots , x _ { n } \right) = x _ { 1 } + \cdots + x _ { n } - 1 .$$
We denote by $H _ { n }$ the set $H _ { n } = \left\{ \left( x _ { 1 } , \ldots , x _ { n } \right) \in \mathbb { R } ^ { n } \mid x _ { 1 } + \cdots + x _ { n } = 1 \right\}$. We denote by $M _ { n }$ the maximum of $F _ { n }$ on $\overline { U _ { n } } \cap H _ { n }$ and we denote by $( a _ { 1 } , \ldots , a _ { n } )$ a point of $U _ { n } \cap H _ { n }$ at which it is attained. For $k$ between 1 and $n$, we denote $\gamma _ { k } = \left( a _ { 1 } a _ { 2 } \cdots a _ { k } \right) ^ { 1 / k }$.
Prove that there exists a real number $\lambda > 0$ such that
$$\left\{ \begin{aligned} \gamma _ { 1 } + \frac { \gamma _ { 2 } } { 2 } + \cdots + \frac { \gamma _ { n } } { n } & = \lambda a _ { 1 } \\ \frac { \gamma _ { 2 } } { 2 } + \cdots + \frac { \gamma _ { n } } { n } & = \lambda a _ { 2 } \\ & \vdots \\ \frac { \gamma _ { n } } { n } & = \lambda a _ { n } \\ a _ { 1 } + a _ { 2 } + \cdots + a _ { n } & = 1 \end{aligned} \right.$$

Let $n$ be in $\mathbb { N } ^ { * }$. We denote by $U _ { n }$ the open set $\left( \mathbb { R } _ { + } ^ { * } \right) ^ { n }$. Its closure, denoted $\overline { U _ { n } }$, is $\left( \mathbb { R } _ { + } \right) ^ { n }$. We consider the map $F _ { n }$ from $\overline { U _ { n } }$ to $\mathbb { R }$, defined by
$$\forall \left( x _ { 1 } , \ldots , x _ { n } \right) \in \overline { U _ { n } } , \quad F _ { n } \left( x _ { 1 } , \ldots , x _ { n } \right) = x _ { 1 } + \left( x _ { 1 } x _ { 2 } \right) ^ { 1 / 2 } + \left( x _ { 1 } x _ { 2 } x _ { 3 } \right) ^ { 1 / 3 } + \cdots + \left( x _ { 1 } \cdots x _ { n } \right) ^ { 1 / n } .$$
We denote by $M _ { n }$ the maximum of $F _ { n }$ on $\overline { U _ { n } } \cap H _ { n }$ and we denote by $( a _ { 1 } , \ldots , a _ { n } )$ a point of $U _ { n } \cap H _ { n }$ at which it is attained. For $k$ between 1 and $n$, we denote $\gamma _ { k } = \left( a _ { 1 } a _ { 2 } \cdots a _ { k } \right) ^ { 1 / k }$. There exists a real number $\lambda > 0$ satisfying the system in Q17.
Deduce that:
a) $\lambda = \gamma _ { 1 } + \gamma _ { 2 } + \cdots + \gamma _ { n } = M _ { n }$;
b) for all $k$ in $\llbracket 1 , n \rrbracket , \gamma _ { k } = \lambda \omega _ { k } a _ { k }$, where
$$\left\{ \begin{array} { l } \omega _ { k } = k \left( 1 - \frac { a _ { k + 1 } } { a _ { k } } \right) \text { if } k \in \llbracket 1 , n - 1 \rrbracket \\ \omega _ { n } = n \end{array} \right.$$

Let $a \in E$ and let $U$ be an open set of $E$ containing $a$. Let $f : U \rightarrow E$ be an application of class $\mathscr{C}^1$ on $U$ such that $df(a) = \operatorname{Id}_E$. We fix a real number $r > 0$ such that $\overline{B(a,r)} \subset U$ and $$\forall x_1, x_2 \in \overline{B(a,r)}, \quad \left\|f(x_1) - f(x_2)\right\| \geqslant \frac{1}{2} \left\|x_1 - x_2\right\|.$$ Let $y_0 \in E$ such that $\left\|y_0 - f(a)\right\| \leqslant \frac{r}{4}$.
Show that the application $$\begin{array}{ccc} \overline{B(a,r)} & \longrightarrow & \mathbb{R} \\ x & \longmapsto & \left\|y_0 - f(x)\right\|^2 \end{array}$$ admits a minimum attained at a point $x_0$ of $B(a,r)$.

