tag:blogger.com,1999:blog-63001658778292633842024-03-05T21:11:05.429+01:00La math-à-outilsCe blog relate des coups de pouce et des aides issues de mon enseignement des mathématiques dans le supérieurSylvain Goléniahttp://www.blogger.com/profile/05638350035551044588noreply@blogger.comBlogger144125tag:blogger.com,1999:blog-6300165877829263384.post-1714203800128070102021-09-26T17:35:00.000+02:002021-09-26T17:35:11.516+02:00Remboursement d'un crédit<div dir="ltr" style="text-align: left;" trbidi="on">
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<p> On considère le cas d'un crédit de 5 000 euros avec un taux annuel de 2,4% et une mensualité de 100 euros. Le taux mensuel est donc de 0,2%. On pose $u_n$ le montant à rembourser au mois $n$. On a alors</p><p>$$u_{n+1}=\frac{1002}{1000}u_n-100,$$</p><p>avec $u_0=5000$. C'est une suite arithmético-géométrique. On commence par trouver le point fixe.</p><div class="separator" style="clear: both; text-align: center;"><div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjDuwfTYx53-S8IufN77I7NL7LlDDdA9wNPgr5EZPtJurQ0n1Qi-GvA1dCvsRQogfIGb1KwFNO9HL_2YPY1qcoG0J8QHBfzmfk7ug4syqJcS225jBF7GezhNgpKtM_mvFmkSs9UpHU-dfs/s2048/1632665129684.jpg" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="2048" data-original-width="1536" height="640" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjDuwfTYx53-S8IufN77I7NL7LlDDdA9wNPgr5EZPtJurQ0n1Qi-GvA1dCvsRQogfIGb1KwFNO9HL_2YPY1qcoG0J8QHBfzmfk7ug4syqJcS225jBF7GezhNgpKtM_mvFmkSs9UpHU-dfs/w480-h640/1632665129684.jpg" width="480" /></a></div><div class="separator" style="clear: both; text-align: center;"><br /></div><hr color="black" width="75%" />
<br /><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgpfBT-ZBQA4_MRCVaO4wbplhIwAiq3Y_zK9M_mBwaCutTd1bQzNcv0NL_SK9uzTUx8Cx8DF8kbEJaWlnY0B1QsWtmbTGiNkx0KZzcqrCQhQxy_Z6NcCKQtITunpTBRk3-Qi5GFKcxk1lU/s2048/1632665129673.jpg" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="2048" data-original-width="1536" height="640" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgpfBT-ZBQA4_MRCVaO4wbplhIwAiq3Y_zK9M_mBwaCutTd1bQzNcv0NL_SK9uzTUx8Cx8DF8kbEJaWlnY0B1QsWtmbTGiNkx0KZzcqrCQhQxy_Z6NcCKQtITunpTBRk3-Qi5GFKcxk1lU/w480-h640/1632665129673.jpg" width="480" /></a></div><div class="separator" style="clear: both; text-align: center;"><br /></div><div class="separator" style="clear: both; text-align: center;">On obtient donc l'expression de $(u_n)$. On remarque que sa limite est $-\infty$, cela veut dire qu'il existe un $n$ à partir duquel la suite sera négative. Le crédit est remboursable. </div><div class="separator" style="clear: both; text-align: center;"><br /></div><hr color="black" width="75%" />
On cherche à calculer le nombre de mois pour rembourser.</div><div dir="ltr" style="text-align: left;" trbidi="on"><br /><div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEilgpnH6eXNx-tSDa409_yTo67prL83Ax_kQkBUqHg_WNr_i9gi8VOj3jrT49i67P93t9NQPFZ1xEvsWINnJZ25QTcSBKQmA6wzInP7lugKWhnmA9Koh7CRdRxLJkx1zhOoMq7niv9T78s/s2048/1632665129662.jpg" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="2048" data-original-width="1536" height="640" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEilgpnH6eXNx-tSDa409_yTo67prL83Ax_kQkBUqHg_WNr_i9gi8VOj3jrT49i67P93t9NQPFZ1xEvsWINnJZ25QTcSBKQmA6wzInP7lugKWhnmA9Koh7CRdRxLJkx1zhOoMq7niv9T78s/w480-h640/1632665129662.jpg" width="480" /></a></div><div class="separator" style="clear: both; text-align: center;"><br /></div><hr color="black" width="75%" />
<br />On calcule enfin ce que le crédit nous a vraiment coûté. <br /><br /><div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhaB10gKuF2bYrM8dTdJoXhfMRiLrnk8tv1Q5pwFTYKa4tKfbtc9522I86cB6F_Rf4uNNOGU31-Xs70FOORCHzqDSa3sBtuisGR9IH28AyPbkE7fHxQak1iq0ph0Wuhgua8eJ7Hsnaje_4/s2048/1632665129651.jpg" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="2048" data-original-width="1536" height="640" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhaB10gKuF2bYrM8dTdJoXhfMRiLrnk8tv1Q5pwFTYKa4tKfbtc9522I86cB6F_Rf4uNNOGU31-Xs70FOORCHzqDSa3sBtuisGR9IH28AyPbkE7fHxQak1iq0ph0Wuhgua8eJ7Hsnaje_4/w480-h640/1632665129651.jpg" width="480" /></a></div><br /><br /><p></p></div>Sylvain Goléniahttp://www.blogger.com/profile/05638350035551044588noreply@blogger.com0tag:blogger.com,1999:blog-6300165877829263384.post-89579090870157421252021-09-26T17:21:00.001+02:002021-09-26T17:21:35.481+02:00Étude d'une suite arithmético-géométrique<div dir="ltr" style="text-align: left;" trbidi="on">
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<p>On étudie la suite $$u_{n+1}= \frac{2}{10} u_n +8$$ avec $u_0=2$. On commence par trouver le point fixe de la suite. Puis, dans le 1) on construit une suite annexe $(v_n)$ qui est géométrique.</p><div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEg2w5h1JG_1yjGwR9HiPWm8mBMgi9Gx78Kwn83pl555654Rq8XKjY33CvZ2Ub7Os2x7N2nRTgYKNMUsSEUytUGhz-4gIhZkHRmKWrz7kUz4lUXf05KMGTi2KjZNyjqilfpM5EgOq4ZYZeA/s2048/1632665129727.jpg" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="2048" data-original-width="1536" height="640" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEg2w5h1JG_1yjGwR9HiPWm8mBMgi9Gx78Kwn83pl555654Rq8XKjY33CvZ2Ub7Os2x7N2nRTgYKNMUsSEUytUGhz-4gIhZkHRmKWrz7kUz4lUXf05KMGTi2KjZNyjqilfpM5EgOq4ZYZeA/w480-h640/1632665129727.jpg" width="480" /></a></div><div class="separator" style="clear: both; text-align: center;"><br /></div><div class="separator" style="clear: both; text-align: center;"><hr color="black" style="text-align: left;" width="75%" /></div><br /><div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEh9O80FzN6Z73UYf-uJSSL9zwZQb1qVuowt7_w_supehHd9f4afWUWV1ndxFhW0MT3IAh3By0wbtewGW20oOshOaVDMvsiJbdeOfwU7GQa1z_X1AZPhDSvyDNbwp7cHcog54xwcs9tJOZs/s2048/1632665129716.jpg" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="2048" data-original-width="1536" height="640" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEh9O80FzN6Z73UYf-uJSSL9zwZQb1qVuowt7_w_supehHd9f4afWUWV1ndxFhW0MT3IAh3By0wbtewGW20oOshOaVDMvsiJbdeOfwU7GQa1z_X1AZPhDSvyDNbwp7cHcog54xwcs9tJOZs/w480-h640/1632665129716.jpg" width="480" /></a></div><div class="separator" style="clear: both; text-align: center;"><br /></div><hr color="black" width="75%" />
