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| Author | SHA1 | Date | |
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| 15e0fec150 | |||
| dbc3f60254 | |||
| d20cd149cd | |||
| bdf6abcc39 | |||
| b517b6318c |
18
Dockerfile
18
Dockerfile
@ -1,14 +1,14 @@
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FROM alpine:3.21.3
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FROM alpine:3.22.0
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RUN apk add --no-cache \
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make=4.4.1-r2 \
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graphviz=12.2.0-r0 \
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texlive-xetex=20240210.69778-r8 \
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texmf-dist-langfrench=2024.0-r6 \
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texmf-dist-latexextra=2024.0-r6 \
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texmf-dist-bibtexextra=2024.0-r6 \
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texmf-dist-mathscience=2024.0-r6 \
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texmf-dist-publishers=2024.0-r6 \
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make=4.4.1-r3 \
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graphviz=12.2.1-r0 \
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texlive-xetex=20240210.69778-r10 \
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texmf-dist-langfrench=2024.0-r7 \
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texmf-dist-latexextra=2024.0-r7 \
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texmf-dist-bibtexextra=2024.0-r7 \
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texmf-dist-mathscience=2024.0-r7 \
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texmf-dist-publishers=2024.0-r7 \
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font-freefont=20120503-r4 \
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&& rm -rf /var/cache/apk/*
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@ -545,110 +545,3 @@
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% TODO Complete proof
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\end{proof}
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\langsubsubsection{Exercices}{Exercises}
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\begin{exercise_sq}
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Soit $T := \{-1, 1\}$ ainsi que $\function{f}{\Z}{T} \functiondef{x}{\begin{cases} 1 & \text{\lang{Pair}{Even} } x \\ -1 & \text{\lang{Impair}{Odd} } x \end{cases}}$
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\begin{enumerate}[(a)]
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\item{Montrer que $\forall (x, y) \in \Z^2, f(x + y) = f(x)f(y)$, que cela dit sur les entiers relatifs ?}
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\item{Est-ce que $\forall (x, y) \in \Z^2, f(xy) = f(x)f(y)$ est aussi vrai ?}
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\end{enumerate}
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\end{exercise_sq}
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\begin{proof}
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Soit $T := \{-1, 1\}$ ainsi que $\function{f}{\Z}{T} \functiondef{x}{\begin{cases} 1 & \text{\lang{Pair}{Even} } x \\ -1 & \text{\lang{Impair}{Odd} } x \end{cases}}$.
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\begin{enumerate}[(a)]
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\item{Soit $(x, y) \in \Z^2$
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Comme tout entier est soit pair ou impair, on peut donc faire une disjonction en 4 cas
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\begin{tabular}{c|c|c|c}
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$x$ & $y$ & $f(x + y)$ & $f(x)f(y)$ \\
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\hline
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$\text{\lang{Pair}{Even} } x$ & $\text{\lang{Pair}{Even} } y$ & $f(2x' + 2y') = f(2(x' + y')) = 1$ & $1 \cdot 1 = 1$ \\
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\hline
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$\text{\lang{Pair}{Even} } x$ & $\text{\lang{Impair}{Odd} } y$ & $f(2x' + 2y' + 1) = f(2(x' + y') + 1) = -1$ & $1 \cdot -1 = -1$ \\
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\hline
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$\text{\lang{Impair}{Odd} } x$ & $\text{\lang{Pair}{Even} } y$ & $f(2x' + 1 + 2y') = f(2(x' + y') + 1) = -1$ & $-1 \cdot 1 = -1$ \\
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\hline
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$\text{\lang{Impair}{Odd} } x$ & $\text{\lang{Impair}{Odd} } y$ & $f(2x' + 1 + 2y' + 1) = f(2(x' + y' + 1)) = 1$ & $-1 \cdot -1 = 1$
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\end{tabular}
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On a donc $\forall (x, y) \in \Z^2, f(x + y) = f(x)f(y)$, $f$ est donc un homomorphisme.}
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\item{Soit $(x, y) \in \Z^2$, dans le cas ou $x$ est pair et $y$ est impair, on remarque que $f(xy) = f(2x'(2y' + 1)) = f(2(2x'y' + x')) = 1$ alors que $f(x)f(y) = 1 \cdot -1 = -1$.
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}
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\end{enumerate}
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\end{proof}
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\begin{exercise_sq}
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Soit $G$ un groupe d'ordre 4. Supposons que $G$ n'est pas isomorphe au groupe $\Z/4\Z$. Montrer que $G$ est isomorphe à $\Z/2\Z \cartesianProduct \Z/2\Z$.
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\end{exercise_sq}
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\begin{proof}
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\lipsum[2]
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% TODO: Complete proof
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\end{proof}
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\begin{exercise_sq}
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\begin{enumerate}[(a)]
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\item{Montrer qu'un élément $\bar{a}$ de $(\Z/n\Z, +)$ est générateur si et seulement s'il existe $b \in \Z$ tel que $\bar{a} \cdot \bar{b} = \bar{1}$.}
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\item{Soit $(G, \composes)$ un groupe, Montrer qu'un morphisme de groupe $$\function{\varphi}{(\Z/n\Z, +)}{(G, \composes)}$$ est déterminé par $\varphi(\bar{1})$.}
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\item{Soit $\function{\varphi}{(\Z/n\Z, +)}{(\Z/n\Z, +)}$ un morphisme. Montrer que $\varphi$ est un isomorphisme si et seulement si $\varphi(\bar{1})$ est générateur.}
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\end{enumerate}
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\end{exercise_sq}
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\begin{proof}
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% TODO: Complete proof
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\begin{enumerate}[(a)]
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\item{\impliespart
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Soit $(\Z/n\Z, +) \in \Grp$ et $\bar{a}$ générateur de $\Z/n\Z$, il existe donc $b \in \N^*$ tel que $\sum\limits_{i = 1}^b \bar{a} = 1$, or comme $\card{\generator{\bar{a}}} = n \implies b \le n$.
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$\Limpliespart$
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}
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\end{enumerate}
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\end{proof}
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\begin{exercise_sq}
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\begin{enumerate}[(a)]
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\item{Montrer que l'ensemble $G$ des matrices défini par $$G := \left\{ \begin{pmatrix} 1 & x & y \\ 0 & 1 & z \\ 0 & 0 & 1 \end{pmatrix} \suchthat x, y, z \in \R \right\}$$ est un sous-groupe de $GL_3(\R)$.}
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\item{Calculer le centre de $G$, c'est-à-dire $$Z(G) = \{ g \in G \suchthat gh = hg \forall h \in G \}$$}
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\end{enumerate}
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\end{exercise_sq}
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\begin{proof}
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\begin{enumerate}[(a)]
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\item{Tous les éléments de $G$ sont des matrices triangulaires supérieures qui sont inversibles, il suffit donc de vérifier chaque axiome d'un sous-groupe.
