{"id":7843,"date":"2022-07-13T11:03:42","date_gmt":"2022-07-13T09:03:42","guid":{"rendered":"https:\/\/www.mathweb.fr\/euclide\/?p=7843"},"modified":"2022-07-13T11:03:44","modified_gmt":"2022-07-13T09:03:44","slug":"calcul-dune-somme-infinie","status":"publish","type":"post","link":"https:\/\/www.mathweb.fr\/euclide\/2022\/07\/13\/calcul-dune-somme-infinie\/","title":{"rendered":"Calcul d&#8217;une somme infinie"},"content":{"rendered":"\n<p class=\"wp-block-paragraph\">Calcul d&#8217;une somme infinie: Cliff Pickover a publi\u00e9 un tweet le 13 juillet 2022 qui m&#8217;inspira:<\/p>\n\n\n\n<figure class=\"wp-block-embed is-type-rich is-provider-twitter wp-block-embed-twitter\"><div class=\"wp-block-embed__wrapper\">\n<blockquote class=\"twitter-tweet\" data-width=\"550\" data-dnt=\"true\"><p lang=\"en\" dir=\"ltr\">Shiver in ecstasy.  Mathematics. <a href=\"https:\/\/t.co\/pwAwtDdcTY\">pic.twitter.com\/pwAwtDdcTY<\/a><\/p>&mdash; Cliff Pickover (@pickover) <a href=\"https:\/\/twitter.com\/pickover\/status\/1546986063931006977?ref_src=twsrc%5Etfw\">July 12, 2022<\/a><\/blockquote><script async src=\"https:\/\/platform.twitter.com\/widgets.js\" charset=\"utf-8\"><\/script>\n<\/div><\/figure>\n\n\n\n<p class=\"wp-block-paragraph\">Ce tweet stipule que:$$\\sum_{n\\geq0}\\frac{n^3}{2^n}=26.$$<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">J&#8217;avais envie de vous exposer la preuve donn\u00e9e par l&#8217;un de ses followers, mais de fa\u00e7on plus p\u00e9dagogique et plus pr\u00e9sentable&#8230;<\/p>\n\n\n\n<!--more-->\n\n\n\n<div id=\"ez-toc-container\" class=\"ez-toc-v2_0_84 counter-hierarchy ez-toc-counter ez-toc-white ez-toc-container-direction\">\n<div class=\"ez-toc-title-container\">\n<p class=\"ez-toc-title\" style=\"cursor:inherit\">Au menu sur cette page...<\/p>\n<span class=\"ez-toc-title-toggle\"><a href=\"#\" class=\"ez-toc-pull-right ez-toc-btn ez-toc-btn-xs ez-toc-btn-default ez-toc-toggle\" aria-label=\"Toggle Table of Content\"><span class=\"ez-toc-js-icon-con\"><span class=\"\"><span class=\"eztoc-hide\" style=\"display:none;\">Toggle<\/span><span class=\"ez-toc-icon-toggle-span\"><svg style=\"fill: #999;color:#999\" xmlns=\"http:\/\/www.w3.org\/2000\/svg\" class=\"list-377408\" width=\"20px\" height=\"20px\" viewBox=\"0 0 24 24\" fill=\"none\"><path d=\"M6 6H4v2h2V6zm14 0H8v2h12V6zM4 11h2v2H4v-2zm16 0H8v2h12v-2zM4 16h2v2H4v-2zm16 0H8v2h12v-2z\" fill=\"currentColor\"><\/path><\/svg><svg style=\"fill: #999;color:#999\" class=\"arrow-unsorted-368013\" xmlns=\"http:\/\/www.w3.org\/2000\/svg\" width=\"10px\" height=\"10px\" viewBox=\"0 0 24 24\" version=\"1.2\" baseProfile=\"tiny\"><path d=\"M18.2 9.3l-6.2-6.3-6.2 6.3c-.2.2-.3.4-.3.7s.1.5.3.7c.2.2.4.3.7.3h11c.3 0 .5-.1.7-.3.2-.2.3-.5.3-.7s-.1-.5-.3-.7zM5.8 14.7l6.2 6.3 6.2-6.3c.2-.2.3-.5.3-.7s-.1-.5-.3-.7c-.2-.2-.4-.3-.7-.3h-11c-.3 0-.5.1-.7.3-.2.2-.3.5-.3.7s.1.5.3.7z\"\/><\/svg><\/span><\/span><\/span><\/a><\/span><\/div>\n<nav><ul class='ez-toc-list ez-toc-list-level-1 ' ><li class='ez-toc-page-1 ez-toc-heading-level-2'><a class=\"ez-toc-link ez-toc-heading-1\" href=\"https:\/\/www.mathweb.fr\/euclide\/2022\/07\/13\/calcul-dune-somme-infinie\/#Calcul_dune_somme_infinie_demonstration_mathematique\" >Calcul d&#8217;une somme infinie: d\u00e9monstration math\u00e9matique<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-2'><a class=\"ez-toc-link ez-toc-heading-2\" href=\"https:\/\/www.mathweb.fr\/euclide\/2022\/07\/13\/calcul-dune-somme-infinie\/#Calcul_dune_somme_infinie_commande_Python\" >Calcul d&#8217;une somme infinie: commande Python<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-2'><a class=\"ez-toc-link ez-toc-heading-3\" href=\"https:\/\/www.mathweb.fr\/euclide\/2022\/07\/13\/calcul-dune-somme-infinie\/#Aller_plus_loin%E2%80%A6\" >Aller plus loin&#8230;<\/a><\/li><\/ul><\/nav><\/div>\n<h2 class=\"wp-block-heading\"><span class=\"ez-toc-section\" id=\"Calcul_dune_somme_infinie_demonstration_mathematique\"><\/span>Calcul d&#8217;une somme infinie: d\u00e9monstration math\u00e9matique<span class=\"ez-toc-section-end\"><\/span><\/h2>\n\n\n\n<p class=\"wp-block-paragraph\">Posons:$$f(x)=\\sum_{n\\geq0}x^n.