<br />Comme on connait tout sur les suites géométiques, on a l'expression explicite de $v_n$ pour tout $n$. On peut alors déduire l'expression de $(u_n)$.<br /><br /><div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjFWPcMMidvCpnrCz2n-Xvb82JxXW16q6qsgV7oGi46ji_WA7E5BIgNCxIPK9mhYfI_3zfWL_6VU48JHsh_dPaTfQkMiqED9JkxYd09S7ZJgaHLsP3F8JlD85i4XkWz5CY-6Jf7fQSJTuQ/s2048/1632665129705.jpg" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="2048" data-original-width="1536" height="640" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjFWPcMMidvCpnrCz2n-Xvb82JxXW16q6qsgV7oGi46ji_WA7E5BIgNCxIPK9mhYfI_3zfWL_6VU48JHsh_dPaTfQkMiqED9JkxYd09S7ZJgaHLsP3F8JlD85i4XkWz5CY-6Jf7fQSJTuQ/w480-h640/1632665129705.jpg" width="480" /></a></div><div class="separator" style="clear: both; text-align: center;"><br /></div><hr color="black" width="75%" />
<br />Une question moins fréquente est celle du calcul de la somme de $u_n$. La formule générale n'est pas à connaître mais il faut bien retenir la méthode.<br /><br /><div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhnyucfiYKtAfTyQhiuCUclCVJcSERVaekaqIManJGqpAlktaqdtqqGHSQaJtdJ0qXCDGRb0BKuNjAuNv4tAv3S9FgG9tG8HHcf_ZLYSfWRQ-7EvJtA8mjVc8ezAJR8zbVV_cIqg4N7LTQ/s2048/1632665129694.jpg" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="2048" data-original-width="1536" height="640" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhnyucfiYKtAfTyQhiuCUclCVJcSERVaekaqIManJGqpAlktaqdtqqGHSQaJtdJ0qXCDGRb0BKuNjAuNv4tAv3S9FgG9tG8HHcf_ZLYSfWRQ-7EvJtA8mjVc8ezAJR8zbVV_cIqg4N7LTQ/w480-h640/1632665129694.jpg" width="480" /></a></div><br /><p></p></div>Sylvain Goléniahttp://www.blogger.com/profile/05638350035551044588noreply@blogger.com0tag:blogger.com,1999:blog-6300165877829263384.post-69251404125696301052021-09-26T17:04:00.006+02:002021-09-26T17:04:53.824+02:00Application des suites géométriques - flocon de Kock et triangle de Siepinski<p>On calcule la longueur du flocon de Koch à l'aide suite géométrique. On peut se référer à <a href="https://fr.wikipedia.org/wiki/Flocon_de_Koch">wikipédia</a> pour plus de contexte et aller plus loin. </p><div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjgiUT_U2tSa5_3mu6EKwbEtgjoXA9b7tImsAxevrrhnl_3nkaLhP3c5EvsY_92pIGyOsCA5Yl4oRgIaajFXVqfGpJnxfrrwy2bh-VEz3hma1CP4xkwVY3XLnZAAoMdQ2-UCuCN9sgu5gQ/s2048/1632665129773.jpg" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="2048" data-original-width="1536" height="640" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjgiUT_U2tSa5_3mu6EKwbEtgjoXA9b7tImsAxevrrhnl_3nkaLhP3c5EvsY_92pIGyOsCA5Yl4oRgIaajFXVqfGpJnxfrrwy2bh-VEz3hma1CP4xkwVY3XLnZAAoMdQ2-UCuCN9sgu5gQ/w480-h640/1632665129773.jpg" width="480" /></a></div><div class="separator" style="clear: both; text-align: center;"><br /></div>
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<div class="separator" style="clear: both; text-align: center;"><br /></div><div class="separator" style="clear: both; text-align: center;">Pour le triangle Sierpinski, on pourra lire le <a href="https://fr.wikipedia.org/wiki/Triangle_de_Sierpiński">wikipédia</a>.</div><div class="separator" style="clear: both; text-align: center;"><br /></div><div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiw5l0LgD_EQkbDgRrUDl_Y1dMsuyZTwsoER9NK9XtXm4g4OWhgPkxB2OgHjLUic30H9kCYJjcWHbj-Q0hfIbi9SxRVko9wJWGoLo905YVenqvd295EdqYyuNGZpnUeoBPWl0vGV9IbJQI/s2048/1632665129738.jpg" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="2048" data-original-width="1536" height="640" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiw5l0LgD_EQkbDgRrUDl_Y1dMsuyZTwsoER9NK9XtXm4g4OWhgPkxB2OgHjLUic30H9kCYJjcWHbj-Q0hfIbi9SxRVko9wJWGoLo905YVenqvd295EdqYyuNGZpnUeoBPWl0vGV9IbJQI/w480-h640/1632665129738.jpg" width="480" /></a></div><br /><p></p>Sylvain Goléniahttp://www.blogger.com/profile/05638350035551044588noreply@blogger.com0tag:blogger.com,1999:blog-6300165877829263384.post-62066357539456305862019-12-17T11:10:00.001+01:002019-12-17T11:10:54.914+01:00Non surjectivité de la transformé de Fourier dans L1<div dir="ltr" style="text-align: left;" trbidi="on">
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Pour montrer ce résultat on propose une solution "à la main". Elle ne repose pas sur l'utilisation de le principe de la borne uniforme qui n'est plus au programme de l'agrégation. Cet exercice est tiré du site <a href="http://www.bibmath.net/ressources/index.php?action=affiche&quoi=bde/analyse/integration/integralesimpropres&type=fexo">bibmath</a>.<br />
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La convention prise pour la transformé de Fourier ici est $F(f)(x) = \frac{1}{\sqrt{2\pi}}\int_{\mathbb{R}} f(t) e^{-ixt}dt$.<br />
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Le 1) est clair. Pour le 2), c'est classique. On commence par montrer l'existence de l'intégral. Attention la fonction n'est pas $L^1$, on a affaire à un intégrale impropre. </div>
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Pour montrer la bornitude et la continuité on procède par exemple en découpant l'intégrale en deux et en reportant la difficulté sur un cas plus simple, celui d'une intégrale d'une fonction continue sur une compact. On conclue grâce au théorème fondamentale de l'analyse (version Riemann en fait).</div>
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Une alternative pour la continuité est d'utiliser la convergence dominée. </div>
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Pour la question 3) on fait un Fubini. Pour cela on utilise Tonelli (cadre du bas) pour montrer que la fonction (de deux variables) est bien dans $L^1$. </div>
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4) On pousse maintenant $R\to \infty$. Le membre de gauche est une intégrale impropre et en générale pas une intégrale de Lebesgue !</div>
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On justifie alors la convergence dominée en faisant attention de ne pas faire apparaître le $R$. </div>
