Suppose that $S _ { 0 } > 0$. Show that the function $S$ of the solution triplet $( S , I , R )$ of $(F)$ satisfies the relation $$\left( - \frac { S ^ { \prime } } { S } \right) ^ { \prime } = - S ^ { \prime } + \frac { S ^ { \prime } } { S }$$ The system $(F)$ is: $$( F ) : \left\{ \begin{array} { l } S ^ { \prime } ( x ) = - I ( x ) S ( x ) \\ I ^ { \prime } ( x ) = I ( x ) S ( x ) - I ( x ) \\ R ^ { \prime } ( x ) = I ( x ) \\ S ( 0 ) = S _ { 0 } , \quad I ( 0 ) = I _ { 0 } , \quad R ( 0 ) = R _ { 0 } \end{array} \right.$$
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 $n \rightarrow \infty$.
With $S _ { 0 } = 1 / 2$, $I _ { 0 } = 1 / 2$, $R _ { 0 } = 0$, and $h : \mathbf { R } _ { + } \rightarrow \mathbf { R }$ defined by $$\forall x \in \mathbf { R } _ { + } \quad h ( x ) = \ln \left( \frac { S ( x ) } { S _ { 0 } } \right) = \ln ( 2 S ( x ) ),$$ show that $h$ is a solution of the Cauchy problem $(C)$: $$( C ) : \left\{ \begin{array} { l } y ^ { \prime } ( x ) + y ( x ) + 1 = \frac { 1 } { 2 } \mathrm { e } ^ { y ( x ) } \\ y ( 0 ) = 0 \end{array} \right.$$ using the relation established in question 18.
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 $$\left|\frac{1}{2}(x_1 + x_2) - x_0\right|^2 < \frac{1}{2}|x_1 - x_0|^2 + \frac{1}{2}|x_2 - x_0|^2 \quad \text{if } x_1 \neq x_2$$
Show that $S _ { N }$ converges uniformly to $S$ on $\mathbf { R } _ { + }$ when $N \rightarrow + \infty$ and that $$\left\| S _ { N } - S \right\| _ { \infty , \mathbf { R } _ { + } } \leq \frac { M \mathrm { e } ^ { 2 M } } { 2 ^ { N + 1 } }$$ where $S _ { N } ( x ) = S _ { 0 } \mathrm { e } ^ { y _ { N } ( x ) } = \frac { 1 } { 2 } \exp \left( \sum _ { n = 0 } ^ { N } a _ { n } \mathrm { e } ^ { - \lambda _ { n } x } \right)$, $S(x) = S_0 e^{y(x)} = \frac{1}{2} e^{y(x)}$, and $\left\| y_N - y \right\|_{\infty, \mathbf{R}_+} \leq \frac{M}{2^N}$.
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 $t \rightarrow 0^+$.
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 $F_{x_0}(x) := \frac{1}{2}|x - x_0|^2 + \tau f(x)$. In this question only, we suppose that $f \in \mathcal{C}^1(\mathbb{R})$. Show that $x_1 := p_f(x_0)$ satisfies $$x_1 = x_0 - \tau f'(x_1)$$
We consider a convex function $f \in \mathcal{C}(\mathbb{R})$, admitting a minimizer $x_* \in \mathbb{R}$, and $\tau > 0$. The sequence $(x_n)_{n \in \mathbb{N}}$ is defined by $x_{n+1} := p_f(x_n)$, where $p_f(x_0)$ is the unique minimizer of $F_{x_0}(x) := \frac{1}{2}|x - x_0|^2 + \tau f(x)$. In this question only, we set $f(x) := |x|$ for all $x \in \mathbb{R}$. a) Show that $$p_f(x) = \begin{cases} x - \tau & \text{if } x \geq \tau \\ x + \tau & \text{if } x \leq -\tau \\ 0 & \text{otherwise} \end{cases}$$ b) Deduce that $x_n \rightarrow 0$ as $n \rightarrow \infty$, for all $x_0 \in \mathbb{R}$ and $\tau > 0$.
We consider a convex function $f \in \mathcal{C}(\mathbb{R})$, admitting a minimizer $x_* \in \mathbb{R}$, and $\tau > 0$. The sequence $(x_n)_{n \in \mathbb{N}}$ is defined by $x_{n+1} := p_f(x_n)$. Show that $\frac{1}{2}|x_1 - x_0|^2 + \tau f(x_1) \leq \tau f(x_0)$. Deduce that for all integers $N > M \geq 0$ $$\frac{1}{2}\sum_{M < n \leq N}|x_n - x_{n-1}|^2 \leq \tau\left(f(x_M) - f(x_N)\right)$$ Deduce that $|x_{n+1} - x_n| \rightarrow 0$ as $n \rightarrow \infty$.
