{"id":5992,"date":"2021-04-13T18:14:43","date_gmt":"2021-04-13T16:14:43","guid":{"rendered":"https:\/\/www.mathweb.fr\/euclide\/?p=5992"},"modified":"2021-04-18T10:43:20","modified_gmt":"2021-04-18T08:43:20","slug":"une-enigme-pour-moi-help-me-please","status":"publish","type":"post","link":"https:\/\/www.mathweb.fr\/euclide\/2021\/04\/13\/une-enigme-pour-moi-help-me-please\/","title":{"rendered":"Une \u00e9nigme pour moi&#8230;"},"content":{"rendered":"\n<p class=\"wp-block-paragraph\">Bonjour les gens ! En effectuant quelques recherches pour <a href=\"https:\/\/www.mathweb.fr\/euclide\/produit\/python-en-mathematiques-au-lycee\/\" target=\"_blank\" rel=\"noreferrer noopener\">mon prochain livre \u00e0 para\u00eetre uniquement sur mathweb.fr<\/a>, je suis tomb\u00e9 sur le probl\u00e8me suivant:<\/p>\n\n\n\n<blockquote class=\"wp-block-quote is-layout-flow wp-block-quote-is-layout-flow\"><p>On choisit au hasard un nombre entre 0 et 1, et on r\u00e9p\u00e8te cela jusqu&rsquo;\u00e0 ce que la somme des nombres choisis d\u00e9passe 1.<\/p><p>Quelle est le nombre moyen de nombres choisis ?<\/p><\/blockquote>\n\n\n\n<!--more-->\n\n\n\n<p class=\"wp-block-paragraph\">Le probl\u00e8me peut s&rsquo;impl\u00e9menter en Python \u00e0 l&rsquo;aide de la fonction suivante:<\/p>\n\n\n\n<pre class=\"EnlighterJSRAW\" data-enlighter-language=\"python\" data-enlighter-theme=\"dracula\" data-enlighter-highlight=\"\" data-enlighter-linenumbers=\"\" data-enlighter-lineoffset=\"\" data-enlighter-title=\"\" data-enlighter-group=\"\">from random import random\n\ndef moyenne(n):\n    T = 0\n    for k in range(n):\n        s, i = 0, 0\n\n        while s &lt; 1:\n            s += random()\n            i += 1\n            \n        T += i\n    \n    return T \/ n\n\nprint( moyenne( 10**9 ) )\n<\/pre>\n\n\n\n<p class=\"wp-block-paragraph\">On obtient alors:<\/p>\n\n\n\n<pre class=\"wp-block-code\"><code>&gt;&gt;&gt; moyenne(10**7)\n2.7186881\n&gt;&gt;&gt; moyenne(10**8)\n2.71829368\n&gt;&gt;&gt; moyenne(10**9)\n2.718286834<\/code><\/pre>\n\n\n\n<p class=\"wp-block-paragraph\">La moyenne semble \u00eatre \u00e9gale \u00e0 \u00ab\u00a0e\u00a0\u00bb (ce n&rsquo;est qu&rsquo;une conjecture).<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Le probl\u00e8me est que je n&rsquo;arrivais pas \u00e0 \u00e9tablir la moindre formule&#8230;<\/p>\n\n\n\n<h2 class=\"wp-block-heading\">La solution de Philippe Rackette<\/h2>\n\n\n\n<p class=\"wp-block-paragraph\">Fort heureusement, il y a parmi les visiteurs et abonn\u00e9\u00b7e\u00b7s de ce site de bien meilleurs probabilistes que moi ! C&rsquo;est ainsi que l&rsquo;un d&rsquo;eux m&rsquo;a propos\u00e9 la solution suivante (le lien vers le PDF a \u00e9t\u00e9 mis en commentaire, mais il va sans doute arriver un jour o\u00f9 le fichier correspondant sera effac\u00e9, donc j&rsquo;ai pr\u00e9f\u00e9r\u00e9 le retranscrire ici.