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\begin{itemize}
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\item{Soit $A \in G$ tel que $x = y = z = 0 \implies A = \Identity_3 \implies \Identity_3 \in G$}
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\item{Soit $(A, B) \in G^2$ ainsi que $(a, b, c, x, y, z) \in \R^6$ tel que
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$A = \begin{pmatrix} 1 & a & b \\ 0 & 1 & c \\ 0 & 0 & 1 \end{pmatrix}$ et
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$B = \begin{pmatrix} 1 & x & y \\ 0 & 1 & z \\ 0 & 0 & 1 \end{pmatrix}$
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$AB = \begin{pmatrix} 1 & x + a & y + az + b \\ 0 & 1 & z + c \\ 0 & 0 & 1 \end{pmatrix}$,
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hors $(x + a, y + az + b, z + c) \in \R^3 \implies AB \in G$}
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\item{Soit $A \in G, \exists! \inv{A} \in GL_3(\R)$ ainsi que $(a, b, c) \in \R^3$ tel que
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$A = \begin{pmatrix} 1 & a & b \\ 0 & 1 & c \\ 0 & 0 & 1 \end{pmatrix}$ comme
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$A \inv{A} = \Identity_G \implies \inv{A} =
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\begin{pmatrix} 1 & -a & ac - b \\ 0 & 1 & -c \\ 0 & 0 & 1 \end{pmatrix}$,
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hors $(-a, ac - b, -c) \in \R^3 \implies \inv{A} \in G$}
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\end{itemize}
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$G$ est de ce fait un sous-groupe de $GL_3(\R)$.
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}
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\item{Soit $(A, B) \in G^2$ ainsi que $(a, b, c, x, y, z) \in \R^6$ tel que
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$A = \begin{pmatrix} 1 & a & b \\ 0 & 1 & c \\ 0 & 0 & 1 \end{pmatrix}$ et
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$B = \begin{pmatrix} 1 & x & y \\ 0 & 1 & z \\ 0 & 0 & 1 \end{pmatrix}$
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$AB = BA \equivalence \begin{pmatrix} 1 & x + a & y + az + b \\ 0 & 1 & z + c \\ 0 & 0 & 1 \end{pmatrix} =
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\begin{pmatrix} 1 & a + x & b + cx + y \\ 0 & 1 & c + z \\ 0 & 0 & 1 \end{pmatrix} \equivalence az = cx$
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Pour un $A$ fixé, la seule manière de rendre le produit commutatif pour tout $B$ est de mettre $a = c = 0$
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$\implies Z(G) = \left\{ \begin{pmatrix} 1 & 0 & b \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} \suchthat b \in \R \right\}$
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}
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\end{enumerate}
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\end{proof}
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183
contents/group_theory_exo.tex
Normal file
183
contents/group_theory_exo.tex
Normal file
@ -0,0 +1,183 @@
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\langsubsubsection{Exercices}{Exercises}
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\begin{exercise_sq}[TD2 EX1]
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Est-ce que les groupes suivants sont isomorphes ?
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\begin{enumerate}[(a)]
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\item{$(\Z/4\Z, +)$ et $(\Z/2\Z \cartesianProduct \Z/2\Z, +)$}
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\item{$(\{1, -1, i, -i\}, \cdot)$ et $(\Z/4\Z, +)$}
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\item{$(S_3, \composes)$ et $(\Z/6\Z, +)$}
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\end{enumerate}
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\end{exercise_sq}
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\begin{proof}
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\begin{enumerate}[(a)]
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\item{$\forall x \in ((\Z/2\Z)^2, +), x + x = 0 \implies \card{\generator{x}} \le 2$ alors que $\bar{1} \in (\Z4/\Z, +), \card{\generator{\bar{1}}} = 4$. Comme les isomorphismes préservent l'ordre des éléments, on en conclut que $((\Z/2\Z)^2, +) \not \isomorphic (\Z/4\Z, +)$}
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\item{Posons $\function{f}{\Z/4\Z}{\{1, i, -1, -i\}} \functiondef{x}{e^{i \frac{x \pi}{2}}}$ or $\forall (x, y) \in (\Z/4\Z)^2, f(x) \cdot f(y) = e^{i \frac{x \pi}{2}} \cdot e^{i \frac{y \pi}{2}} = e^{i \frac{(x + y) \pi}{2}} = f(x + y) \implies f \in Hom(\Z/4\Z, \{1, i, -1, -i\})$ de plus $\inv{f}(y) = -\frac{2i}{\pi} \log(y)$ ce qui permet de conclure $(\{1, -1, i, -i\}, \cdot) \isomorphic (\Z/4\Z, +)$}
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\item{Soit $ (f, g) \in (S_3, \composes)^2$ tel que $f = \begin{bmatrix} 1 & 3 & 2 \end{bmatrix}$ et $g = \begin{bmatrix} 2 & 1 & 3 \end{bmatrix}$.
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Observons que $f \composes g = \begin{bmatrix} 2 & 3 & 1 \end{bmatrix}$ ainsi que $g \composes f = \begin{bmatrix} 3 & 1 & 2\end{bmatrix}$.
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Comme $f \composes g \ne g \composes f \implies (S_3, \composes) \notin \Ab$. Sachant que $(\Z/6\Z, +) \in \Ab$ on en conclut que $(S_3, \composes) \not \isomorphic (\Z/6\Z, +)$
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}
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\end{enumerate}
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\end{proof}
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\begin{exercise_sq}[TD2 EX4]
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On considère le groupe (voir Feuille 1)
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$$G = \{ \function{f_{a,b}}{\R}{\R} \suchthat f_{a,b}(x) = ax + b, a \in \R^*, b \in \R \}$$
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dont la loi de groupe est la composition des fonctions. On pose
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$$H := \{ \function{f_b}{\R}{\R} \suchthat f_b(x) = x + b, b \in \R \}$$
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\begin{enumerate}[(a)]
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\item{Montrer que $H \subset G$ est un sous-groupe.}
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\item{Montrer que $$f_{a,b} H = f_{c,d} H$$ si et seulement si $a = c$.}
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\end{enumerate}
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\end{exercise_sq}
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\begin{proof}
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Soit $G = \{ \function{f_{a,b}}{\R}{\R} \suchthat f_{a,b}(x) = ax + b, a \in \R^*, b \in \R \}$
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ainsi que $H := \{ \function{f_b}{\R}{\R} \suchthat f_b(x) = x + b, b \in \R \}$
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\begin{enumerate}[(a)]
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\item{
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\begin{itemize}
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\item{$\Identity_G = f_{1, 0} \implies \Identity_G \in H$}
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\item{$\forall (f_{1, b}, f_{1, d}) \in H^2, \forall x \in \R, f_{1, b} \composes f_{1, d} = (x + b) + d = f_{1, b + d} \implies f_{1, b} \composes f_{1, d} \in H$}
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\item{$\forall f_{1, b} \in H, \exists! f_{c, d} \in G, f_{1, b} \composes f_{c, d} = \Identity_G \implies \forall x \in \R, f_{1, b} \composes f_{c, d} = (cx + d) + b \implies f_{c, d} = f_{1, -b} \implies \inv{f_{1, b}} = f_{c, d} \in H$}
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\end{itemize}
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}
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\item{
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\impliespart
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Soit $(f_{a, b}, f_{c, d}) \in G^2$ tel que $f_{a,b} H = f_{c,d} H
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\equivalence \forall f_{1, e} \in H, \inv{f_{c, d} \composes f_{a, b} \composes f_{1, e} \in H}
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\equivalence \frac{1}{c} (a(x + e) + b) - \frac{d}{c} \in H
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\equivalence \frac{a}{c} x + (\frac{ae + b - d}{c}) \in H
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\implies a = c$
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\Limpliespart
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Soit $(f_{a, b}, f_{c, d}) \in G^2$ tel que $a = c$.