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5) On peut enfin conclure. On suppose qu'il existe $g\in C_0(\mathbb{R})$ qui soit la transformée de Fourier de $f$. La fonction $g$ est nécessairement impaire car $f$ l'est. On choisit $g(x)= 1/\ln(x)$ pour $x\geq 2$ et on la prolonge en une fonction impaire et continue (sur $[-2,2]$ il suffit de compléter par une droite). Comme $\int_1^\infty \frac{g(t)}{t}dt$ n'est pas convergente, on en déduit qu'il n'existe pas de $f\in L^1$ impaire telle que $\hat f=g$. La transformée de Fourier n'est donc pas surjective de $L^1$ dans $C_0$. </div>
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Sylvain Goléniahttp://www.blogger.com/profile/05638350035551044588noreply@blogger.com1tag:blogger.com,1999:blog-6300165877829263384.post-44709149897942402262019-12-17T10:51:00.000+01:002019-12-17T10:51:36.784+01:00Equation différentielle linéaire d'ordre 1 - partie 1<div dir="ltr" style="text-align: left;" trbidi="on">
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Tout d'abord il faut faire attention lors du travail de mémorisation de ne pas tromper de signe. Suivant les auteurs on se ramène à $y'+a(x)y=0$ ou à $y'=b(x)y$. Cela donne comme solution $y= c e^{-\int a(x)dx}$ et $y= c e^{\int b(x)dx}$ avec $c\in\mathbb{R}$.</div>
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Ensuite on voit que si on choisit une primitive différente, cela change la valeur du $c$. Cela n'a donc pas d'impact sur la solution de l'équation différentielle.</div>
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On commence facile. Ici on a une condition initiale. Elle permet de déterminer la constante de façon unique.</div>
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On trace $x\mapsto ce^{-x}$ et l'on fait varier la constante de -5 à 5. On voit ici que le choix d'un $c$ détermine de façon unique la valeur de la fonction en $0$ et vice versa. Plus généralement le théorème de Cauchy-Lipschitz montre que cette correspondance est une bijection. De plus, les courbes $x\mapsto ce^{-x}$ et $x\mapsto c'e^{-x}$ se coupent si et seulement si $c=c'$.</div>
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Ici pour se ramener à la forme du cours, on élimine le $2$ qui est devant le $y'$.</div>
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On passe maintenant à une équation avec second membre que l'on note $(E)$. </div>
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On commence par l'étude de l'équation sans second membre $(ESSM)$ que l'on appelle aussi équation homogène $(EH)$. On la résout. </div>
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On cherche ensuite une solution particulière de $(E)$. Ici la fonction constante $3$ est solution évidente. </div>
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La solution générale de $(E)$ est donnée par la solution particulière de $(E)$ + la solution générale de $(ESSM)$.</div>
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On peut se demander comment on aurait pu trouver le $3$. Une règle (de type cuisine) est que si le second membre est un polynôme alors on cherche un polynôme comme solution particulière. On teste alors les polynômes de degré $0$, puis de degré $1$ si cela ne marche pas, puis $2$...</div>
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On met cela en pratique. Ici la solution particulière est une constante encore.</div>
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On passe à autre exemple. On divise par $x$ car il est devant le $y'$. Attention en faisant cela on se "condamne" à résoudre pour $x>0$ ou $x<0$. La solution de l'équation homogène est $y_H(x)= c|x|$. Ici il faut fair très attention. Il n'y a pas un seul $c$ mais deux : $c_+$ et $c_-$, un pour $x<0$ et un pour $x>0$. Par exemple, avec un certain choix on voit que $x\mapsto x$ est solution sur $\mathbb{R}^*$ et que $x\mapsto |x|$ aussi. </div>
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Pour rendre le choix du $c$ unique, on faire un recollement et chercher une solution de classe $C^1$ sur $\mathbb{R}$. Cela est possible ici si $c_+= -c_-$. En d'autres termes $x\mapsto cx$, avec $c\in \mathbb{R}$ est l'ensemble des solutions de classe $C^1$ de $(E)$ dans $\mathbb{R}$. </div>
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On passe ensuite à la solution particulière. On voit rapidement que l'on n'a pas de solution avec des polynôme de degré inférieur à $2$. On essaye un de degré $3$. En procédant pas identification on obtient $y_p(x)= \frac{1}{2} x^3$. </div>
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La solution générale de $(E)$ est de nouveau de la forme solution particulière de $(E)$ + solution générale de l'équation sans second membre.</div>
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Sylvain Goléniahttp://www.blogger.com/profile/05638350035551044588noreply@blogger.com0tag:blogger.com,1999:blog-6300165877829263384.post-30969192223800788052019-12-17T09:47:00.002+01:002019-12-17T09:47:52.169+01:00Quelques décompositions en éléments simples<div dir="ltr" style="text-align: left;" trbidi="on">
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On procède à changement de variable affine.</div>
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Pour traiter les quotients, quand le dénominateur est de degré 2 et que le Delta est strictement négatif, on procède comme suit :</div>
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1) Si le degré numérateur soit de degré supérieur à 2, on fait une division euclidienne.</div>
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2) On fait du $u'/u$. Le but est de faire disparaître le terme en $t$. Cela donnera un terme en $\ln$.</div>
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3) On utilise la forme canonique du dénominateur et on fait un changement affine pour se ramener à du $1/(1+x^2)$ qui donnera de l'$\arctan$.</div>