We consider a convex function $f \in \mathcal{C}(\mathbb{R})$, admitting a minimizer $x_* \in \mathbb{R}$, and $\tau > 0$. The sequence $(x_n)_{n \in \mathbb{N}}$ is defined by $x_{n+1} := p_f(x_n)$. Show that for all $M, N \in \mathbb{N}$ $$|x_N - x_M| \leq \sqrt{2\tau|N-M|}\sqrt{\left|f(x_M) - f(x_N)\right|}$$
We assume in this part that all roots of $p$ are stable and have multiplicity 1 and we denote by $h = Xp'$ (where $p'$ is the derivative polynomial of $p$) and $h_0$ the reciprocal polynomial of $h$. We recall that, according to question 3, there exists a real number $\lambda \in \{-1, 1\}$ such that $p = \lambda p_0$. For every real number $r > 0$, we denote by $F(r) = J(p(rX))$. Deduce, using the results of question 4, that $$\frac{n}{2(r-1)} F(r) \underset{r \rightarrow 1}{=} J(h) + o(1)$$
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 $F_{x_0}(x) := \frac{1}{2}|x - x_0|^2 + \tau f(x)$. Let $x \in \mathbb{R}$ and $\tilde{x} := p_f(x)$. Show that for all $v \in \mathbb{R}$ and $t \in \mathbb{R}$ $$\tau f(\tilde{x}) + \frac{1}{2}|\tilde{x} - x|^2 \leq \tau f(\tilde{x} + tv) + \frac{1}{2}|\tilde{x} + tv - x|^2$$ Let also $y \in \mathbb{R}$, and $\tilde{y} := p_f(y)$. Deduce that $$2\tau(f(\tilde{x}) + f(\tilde{y}) - f(\tilde{x} + tv) - f(\tilde{y} - tv)) \leq |\tilde{x} + tv - x|^2 + |\tilde{y} - tv - y|^2 - |\tilde{x} - x|^2 - |\tilde{y} - y|^2$$
We assume in this part that all roots of $p$ are stable and have multiplicity 1 and we denote by $h = Xp'$ (where $p'$ is the derivative polynomial of $p$) and $h_0$ the reciprocal polynomial of $h$. We recall that, according to question 3, there exists a real number $\lambda \in \{-1, 1\}$ such that $p = \lambda p_0$. We admit that the map defined on $S_n(\mathbf{R})$ with values in $\mathbf{R}^n$ which associates to a symmetric matrix the $n$-tuple of its real eigenvalues counted with their multiplicities, arranged in decreasing order, is continuous. Deduce that $\sigma(p) = n - 1 - \pi(J(p'))$.
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 $F_{x_0}(x) := \frac{1}{2}|x - x_0|^2 + \tau f(x)$. Let $x, y \in \mathbb{R}$, $\tilde{x} := p_f(x)$, $\tilde{y} := p_f(y)$. Show that the right-hand side in inequality $$2\tau(f(\tilde{x}) + f(\tilde{y}) - f(\tilde{x} + tv) - f(\tilde{y} - tv)) \leq |\tilde{x} + tv - x|^2 + |\tilde{y} - tv - y|^2 - |\tilde{x} - x|^2 - |\tilde{y} - y|^2$$ admits the asymptotic expansion $2tv(\tilde{x} - x + y - \tilde{y}) + o(t)$ as $t \rightarrow 0$.
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 $F_{x_0}(x) := \frac{1}{2}|x - x_0|^2 + \tau f(x)$. Let $x, y \in \mathbb{R}$, $\tilde{x} := p_f(x)$, $\tilde{y} := p_f(y)$. We choose $v := \tilde{y} - \tilde{x}$ in inequality $$2\tau(f(\tilde{x}) + f(\tilde{y}) - f(\tilde{x} + tv) - f(\tilde{y} - tv)) \leq |\tilde{x} + tv - x|^2 + |\tilde{y} - tv - y|^2 - |\tilde{x} - x|^2 - |\tilde{y} - y|^2$$ Show that the left-hand side is positive for all $t \in [0,1]$. Deduce that $$|\tilde{x} - \tilde{y}|^2 \leq (x-y)(\tilde{x} - \tilde{y}).$$
We consider a convex function $f \in \mathcal{C}(\mathbb{R})$, admitting a minimizer $x_* \in \mathbb{R}$, and $\tau > 0$. The sequence $(x_n)_{n \in \mathbb{N}}$ is defined by $x_{n+1} := p_f(x_n)$. Show that $\left|p_f(x) - p_f(y)\right| \leq |x-y|$ for all $x, y \in \mathbb{R}$. Deduce that the sequence $\left(\left|x_n - x_*\right|\right)_{n \in \mathbb{N}}$ is decreasing.
