{"id":11482,"date":"2025-10-08T11:38:17","date_gmt":"2025-10-08T09:38:17","guid":{"rendered":"https:\/\/www.mathweb.fr\/euclide\/?p=11482"},"modified":"2025-10-08T11:40:09","modified_gmt":"2025-10-08T09:40:09","slug":"suites-homographiques-comment-les-etudier","status":"publish","type":"post","link":"https:\/\/www.mathweb.fr\/euclide\/2025\/10\/08\/suites-homographiques-comment-les-etudier\/","title":{"rendered":"Suites homographiques: comment les \u00e9tudier ?"},"content":{"rendered":"\n<p class=\"wp-block-paragraph\">Suites homographiques: quels sont leurs secrets ? Comment les \u00e9tudier ? Nous allons voir dans cet article comment trouver une suite auxiliaire qui pourra nous permettre de trouver son terme g\u00e9n\u00e9ral.<\/p>\n\n\n\n<!--more-->\n\n\n\n<div id=\"ez-toc-container\" class=\"ez-toc-v2_0_88 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\/2025\/10\/08\/suites-homographiques-comment-les-etudier\/#Suites_homographiques_definitions\" >Suites homographiques: d\u00e9finitions<\/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\/2025\/10\/08\/suites-homographiques-comment-les-etudier\/#Definition_dune_suite_homographique\" >D\u00e9finition d&rsquo;une suite homographique<\/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\/2025\/10\/08\/suites-homographiques-comment-les-etudier\/#Polynome_caracteristique_des_suites_homographiques\" >Polyn\u00f4me caract\u00e9ristique des suites homographiques<\/a><\/li><\/ul><\/li><li class='ez-toc-page-1 ez-toc-heading-level-2'><a class=\"ez-toc-link ez-toc-heading-4\" href=\"https:\/\/www.mathweb.fr\/euclide\/2025\/10\/08\/suites-homographiques-comment-les-etudier\/#Suites_homographique_introduction_dune_suite_auxiliaire\" >Suites homographique: introduction d&rsquo;une suite auxiliaire<\/a><ul class='ez-toc-list-level-3' ><li class='ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-5\" href=\"https:\/\/www.mathweb.fr\/euclide\/2025\/10\/08\/suites-homographiques-comment-les-etudier\/#Cas_ou_le_polynome_caracteristique_admet_deux_racines_reelles\" >Cas o\u00f9 le polyn\u00f4me caract\u00e9ristique admet deux racines r\u00e9elles<\/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\/2025\/10\/08\/suites-homographiques-comment-les-etudier\/#Cas_ou_le_polynome_caracteristique_admet_une_unique_racine_reelle\" >Cas o\u00f9 le polyn\u00f4me caract\u00e9ristique admet une unique racine r\u00e9elle<\/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\/2025\/10\/08\/suites-homographiques-comment-les-etudier\/#Cas_ou_le_polynome_caracteristique_dadmet_aucune_racine_reelle\" >Cas o\u00f9 le polyn\u00f4me caract\u00e9ristique d&rsquo;admet aucune racine r\u00e9elle<\/a><\/li><\/ul><\/li><\/ul><\/nav><\/div>\n<h2 class=\"wp-block-heading\"><span class=\"ez-toc-section\" id=\"Suites_homographiques_definitions\"><\/span>Suites homographiques: d\u00e9finitions<span class=\"ez-toc-section-end\"><\/span><\/h2>\n\n\n\n<h3 class=\"wp-block-heading\"><span class=\"ez-toc-section\" id=\"Definition_dune_suite_homographique\"><\/span>D\u00e9finition d&rsquo;une suite homographique<span class=\"ez-toc-section-end\"><\/span><\/h3>\n\n\n\n<p class=\"wp-block-paragraph\">Avant tout, il faut savoir qu&rsquo;une fonction homographique est une fonction de la forme:$$f(x)=\\frac{ax+b}{cx+d}\\quad,\\quad c\\neq0.