{"id":5238,"date":"2020-12-18T14:15:10","date_gmt":"2020-12-18T13:15:10","guid":{"rendered":"https:\/\/www.mathweb.fr\/euclide\/?p=5238"},"modified":"2023-06-11T10:04:55","modified_gmt":"2023-06-11T08:04:55","slug":"la-suite-de-heron-etude-mathematique-et-implementation-en-python","status":"publish","type":"post","link":"https:\/\/www.mathweb.fr\/euclide\/2020\/12\/18\/la-suite-de-heron-etude-mathematique-et-implementation-en-python\/","title":{"rendered":"La suite de H\u00e9ron, \u00e9tude math\u00e9matique et impl\u00e9mentation en python"},"content":{"rendered":"\n<p>La suite de H\u00e9ron est une suite permettant de trouver une valeur approch\u00e9e d&#8217;une racine carr\u00e9e. <\/p>\n\n\n\n<p>Elle tire son nom du math\u00e9maticien <a href=\"https:\/\/fr.wikipedia.org\/wiki\/H%C3%A9ron_d%27Alexandrie\" target=\"_blank\" rel=\"noreferrer noopener\">H\u00e9ron d&#8217;Alexandrie<\/a>.<\/p>\n\n\n\n<!--more-->\n\n\n<div class=\"wp-block-image is-style-rounded\">\n<figure class=\"aligncenter size-large\"><img loading=\"lazy\" decoding=\"async\" width=\"220\" height=\"309\" src=\"https:\/\/www.mathweb.fr\/euclide\/wp-content\/uploads\/2020\/12\/Heron_Alexandrie.png\" alt=\"suite de h\u00e9ron\" class=\"wp-image-5239\" srcset=\"https:\/\/www.mathweb.fr\/euclide\/wp-content\/uploads\/2020\/12\/Heron_Alexandrie.png 220w, https:\/\/www.mathweb.fr\/euclide\/wp-content\/uploads\/2020\/12\/Heron_Alexandrie-214x300.png 214w\" sizes=\"auto, (max-width: 220px) 100vw, 220px\" \/><figcaption class=\"wp-element-caption\">H\u00e9ron d&#8217;Alexandrie<\/figcaption><\/figure>\n<\/div>\n\n\n<div id=\"ez-toc-container\" class=\"ez-toc-v2_0_82_2 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\/2020\/12\/18\/la-suite-de-heron-etude-mathematique-et-implementation-en-python\/#Suite_de_Heron_etude_mathematique\" >Suite de H\u00e9ron: \u00e9tude math\u00e9matique<\/a><ul class='ez-toc-list-level-3' ><li class='ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-2\" href=\"https:\/\/www.mathweb.fr\/euclide\/2020\/12\/18\/la-suite-de-heron-etude-mathematique-et-implementation-en-python\/#Fonction_associee_a_la_suite_de_Heron\" >Fonction associ\u00e9e \u00e0 la suite de H\u00e9ron<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-3\" href=\"https:\/\/www.mathweb.fr\/euclide\/2020\/12\/18\/la-suite-de-heron-etude-mathematique-et-implementation-en-python\/#Tous_les_termes_de_la_suite_sont_positifs\" >Tous les termes de la suite sont positifs<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-4\" href=\"https:\/\/www.mathweb.fr\/euclide\/2020\/12\/18\/la-suite-de-heron-etude-mathematique-et-implementation-en-python\/#La_suite_de_Heron_est_minoree_par_sqrta\" >La suite de H\u00e9ron est minor\u00e9e par \\(\\sqrt{a}\\)<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-5\" href=\"https:\/\/www.mathweb.fr\/euclide\/2020\/12\/18\/la-suite-de-heron-etude-mathematique-et-implementation-en-python\/#La_suite_est_decroissante\" >La suite est d\u00e9croissante<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-6\" href=\"https:\/\/www.mathweb.fr\/euclide\/2020\/12\/18\/la-suite-de-heron-etude-mathematique-et-implementation-en-python\/#La_suite_est_convergente\" >La suite est convergente<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-7\" href=\"https:\/\/www.mathweb.fr\/euclide\/2020\/12\/18\/la-suite-de-heron-etude-mathematique-et-implementation-en-python\/#Vitesse_de_convergence_de_la_suite_de_Heron\" >Vitesse de convergence de la suite de H\u00e9ron<\/a><\/li><\/ul><\/li><li class='ez-toc-page-1 ez-toc-heading-level-2'><a class=\"ez-toc-link ez-toc-heading-8\" href=\"https:\/\/www.mathweb.fr\/euclide\/2020\/12\/18\/la-suite-de-heron-etude-mathematique-et-implementation-en-python\/#Suite_de_Heron_du_cote_de_Python\" >Suite de H\u00e9ron: du c\u00f4t\u00e9 de Python<\/a><\/li><\/ul><\/nav><\/div>\n<h2 class=\"wp-block-heading\"><span class=\"ez-toc-section\" id=\"Suite_de_Heron_etude_mathematique\"><\/span>Suite de H\u00e9ron: \u00e9tude math\u00e9matique<span class=\"ez-toc-section-end\"><\/span><\/h2>\n\n\n\n<p>On consid\u00e8re la suite \\((u_n)_{n\\in\\mathbb{N}}\\) d\u00e9finie par son premier terme \\(u_0 > 0\\) et par la relation de r\u00e9currence suivante :$$\\forall n\\in\\mathbb{N},\\quad u_{n+1}=\\frac{1}{2}\\left(u_n+\\frac{a}{u_n}\\right)$$o\u00f9 \\(a\\) est un r\u00e9el strictement plus grand que 0 (le cas o\u00f9 il est \u00e9gal \u00e0 0 ne nous importe peu car la suite devient g\u00e9om\u00e9trique de raison \\(\\frac{1}{2}\\) et converge donc vers 0).<\/p>\n\n\n\n<p>Cette suite est appel\u00e9e une suite de H\u00e9ron de param\u00e8tre <em>a<\/em>.<\/p>\n\n\n\n<h3 class=\"wp-block-heading\"><span class=\"ez-toc-section\" id=\"Fonction_associee_a_la_suite_de_Heron\"><\/span>Fonction associ\u00e9e \u00e0 la suite de H\u00e9ron<span class=\"ez-toc-section-end\"><\/span><\/h3>\n\n\n\n<p>Imm\u00e9diatement, on peut constater que \\(u_{n+1} = f(u_n)\\), avec:$$f(x)=\\frac{1}{2}\\left(x+\\frac{a}{x}\\right)$$que l&#8217;on peut d\u00e9finir sur \\(]0;+\\infty[\\).<\/p>\n\n\n\n<p>Sa d\u00e9riv\u00e9e est alors:$$f'(x)=\\frac{1}{2}\\left(1-\\frac{a}{x^2}\\right)$$que l&#8217;on peut aussi \u00e9crire:$$f'(x)=\\frac{x^2-a}{2x^2}.