<\/span><\/h3>\n\n\n\n\u00c0 travers mon activit\u00e9 d’enseignement \u00e0 distance, je suis quelques fois confront\u00e9 \u00e0 des situations qui me surprennent au niveau p\u00e9dagogique. C’est notamment le cas lorsque je vois que certains enseignants d\u00e9finissent une fonction convexe<\/em> \\(f\\) sur un intervalle I comme \u00e9tant une fonction v\u00e9rifiant l’in\u00e9galit\u00e9 suivante:$$\\forall\\ (x,y)\\in I\\times I,\\ \\forall t\\in[0;1],\\ f\\big(tx+(1-t)y\\big) \\leq tf(x)+(1-t)f(y).$$C’est une in\u00e9galit\u00e9 barycentrique<\/em>. Or, la notion de barycentre ayant disparue des programmes du lyc\u00e9e depuis moultes ann\u00e9es, il est \u00e9tonnant de voir cette d\u00e9finition en Terminale car elle n’a de sens que si les \u00e9l\u00e8ves la comprennent.<\/p>\n\n\n\nNous allons donc tenter de la comprendre.<\/p>\n\n\n\n
<\/span>Ecriture barycentrique d’un point sur un segment<\/span><\/h3>\n\n\n\nPrenons un segment [AB] (par exemple, un b\u00e2ton). Suspendons \u00e0 l’extr\u00e9mit\u00e9 A un poids d’une masse \\(m_1\\) et \u00e0 l’autre extr\u00e9mit\u00e9, un poids d’une masse \\(m_2\\). <\/p>\n\n\n
\n
\n\t\t\t\n\t\t\t\tLe petit bout de bois (M) peut se d\u00e9placer entre A et B. On peut alors \u00e9crire:$$t\\vec{MA}+(1-t)\\vec{MB}=\\vec{0}\\quad,\\quad t\\in[0;1].$$En effet, nous avons les \u00e9quivalences suivantes:$$\\begin{align} t\\vec{MA}+(1-t)\\vec{MB}=\\vec{0} & \\iff t\\vec{MA}+(1-t)\\big(\\vec{MA}+\\vec{AB}\\big)=\\vec{0} \\\\ & \\iff t\\vec{MA}+\\vec{MA}+\\vec{AB}-t\\vec{MA}-t\\vec{AB}=\\vec{0} \\\\ & \\iff \\vec{MA}=(1-t)\\vec{BA}\\end{align}$$<\/p>\n\n\n\n
t<\/em> \u00e9tant compris entre 0 et 1 (compris), \\((1-t)\\vec{BA}\\) va varier de \\(\\vec{BA}\\) \u00e0 \\(\\vec{0}\\). Donc \\(\\vec{MA}\\) va lui aussi varier de \\(\\vec{BA}\\) (M = B) \u00e0 \\(\\vec{0}\\) (M = A). Le point M bouge donc de B \u00e0 A.<\/p>\n\n\n\nL’\u00e9criture \\( t\\vec{MA}+(1-t)\\vec{MB}=\\vec{0}\\quad,\\quad t\\in[0;1] \\) est appel\u00e9e \u00e9criture barycentrique<\/em> de M.<\/p>\n\n\n\nLe barycentre<\/em> du b\u00e2ton est le point M sur lequel on peut \u00ab\u00a0poser le doigt\u00a0\u00bb de sorte qu’il y ait un parfait \u00e9quilibre. Dans ce cas, il faut que \\(m_1\\vec{MA} + m_2\\vec{MB} = \\vec{0}\\). <\/p>\n\n\n\n<\/span>Lien avec la convexit\u00e9 d’une fonction<\/span><\/h3>\n\n\n\nDans la d\u00e9finition de la convexit\u00e9: $$\\forall\\ (x,y)\\in I\\times I,\\ \\forall t\\in[0;1],\\ f\\big(tx+(1-t)y\\big) \\leq tf(x)+(1-t)f(y)$$<\/p>\n\n\n\n
\n\\( tx+(1-t)y \\) repr\u00e9sente un nombre compris entre x<\/em> et y<\/em>, donc \\( f\\big(tx+(1-t)y\\big) \\) est l’image de ce nombre par la fonction f<\/em>;<\/li>\n\n\n\n\\( tf(x)+(1-t)f(y) \\) repr\u00e9sente un nombre dans l’intervalle [ f(x)<\/em> ; f(y)<\/em> ].<\/li>\n<\/ul>\n\n\n\nAinsi, cette in\u00e9galit\u00e9 repr\u00e9sente la situation suivante:<\/p>\n\n\n
\n
\n\t\t\t\n\t\t\t\tOn a \\( M\\big( tx+(1-t)y\\ ;\\ f(tx+(1-t)y) \\big) \\) qui appartient \u00e0 la courbe repr\u00e9sentative de f<\/em> et \\( N\\big( tx+(1-t)y\\ ;\\ tf(x)+(1-t)f(y) \\big) \\) qui est sur le segment [AB] (son abscisse et son ordonn\u00e9e sont \u00e9crits de mani\u00e8re barycentrique).<\/p>\n\n\n\nL’in\u00e9galit\u00e9 de convexit\u00e9 signifie alors que M est toujours au-dessous de N sur l’intervalle consid\u00e9r\u00e9 et donc que la courbe repr\u00e9sentative de f<\/em> est toujours en dessous de la s\u00e9cante (la corde) [AB].