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On observe que $f_{a, b} \composes f_{1, d - b} = a(x + d - b) + b = f_{a, d}$
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ainsi que $f_{c, d} \composes f_{1, b - d} = c(x + b - d) + d = f_{a, b}$.
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Or $f_{1, d - b}$ et $f_{1, b - d}$ sont dans $H$ ce qui montre que $f_{a,b} H = f_{c,d} H$.
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}
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\end{enumerate}
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\end{proof}
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\begin{exercise_sq}
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Soit $T := \{-1, 1\}$ ainsi que $\function{f}{\Z}{T} \functiondef{x}{\begin{cases} 1 & \text{\lang{Pair}{Even} } x \\ -1 & \text{\lang{Impair}{Odd} } x \end{cases}}$
|
||||
\begin{enumerate}[(a)]
|
||||
\item{Montrer que $\forall (x, y) \in \Z^2, f(x + y) = f(x)f(y)$, que cela dit sur les entiers relatifs ?}
|
||||
\item{Est-ce que $\forall (x, y) \in \Z^2, f(xy) = f(x)f(y)$ est aussi vrai ?}
|
||||
\end{enumerate}
|
||||
\end{exercise_sq}
|
||||
|
||||
\begin{proof}
|
||||
Soit $T := \{-1, 1\}$ ainsi que $\function{f}{\Z}{T} \functiondef{x}{\begin{cases} 1 & \text{\lang{Pair}{Even} } x \\ -1 & \text{\lang{Impair}{Odd} } x \end{cases}}$.
|
||||
\begin{enumerate}[(a)]
|
||||
\item{Soit $(x, y) \in \Z^2$
|
||||
|
||||
Comme tout entier est soit pair ou impair, on peut donc faire une disjonction en 4 cas
|
||||
|
||||
\begin{tabular}{c|c|c|c}
|
||||
$x$ & $y$ & $f(x + y)$ & $f(x)f(y)$ \\
|
||||
\hline
|
||||
$\text{\lang{Pair}{Even} } x$ & $\text{\lang{Pair}{Even} } y$ & $f(2x' + 2y') = f(2(x' + y')) = 1$ & $1 \cdot 1 = 1$ \\
|
||||
\hline
|
||||
$\text{\lang{Pair}{Even} } x$ & $\text{\lang{Impair}{Odd} } y$ & $f(2x' + 2y' + 1) = f(2(x' + y') + 1) = -1$ & $1 \cdot -1 = -1$ \\
|
||||
\hline
|
||||
$\text{\lang{Impair}{Odd} } x$ & $\text{\lang{Pair}{Even} } y$ & $f(2x' + 1 + 2y') = f(2(x' + y') + 1) = -1$ & $-1 \cdot 1 = -1$ \\
|
||||
\hline
|
||||
$\text{\lang{Impair}{Odd} } x$ & $\text{\lang{Impair}{Odd} } y$ & $f(2x' + 1 + 2y' + 1) = f(2(x' + y' + 1)) = 1$ & $-1 \cdot -1 = 1$
|
||||
\end{tabular}
|
||||
|
||||
On a donc $\forall (x, y) \in \Z^2, f(x + y) = f(x)f(y)$, $f$ est donc un homomorphisme.}
|
||||
|
||||
\item{Soit $(x, y) \in \Z^2$, dans le cas ou $x$ est pair et $y$ est impair, on remarque que $f(xy) = f(2x'(2y' + 1)) = f(2(2x'y' + x')) = 1$ alors que $f(x)f(y) = 1 \cdot -1 = -1$.
|
||||
}
|
||||
\end{enumerate}
|
||||
\end{proof}
|
||||
|
||||
\begin{exercise_sq}
|
||||
Soit $G$ un groupe d'ordre 4. Supposons que $G$ n'est pas isomorphe au groupe $\Z/4\Z$. Montrer que $G$ est isomorphe à $\Z/2\Z \cartesianProduct \Z/2\Z$.
|
||||
\end{exercise_sq}
|
||||
|
||||
\begin{proof}
|
||||
\lipsum[2]
|
||||
% TODO: Complete proof
|
||||
\end{proof}
|
||||
|
||||
\begin{exercise_sq}
|
||||
\begin{enumerate}[(a)]
|
||||
\item{Montrer qu'un élément $\bar{a}$ de $(\Z/n\Z, +)$ est générateur si et seulement s'il existe $b \in \Z$ tel que $\bar{a} \cdot \bar{b} = \bar{1}$.}
|
||||
\item{Soit $(G, \composes)$ un groupe, Montrer qu'un morphisme de groupe $$\function{\varphi}{(\Z/n\Z, +)}{(G, \composes)}$$ est déterminé par $\varphi(\bar{1})$.}
|
||||
\item{Soit $\function{\varphi}{(\Z/n\Z, +)}{(\Z/n\Z, +)}$ un morphisme. Montrer que $\varphi$ est un isomorphisme si et seulement si $\varphi(\bar{1})$ est générateur.}
|
||||
\end{enumerate}
|
||||
\end{exercise_sq}
|
||||
|
||||
\begin{proof}
|
||||
% TODO: Complete proof
|
||||
\begin{enumerate}[(a)]
|
||||
\item{\impliespart
|
||||
|
||||
Soit $(\Z/n\Z, +) \in \Grp$ et $\bar{a}$ générateur de
|
||||
$\Z/n\Z \implies \forall \bar{k} \in \Z/n\Z, \exists \bar{b} \in \Z/n\Z, \bar{a} \cdot \bar{b} = \sum\limits_{i = 1}^{\bar{b}} \bar{a} = \bar{k}$,
|
||||
il existe donc en particulier $\bar{b} \in \Z/n\Z$ tel que $\bar{a} \cdot \bar{b} = \sum\limits_{i = 1}^{\bar{b}} \bar{a} = \bar{1}$.