$$<\/p>\n\n\n\n<p>L&#8217;expression \\(x^2-a\\) s&#8217;annule pour \\(x=-\\sqrt{a}\\) et pour \\(x=\\sqrt{a}\\). On a alors le tableau de variations suivant :<\/p>\n\n\n<div class=\"wp-block-image\">\n<figure class=\"aligncenter size-large\"><img loading=\"lazy\" decoding=\"async\" width=\"573\" height=\"149\" src=\"https:\/\/www.mathweb.fr\/euclide\/wp-content\/uploads\/2020\/12\/tableau-variations-suite-heron.jpg\" alt=\"tableau de variations de la fonction associ\u00e9e \u00e0 la suite de h\u00e9ron\" class=\"wp-image-5243\" srcset=\"https:\/\/www.mathweb.fr\/euclide\/wp-content\/uploads\/2020\/12\/tableau-variations-suite-heron.jpg 573w, https:\/\/www.mathweb.fr\/euclide\/wp-content\/uploads\/2020\/12\/tableau-variations-suite-heron-300x78.jpg 300w\" sizes=\"auto, (max-width: 573px) 100vw, 573px\" \/><figcaption class=\"wp-element-caption\">Tableau de variations de la fonction associ\u00e9e \u00e0 la suite de H\u00e9ron de param\u00e8tre <em>a<\/em><\/figcaption><\/figure>\n<\/div>\n\n\n<p><em>f<\/em> admet donc un minimum pour \\(x=\\sqrt{a}\\) qui vaut \\(\\sqrt{a}\\).<\/p>\n\n\n\n<blockquote class=\"wp-block-quote is-layout-flow wp-block-quote-is-layout-flow\">\n<p>Pour tout r\u00e9el <em>x<\/em> &gt; 0, \\(f(x) \\geqslant \\sqrt{a}\\).<\/p>\n<\/blockquote>\n\n\n\n<h3 class=\"wp-block-heading\"><span class=\"ez-toc-section\" id=\"Tous_les_termes_de_la_suite_sont_positifs\"><\/span>Tous les termes de la suite sont positifs<span class=\"ez-toc-section-end\"><\/span><\/h3>\n\n\n\n<p>Ce r\u00e9sultat est presque imm\u00e9diat. En effet, $$u_0&gt;0$$ donc $$\\frac{1}{2}\\left(u_0 + \\frac{a}{u_0}\\right)&gt;0$$donc:$$u_1&gt;0.$$<\/p>\n\n\n\n<p>De plus, si on suppose que pour un entier <em>k<\/em> fix\u00e9, \\(u_k&gt;0\\),$$\\frac{1}{2}\\left(u_k + \\frac{a}{u_k}\\right)&gt;0$$donc:$$u_{k+1}&gt;0.$$<\/p>\n\n\n\n<p>D&#8217;apr\u00e8s le principe de r\u00e9currence, on peut conclure que pour tout entier naturel <em>n<\/em>, \\(u_n&gt;0\\).<\/p>\n\n\n\n<h3 class=\"wp-block-heading\"><span class=\"ez-toc-section\" id=\"La_suite_de_Heron_est_minoree_par_sqrta\"><\/span>La suite de H\u00e9ron est minor\u00e9e par \\(\\sqrt{a}\\)<span class=\"ez-toc-section-end\"><\/span><\/h3>\n\n\n\n<p>Nous venons en effet de d\u00e9montrer que tous les termes de la suite sont strictement positifs donc pour tout entier naturel <em>n<\/em>, \\(f(u_n) \\geqslant \\sqrt{a}\\) d&#8217;apr\u00e8s les variations de la fonction <em>f<\/em>.<\/p>\n\n\n\n<h3 class=\"wp-block-heading\"><span class=\"ez-toc-section\" id=\"La_suite_est_decroissante\"><\/span>La suite est d\u00e9croissante<span class=\"ez-toc-section-end\"><\/span><\/h3>\n\n\n\n<p>En effet, on a:$$\\begin{align}u_{n+1}-u_n &amp; = \\frac{1}{2}\\left(u_n+\\frac{a}{u_n}\\right)-u_n\\\\&amp;=\\frac{1}{2}\\left(u_n+\\frac{a}{u_n}\\right)-\\frac{1}{2}\\times2u_n\\\\&amp;=\\frac{1}{2}\\left(u_n+\\frac{a}{u_n}-2u_n\\right) \\\\&amp;=\\frac{1}{2}\\left(\\frac{a-u_n^2}{u_n}\\right)\\end{align}$$<\/p>\n\n\n\n<p>Or, nous avons vu pr\u00e9c\u00e9demment que pour tout entier naturel <em>n<\/em> > 0, \\(u_n\\geqslant\\sqrt{a}\\), donc que \\(u_n^2 \\geqslant a\\), ce qui nous assure que \\(u_{n+1}-u_n \\leqslant 0\\) pour <em>n<\/em> > 0.