<\/p>\n\n\n\n% code LaTeX pour construire l'illustration ci-dessus\n\\documentclass{standalone}\n\\usepackage{tikz}\n\n\\begin{document}\n\n\\begin{tikzpicture}[>=latex]E\n\\draw[->,thick] (-1,0) -- (5,0);\n\\draw[->,thick] (0,-1) -- (0,5);\n\\draw[thick,purple] (0.5,1) to[bend right=40] (4,4);\n\\draw[orange,dotted] (0.5,1) -- (0.5,0) node[below] {$x$};\n\\draw[orange,dotted] (4,4) -- (4,0) node[below] {$y$};\n\\draw[orange,dotted] (0.5,1) -- (0,1) node[left] {$f(x)$};\n\\draw[orange,dotted] (4,4) -- (0,4) node[left] {$f(y)$};\n\\draw[green!50!black] (0.5,1) -- (4,4);\n\\draw[blue!50!black,dashed] (2,0) node[below] {\\tiny $tx+(1-t)y$} -- (2,1.32) -- (0,1.32) node[left] {\\tiny $f(tx+(1-t)y)$};\n\\draw[blue!50!black,dotted] (2,1.32) -- (2,2.3) node[above] {$N$} -- (0,2.3) node[left] {\\tiny $tf(x)+(1-t)f(y)$};\n\\fill[blue!50!black] (2,2.3) circle(1pt);\n\\fill[blue!50!black] (2,1.31) circle(1pt) node[below right] {$M$};\n\\fill (0.5,1) node[below left] {$A$};\n\\fill (4,4) node[above right] {$B$};\n\\end{tikzpicture}\n\n\\end{document}<\/pre>\n\n\n\n<\/span>In\u00e9galit\u00e9 de convexit\u00e9 en Terminale: une d\u00e9finition \u00e9quivalente<\/span><\/h2>\n\n\n\nLe programme officiel stipule:<\/p>\n\n\n\n
\nFonction convexe sur un intervalle : d\u00e9finition par la position relative de la courbe repr\u00e9sentative et des s\u00e9cantes. Pour une fonction deux fois d\u00e9rivable, \u00e9quivalence admise avec la position par rapport aux tangentes, la croissance de \u0192\u2019, la positivit\u00e9 de \u0192\u2019\u2019.<\/p>\n<\/blockquote>\n\n\n\n
L’enseignant(e) est donc libre de d\u00e9finir une fonction convexe \u00e0 l’aide de l’in\u00e9galit\u00e9 pr\u00e9c\u00e9dente ou de mani\u00e8re purement graphique.<\/p>\n\n\n\n
Mais tr\u00e8s souvent, on retient la propri\u00e9t\u00e9 suivante:<\/p>\n\n\n\n
\nSoit f<\/em> une fonction deux fois d\u00e9rivable.<\/p>\n\n\n\nf\u00a0\u00bb(x) > 0 sur un intervalle I \u00e9quivaut \u00e0 dire que f est convexe sur I.<\/p>\n<\/blockquote>\n\n\n\n
C’est en effet ce dernier r\u00e9sultat qui est, en pratique, utilis\u00e9 car toutes les fonctions que nous \u00e9tudions au lyc\u00e9e sont deux fois d\u00e9rivables sur un intervalle judicieusement choisi.<\/p>\n\n\n\n
<\/span>D\u00e9montrer la convexit\u00e9 de la fonction carr\u00e9e \u00e0 l’aide de l’in\u00e9galit\u00e9<\/span><\/h2>\n\n\n\nPour d\u00e9montrer que la fonction f<\/em> telle que \\(f(x)=x^2\\) est convexe sur \\(\\mathbb{R}\\), nous allons prendre deux r\u00e9els x<\/em> et y<\/em>, avec x<\/em> < y<\/em>. On a alors: $$\\begin{align}f(tx+(1-t)y) & = (tx+(1-t)y)^2\\\\ & = t^2x^2 + (1-t)^2y^2 + 2t(1-t)xy \\end{align}$$ et $$ tf(x) + (1-t)f(y) = tx^2+(1-t)y^2.$$<\/p>\n\n\n\nAinsi, la diff\u00e9rence est:$$ \\begin{align} & f(tx+(1-t)y) -\\big(tf(x) + (1-t)f(y)\\big)\\\\ = & t^2x^2 + (1-t)^2y^2 + 2t(1-t)xy – tx^2-(1-t)y^2 \\\\ = & t(t-1)x^2+(1-t)(1-t-1)y^2+2t(1-t)xy\\\\=&t(t-1)x^2+t(t-1)y^2-2t(t-1)xy\\\\=&t(t-1)(x^2+y^2-2xy)\\\\=&t(t-1)(x-y)^2\\end{align}$$<\/p>\n\n\n\n
\\(t\\in[0;1]\\) donc \\(t-1\\leq0\\) et donc cette derni\u00e8re diff\u00e9rence est n\u00e9gative. Ce qui signifie donc que:$$\\forall\\ (x,y)\\in \\mathbb{R}^2,\\ \\forall t\\in[0;1],\\ f\\big(tx+(1-t)y\\big) \\leq tf(x)+(1-t)f(y).$$<\/p>\n\n\n\n
La fonction carr\u00e9 est donc convexe sur \\(\\mathbb{R}\\).<\/p>\n","protected":false},"excerpt":{"rendered":"
L’in\u00e9galit\u00e9 de convexit\u00e9 en Terminale n’est pas explicitement au programme. Cependant, certains enseignants d\u00e9cident de l’utiliser pour d\u00e9finir la convexit\u00e9 d’une fonction sur un intervalle.<\/p>\n","protected":false},"author":1,"featured_media":11551,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[21,6],"tags":[],"class_list":["post-11539","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-enseignement","category-mathematiques"],"yoast_head":"\n
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