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Ici le dénominateur a un discriminant strictement positif. On calcule les racines et on invoque la décomposition en éléments simples. Pour calculer les coefficients, on peut procéder par identification et résoudre un système. </div>
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On intègre tranquillement. Attention une primitive de $1/(x-a)$ est $\ln(|x-a|)$. Il ne faut pas oublier la valeur absolue !! Ici, on a supposé que $x\geq 0$. </div>
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Dans le c), on demande de calculer la limite de $F$ en $+\infty$. Dans le $d)$, on propose de calculer cette intégrale. On pose le changement de variable et on se ramène au cas précédent.</div>
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On finit le calcul.</div>
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Pour cette décomposition, on a $(x-1)^2$ au dénominateur. La forme de la décomposition en éléments simples n'est pas la même que pour l'exercice précédent. Il y a un terme supplémentaire. </div>
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Ensuite on propose de calculer les constantes grâce à des astuces et non en procédant par identification. </div>
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On continue pour les autres constantes.</div>
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On intègre !</div>
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Sylvain Goléniahttp://www.blogger.com/profile/05638350035551044588noreply@blogger.com0tag:blogger.com,1999:blog-6300165877829263384.post-35260906553063541362019-12-16T18:19:00.000+01:002019-12-16T18:19:40.941+01:00Quelques exercices d'intégration - changement de variables<div dir="ltr" style="text-align: left;" trbidi="on">
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On rappelle tout d'abord le théorème de changement de variable pour une intégrale. </div>
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On commence avec un exemple élémentaire. Il y a 3 étapes à respecter. </div>
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1) la nouvelle variable, </div>
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2) le calcul de nouvel élément différentiel </div>
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3) Les nouvelles bornes de l'intégrale</div>
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Si l'on a besoin d'avoir une primitive alors il ne faut pas oublier de revenir dans la variable initiale (voir le terme entre crochet dans la dernière ligne)</div>
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Ici on utilise le fait que $\sin:[-\pi/2, \pi/2]\to [-1,1]$ est une bijection qui est de classe $C^1$. Une fois le changement de variable réalisé, on tombe sur l'intégrale de $\cos^2(u)$. Pour l'intégrer on peut ou bien faire un peu trigonométrie ou bien faire une linéarisation (carré de gauche).</div>
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Cette primitive nous permet de calculer l'aire d'un demi-cercle. </div>
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On recommence cet exercice en se basant sur le fait que $\cos:[0, \pi] \to [-1,1]$ est une bijection de classe $C^1$. Encore une fois on refait une linéarisation.</div>
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On finit le calcul.</div>
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On passe à un autre type d'exercice. On commence par une forme canonique.</div>
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On fait ensuite apparaître du $1/(1+t^2)$ qui va nous donne de l'arctangente. Puis nous effectuons le changement de variable. </div>
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Sylvain Goléniahttp://www.blogger.com/profile/05638350035551044588noreply@blogger.com0tag:blogger.com,1999:blog-6300165877829263384.post-74225103136970607092019-12-16T18:02:00.002+01:002019-12-16T18:03:43.781+01:00Intégrales généralisés partie 1<div dir="ltr" style="text-align: left;" trbidi="on">
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Ici on a que $\ln$ n'est pas continue en $0$. On regarde l'existence de la limite $\int_\epsilon^1 \ln(x) dx$ quand $\epsilon\to 0^+$. Quand <b>elle existe et qu'elle est finie</b> on dit que $\int_0^1 \ln(x) dx$ converge, sinon on dit l'intégrale diverge.</div>
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L'intégrale est dite "généralisée" ou "impropre".</div>
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Il est rare de pouvoir calculer les intégrales explicitement. On utilise alors volontiers des théorèmes de comparaison.</div>
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Attention, ici on utilise de façon cruciale l'hypothèse de positivité des fonctions.</div>
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On essaye une première fois le critère de comparaison. Le premier point est qu'on peut toujours séparer le comportement sur un compact du comportement en l'infini. </div>
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L'idée ici est d'améliorer l'efficacité du critère. Pour cela on va comparer avec le critère de Riemann. Encore une fois, on peut "oublier" ce qui se passe sur un compact. </div>
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On combine le critère de Riemann avec le critère de comparaison.</div>
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On donne un autre résultat important. C'est le critère d'absolue convergence. Sa démonstration repose sur la complétude de $\mathbb{R}$. </div>
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On donne une application directe ici. </div>
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Sylvain Goléniahttp://www.blogger.com/profile/05638350035551044588noreply@blogger.com0tag:blogger.com,1999:blog-6300165877829263384.post-44800565334438744962019-12-16T11:35:00.000+01:002019-12-16T11:36:25.300+01:00Exercices types sur arccos, arcsin et arctan<div dir="ltr" style="text-align: left;" trbidi="on">