We are given $f \in \mathcal{C}^1(\mathbb{R})$, convex, admitting a minimizer $x_* \in \mathbb{R}$, with $f'$ being $L$-Lipschitzian, and $0 < \tau < 2/L$. The sequence $(x_n)_{n \in \mathbb{N}}$ is defined by $x_{n+1} := x_n - \tau f'(x_n)$. Establish an upper bound for the sequence with general term $a_n := f(x_n) - f(x_*)$. Conclude that $f(x_n) \rightarrow f(x_*)$ when $n \rightarrow \infty$.

We are given $f \in \mathcal{C}^1(\mathbb{R})$, convex, admitting a minimizer $x_* \in \mathbb{R}$, with $f'$ being $L$-Lipschitzian, and $0 < \tau < 2/L$. The sequence $(x_n)_{n \in \mathbb{N}}$ is defined by $x_{n+1} := x_n - \tau f'(x_n)$. Show that $\frac{\tau}{2}(2 - \tau L)\sum_{0 \leq i < n}\left|f'(x_i)\right|^2 \leq \left(f(x_0) - f(x_n)\right)$ for all $n \in \mathbb{N}^*$. Deduce that $f'(x_n) \rightarrow 0$ when $n \rightarrow \infty$.

We are given $f \in \mathcal{C}^1(\mathbb{R})$, convex, admitting a minimizer $x_* \in \mathbb{R}$, with $f'$ being $L$-Lipschitzian, and $0 < \tau < 2/L$. The sequence $(x_n)_{n \in \mathbb{N}}$ is defined by $x_{n+1} := x_n - \tau f'(x_n)$. a) Show that the sequence $\left(x_n\right)_{n \in \mathbb{N}}$ admits a convergent subsequence. We denote by $\varphi : \mathbb{N} \rightarrow \mathbb{N}$ the extractor and $x_{**}$ the corresponding limit, so that $x_{\varphi(n)} \rightarrow x_{**}$ when $n \rightarrow \infty$. Hint. We may use without proof the Bolzano-Weierstrass theorem: from any bounded sequence in $\mathbb{R}$, we can extract a convergent subsequence. b) Show that $f'(x_{**}) = 0$. c) Deduce that $x_{**}$ is a minimizer of $f$, then that $\left|x_n - x_{**}\right| \rightarrow 0$ when $n \rightarrow \infty$.

We consider a convex function $f \in \mathcal{C}(\mathbb{R})$, admitting a minimizer $x_* \in \mathbb{R}$. Let also $\tau > 0$. Show that the function $F_{x_0}(x) := \frac{1}{2}|x - x_0|^2 + \tau f(x)$ admits a unique minimizer on $\mathbb{R}$, which we will denote $p_f(x_0)$. Hint: We may consider minimizers $x_1$ and $x_2$ of $F_{x_0}$, and note that $$\left|\frac{1}{2}(x_1 + x_2) - x_0\right|^2 < \frac{1}{2}|x_1 - x_0|^2 + \frac{1}{2}|x_2 - x_0|^2 \quad \text{if } x_1 \neq x_2$$

We consider a convex function $f \in \mathcal{C}(\mathbb{R})$, admitting a minimizer $x_* \in \mathbb{R}$, and $\tau > 0$. The operator $p_f$ is defined as the unique minimizer of $F_{x_0}(x) := \frac{1}{2}|x - x_0|^2 + \tau f(x)$. Show that $x_0 \in \mathbb{R}$ is a minimizer of $f$ if and only if $p_f(x_0) = x_0$. Hint: consider the quantity $F_{x_0}((1-t)x_0 + tx_*)$ when $t \rightarrow 0^+$.