|
||||
Or $\bar{b} = \{ b \cdot n \suchthat n \in \Z \}$, en conséquence, il existe $b \in \Z$ tel que $\bar{a} \cdot \bar{b} = \bar{1}$.
|
||||
|
||||
\Limpliespart
|
||||
|
||||
Soit $b \in \Z$ tel que $\bar{a} \cdot \bar{b} = \bar{1} \implies \forall k \in \Z, (k \cdot \bar{b}) \cdot \bar{a} = k \cdot (\bar{a} \cdot \bar{b}) \equiv k \mod n \implies \forall $
|
||||
}
|
||||
\item{Soit $x \in \Z/n\Z, \varphi(x) = \varphi(\sum\limits_{i = 1}^x \bar{1}) = \composes\limits_{i = 1}^x \varphi(\bar{1})$}
|
||||
\item{\impliespart
|
||||
asdasd
|
||||
|
||||
\Limpliespart
|
||||
|
||||
asdasd
|
||||
}
|
||||
\end{enumerate}
|
||||
\end{proof}
|
||||
|
||||
\begin{exercise_sq}
|
||||
\begin{enumerate}[(a)]
|
||||
\item{Montrer que l'ensemble $G$ des matrices défini par $$G := \left\{ \begin{pmatrix} 1 & x & y \\ 0 & 1 & z \\ 0 & 0 & 1 \end{pmatrix} \suchthat x, y, z \in \R \right\}$$ est un sous-groupe de $GL_3(\R)$.}
|
||||
\item{Calculer le centre de $G$, c'est-à-dire $$Z(G) = \{ g \in G \suchthat gh = hg \forall h \in G \}$$}
|
||||
\end{enumerate}
|
||||
\end{exercise_sq}
|
||||
|
||||
\begin{proof}
|
||||
\begin{enumerate}[(a)]
|
||||
\item{Tous les éléments de $G$ sont des matrices triangulaires supérieures qui sont inversibles, il suffit donc de vérifier chaque axiome d'un sous-groupe.
|
||||
|
||||
\begin{itemize}
|
||||
\item{Soit $A \in G$ tel que $x = y = z = 0 \implies A = \Identity_3 \implies \Identity_3 \in G$}
|
||||
\item{Soit $(A, B) \in G^2$ ainsi que $(a, b, c, x, y, z) \in \R^6$ tel que
|
||||
$A = \begin{pmatrix} 1 & a & b \\ 0 & 1 & c \\ 0 & 0 & 1 \end{pmatrix}$ et
|
||||
$B = \begin{pmatrix} 1 & x & y \\ 0 & 1 & z \\ 0 & 0 & 1 \end{pmatrix}$
|
||||
|
||||
$AB = \begin{pmatrix} 1 & x + a & y + az + b \\ 0 & 1 & z + c \\ 0 & 0 & 1 \end{pmatrix}$,
|
||||
hors $(x + a, y + az + b, z + c) \in \R^3 \implies AB \in G$}
|
||||
\item{Soit $A \in G, \exists! \inv{A} \in GL_3(\R)$ ainsi que $(a, b, c) \in \R^3$ tel que
|
||||
$A = \begin{pmatrix} 1 & a & b \\ 0 & 1 & c \\ 0 & 0 & 1 \end{pmatrix}$ comme
|
||||
$A \inv{A} = \Identity_G \implies \inv{A} =
|
||||
\begin{pmatrix} 1 & -a & ac - b \\ 0 & 1 & -c \\ 0 & 0 & 1 \end{pmatrix}$,
|
||||
hors $(-a, ac - b, -c) \in \R^3 \implies \inv{A} \in G$}
|
||||
\end{itemize}
|
||||
|
||||
$G$ est de ce fait un sous-groupe de $GL_3(\R)$.
|
||||
}
|
||||
\item{Soit $(A, B) \in G^2$ ainsi que $(a, b, c, x, y, z) \in \R^6$ tel que
|
||||
$A = \begin{pmatrix} 1 & a & b \\ 0 & 1 & c \\ 0 & 0 & 1 \end{pmatrix}$ et
|
||||
$B = \begin{pmatrix} 1 & x & y \\ 0 & 1 & z \\ 0 & 0 & 1 \end{pmatrix}$
|
||||
|
||||
$AB = BA \equivalence \begin{pmatrix} 1 & x + a & y + az + b \\ 0 & 1 & z + c \\ 0 & 0 & 1 \end{pmatrix} =
|
||||
\begin{pmatrix} 1 & a + x & b + cx + y \\ 0 & 1 & c + z \\ 0 & 0 & 1 \end{pmatrix} \equivalence az = cx$
|
||||
|
||||
Pour un $A$ fixé, la seule manière de rendre le produit commutatif pour tout $B$ est que $a = c = 0$
|
||||
|
||||
$\implies Z(G) = \left\{ \begin{pmatrix} 1 & 0 & b \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} \suchthat b \in \R \right\}$
|
||||
}
|
||||
\end{enumerate}
|
||||
\end{proof}
|
||||
|
||||
@ -18,6 +18,37 @@
|
||||
Soit $(R, +, \star) \in \Ring$ un ensemble $S \subseteq R$ est un \textbf{sous-anneau} si $(S, +, \star)$ est un anneau.
|
||||
\end{definition_sq}
|
||||
|
||||
\begin{definition_sq} \label{definition:ideal}
|
||||
Soit $(R, +, \star) \in \Ring$ un ensemble $I \subseteq R$ est un \textbf{idéal} si $(I, +)$ est un groupe et $\forall x \in I, \forall y \in R, \{ x \star y, y \star x \} \subset I$
|
||||
\end{definition_sq}
|
||||
|
||||
\begin{theorem_sq}
|
||||
Soit $(R, +, \star)$, $(S, +, \star)$ ainsi que $\function{f}{(R, +, \star)}{(S, +, \star)}$ un homomorphisme.
|
||||
|
||||
\begin{itemize}
|
||||
\item{$\ker f \subset R$ est un idéal}
|
||||
\item{$im f \subset S$ est un sous-anneau}
|
||||
\end{itemize}
|
||||
\end{theorem_sq}
|
||||
|
||||
\begin{proof}
|
||||
\lipsum[2]
|
||||
% TODO Complete proof
|
||||
\end{proof}
|
||||
|
||||
\begin{theorem_sq}
|
||||
Soit $(R, +, \star)$, $(S, +, \star)$ ainsi que $\function{f}{(R, +, \star)}{(S, +, \star)}$ est un monomorphisme si et seulement si $\ker f = \{ 0 \}$.