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Voici quelques exercices types pour les fonctions réciproques trigonométriques</div>
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Quand on veut démontrer qu'une fonction (ici le membre de gauche) est constante, il suffit de montrer que sa dérivée est nulle. C'est ce qu'on fait. Pour trouver la valeur de la dérivée, on évalue en un point.</div>
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Cet exercice est l'analogue direct en version $\arctan$.</div>
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Pour trouver la constante, on peut par exemple évaluer en $\pi/4$ ou bien regarder la limite quand $x\to \infty$. </div>
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On voit que les deux compositions n'ont pas le même domaine de définition ! L'une n'est définie que sur $[-1,1]$ alors que l'autre l'est sur $\mathbb{R}$. On travaille sur des intervalles de longueur $\pi$ pour trouver les bonnes formulations.</div>
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C'est une intégrale classique. On passe en forme canonique.</div>
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Il y a juste un changement de variable affine à faire et on retombe sur du $\arcsin$. On aurait pu choisir du $\arccos$ pour intégrer aussi.</div>
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Enfin on remarque que quand on change le signe sous la racine, on sort du contexte des fonctions trigonométriques inverses. On passe en hyperbolique.</div>
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Sylvain Goléniahttp://www.blogger.com/profile/05638350035551044588noreply@blogger.com0tag:blogger.com,1999:blog-6300165877829263384.post-65929253008728307492019-12-15T15:47:00.002+01:002019-12-16T11:16:07.391+01:00Le paradoxe de l'hôtel infini par Jeff Dekofsky<div dir="ltr" style="text-align: left;" trbidi="on">
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Cette vidéo traite de la notion de dénombrabilité. Les sous-titres en français sont disponibles. </div>
Sylvain Goléniahttp://www.blogger.com/profile/05638350035551044588noreply@blogger.com0tag:blogger.com,1999:blog-6300165877829263384.post-38033376855433451702019-12-13T20:45:00.002+01:002019-12-16T11:36:04.734+01:00Les fonctions réciproques : argch et argsh<div dir="ltr" style="text-align: left;" trbidi="on">
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Les propriétés générales des fonctions $\cosh$ et $\sinh$ sont laissées en exercice. </div>
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Ici on montre que $\cosh$ est une bijection de $[0, \infty[$ dans $[1, \infty[$. Il faut bien faire attention au fait que le point $0$ suive de la continuité de $\cosh$, attention on a $\cosh'(0)=0$.</div>
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On peut maintenant définir la fonction réciproque. La continuité est automatique. Pour la dérivabilité, on fait bien attention que la dérivée de la fonction réciproque existe quand on a pas une tangente plate pour la fonction. Géométriquement, on voit que cela implique d'avoir une tangente verticale et donc exclus le point du domaine de définition de la dérivée. </div>
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Maintenant qu'on a prouvé que la fonction est dérivable, on passe au calcul de sa dérivée. </div>
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Sylvain Goléniahttp://www.blogger.com/profile/05638350035551044588noreply@blogger.com0tag:blogger.com,1999:blog-6300165877829263384.post-69433622012349958322019-12-13T20:34:00.002+01:002019-12-16T11:15:10.289+01:00Décomposition en éléments simples - partie 2<div dir="ltr" style="text-align: left;" trbidi="on">
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La première étape est de faire baisser le degré du polynôme du dénominateur.</div>
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Ici une erreur est faite. En effet l'étape $u'/u$ n'est primordiale que dans le cas où le discriminant du dénominateur est négatif ou nul. Il aurait fallut directement chercher $a$ et $b$ tel que le quotient soit égal à </div>
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\[\frac{a}{x- \frac{-1-\sqrt{5}}{2}}+ \frac{b}{x- \frac{-1+\sqrt{5}}{2}}\]</div>
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On va maintenant expliquer une méthode pour trouver $a$ et $b$. La méthode la plus simple est de mettre tout sur le même dénominateur et d'identifier les termes avec le membres de gauche. Nous présenterons une autre astuce.</div>
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Ici cette astuce marche pour tous les termes de type $\frac{1}{x-\lambda}$ quand il n'y a pas de $\frac{1}{(x-\lambda)^2}$.</div>
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On recommence. Cette fois on voit que le $\Delta$ de dénominateur est strictement négatif.</div>
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L'étape $u'/u$ est primordiale.</div>
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Comme $\Delta<0$, on vise d'obtenir un dénominateur sous le forme $X^2+1$. En l'intégrant nous aurons alors de l'arctangente. </div>
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On se focalise maintenant sur un autre terme apparaissant des la décomposition en éléments simples. </div>
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On rappelle au passage la méthode de linéarisation. </div>
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La conclusion est simple.</div>
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Sylvain Goléniahttp://www.blogger.com/profile/05638350035551044588noreply@blogger.com0tag:blogger.com,1999:blog-6300165877829263384.post-68346444125044336772019-12-11T17:54:00.000+01:002019-12-16T11:15:10.279+01:00Décomposition en éléments simples - partie 1<div dir="ltr" style="text-align: left;" trbidi="on">
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On se propose de revoir la décomposition en éléments simples. Un rappel concis de cours et des exemples corrigés se trouve <a href="http://maths.cnam.fr/IMG/pdf/Ana.4_integration_fract_rat.pdf">ici</a>. On commence par les cas les plus simples.<br />