$$<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">La suite \\((u_n)\\) est dite <em>homographique<\/em> si elle est d\u00e9finie par:$$\\begin{cases}u_0\\\\u_{n+1}=\\frac{au_n+b}{cu_n+d}=f(u_n)\\end{cases}$$<\/p>\n\n\n\n<h3 class=\"wp-block-heading\"><span class=\"ez-toc-section\" id=\"Polynome_caracteristique_des_suites_homographiques\"><\/span>Polyn\u00f4me caract\u00e9ristique des suites homographiques<span class=\"ez-toc-section-end\"><\/span><\/h3>\n\n\n\n<p class=\"wp-block-paragraph\">En supposant que \\((u_n)\\) converge, et en utilisant le th\u00e9or\u00e8me du point fixe, sa limite \\(x\\) v\u00e9rifie l&rsquo;\u00e9quation:$$\\begin{align}f(x)=x &amp; \\iff x(cx+d) = ax+b \\\\ cx^2 + (d-a)x &#8211; b = 0  \\end{align}$$<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">On d\u00e9finit alors le <em>polyn\u00f4me caract\u00e9ristique<\/em> de la suite par le polyn\u00f4me:$$P(x)=cx^2+(d-a)x &#8211; b.$$<\/p>\n\n\n\n<h2 class=\"wp-block-heading\"><span class=\"ez-toc-section\" id=\"Suites_homographique_introduction_dune_suite_auxiliaire\"><\/span>Suites homographique: introduction d&rsquo;une suite auxiliaire<span class=\"ez-toc-section-end\"><\/span><\/h2>\n\n\n\n<h3 class=\"wp-block-heading\"><span class=\"ez-toc-section\" id=\"Cas_ou_le_polynome_caracteristique_admet_deux_racines_reelles\"><\/span>Cas o\u00f9 le polyn\u00f4me caract\u00e9ristique admet deux racines r\u00e9elles<span class=\"ez-toc-section-end\"><\/span><\/h3>\n\n\n\n<p class=\"wp-block-paragraph\">Notons \\(\\alpha\\) et \\(\\beta\\) les deux racines du polyn\u00f4me caract\u00e9ristique, et consid\u00e9rons la suite \\((v_n)\\) d\u00e9finie pour tout entier naturel \\(n\\) par: $$v_n = \\frac{u_n-\\alpha}{u_n-\\beta}.$$Alors:<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">$$\\begin{aligned}v_{n+1} &amp; = \\frac{u_{n+1}-\\alpha}{u_{n+1}-\\beta} \\\\ &amp; = \\frac{\\frac{au_n+b}{cu_n+d}-\\frac{a\\alpha+b}{c\\alpha+d}}{\\frac{au_n+b}{cu_n+d} &#8211; \\frac{a\\beta+b}{c\\beta+d}}\\text{ car }\\alpha=f(\\alpha)\\text{ et }\\beta=f(\\beta) \\\\ &amp; = \\frac{ \\frac{\\left(au_n+b\\right)\\left(c\\alpha+d\\right)-\\left(cu_n+d\\right)(a\\alpha+b)}{(cu_n+d)(c\\alpha+d)}}{\\frac{\\left(au_n+b\\right)(c\\beta+d)-\\left(cu_n+d\\right)(a\\beta+b)}{(cu_n+d)(c\\beta+d)}} \\\\ &amp; =  \\frac{\\left(au_n+b\\right)\\left(c\\alpha+d\\right)-\\left(cu_n+d\\right)(a\\alpha+b)}{(cu_n+d)(c\\alpha+d)} \\times \\frac{(cu_n+d)(c\\beta+d)}{\\left(au_n+b\\right)(c\\beta+d)-\\left(cu_n+d\\right)(a\\beta+b)} \\\\ &amp; = \\frac{ac\\alpha u_n + adu_n + bc\\alpha + bd &#8211; ac\\alpha u_n &#8211; bcu_n &#8211; ad\\alpha &#8211; bd}{ac\\beta u_n + adu_n + bc\\beta + bd &#8211; ac\\beta u_n &#8211; bcu_n &#8211; ad\\beta &#8211; bd} \\times \\frac{c\\beta + d}{c\\alpha + d} \\\\ &amp; = \\frac{\\left(u_n &#8211; \\alpha\\right)(ad-bc)}{\\left(u_n &#8211; \\beta\\right)(ad-bc)}\\times\\frac{c\\beta+d}{c\\alpha+d}\\\\ &amp; = \\frac{c\\beta+d}{c\\alpha+d}v_n.