17. $f ( x )$ is a differentiable function and $g ( x )$ is a double differentiable function such that $| f ( x ) | < 1$ and $f ^ { \prime } ( x ) = g ( x )$. If $f ^ { 2 } ( 0 ) + g ^ { 2 } ( 0 ) = 0$. Prove that there exists some $c \hat { I } ( - 3,3 )$ such that $\mathrm { g } ( \mathrm { c } ) . \mathrm { gn } ( \mathrm { c } ) < 0$.

27. Let $f$ be twice differentiable function satisfying $f ( 1 ) = 1 , f ( 2 ) = 4 , f ( 3 ) = 9$, then :
(a) $f ^ { \prime } ( x ) = 2 , \forall x \hat { I } ( R )$
(b) $f ^ { \prime } ( x ) = 5 = f ^ { \prime \prime } ( x )$, for some $x \hat { I } ( 1,3 )$
(c) There exists at least one $x \hat { I } ( 1,3 )$ such that $f ^ { \prime } ( x ) = 2$
(d) none of these

Let $f , g : [ - 1,2 ] \rightarrow \mathbb { R }$ be continuous functions which are twice differentiable on the interval $( - 1,2 )$. Let the values of $f$ and $g$ at the points $- 1,0$ and 2 be as given in the following table:

	$x = - 1$	$x = 0$	$x = 2$
$f ( x )$	3	6	0
$g ( x )$	0	1	- 1

In each of the intervals $( - 1,0 )$ and $( 0,2 )$ the function $( f - 3 g ) ^ { \prime \prime }$ never vanishes. Then the correct statement(s) is(are)
(A) $\quad f ^ { \prime } ( x ) - 3 g ^ { \prime } ( x ) = 0$ has exactly three solutions in $( - 1,0 ) \cup ( 0,2 )$
(B) $f ^ { \prime } ( x ) - 3 g ^ { \prime } ( x ) = 0$ has exactly one solution in $( - 1,0 )$
(C) $f ^ { \prime } ( x ) - 3 g ^ { \prime } ( x ) = 0$ has exactly one solution in $( 0,2 )$
(D) $f ^ { \prime } ( x ) - 3 g ^ { \prime } ( x ) = 0$ has exactly two solutions in ( $- 1,0$ ) and exactly two solutions in ( 0,2 )

For every twice differentiable function $f : \mathbb { R } \rightarrow [ - 2,2 ]$ with $( f ( 0 ) ) ^ { 2 } + \left( f ^ { \prime } ( 0 ) \right) ^ { 2 } = 85$, which of the following statement(s) is (are) TRUE?
(A) There exist $r , s \in \mathbb { R }$, where $r < s$, such that $f$ is one-one on the open interval ( $r , s$ )
(B) There exists $x _ { 0 } \in ( - 4,0 )$ such that $\left| f ^ { \prime } \left( x _ { 0 } \right) \right| \leq 1$
(C) $\lim _ { x \rightarrow \infty } f ( x ) = 1$
(D) There exists $\alpha \in ( - 4,4 )$ such that $f ( \alpha ) + f ^ { \prime \prime } ( \alpha ) = 0$ and $f ^ { \prime } ( \alpha ) \neq 0$

Let $S$ be the set of all twice differentiable functions $f$ from $\mathbb { R }$ to $\mathbb { R }$ such that $\frac { d ^ { 2 } f } { d x ^ { 2 } } ( x ) > 0$ for all $x \in ( - 1,1 )$. For $f \in S$, let $X _ { f }$ be the number of points $x \in ( - 1,1 )$ for which $f ( x ) = x$. Then which of the following statements is(are) true?
(A) There exists a function $f \in S$ such that $X _ { f } = 0$
(B) For every function $f \in S$, we have $X _ { f } \leq 2$
(C) There exists a function $f \in S$ such that $X _ { f } = 2$
(D) There does NOT exist any function $f$ in $S$ such that $X _ { f } = 1$

A polynomial $P(x)$ with real coefficients and of degree four satisfies the inequality
$$P(x) \geq x$$
for every real number $x$.
$$\begin{aligned} & P(1) = 1 \\ & P(2) = 4 \\ & P(3) = 3 \end{aligned}$$
according to, $\mathbf{P(4)}$ is equal to what?