|
||||
\end{theorem_sq}
|
||||
|
||||
\begin{proof}
|
||||
\lipsum[2]
|
||||
% TODO Complete proof
|
||||
\end{proof}
|
||||
|
||||
\begin{definition_sq}
|
||||
Soit $(R, +, \star)$ et $I \subset R$ un idéal. On définit \textbf{l'anneau quotient} $\function{q}{R}{R/I}$ le quotient du groupe abélien $(R, +)$ par le sous-groupe $I$.
|
||||
\end{definition_sq}
|
||||
|
||||
\begin{definition_sq} \label{definition:ring_unit}
|
||||
Soit $(R, +, \star) \in \Ring$ un élément $x \in R$ est dit \textbf{inversible} (on dit aussi que $x$ est une \textbf{unité}) s'il existe $y \in R$ tel que $x \star y = y \star x = \Identity_\star$
|
||||
|
||||
@ -36,6 +67,10 @@
|
||||
$$\phi(\Identity_{\cartesianProduct_R}) = \Identity_{\cartesianProduct_S}$$
|
||||
\end{definition_sq}
|
||||
|
||||
\begin{definition_sq} \label{definition:ring_morphism_kernel}
|
||||
Soit $(R, +, \star)$ et $(S, +, \star)$ ainsi que d'un morphisme d'anneau $\function{\phi}{R}{S}$. On appelle \textbf{noyau de $\phi$} l'ensemble $\ker(\phi) := \{ x \in R \suchthat \phi(x) = \Identity_{+_S} \}$.
|
||||
\end{definition_sq}
|
||||
|
||||
\begin{theorem_sq}
|
||||
Soit $(R, +, \cartesianProduct)$ et $(S, +, \cartesianProduct)$ deux anneaux ainsi que l'homomorphisme $\function{\phi}{R}{S}$
|
||||
$$\phi(\Identity_{+_R}) = \Identity_{+_S}$$
|
||||
|
||||
107
contents/topology_exo.tex
Normal file
107
contents/topology_exo.tex
Normal file
@ -0,0 +1,107 @@
|
||||
\langsubsubsection{Exercices}{Exercises}
|
||||
|
||||
\begin{exercise_sq}[TD3 EX1]
|
||||
Soit $E$ un $\R$-espace vectoriel de dimension finie muni de deux normes $N_1, N_2$.
|
||||
|
||||
\begin{enumerate}[(a)]
|
||||
\item{Montrer que les boules-unité $B_1, B_2$ pour $N_1, N_2$ sont homéomorphes. En déduire que si l'une est compacte, alors de même l'autre. Elles sont donc toutes compactes étant donné que la boule euclidienne l'est.}
|
||||
\item{Montrer que la norme $\function{N_2}{E}{\R}_+$ restreinte à la boules-unité $B_1$ est majoré par un réel $\lambda$. En déduire que pour tout $x \in E$ on a $\lambda N_1(x) \le N_2(x)$.}
|
||||
\item{En déduire que les normes $N_1$ et $N_2$ sont équivalentes. En particulier, la "bornitude" d'une partie de $E$ ne dépend pas du choix de la norme.}
|
||||
\end{enumerate}
|
||||
\end{exercise_sq}
|
||||
|
||||
\begin{proof}
|
||||
\lipsum[2]
|
||||
% TODO Complete proof
|
||||
\end{proof}
|
||||
|
||||
\begin{exercise_sq}[TD3 EX2]
|
||||
Parmi les parties suivantes de R2, lesquelles sont compactes ?
|
||||
|
||||
\begin{enumerate}[(a)]
|
||||
\item{$H_a = \{ (x, y) \in \R^2 \suchthat xy = 1, \abs{x + y} \le a \}$ pour $2 \le a \le +\infty$}
|
||||
\item{$S_b = \{ (x, y) \in \R^2 \suchthat −b \abs{x} \le y \le 1 −x^2 \}$ pour $b \in \R_+$}
|
||||
\item{$P = \{ (0, 0) \} \union \Union\limits_{n \in \N^*} \{ \frac{1}{n} \} \cartesianProduct [0, \frac{1}{n}]$}
|
||||
\item{$S = \{ (0, 0) \} \union \{ (x, x \sin(\frac{1}{x})) \suchthat 0 < x \le 1 \}$}
|
||||
\item{$D = \{ (x, y) \in \R^2 \suchthat x^2 + y^2 \le 1 \}, D_\Q = D \intersection \Q^2, D_\Z = D \intersection \Z^2$}
|
||||
\item{Donner trois raisons du fait que $]0, 1] \subset \R$ n'est pas compact.}
|
||||
\end{enumerate}
|
||||
\end{exercise_sq}
|
||||
|
||||
\begin{proof}
|
||||
\lipsum[2]
|
||||
% TODO Complete proof
|
||||
\end{proof}
|
||||
|
||||
\begin{exercise_sq}[TD3 EX3]
|
||||
Soit $A$ un compact de $(R^n,d)$ et $\function{\phi}{A}{A}$ une application contractante.
|
||||
|
||||
\begin{enumerate}[(a)]
|
||||
\item{Montrer que $A \cartesianProduct A$ est un fermé de $R^{2n}$.}
|
||||
\item{Montrer que $A \cartesianProduct A$ est un compact de $R^{2n}$.}
|
||||
\item{On suppose que $A$ n'est pas singleton. Montrer que $\function{\phi}{A}{A}$ ne peut pas être surjective.
|
||||
On pourra considérer les antécédents de deux points $(x_0, y_0) \in A^2$ tels que $d(x_0, y_0) = diam(A) = sup_{(x, y) \in A} d(x, y)$}
|
||||
\item{On note $A= A_0$ et $A_{n + 1} = \phi(A_n)$. Que peut-on dire sur $\lim_{n \to \infty} diam(A_n)$ et sur $\Intersection_{n \ge 0} A_n$ ?}
|
||||
\end{enumerate}
|
||||
\end{exercise_sq}
|
||||
|
||||
\begin{proof}
|
||||
\lipsum[2]
|
||||
% TODO Complete proof
|
||||
\end{proof}
|
||||
|
||||
\begin{exercise_sq}[TD3 EX4]
|
||||
On se place dans $\R^n$ muni de la distance euclidienne.