\\end{aligned}$$<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Le calcul est un peu fastidieux, mais on d\u00e9montre ainsi que la suite \\((v_n)\\) est n\u00e9cessairement g\u00e9om\u00e9trique de raison \\(q=\\frac{c\\beta+d}{c\\alpha+d}\\).<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Comment choisir l&rsquo;ordre des racines? Que mettre \u00e0 la place de \\(\\alpha\\) et de \\(\\beta\\)? Si on souhaite que \\((u_n)\\) converge alors il faut que \\((v_n)\\) converge aussi. Et comme \\((v_n)\\) est g\u00e9om\u00e9trique, elle ne peut converger que vers 0, donc il est n\u00e9cessaire que |q|&lt;1. Le choix de \\(\\alpha\\) et \\(\\beta\\) se fait donc sur le fait que \\(\\left|\\frac{c\\beta+d}{c\\alpha+d}\\right|&lt;1\\).<\/p>\n\n\n\n<blockquote class=\"wp-block-quote is-layout-flow wp-block-quote-is-layout-flow\">\n<p class=\"wp-block-paragraph\">Par exemple, si pour tout entier naturel n:$$\\begin{cases}u_0=1\\\\u_{n+1}=\\frac{2u_n+1}{u_n+1}\\end{cases}$$alors son polyn\u00f4me caract\u00e9ristique est:$$P(x)=x^2-x-1$$qui admet pour racines \\(\\varphi=\\frac{1+\\sqrt5}{2}\\) et \\(\\overline{\\varphi}=\\frac{1-\\sqrt5}{2}\\). Comme \\(c\\varphi+d=\\frac{3+\\sqrt5}{2}\\) et \\(c\\overline{\\varphi}+d=\\frac{3-\\sqrt5}{2}&lt;c\\varphi+d\\), on doit prendre \\(\\alpha=\\varphi\\) et \\(\\beta=\\overline{\\varphi}\\) pour que |q|&lt;1. Dans ce cas, \\((u_n)\\) converge vers \\(\\varphi\\).<\/p>\n<\/blockquote>\n\n\n\n<h3 class=\"wp-block-heading\"><span class=\"ez-toc-section\" id=\"Cas_ou_le_polynome_caracteristique_admet_une_unique_racine_reelle\"><\/span>Cas o\u00f9 le polyn\u00f4me caract\u00e9ristique admet une unique racine r\u00e9elle<span class=\"ez-toc-section-end\"><\/span><\/h3>\n\n\n\n<p class=\"wp-block-paragraph\">Notons \\(\\alpha\\) cette racine et posons pour tout entier naturel \\(n\\):$$w_n=\\frac{1}{u_n-\\alpha}.$$Alors,<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">$$\\begin{align}w_{n+1} &amp; = \\frac{1}{u_{n+1}-\\alpha} \\\\ &amp; = \\frac{1}{\\tfrac{au_n+b}{cu_n+d}-\\tfrac{a\\alpha+b}{c\\alpha+d}} \\\\ &amp; = \\frac{\\left(cu_n+d\\right)(c\\alpha+d)}{\\left(au_n+b\\right)(c\\alpha+d)-\\left(cu_n+d\\right)(a\\alpha+b)} \\\\ &amp; = \\frac{(cu_n+d)(c\\alpha+d)}{(u_n-\\alpha)(ad-bc)}.\\end{align}$$<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Or, \\(\\alpha\\) \u00e9tant une racine double de P, ce dernier admet un discriminant nul. Donc :<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">$$\\begin{align}(d-a)^2+4bc=0 &amp; \\iff ad-bc =ad-\\frac{(d-a)^2}{4}\\\\&amp; \\iff ad-bc=\\frac{(a+d)^2}{4}.