$$<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">C&#8217;est une somme g\u00e9om\u00e9trique qui converge lorsque 0 &lt; <em>x<\/em> &lt; 1. Dans ce cas:$$f(x)=\\frac{1}{1-x}\\quad,\\quad 0&lt;x&lt;1.$$On a alors les d\u00e9riv\u00e9es successives suivantes:$$\\begin{array}{l}f'(x)=\\frac{1}{(1-x)^2}\\\\f&#8221;(x)=\\frac{2}{(1-x)^3}\\\\f^{(3)}(x)=\\frac{6}{(1-x)^4}\\end{array}$$<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Si on part maintenant du d\u00e9veloppement en s\u00e9rie de f(x), on a:$$f'(x)=\\sum_{n\\geq0}nx^{n-1}.$$<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">En multipliant par \\(x\\) \u00e0 droite et \u00e0 gauche, on a alors:$$xf'(x)=\\sum_{n\\geq0}nx^n.$$<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">En d\u00e9rivant \u00e0 droite et \u00e0 gauche, on obtient:$$f'(x)+xf&#8221;(x)=\\sum_{n\\geq0}n^2x^{n-1}.$$<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">En multipliant par \\(x\\) \u00e0 droite et \u00e0 gauche, on a:$$xf'(x)+x^2f&#8221;(x)=\\sum_{n\\geq0}n^2x^n.$$<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Le but, en multipliant par \\(x\\), est d&#8217;obtenir une somme infinie o\u00f9 est pr\u00e9sent \\(x^n\\) et non \\(x^{n-1}\\).<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">En d\u00e9rivant les deux membres de cette derni\u00e8re \u00e9galit\u00e9, on arrive \u00e0:$$f'(x)+xf&#8221;(x)+2xf&#8221;(x)+x^2f^{(3)}(x)=\\sum_{n\\geq0}n^3x^{n-1}.$$<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">En multipliant par \\(x\\) \u00e0 droite et \u00e0 gauche, on a:$$xf'(x)+3x^2f&#8221;(x)+x^3f^{(3)}(x)=\\sum_{n\\geq0}n^3x^n.$$<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">On remplace maintenant \\(f'(x)\\), \\(f&#8221;(x)\\) et \\(f^{(3)}(x)\\) par leur expression, ce qui donne:$$\\frac{x}{(1-x)^2}+\\frac{6x^2}{(1-x)^3}+\\frac{6x^3}{(1-x)^4}=\\sum_{n\\geq0}n^3x^n.$$<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">En prenant \\(x=\\frac{1}{2}\\), on obtient:$$2+12+12=26=\\sum_{n\\geq0}n^3x^n.$$<\/p>\n\n\n\n<h2 class=\"wp-block-heading\"><span class=\"ez-toc-section\" id=\"Calcul_dune_somme_infinie_commande_Python\"><\/span>Calcul d&#8217;une somme infinie: commande Python<span class=\"ez-toc-section-end\"><\/span><\/h2>\n\n\n\n<p class=\"wp-block-paragraph\">On peut v\u00e9rifier ais\u00e9ment cette valeur \u00e0 l&#8217;aide de Python:<\/p>\n\n\n\n<pre class=\"EnlighterJSRAW\" data-enlighter-language=\"python\" data-enlighter-theme=\"dracula\" data-enlighter-highlight=\"\" data-enlighter-linenumbers=\"false\" data-enlighter-lineoffset=\"\" data-enlighter-title=\"\" data-enlighter-group=\"\">sum( (n**3\/2**n) for n in range(68) )<\/pre>\n\n\n\n<p class=\"wp-block-paragraph\">Pourquoi s&#8217;arr\u00eater \u00e0 68 ? Parce que si l&#8217;on s&#8217;arr\u00eate \u00e0 67, cela affiche:<\/p>\n\n\n\n<pre class=\"wp-block-code\"><code>25.999999999999996<\/code><\/pre>\n\n\n\n<p class=\"wp-block-paragraph\">et qu&#8217;il est inutile d&#8217;aller au-del\u00e0 de 68.