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Ici on utilise le symbole intégrale pour vouloir dire "primitive". Il faut manier cet abus avec parcimonie. En effet, les constantes d'intégration dépendent de l'intervalle maximale de continuité. Pour $1/x$, la fonction n'est pas continue en $0$. On peut intégrer à droite, et avoir une constante, et intégrer à gauche, et avoir une autre constante. Ici $0$ agit comme un mur. Les deux constantes ne se voient pas et sont indépendantes. </div>
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On présente ici une technique possible pour calculer les coefficients recherchés dans la décomposition. </div>
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Attention ! Quand on intègre un quotient de type $1/(x-a)$, il ne faut pas oublier la valeur absolue dans le $\ln(|x-a|)$.</div>
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Sylvain Goléniahttp://www.blogger.com/profile/05638350035551044588noreply@blogger.com0tag:blogger.com,1999:blog-6300165877829263384.post-44683846236408379452019-12-11T08:21:00.000+01:002019-12-11T17:43:06.105+01:00Colle de maths - les diplômés du dernier rang<div dir="ltr" style="text-align: left;" trbidi="on">
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Extrait du film des diplômés du dernier rang de <a class="blue-link" href="http://www.allocine.fr/personne/fichepersonne_gen_cpersonne=3342.html" style="box-sizing: border-box; color: #005ea8; cursor: pointer; font-family: Arial, sans-serif; font-size: 13px; outline: none; text-decoration: none;" title="Christian Gion">Christian Gion</a>. Michel Galabru y incarne un professeur de mathématiques.</div>
Sylvain Goléniahttp://www.blogger.com/profile/05638350035551044588noreply@blogger.com0tag:blogger.com,1999:blog-6300165877829263384.post-267706841172087632019-12-09T08:41:00.001+01:002019-12-09T08:41:47.631+01:00Pourquoi le zéro n'a pas toujours compté dans l'Histoire ?<div dir="ltr" style="text-align: left;" trbidi="on">
Non "2 x 5" était loin d'être égal à 10 durant les grandes époques reculées. Ce n'est que très tardivement que le zéro est devenu la quantité nulle que nous lui prêtons aujourd'hui. Alors qu'il est difficile d’imaginer les mathématiques sans le zéro, celui-ci a d'abord commencé par être exclu philosophiquement. [...]<br /><br />La suite sur <a href="https://www.franceinter.fr/culture/mathematiques-pourquoi-le-zero-n-a-pas-toujours-compte-dans-l-histoire?fbclid=IwAR3lL-UV7kY0rcRixFFV8U0shFf9EcCDT-hhiQiYLBZDysynBPIEfcL5DxI">franceinter.fr</a>. </div>
Sylvain Goléniahttp://www.blogger.com/profile/05638350035551044588noreply@blogger.com0tag:blogger.com,1999:blog-6300165877829263384.post-42369582406904834022019-12-02T11:33:00.000+01:002019-12-02T11:33:05.793+01:00Quelques limites<div dir="ltr" style="text-align: left;" trbidi="on">
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On commence facile. La règle de survie est de factoriser par le plus grand terme. Ici c'est $x$.</div>
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Ici on repère le plus grand terme qui se cache cette fois à l'intérieur du $\ln$. C'est une nouvelle fois $x$. </div>
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Même règle de survie. On factorise en haut et en bas par les plus grands termes. Attention pour justifier que $\ln(x)/x$ tend vers $0$ en $+\infty$, on invoque les croissances comparées (C.C. sur le tableau).</div>
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Attention ici on a une forme indéterminée du type $1^\infty$. La difficulté est de repérer que l'on a affaire à une forme inderterminée de type quotient différentielle : $\frac{f(x)-f(a)}{x-a}$ tend vers $f'(a)$ quand $x\to a$ et $f$ est dérivable en $a$. </div>
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Pour celui-là, pour être plus à l'aise (pas une obligation), on se place en $+\infty$ grâce à un changement de variable. On utilise une nouvelle fois la croissance comparée. </div>
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La difficulté ici est que le cours sur les croissances comparées est formulée avec du $X^\alpha e^X$. Il faut donc faire sortir le $X= 1+\sqrt{x}$ et trouver le bon $\alpha$. </div>
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On fait tout d'abord un essai non concluant (avec $\alpha=1$), il nous ramène à une forme indéterminée. </div>
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On recommence avec $\alpha=200$. L'idée est que $X^{200}$ est du même ordre que $x^{100}$ à l'infini. </div>
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c'est la justification de la limite du dessus.</div>
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Pour finir on a de nouveau la même difficulté, il faut faire apparaître du $X^\alpha\ln(X)$. </div>
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On se place en $+\infty$ par commodité. Ici $\alpha =1$ convient. </div>
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Sylvain Goléniahttp://www.blogger.com/profile/05638350035551044588noreply@blogger.com0tag:blogger.com,1999:blog-6300165877829263384.post-1563180058586584282019-11-29T19:29:00.000+01:002019-11-29T19:29:28.188+01:00Intégration de la partie entière<script type="text/javascript" src="http://cdn.mathjax.org/mathjax/latest/MathJax.js">
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On se propose d'intégrer la partie entière.<br />
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Elle est continue par morceaux, donc Riemann intégrable. On calcule les premiers termes.</div>
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On donne la formule pour $x\in [k,k+1]$ avec $k\geq 1$. On laisse le cas, $k\leq 0$ en exercice.</div>