|
||||
|
||||
\begin{enumerate}[(a)]
|
||||
\item{Montrer que la somme $K + L = \{ x + y \in \R^n \suchthat x \in K, y \in L \}$ de deux parties compactes $K$, $L$ est compacte.}
|
||||
\item{Montrer que l'intersection de deux parties compactes est compacte. Montrer que la réunion finie de parties compactes est compacte.}
|
||||
\item{Montrer que pour deux compacts $K$, $L$ disjoints la distance $d(K, L) = inf_{(x, y) \in K \cartesianProduct L} d(x, y)$ est strictement positive.
|
||||
En déduire l’existence de deux ouverts $U$, $V$ disjoints tels que $K \subset U$ et $L \subset V$.}
|
||||
\item{Montrer que l'intersection $K = \Intersection_{n \ge 0} K_n$ d'une famille décroissante de parties compactes non vides est compacte non vide.
|
||||
Montrer que si $K \subset U$ pour un ouvert $U$ alors il existe $n \in \N$ tel que $K_n \subset U$.}
|
||||
\end{enumerate}
|
||||
\end{exercise_sq}
|
||||
|
||||
\begin{proof}
|
||||
\lipsum[2]
|
||||
% TODO Complete proof
|
||||
\end{proof}
|
||||
|
||||
\begin{exercise_sq}[TD3 EX5]
|
||||
Montrer que toute suite de points bornée de $\R^n$ possède une sous-suite qui converge (théorème de Bolzano-Weierstrass).
|
||||
En déduire que toute suite \suite{x} n’admettant pas de sous-suite convergente, diverge dans le sens suivant : $\lim_{n \to \infty} \norm{x_n} = \infty$.
|
||||
\end{exercise_sq}
|
||||
|
||||
\begin{proof}
|
||||
\lipsum[2]
|
||||
% TODO Complete proof
|
||||
\end{proof}
|
||||
|
||||
\begin{exercise_sq}[TD3 EX6]
|
||||
Soit un espace métrique $(E, d)$, nous allons montrer l'équivalence entre les propositions suivantes :
|
||||
|
||||
\begin{enumerate}
|
||||
\item{De tout recouvrement ouvert de $E$ on peut extraire un recouvrement fini (la propriété de Borel-Lebesgue).}
|
||||
\item{$E$ est compact (i.e. toute suite admet des valeurs d’adhérence).}
|
||||
\item{$E$ est pré-compact (3a) et complet (3b).}
|
||||
\item{$E$ est pré-compact et pour tout recouvrement ouvert de $E$ il existe $\epsilon > 0$ tel que toute $\epsilon$-boule de $E$ est contenue dans un des ouverts du recouvrement.}
|
||||
\end{enumerate}
|
||||
|
||||
\begin{enumerate}[(a)]
|
||||
\item{Montrer que l'ensemble des valeurs d’adhérence d'une suite \suite{x} s'identifie à $\Intersection\limits_{N \ge 0} \overline{X_N}$, où $X_N = \Union\limits_{n \ge N} \{ x_n \}$.}
|
||||
\item{Montrer que la propriété de Borel-Lebesgue implique que pour toute suite décroissante de fermés dont l’intersection est vide les termes de la suite
|
||||
sont vides à partir d’un certain rang. En déduire que l’intersection de (a) est non-vide, donc (1) $\implies$ (2).
|
||||
On a vu en cours que (2) $\implies$ (3b), On admettra ici que (2) $\implies$ (3a) complétant ainsi (2) $\implies$ (3). En cours, on a vu (3) $\implies$ (2).}
|
||||
\item{Pour (3) $\implies$ (4) on raisonne par l’absurde : on suppose qu'il existe un recouvrement $(U_i)_{i \in I}$ de $E$ mettant en défaut (4), autrement dit : pour $\epsilon_n = \frac{1}{2^n}$
|
||||
il existe une boule $B(x_n, \epsilon_n)$ contenue dans aucun des ouverts du recouvrement.
|
||||
La suite des centres $(x_n)$ admet alors (par (3) $\implies$ (2)) une sous-suite qui converge vers $x \in E$. Montrer qu'un ouvert du recouvrement de $E$ contenant $x$ contient forcément des boules $B(x_n, \epsilon_n)$ en contradiction avec l'hypothèse.}
|
||||
\item{Montrer (4) $\implies$ (1).}
|
||||
\end{enumerate}
|
||||
\end{exercise_sq}
|
||||
|
||||
\begin{proof}
|
||||
\lipsum[2]
|
||||
% TODO Complete proof
|
||||
\end{proof}
|
||||
|
||||
2
main.tex
2
main.tex
@ -73,6 +73,7 @@ Et de manière honteusement démagogique, je vous remercie tous lecteurs de ce c
|
||||
\input{contents/combinatorics}
|
||||
\input{contents/algebra}
|
||||
\input{contents/group_theory}
|
||||
\input{contents/group_theory_exo}
|
||||
\input{contents/ring_theory}
|
||||
\input{contents/algebra_dm1}
|
||||
\input{contents/algebra_dm2}
|
||||
@ -84,6 +85,7 @@ Et de manière honteusement démagogique, je vous remercie tous lecteurs de ce c
|
||||
\input{contents/suites}
|
||||
\input{contents/fourier}
|
||||
\input{contents/topology}
|
||||
\input{contents/topology_exo}
|
||||
\input{contents/topology_dm1}
|
||||
\input{contents/dynamic_systems}
|
||||
\input{contents/category_theory}
|
||||
|
||||
@ -40,7 +40,7 @@
|
||||
}
|
||||
@online{wimmics_website,
|
||||
author = {Wimmics},
|
||||
title = {Wimmics – Bridging social semantics and formal semantics on the web},
|
||||
title = {Wimmics - Bridging social semantics and formal semantics on the web},
|
||||
url = {https://team.inria.fr/wimmics}
|
||||
}
|
||||
@online{tyrex_website,
|
||||
@ -193,13 +193,13 @@
|
||||
url = {https://www.i3s.unice.fr}
|
||||
}
|
||||
@online{rdf2rdf_website,
|
||||
title = {RDF2RDF’s official website},
|
||||
title = {RDF2RDF's official website},
|
||||
url = {http://www.l3s.de/~minack/rdf2rdf}
|
||||
}
|
||||
@online{team_github,
|
||||
author = {Damien Graux and Pierre Saunders},
|
||||
year = {2021},
|
||||
title = {Team’s GitHub},
|
||||
title = {Team's GitHub},
|
||||
url = {https://github.com/SemanticWebBenchmarker}
|
||||
}
|
||||
@manual{unix_standard,
|
||||
@ -207,20 +207,20 @@
|
||||
url = {https://www.opengroup.org/membership/forums/platform/unix}
|
||||
}
|
||||
@online{virtuoso_website,
|
||||
title = {Virtuoso’s official website},
|
||||
title = {Virtuoso's official website},
|
||||
url = {https://virtuoso.openlinksw.com}