<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Posons \\(S_{k+1} = P(X&gt;k+1)\\) pour tout entier naturel \\(k\\). Alors,$$S_{k+1}=\\int_{x_1+\\cdots+x_{k+1}&lt;1,\\ x_i \\geqslant0} 1\\text{d}x_1\\cdots \\text{d}x_{k+1}$$En effet, la densit\u00e9 de probabilit\u00e9 est homog\u00e8ne et la probabilit\u00e9 qu&rsquo;il faille plus de <em>k<\/em>+1 nombres n\u00e9cessite que la somme des <em>k<\/em>+1 premiers nombres soit strictement inf\u00e9rieure \u00e0 1.<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">On peur alors \u00e9crire pour <em>k<\/em> non nul:$$S_{k+1}=\\int_{x_{k+1}=0}^{x_{k+1}=1^-}\\left( \\int_{x_1+\\ \\cdots\\ +x_k&lt;1-x_{k+1},\\ x_i\\geqslant0}1\\text{d}x_1\\cdots\\text{d}x_k \\right)\\text{d}x_{k+1}$$<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">En effectuant les changements de variables:$$u_i = \\frac{1}{1-x_{k+1}}x_i$$pour <em>i<\/em> variant de 1 \u00e0 <em>k<\/em>, on obtient:$$S_{k+1}=\\int_{x_{k+1}=0}^{x_{k+1}=1^-}\\left( \\int_{u_1+\\ \\cdots\\ +u_k&lt;1,\\ u_i\\geqslant0}(1-x_{k+1})^k\\text{d}u_1\\cdots\\text{d}u_k \\right)\\text{d}x_{k+1}$$que l&rsquo;on peut aussi \u00e9crire: $$S_{k+1}=\\int_{x_{k+1}=0}^{x_{k+1}=1^-}\\left( \\int_{u_1+\\ \\cdots\\ +u_k&lt;1,\\ u_i\\geqslant0}1\\text{d}u_1\\cdots\\text{d}u_k \\right)(1-x_{k+1})^k\\text{d}x_{k+1}$$c&rsquo;est-\u00e0-dire:$$S_{k+1}=\\int_{x_{k+1}=0}^{x_{k+1}=1^-}S_k(1-x_{k+1})^k\\text{d}x_{k+1}$$<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Comme \\(S_k\\) est ind\u00e9pendant de \\(x_{k+1}\\), on a alors:$$S_{k+1}=S_k\\int_{x_{k+1}=0}^{x_{k+1}=1^-}(1-x_{k+1})^k\\text{d}x_{k+1}$$<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Or,$$\\int_0^1(1-x)^k\\text{d}x=\\left[-\\frac{(1-x)^{k+1}}{k+1}\\right]_0^1=\\frac{1}{k+1}$$ d&rsquo;o\u00f9:$$S_{k+1}=\\frac{1}{k+1}S_k.$$<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">De plus, \\(S_1=1\\) car l&rsquo;\u00e9v\u00e9nement \u00ab\u00a0plus d&rsquo;un nombre est n\u00e9cessaire pour obtenir une somme strictement plus grande que 1\u00a0\u00bb est certain. On a alors:$$\\begin{align}S_k&amp;=\\frac{1}{k}S_{k-1}\\\\&amp; = \\frac{1}{k}\\times\\frac{1}{k-1}S_{k-2}\\\\&amp;=\\cdots\\\\&amp;=\\frac{1}{k}\\times\\frac{1}{k-1}\\times\\cdots\\times\\frac{1}{k-i}S_{k-i-1}\\\\&amp;=\\cdots\\\\&amp;=\\frac{1}{k}\\times\\frac{1}{k-1}\\times\\cdots\\times\\frac{1}{k-(k-2)}S_{k-(k-2)-1}\\\\&amp;=\\frac{1}{k}\\times\\frac{1}{k-1}\\times\\cdots\\times\\frac{1}{2}S_{1}\\\\&amp; = \\frac{1}{k!}\\end {align}$$<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Ainsi,$$P(X\\leqslant k)=1-P(X&gt;k)=1-S_k=1-\\frac{1}{k!}$$<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Par suite, on a:$$\\begin{align}P(X=k) &amp; =P(X\\leqslant k)-P(X\\leqslant k-1)\\\\&amp;=\\left(1-\\frac{1}{k!