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Sylvain Goléniahttp://www.blogger.com/profile/05638350035551044588noreply@blogger.com0tag:blogger.com,1999:blog-6300165877829263384.post-50642617461085815242019-11-29T19:25:00.000+01:002019-11-29T19:25:56.172+01:00Fonction inverse et intégration<div dir="ltr" style="text-align: left;" trbidi="on">
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On justifie tout d'abord que l'on peut bien inverser la fonction $\cos$ sur $[0,\pi]$ et que la dérivé de l'inverse existe sur $]-1,1[$.</div>
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On calcule sa dérivée. </div>
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On peut alors calculer notre intégrale à l'aide d'un petit changement de variable. </div>
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On aurait pu choisir d'utiliser $\arcsin'(x) = 1/\sqrt{1-x^2}$. </div>
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Sylvain Goléniahttp://www.blogger.com/profile/05638350035551044588noreply@blogger.com0tag:blogger.com,1999:blog-6300165877829263384.post-23177611260012892762019-11-29T19:20:00.001+01:002019-11-29T19:27:10.544+01:00Quelques intégrales<div dir="ltr" style="text-align: left;" trbidi="on">
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On se propose de calculer quelques intégrales.<br />
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Premier exercice type d'intégration par partie. L'idée générale est de dériver le terme dont on veut se débarrasser (ici le $ln$). </div>
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Ici même stratégie, on dérive l'arctangente car on ne "sait" pas l'intégrer.</div>
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Idem ici, on dérive le $\ln$.</div>
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Cet exercice est un grand classique de l'intégration par partie. On propose une preuve alternative qui utilise les complexes.</div>
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On passe maintenant au changement de variable (ici on sait pas trop quoi dériver facilement donc l'ipp est à proscrire). Comme on "aime" pas le $\ln$ on le fait disparaitre grâce à un changement de variables.</div>
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ici c'est direct. </div>
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Ici l'$\exp$ nous dérange au dénominateur. On utilise donc un changement de variable pour le retirer. On se retrouve à étudier une fraction rationnelle et on tombe dans la théorie des décompositions en éléments simples. </div>
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Attention pour intégrer les termes de type $u'/u$. Une primitive est $\ln(|u|)$. Il ne faut jamais oublier la valeur absolue !</div>
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Voici une méthode plus astucieuse...</div>
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Sylvain Goléniahttp://www.blogger.com/profile/05638350035551044588noreply@blogger.com0tag:blogger.com,1999:blog-6300165877829263384.post-81599675890141977372019-11-29T18:55:00.002+01:002019-11-29T18:55:46.475+01:00Egalité des modules<div dir="ltr" style="text-align: left;" trbidi="on">
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On se propose de résoudre l'exercice suivant : Résoudre $|z-i| = |z+3|$.<br />
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<span style="text-align: left;">Nous procédons d'abord par une méthode géométrique.</span></div>
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<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhS3ZQZoK1vnshqe6MH0jMc3I9WHzkFQ-CAI5pNM8VTs57D-GkTx79oMugwvdhegN2wqMqfDyMIwuJW9hEt5srH27BGwCWLlKa8NboL-8Lc90rITwAKbAh8mErXPa413vb8ktGHTGEqc4g/s1600/IMG_20191107_120134.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="1600" data-original-width="1200" height="640" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhS3ZQZoK1vnshqe6MH0jMc3I9WHzkFQ-CAI5pNM8VTs57D-GkTx79oMugwvdhegN2wqMqfDyMIwuJW9hEt5srH27BGwCWLlKa8NboL-8Lc90rITwAKbAh8mErXPa413vb8ktGHTGEqc4g/s640/IMG_20191107_120134.jpg" width="480" /></a></div>
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Voici une méthode analytique.</div>
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Sylvain Goléniahttp://www.blogger.com/profile/05638350035551044588noreply@blogger.com0tag:blogger.com,1999:blog-6300165877829263384.post-33885087881812902402019-11-29T18:47:00.000+01:002019-11-29T18:50:15.325+01:00tan(a+b)<div dir="ltr" style="text-align: left;" trbidi="on">
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La formule $\tan(a+b)$ est aussi souvent méconnue. On la redémontre.<br />
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Alors que les deux premières lignes sont valides lorsque $a+b \neq \pi/2 [\pi]$, on suppose aussi que $a\neq \pi/2 [\pi]$ et $b \neq \pi/2 [\pi]$ pour écrire les deux dernières.</div>
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Sylvain Goléniahttp://www.blogger.com/profile/05638350035551044588noreply@blogger.com0tag:blogger.com,1999:blog-6300165877829263384.post-88254364607134690392019-11-29T18:31:00.000+01:002019-11-29T18:49:54.472+01:00Quelques formules trigonométriques<div dir="ltr" style="text-align: left;" trbidi="on">
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</script>On commence par une formule bien connue :<br />
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<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgCty8_QExRUnJ3Iyhlpx38gVLi_FbMGVcB5ZoDeb76zYVLJhcuiTuJ6HCx_M9X-C1QAHpJ4Zxk5LHtQx2R1N4aIlMOmQkmJXRyTFgvxl_LIjVLDq7EqrHmA33ly2BePsinGrAh2yNW0WA/s1600/IMG_20191107_115003.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="1600" data-original-width="1200" height="640" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgCty8_QExRUnJ3Iyhlpx38gVLi_FbMGVcB5ZoDeb76zYVLJhcuiTuJ6HCx_M9X-C1QAHpJ4Zxk5LHtQx2R1N4aIlMOmQkmJXRyTFgvxl_LIjVLDq7EqrHmA33ly2BePsinGrAh2yNW0WA/s640/IMG_20191107_115003.jpg" width="480" /></a></div>