|
||||
}
|
||||
@online{dbpedia_website,
|
||||
title = {Dbpedia’s official website},
|
||||
title = {Dbpedia's official website},
|
||||
url = {https://www.dbpedia.org}
|
||||
}
|
||||
@online{lod-cloud_website,
|
||||
title = {The Linked Open Data Cloud’s official website},
|
||||
title = {The Linked Open Data Cloud's official website},
|
||||
url = {https://www.lod-cloud.net}
|
||||
}
|
||||
@online{fuseki_website,
|
||||
author = {Apache},
|
||||
title = {Apache Jena Fueski’s official website},
|
||||
title = {Apache Jena Fueski's official website},
|
||||
url = {https://jena.apache.org/documentation/fuseki2}
|
||||
}
|
||||
@manual{sparql_standard,
|
||||
@ -229,11 +229,11 @@
|
||||
}
|
||||
@online{tdb2_website,
|
||||
author = {Apache},
|
||||
title = {TDB2’s official website},
|
||||
title = {TDB2's official website},
|
||||
url = {https://jena.apache.org/documentation/tdb2}
|
||||
}
|
||||
@online{watdiv_website,
|
||||
title = {WatDiv’s official website},
|
||||
title = {WatDiv's official website},
|
||||
url = {https://dsg.uwaterloo.ca/watdiv}
|
||||
}
|
||||
@online{sp2bench_website,
|
||||
@ -282,7 +282,7 @@
|
||||
url = {https://github.com/SemanticWebBenchmarker/4store}
|
||||
}
|
||||
@online{wikipedia_cayley_dickson,
|
||||
title = {Cayley–Dickson construction},
|
||||
title = {Cayley-Dickson construction},
|
||||
url = {https://en.wikipedia.org/wiki/Cayley-Dickson\_construction}
|
||||
}
|
||||
@online{wikipedia_complex_number,
|
||||
|
||||
@ -7,7 +7,7 @@
|
||||
address = {Red Hook, NY, USA},
|
||||
abstract = {Generative Adversarial Networks (GANs) are powerful generative models, but suffer from training instability. The recently proposed Wasserstein GAN (WGAN) makes progress toward stable training of GANs, but sometimes can still generate only poor samples or fail to converge. We find that these problems are often due to the use of weight clipping in WGAN to enforce a Lipschitz constraint on the critic, which can lead to undesired behavior. We propose an alternative to clipping weights: penalize the norm of gradient of the critic with respect to its input. Our proposed method performs better than standard WGAN and enables stable training of a wide variety of GAN architectures with almost no hyperparameter tuning, including 101-layer ResNets and language models with continuous generators. We also achieve high quality generations on CIFAR-10 and LSUN bedrooms.},
|
||||
booktitle = {Proceedings of the 31st International Conference on Neural Information Processing Systems},
|
||||
pages = {5769–5779},
|
||||
pages = {5769-5779},
|
||||
numpages = {11},
|
||||
location = {Long Beach, California, USA},
|
||||
series = {NIPS'17},
|
||||
@ -33,7 +33,8 @@
|
||||
year = {1983},
|
||||
volume = {269},
|
||||
pages = {543-547},
|
||||
booktitle = {Doklady ANSSSR (translated as Soviet.Math.Docl.)}
|
||||
booktitle = {Doklady ANSSSR (translated as Soviet.Math.Docl.)},
|
||||
journal = {Dokl Akad Nauk SSSR}
|
||||
}
|
||||
@article{adagrad_paper,
|
||||
author = {John Duchi and Elad Hazan and Yoram Singer},
|
||||
@ -117,7 +118,7 @@ we aim to support the design and implementation of more diverse benchmarks. Appl
|
||||
developers can use our result to analyze their data and queries and choose a data
|
||||
management system.},
|
||||
booktitle = {The World Wide Web Conference},
|
||||
pages = {1623–1633},
|
||||
pages = {1623-1633},
|
||||
numpages = {11},
|
||||
location = {San Francisco, CA, USA},
|
||||
series = {WWW '19}
|
||||
@ -160,7 +161,7 @@ benchmark queries that include pattern matching and long join paths in the under
|
||||
data graphs.},
|
||||
journal = {Proc. VLDB Endow.},
|
||||
month = aug,
|
||||
pages = {647–659},
|
||||
pages = {647-659},
|
||||
numpages = {13}
|
||||
}
|
||||
@article{lubm_article,
|
||||
@ -252,7 +253,7 @@ the desired benchmark datasets. To our knowledge, this is the first methodologic
|
||||
study of RDF benchmarks, as well as the first attempt on generating RDF benchmarks
|
||||
in a principled way.},
|
||||
booktitle = {Proceedings of the 2011 ACM SIGMOD International Conference on Management of Data},
|
||||
pages = {145–156},
|
||||
pages = {145-156},
|
||||
numpages = {12},
|
||||
keywords = {RDF, benchmark},
|
||||
location = {Athens, Greece},
|
||||
@ -298,33 +299,34 @@ in a principled way.},
|
||||
bibsource = {dblp computer science bibliography, https://dblp.org}
|
||||
}
|
||||
@article{transfer_learning_survey,
|
||||
author = {Fuzhen Zhuang and
|
||||
Zhiyuan Qi and
|
||||
Keyu Duan and
|
||||
Dongbo Xi and
|
||||
Yongchun Zhu and
|
||||
Hengshu Zhu and
|
||||
Hui Xiong and
|
||||
Qing He},
|
||||
title = {A Comprehensive Survey on Transfer Learning},
|
||||
journal = {CoRR},
|
||||
volume = {abs/1911.02685},
|
||||
year = {2019},
|
||||
url = {http://arxiv.org/abs/1911.02685},
|
||||
author = {Fuzhen Zhuang and
|
||||
Zhiyuan Qi and
|
||||
Keyu Duan and
|
||||
Dongbo Xi and
|
||||
Yongchun Zhu and
|
||||
Hengshu Zhu and
|
||||
Hui Xiong and
|
||||
Qing He},
|
||||
title = {A Comprehensive Survey on Transfer Learning},
|
||||
journal = {CoRR},
|
||||
volume = {abs/1911.02685},
|
||||
year = {2019},
|
||||
url = {http://arxiv.org/abs/1911.02685},
|
||||
eprinttype = {arXiv},
|
||||
eprint = {1911.02685},
|
||||
timestamp = {Sat, 29 Aug 2020 18:19:14 +0200},
|
||||
biburl = {https://dblp.org/rec/journals/corr/abs-1911-02685.bib},
|
||||