We consider a convex function $f \in \mathcal{C}(\mathbb{R})$, admitting a minimizer $x_* \in \mathbb{R}$, and $\tau > 0$. The sequence $(x_n)_{n \in \mathbb{N}}$ is defined by $x_{n+1} := p_f(x_n)$. We have a convergent subsequence $x_{\varphi(n)} \rightarrow x_{**}$ as $n \rightarrow \infty$, where $\varphi : \mathbb{N} \rightarrow \mathbb{N}$ is strictly increasing and $x_{**} \in \mathbb{R}$. Show that $x_{\varphi(n)+1} \rightarrow x_{**}$ as $n \rightarrow \infty$, then deduce that $p_f(x_{**}) = x_{**}$.
We consider a convex function $f \in \mathcal{C}(\mathbb{R})$, admitting a minimizer $x_* \in \mathbb{R}$, and $\tau > 0$. The sequence $(x_n)_{n \in \mathbb{N}}$ is defined by $x_{n+1} := p_f(x_n)$. We have established that $p_f(x_{**}) = x_{**}$ for some $x_{**} \in \mathbb{R}$. Conclude that $x_{**}$ is a minimizer of $f$, and that $x_n \rightarrow x_{**}$ as $n \rightarrow \infty$.
We fix an integer $d \in \mathbb{N}^*$, and we equip $\mathbb{R}^d$ with the usual inner product denoted $\langle \cdot, \cdot \rangle$ and the associated Euclidean norm $\|\cdot\|$. We denote $C := \{x \in \mathbb{R}^d \mid \|x\| \leq 1\}$ the closed unit ball of $\mathbb{R}^d$, and we are given $f \in \mathcal{C}^1(\mathbb{R}^d)$. Show that $f$ admits a minimizer on $C$, which we denote $x_*$ in the following questions.
We fix an integer $d \in \mathbb{N}^*$, and we equip $\mathbb{R}^d$ with the usual inner product denoted $\langle \cdot, \cdot \rangle$ and the associated Euclidean norm $\|\cdot\|$. We denote $C := \{x \in \mathbb{R}^d \mid \|x\| \leq 1\}$ the closed unit ball of $\mathbb{R}^d$, and we are given $f \in \mathcal{C}^1(\mathbb{R}^d)$. Let $x_*$ be a minimizer of $f$ on $C$. Suppose in this question that $\|x_*\| < 1$. Show that $\nabla f(x_*) = 0$.
We fix an integer $d \in \mathbb{N}^*$, and we equip $\mathbb{R}^d$ with the usual inner product denoted $\langle \cdot, \cdot \rangle$ and the associated Euclidean norm $\|\cdot\|$. We denote $C := \{x \in \mathbb{R}^d \mid \|x\| \leq 1\}$ the closed unit ball of $\mathbb{R}^d$, and we are given $f \in \mathcal{C}^1(\mathbb{R}^d)$. Let $x_*$ be a minimizer of $f$ on $C$. Suppose in this question that $\|x_*\| = 1$. The objective is to show that $$\exists \lambda \geq 0,\, \nabla f(x_*) = -\lambda x_*.$$ a) Let $x, y \in \mathbb{R}^d$ such that $x \neq y$ and $\|x\| = \|y\| = 1$. Show that $\langle x, v \rangle > 0$ and $\langle y, v \rangle < 0$, where $v := x - y$. b) Suppose by contradiction that (7) is not satisfied. Show that there exists $v \in \mathbb{R}^d$ such that $\langle v, \nabla f(x_*) \rangle > 0$ and $\langle v, x_* \rangle > 0$. Deduce a contradiction and conclude. Hint: consider the quantities $f(x_* - tv)$ and $\|x_* - tv\|^2$, in the limit $t \rightarrow 0^+$.
114- The function with the rule $f(x) = \begin{cases} |x| + [-x] & ; \ x \notin \mathbb{Z} \\ a & ; \ x \in \mathbb{Z} \end{cases}$, for which value of $a$, is continuous on the set of real numbers? (Note: $[\ ]$ denotes the floor function.) (1) $-1$ (2) $1$ (3) $0$ (4) Always discontinuous
115- The function with the rule $f(x) = \begin{cases} \dfrac{x-[x]}{x^2-x-6} & ; \ x \neq 2 \\ a & ; \ x = 2 \end{cases}$, for which value of $a$, is continuous on the interval $[2,3]$? (1) $\dfrac{1}{11}$ (2) $\dfrac{1}{9}$ (3) $\dfrac{1}{8}$ (4) $\dfrac{1}{6}$
116- The number of discontinuous points of the graph of the function $f(x) = \dfrac{3-\sqrt{x+4}}{1+\sqrt[3]{x+1}} + \dfrac{1}{x+5}$ is which of the following? (1) zero (2) $1$ (3) $2$ (4) $3$