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Il faut prendre un peu de recul sur cette preuve. Si on se place du point de vue du lycée, on ne démontre PAS la formule de $\cos(a+b)$. En effet, au lycée, on démontre tout d'abord cette formule et celle de $\sin(a+b)$ en s'appuyant sur de la géométrie. Ensuite, ces formules permettent de démontrer que l'exponentielle complexe possède bien la propriété du produit : $e^{i(a+b)} = e^{ia}e^{ib}$. D'un point de vue universitaire, on ne s'appuie pas sur la géométrie du plan. Après avoir construit le corps des complexes, on définie l'exponentielle grâce à une série entière de rayon de convergence infini. Puis on montre, grâce au produit de Cauchy que $e^{i(a+b)} = e^{ia}e^{ib}$. On définie alors $\cos$ comme étant sa partie réelle et $\sin$ sa partie imaginaire. On en déduit alors la formule trigonométrique de $\cos(a+b)$, comme sur l'image. </div>
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Du point de vue du CAPES, ce calcul est utile pour se remémorer très rapidement la valeur de $\cos(a+b)$.</div>
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<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjoQRqjjGXGdxiZe5i324WNlIVLOS64IRXd3S9JG2fcTiHVBD1XComGu0eYD0EafNWNmTa8x-OPXNdPeXMPsLYy_L2COfB4SuQ0XsYz9RA7PFqOAdeiTygiy-TJXbvp-ZD6MIqbYpGa41Q/s1600/IMG_20191107_112621.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="1600" data-original-width="1200" height="640" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjoQRqjjGXGdxiZe5i324WNlIVLOS64IRXd3S9JG2fcTiHVBD1XComGu0eYD0EafNWNmTa8x-OPXNdPeXMPsLYy_L2COfB4SuQ0XsYz9RA7PFqOAdeiTygiy-TJXbvp-ZD6MIqbYpGa41Q/s640/IMG_20191107_112621.jpg" width="480" /></a></div>
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<span style="text-align: left;">On regarde maintenant une variante. On se propose de redémontrer la formule (trop souvent oubliée) de $\cos(a)+\cos(b)$.</span></div>
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On utilise la technique du demi-angle. C'est une manière efficace pour factoriser $e^{ia}+e^{ib}$. On utilise aussi que $\cos(a)+ \cos(b)$ est la partie réelle de la somme des exponentielles.</div>
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En prenant la partie imaginaire, on retrouve alors la formule de $\sin(a)+ \sin(b)$.</div>
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On traite ici un cas un peu plus compliqué : $\sin(a)- \cos(b)$. Avec de la trigonométrie, on peut le déduire du cas précédent. Pour le traiter comme avant, on met tout sous la forme d'une partie réelle en remarquant que $\sin(a)= Re(ie^{ia})$. </div>
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Sylvain Goléniahttp://www.blogger.com/profile/05638350035551044588noreply@blogger.com0tag:blogger.com,1999:blog-6300165877829263384.post-54361963031930747722019-11-29T18:22:00.000+01:002019-11-29T18:49:54.468+01:00cos(pi/12) - méthode trigonométrique<div dir="ltr" style="text-align: left;" trbidi="on">
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On se propose maintenant de calculer $\cos(\pi/12)$ en utilisant de la trigonométrie.<br />
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1) Tout d'abord, on remarque que $\cos(\pi/12)>0$ car $0<\pi/12<\pi$. </div>
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2) On se rappelle que $\cos^2(x)= \frac{1+\cos(2x)}{2}$. En posant $x= \pi/12$, on obtient donc :</div>
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\[\cos(\pi/12)= |\cos(\pi/12)|= \sqrt{\frac{1+\cos(2x)}{2}}= \sqrt{\frac{1+\sqrt{3}/2}{2}}.\]</div>
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Sylvain Goléniahttp://www.blogger.com/profile/05638350035551044588noreply@blogger.com0tag:blogger.com,1999:blog-6300165877829263384.post-80760668406138214012019-11-29T18:13:00.001+01:002019-11-29T18:49:41.719+01:00cos(pi/12) - méthode complexe<div dir="ltr" style="text-align: left;" trbidi="on">
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On résout maintenant un exercice classique. Il peut être traiter avec l'aide de relations trigonométrique mais on utilisera une approche complexe.<br />
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On commence par trouver la forme exponentielle de $(1+i\sqrt{3})(1-i)$. Pour cela il ne faut PAS développer mais au contraire traiter chaque terme du produit séparément. </div>
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La forme algébrique est directe.</div>
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Pour déduire, $\cos(\pi/12)$ on identifie les parties réelles. Pour la rédaction, on n'oubliera pas d'écrire que l'écriture algébrique d'un nombre complexe est unique. </div>
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Sylvain Goléniahttp://www.blogger.com/profile/05638350035551044588noreply@blogger.com0tag:blogger.com,1999:blog-6300165877829263384.post-56482916600847624612019-11-29T16:12:00.000+01:002019-11-29T18:49:11.548+01:00Racine n-ième d'un nombre complexe<div dir="ltr" style="text-align: left;" trbidi="on">
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Ici on cherche la racine $n$-ième d'un nombre complexe. Si $n=2$, on peut donner cette exercice au hasard et se reporter au post précédent pour obtenir la forme algébrique. Ici, la méthode repose sur le fait que l'on connait la forme exponentielle du nombre dont on veut prendre la racine $n$-ième.</div>
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D'un point de vue rédactionnel, on n'oublie pas de mentionner l'unicité de l'écriture exponentielle. </div>
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Une erreur courante est le passage de $5\theta$ à $\theta$. Il ne faut pas oublier de diviser les $2k\pi$. </div>
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Pour obtenir les $5$ racines, on prend $5$ "k" qui se suivent. Ici on prend $k=0, 1, 2, 3, 4$.</div>
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On peut aller beaucoup plus loin dans l'analyse et les propriétés de ces solutions. Voir par exemple <a href="http://math.univ-lyon1.fr/capes/IMG/pdf/new.racine.pdf">ici</a>. </div>
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Sylvain Goléniahttp://www.blogger.com/profile/05638350035551044588noreply@blogger.com0