<\/p>\n\n\n\n<h2 class=\"wp-block-heading\"><span class=\"ez-toc-section\" id=\"Aller_plus_loin%E2%80%A6\"><\/span>Aller plus loin&#8230;<span class=\"ez-toc-section-end\"><\/span><\/h2>\n\n\n\n<figure class=\"wp-block-table is-style-stripes\"><table><thead><tr><th class=\"has-text-align-center\" data-align=\"center\">Valeurs de \\(k\\)<\/th><th class=\"has-text-align-center\" data-align=\"center\">Valeur de \\(\\displaystyle\\sum_{n\\geq0}\\frac{n^k}{2^n}\\)<\/th><\/tr><\/thead><tbody><tr><td class=\"has-text-align-center\" data-align=\"center\">0<\/td><td class=\"has-text-align-center\" data-align=\"center\">1<\/td><\/tr><tr><td class=\"has-text-align-center\" data-align=\"center\">1<\/td><td class=\"has-text-align-center\" data-align=\"center\">2<\/td><\/tr><tr><td class=\"has-text-align-center\" data-align=\"center\">2<\/td><td class=\"has-text-align-center\" data-align=\"center\">6<\/td><\/tr><tr><td class=\"has-text-align-center\" data-align=\"center\">3<\/td><td class=\"has-text-align-center\" data-align=\"center\">26<\/td><\/tr><tr><td class=\"has-text-align-center\" data-align=\"center\">4<\/td><td class=\"has-text-align-center\" data-align=\"center\">150<\/td><\/tr><tr><td class=\"has-text-align-center\" data-align=\"center\">5<\/td><td class=\"has-text-align-center\" data-align=\"center\">1082<\/td><\/tr><tr><td class=\"has-text-align-center\" data-align=\"center\">6<\/td><td class=\"has-text-align-center\" data-align=\"center\">9366<\/td><\/tr><tr><td class=\"has-text-align-center\" data-align=\"center\">7<\/td><td class=\"has-text-align-center\" data-align=\"center\">94586<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n<p class=\"wp-block-paragraph\">On &#8220;reconnait&#8221; la s\u00e9quence <a href=\"http:\/\/oeis.org\/A000629\" target=\"_blank\" rel=\"noreferrer noopener\">A000629<\/a> (&#8220;Number of necklaces of partitions of k+1 labeled beads&#8221;). Pour avoir plus d&#8217;informations sur les <em>colliers<\/em> en combinatoire, regardez cette <a href=\"https:\/\/fr.wikipedia.org\/wiki\/Collier_(combinatoire)\" target=\"_blank\" rel=\"noreferrer noopener\">page Wikipedia<\/a>.<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Par exemple, si l&#8217;on dispose du collier form\u00e9 par les 3 \u00e9l\u00e9ments <em>a<\/em>, <em>b<\/em> et <em>c<\/em>, on peut former:<\/p>\n\n\n\n<ul class=\"wp-block-list\"><li>{ {<em>abc<\/em>} } , <\/li><li>{ {<em>ab<\/em>} , {<em>c<\/em>} } ,<\/li><li>{ {<em>ac<\/em>} , {<em>b<\/em>} } ,<\/li><li>{ {<em>bc<\/em>} , {<em>a<\/em>} } ,<\/li><li>{ {<em>a<\/em>} , {<em>b<\/em>} , {<em>c<\/em>} } ,<\/li><li>{ {<em>a<\/em>} , {<em>c<\/em>} , {<em>b<\/em>} }.<\/li><\/ul>\n\n\n\n<p class=\"wp-block-paragraph\">On peut donc, en th\u00e9orie, donner une expression de la valeur de la somme \\(\\displaystyle\\sum_{n\\geq0}\\frac{n^k}{2^k}\\).<\/p>\n","protected":false},"excerpt":{"rendered":"<p>Calcul d&#8217;une somme infinie: Cliff Pickover a publi\u00e9 un tweet le 13 juillet 2022 qui m&#8217;inspira: Ce tweet stipule que:$$\\sum_{n\\geq0}\\frac{n^3}{2^n}=26.$$ J&#8217;avais envie de vous exposer la preuve donn\u00e9e par l&#8217;un de ses followers, mais de fa\u00e7on plus p\u00e9dagogique et plus pr\u00e9sentable&#8230;<\/p>\n","protected":false},"author":1,"featured_media":7852,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[5],"tags":[356,357,359,358],"class_list":["post-7843","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-python","tag-collier","tag-combinatoire","tag-serie","tag-somme"],"yoast_head":"<!-- This site is optimized with the Yoast SEO plugin v27.7 - https:\/\/yoast.com\/product\/yoast-seo-wordpress\/ -->\n<title>Calcul d&#039;une somme infinie - Mathweb.fr<\/title>\n<meta name=\"description\" content=\"Calcul d&#039;une somme infinie: tout est parti d&#039;un tweet de Cliff Pickover qui montre la valeur d&#039;une somme infinie. 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