}\\right)-\\left(1-\\frac{1}{(k-1)!}\\right)\\\\&amp;=\\frac{1}{(k-1)!}-\\frac{1}{k!}\\\\&amp;=\\frac{k-1}{k!} \\end{align}$$<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">On peut alors calculer le nombre moyen que l&rsquo;on cherchait en d\u00e9terminant l&rsquo;esp\u00e9rance de <em>X<\/em>:$$\\begin{align} E(X) &amp; =\\sum_{k\\geqslant1} k \\times P(X=k)\\\\&amp; = \\sum_{k\\geqslant1}k\\times\\frac{k-1}{k!}\\\\&amp;=\\sum_{k\\geqslant1}\\frac{k-1}{(k-1)!}\\\\&amp;=\\sum_{p\\geqslant0}\\frac{1}{p!}\\\\&amp;=\\text{e}.\\end{align}$$<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">L&rsquo;\u00e9nigme est ainsi r\u00e9solue! Me voil\u00e0 rassur\u00e9&#8230; La conjecture de d\u00e9part \u00e9tait donc bonne, mais c&rsquo;est bien mieux de voir pourquoi! <\/p>\n\n\n\n<p class=\"wp-block-paragraph\">En math\u00e9matiques, tout est une question de point de vue: je n&rsquo;avais pas le bon d\u00e8s le d\u00e9part et je ne pouvais donc pas avoir la bonne id\u00e9e !<\/p>\n","protected":false},"excerpt":{"rendered":"<p>Bonjour les gens ! En effectuant quelques recherches pour mon prochain livre \u00e0 para\u00eetre uniquement sur mathweb.fr, je suis tomb\u00e9 sur le probl\u00e8me suivant: On choisit au hasard un nombre entre 0 et 1, et on r\u00e9p\u00e8te cela jusqu&rsquo;\u00e0 ce que la somme des nombres choisis d\u00e9passe 1. Quelle est [&hellip;]<\/p>\n","protected":false},"author":1,"featured_media":6029,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[6],"tags":[],"class_list":["post-5992","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-mathematiques"],"yoast_head":"<!-- This site is optimized with the Yoast SEO plugin v27.9 - https:\/\/yoast.com\/product\/yoast-seo-wordpress\/ -->\n<title>Une \u00e9nigme pour moi... - Mathweb.fr<\/title>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/www.mathweb.fr\/euclide\/2021\/04\/13\/une-enigme-pour-moi-help-me-please\/\" \/>\n<meta property=\"og:locale\" content=\"fr_FR\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"Une \u00e9nigme pour moi... - Mathweb.fr\" \/>\n<meta property=\"og:description\" content=\"Bonjour les gens ! En effectuant quelques recherches pour mon prochain livre \u00e0 para\u00eetre uniquement sur mathweb.fr, je suis tomb\u00e9 sur le probl\u00e8me suivant: On choisit au hasard un nombre entre 0 et 1, et on r\u00e9p\u00e8te cela jusqu&rsquo;\u00e0 ce que la somme des nombres choisis d\u00e9passe 1. 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En effectuant quelques recherches pour mon prochain livre \u00e0 para\u00eetre uniquement sur mathweb.fr, je suis tomb\u00e9 sur le probl\u00e8me suivant: On choisit au hasard un nombre entre 0 et 1, et on r\u00e9p\u00e8te cela jusqu&rsquo;\u00e0 ce que la somme des nombres choisis d\u00e9passe 1. 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