bibsource = {dblp computer science bibliography, https://dblp.org}
|
||||
eprint = {1911.02685},
|
||||
timestamp = {Sat, 29 Aug 2020 18:19:14 +0200},
|
||||
biburl = {https://dblp.org/rec/journals/corr/abs-1911-02685.bib},
|
||||
bibsource = {dblp computer science bibliography, https://dblp.org}
|
||||
}
|
||||
@article{generative_adversarial_nets,
|
||||
doi = {10.48550/ARXIV.1406.2661},
|
||||
url = {https://arxiv.org/abs/1406.2661},
|
||||
author = {Goodfellow, Ian J. and Pouget-Abadie, Jean and Mirza, Mehdi and Xu, Bing and Warde-Farley, David and Ozair, Sherjil and Courville, Aaron and Bengio, Yoshua},
|
||||
keywords = {Machine Learning (stat.ML), Machine Learning (cs.LG), FOS: Computer and information sciences, FOS: Computer and information sciences},
|
||||
title = {Generative Adversarial Networks},
|
||||
doi = {10.48550/ARXIV.1406.2661},
|
||||
url = {https://arxiv.org/abs/1406.2661},
|
||||
author = {Goodfellow, Ian J. and Pouget-Abadie, Jean and Mirza, Mehdi and Xu, Bing and Warde-Farley, David and Ozair, Sherjil and Courville, Aaron and Bengio, Yoshua},
|
||||
keywords = {Machine Learning (stat.ML), Machine Learning (cs.LG), FOS: Computer and information sciences, FOS: Computer and information sciences},
|
||||
title = {Generative Adversarial Networks},
|
||||
publisher = {arXiv},
|
||||
year = {2014},
|
||||
journal = {arXiv},
|
||||
year = {2014},
|
||||
copyright = {arXiv.org perpetual, non-exclusive license}
|
||||
}
|
||||
@article{vae_paper,
|
||||
@ -338,22 +340,22 @@ in a principled way.},
|
||||
copyright = {arXiv.org perpetual, non-exclusive license}
|
||||
}
|
||||
@article{edit_gan_paper,
|
||||
author = {Huan Ling and
|
||||
Karsten Kreis and
|
||||
Daiqing Li and
|
||||
Seung Wook Kim and
|
||||
Antonio Torralba and
|
||||
Sanja Fidler},
|
||||
title = {EditGAN: High-Precision Semantic Image Editing},
|
||||
journal = {CoRR},
|
||||
volume = {abs/2111.03186},
|
||||
year = {2021},
|
||||
url = {https://arxiv.org/abs/2111.03186},
|
||||
author = {Huan Ling and
|
||||
Karsten Kreis and
|
||||
Daiqing Li and
|
||||
Seung Wook Kim and
|
||||
Antonio Torralba and
|
||||
Sanja Fidler},
|
||||
title = {EditGAN: High-Precision Semantic Image Editing},
|
||||
journal = {CoRR},
|
||||
volume = {abs/2111.03186},
|
||||
year = {2021},
|
||||
url = {https://arxiv.org/abs/2111.03186},
|
||||
eprinttype = {arXiv},
|
||||
eprint = {2111.03186},
|
||||
timestamp = {Wed, 10 Nov 2021 16:07:30 +0100},
|
||||
biburl = {https://dblp.org/rec/journals/corr/abs-2111-03186.bib},
|
||||
bibsource = {dblp computer science bibliography, https://dblp.org}
|
||||
eprint = {2111.03186},
|
||||
timestamp = {Wed, 10 Nov 2021 16:07:30 +0100},
|
||||
biburl = {https://dblp.org/rec/journals/corr/abs-2111-03186.bib},
|
||||
bibsource = {dblp computer science bibliography, https://dblp.org}
|
||||
}
|
||||
@misc{dall_e_2_paper,
|
||||
doi = {10.48550/ARXIV.2204.06125},
|
||||
@ -399,26 +401,26 @@ in a principled way.},
|
||||
address = {Cambridge, MA, USA},
|
||||
abstract = {The Bayesian analysis of neural networks is difficult because a simple prior over weights implies a complex prior distribution over functions. In this paper we investigate the use of Gaussian process priors over functions, which permit the predictive Bayesian analysis for fixed values of hyperparameters to be carried out exactly using matrix operations. Two methods, using optimization and averaging (via Hybrid Monte Carlo) over hyperparameters have been tested on a number of challenging problems and have produced excellent results.},
|
||||
booktitle = {Proceedings of the 8th International Conference on Neural Information Processing Systems},
|
||||
pages = {514–520},
|
||||
pages = {514-520},
|
||||
numpages = {7},
|
||||
location = {Denver, Colorado},
|
||||
series = {NIPS'95}
|
||||
}
|
||||
@article{semi-supervised_learning_with_deep_generative_models,
|
||||
author = {Diederik P. Kingma and
|
||||
Danilo Jimenez Rezende and
|
||||
Shakir Mohamed and
|
||||
Max Welling},
|
||||
title = {Semi-Supervised Learning with Deep Generative Models},
|
||||
journal = {CoRR},
|
||||
volume = {abs/1406.5298},
|
||||
year = {2014},
|
||||
url = {http://arxiv.org/abs/1406.5298},
|
||||
author = {Diederik P. Kingma and
|
||||
Danilo Jimenez Rezende and
|
||||
Shakir Mohamed and
|
||||
Max Welling},
|
||||
title = {Semi-Supervised Learning with Deep Generative Models},
|
||||
journal = {CoRR},
|
||||
volume = {abs/1406.5298},
|
||||
year = {2014},
|
||||
url = {http://arxiv.org/abs/1406.5298},
|
||||
eprinttype = {arXiv},
|
||||
eprint = {1406.5298},
|
||||
timestamp = {Mon, 13 Aug 2018 16:47:38 +0200},
|
||||
biburl = {https://dblp.org/rec/journals/corr/KingmaRMW14.bib},
|
||||
bibsource = {dblp computer science bibliography, https://dblp.org}
|
||||
eprint = {1406.5298},
|
||||
timestamp = {Mon, 13 Aug 2018 16:47:38 +0200},
|
||||
biburl = {https://dblp.org/rec/journals/corr/KingmaRMW14.bib},
|
||||
bibsource = {dblp computer science bibliography, https://dblp.org}
|
||||
}
|
||||
@article{every_model_learned_by_gradient_descent_is_approximately_a_kernel_machine,
|
||||
author = {Pedro Domingos},
|
||||
@ -467,5 +469,6 @@ in a principled way.},
|
||||
author = {Edward Jewitt Wheeler},
|
||||
page = {564},
|
||||
volumes = {49},
|
||||
year = {1910}
|
||||
year = {1910},
|
||||
publisher = {New York : Current Literature Pub. Co.}
|
||||
}
|
||||
|
||||
Reference in New Issue
Block a user