<\/p>\n\n\n\n<p>Cette suite est donc d\u00e9croissante \u00e0 partir de <em>n<\/em> = 1.<\/p>\n\n\n\n<h3 class=\"wp-block-heading\"><span class=\"ez-toc-section\" id=\"La_suite_est_convergente\"><\/span>La suite est convergente<span class=\"ez-toc-section-end\"><\/span><\/h3>\n\n\n\n<p>La suite est minor\u00e9e et d\u00e9croissante. D&#8217;apr\u00e8s le th\u00e9or\u00e8me de convergence des suites monotones, elle converge donc. Notons \\(\\ell\\) sa limite.<\/p>\n\n\n\n<p>Comme <em>f<\/em> est une fonction continue, on peut \u00e9crire : $$u_{n+1} = f(u_n) \\Rightarrow \\lim\\limits_{n\\to+\\infty} u_{n+1} = f\\left(\\lim\\limits_{n\\to+\\infty} u_n\\right),$$c&#8217;est-\u00e0-dire:$$\\ell = f(\\ell).$$On doit donc r\u00e9soudre cette derni\u00e8re \u00e9quation pour d\u00e9terminer la valeur de la limite de la suite.<\/p>\n\n\n\n<p>$$\\begin{align}\\ell = f(\\ell) &amp; \\iff \\ell = \\frac{1}{2}\\left(\\ell + \\frac{a}{\\ell}\\right)\\\\&amp;\\iff 2\\ell = \\ell + \\frac{a}{\\ell}\\\\&amp;\\iff \\ell = \\frac{a}{\\ell}\\\\&amp;\\iff \\ell^2=a\\\\&amp;\\iff \\ell=-\\sqrt{a}\\text{ ou }\\ell = \\sqrt{a} \\end{align}$$<\/p>\n\n\n\n<p>Or, tous les \\(u_n\\) sont positifs donc \\(\\ell\\) ne peut pas \u00eatre \u00e9gale \u00e0 \\(\\sqrt{a}\\). Par cons\u00e9quent,$$\\lim\\limits_{n\\to+\\infty} u_n=\\sqrt{a}.$$<\/p>\n\n\n\n<h3 class=\"wp-block-heading\"><span class=\"ez-toc-section\" id=\"Vitesse_de_convergence_de_la_suite_de_Heron\"><\/span>Vitesse de convergence de la suite de H\u00e9ron<span class=\"ez-toc-section-end\"><\/span><\/h3>\n\n\n\n<p class=\"is-style-Paragraph-paragraph\">Effectuons le calcul suivant:$$\\begin{align}u_{n+1}-\\sqrt{a} &amp; = \\frac{1}{2}\\left( u_n + \\frac{a}{u_n} \\right) &#8211; \\sqrt{a} \\\\ &amp; = \\frac{1}{2}\\left( u_n + \\frac{a}{u_n} \\right) &#8211; \\frac{1}{2}\\times2\\sqrt{a}\\\\&amp;=\\frac{1}{2}\\left( u_n + \\frac{a}{u_n} &#8211; 2\\sqrt{a}\\right)\\\\&amp;=\\frac{1}{2}\\left( \\frac{u_n^2 + a &#8211; 2\\sqrt{a} }{u_n} \\right) \\\\&amp; = \\frac{1}{2}\\times\\frac{\\left(u_n-\\sqrt{a}\\right)^2}{u_n} \\end{align}$$<\/p>\n\n\n\n<p class=\"is-style-Paragraph-paragraph\">Or, on sait que \\(u_n \\geqslant \\sqrt{a} \\) pour <em>n<\/em> > 0, donc \\(\\frac{1}{u_n} \\leqslant \\frac{1}{\\sqrt{a}}\\), ce qui donne alors:$$u_{n+1}-\\sqrt{a} \\leqslant  \\frac{\\left(u_n-\\sqrt{a}\\right)^2}{2\\sqrt{a}}$$<\/p>\n\n\n\n<p class=\"is-style-Paragraph-paragraph\">ce qui signifie que d&#8217;un terme \u00e0 l&#8217;autre, le nombre de d\u00e9cimales correctes double. On dit que la convergence est <em>quadratique<\/em>.<\/p>\n\n\n\n<h2 class=\"wp-block-heading\"><span class=\"ez-toc-section\" id=\"Suite_de_Heron_du_cote_de_Python\"><\/span>Suite de H\u00e9ron: du c\u00f4t\u00e9 de Python<span class=\"ez-toc-section-end\"><\/span><\/h2>\n\n\n\n<p class=\"is-style-Paragraph-paragraph\">Voici un exemple de programme Python:<\/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=\"\">def heron(a,p):\n    u = 3 # premier terme quelconque strictement positif\n    delta = 1\n    n = 0\n    while delta >= 10**(-p):\n        prec = u\n        u = 0.5 * (u + a\/u)\n        delta = abs(prec - u)\n        n += 1\n        \n    return u,n\n<\/pre>\n\n\n\n<p class=\"is-style-Paragraph-paragraph\">J&#8217;ai ici impl\u00e9ment\u00e9 une fonction <strong><em>heron(a,p)<\/em><\/strong> qui admet deux arguments : &#8220;<em>a<\/em>&#8221; est le nombre dont on cherche une valeur approch\u00e9e \u00e0 \\(10^{-p}\\) de sa racine carr\u00e9e.<\/p>\n\n\n\n<p class=\"is-style-Paragraph-paragraph\">Il est \u00e0 noter toutefois qu&#8217;il est inutile de mettre de trop grandes valeurs de <em>p<\/em> car Python est assez limit\u00e9 dans les d\u00e9cimales.<\/p>\n\n\n\n<p class=\"is-style-Paragraph-paragraph\">Ce programme donne par exemple:<\/p>\n\n\n\n<pre class=\"wp-block-preformatted\">>>> heron(11,7)\n(3.3166247903554, 5)\n>>> heron(999,10)\n(31.606961258558215, 8)\n>>> heron(0.5,8)\n(0.7071067811865475, 8)<\/pre>\n\n\n\n<p class=\"is-style-Paragraph-paragraph\">Cela signifie que 5 termes ont suffit pour trouver la valeur approch\u00e9e de \\(\\sqrt{11}\\) \u00e0 7 chiffres significatifs, et que 8 ont suffit pour trouver la valeur approch\u00e9e de \\(\\sqrt{999}\\) et \\(\\sqrt{0,5}\\) \u00e0 8 chiffres significatifs.<\/p>\n","protected":false},"excerpt":{"rendered":"<p>La suite de H\u00e9ron est une suite permettant de trouver une valeur approch\u00e9e d&#8217;une racine carr\u00e9e. Elle tire son nom du math\u00e9maticien H\u00e9ron d&#8217;Alexandrie.<\/p>\n","protected":false},"author":1,"featured_media":8409,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[6,5],"tags":[388,108],"class_list":["post-5238","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-mathematiques","category-python","tag-heron","tag-suites"],"yoast_head":"<!-- This site is optimized with the Yoast SEO plugin v27.3 - https:\/\/yoast.com\/product\/yoast-seo-wordpress\/ -->\n<title>La suite de H\u00e9ron, \u00e9tude math\u00e9matique et impl\u00e9mentation en python<\/title>\n<meta name=\"description\" content=\"La suite de H\u00e9ron est une suite permettant de trouver une valeur approch\u00e9e d&#039;une racine carr\u00e9e. 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