\\end{align}$$<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">De plus, comme \\(\\alpha\\) est une racine double de P, on peut la calculer avec les formules vues en 1\u00e8re : $$ \\alpha = \\frac{a-d}{2c}.$$ On a alors : $$c\\alpha+d=\\frac{a+d}{2}.$$ Ainsi :<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">$$\\begin{align}w_{n+1} &amp; = \\frac{cu_n+d}{u_n-\\alpha}\\times\\frac{\\tfrac{a+d}{2}}{\\tfrac{(a+d)^2}{4}}\\\\&amp; = \\frac{cu_n+d}{u_n-\\alpha}\\times\\frac{2}{a+d}\\\\&amp; = \\frac{2}{a+d}\\times\\frac{cu_n-c\\alpha+c\\alpha+d}{u_n-\\alpha}\\\\&amp; = \\frac{2}{a+d}\\times\\left[c+\\frac{c\\alpha+d}{u_n-\\alpha} \\right]\\\\&amp; = \\frac{2c}{a+d}+2\\frac{c\\alpha+d}{a+d}\\times\\frac{1}{u_n-\\alpha}\\\\&amp; = \\frac{2c}{a+d}+2\\frac{c\\alpha+d}{2(c\\alpha+d)}\\times\\frac{1}{u_n-\\alpha}\\\\&amp; = \\frac{2c}{a+d}+\\frac{1}{u_n-a}\\\\w_{n+1} &amp; = w_n+\\frac{2c}{a+d}.\\end{align}$$<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Notons que \\(a+d \\neq 0\\). En effet, \\(c\\alpha+d=\\frac{a+d}{2}\\) et comme \\(\\alpha\\) existe, \\(c\\alpha+d\\neq 0\\). La suite \\((w_n)\\) est donc arithm\u00e9tique de raison non nulle \\(r=\\frac{2c}{a+d}\\).<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">\\((w_n)\\) \u00e9tant arithm\u00e9tique, sa limite est infinie et donc la limite de \\((u_n)\\) est n\u00e9cessairement \u00e9gale \u00e0 \\(\\alpha\\).<\/p>\n\n\n\n<blockquote class=\"wp-block-quote is-layout-flow wp-block-quote-is-layout-flow\">\n<p class=\"wp-block-paragraph\">Par exemple, si pour tout entier naturel n, $$\\begin{cases}u_0=1\\\\u_{n+1}=\\frac{4u_n-1}{4u_n}\\end{cases}$$le polyn\u00f4me caract\u00e9ristique de \\((u_n)\\)est:$$P(x)=(2x-1)^2$$ dont la seule racine r\u00e9elle est \\(\\frac{1}{2}\\). Ainsi, la suite \\((u_n)\\) converge vers \\(\\frac{1}{2}\\).<\/p>\n<\/blockquote>\n\n\n\n<h3 class=\"wp-block-heading\"><span class=\"ez-toc-section\" id=\"Cas_ou_le_polynome_caracteristique_dadmet_aucune_racine_reelle\"><\/span>Cas o\u00f9 le polyn\u00f4me caract\u00e9ristique d&rsquo;admet aucune racine r\u00e9elle<span class=\"ez-toc-section-end\"><\/span><\/h3>\n\n\n\n<p class=\"wp-block-paragraph\">L\u00e0, on sort du champs r\u00e9el pour passer au champs complexe. Dans ce cas, il admet deux racines complexes.<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Ce que nous avons dit dans le premiers cas reste valable: on pose \\(v_n=\\frac{u_n-\\alpha}{u_n-\\beta}\\) mais la suite devient alors une suite g\u00e9om\u00e9trique complexe.<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">On montre alors que pour tout entier naturel <em>n<\/em>, $$u_n=\\frac{\\beta q^n(u_0-\\alpha)-\\alpha(u_0-\\beta)}{q^n(u_0-\\alpha)-(u_0-\\beta)}.$$<\/p>\n\n\n\n<blockquote class=\"wp-block-quote is-layout-flow wp-block-quote-is-layout-flow\">\n<p class=\"wp-block-paragraph\">Par exemple, si pour tout entier naturel n,$$\\begin{cases}u_0\\\\u_{n+1}=\\frac{1}{1-u_n}\\end{cases}$$alors son polyn\u00f4me caract\u00e9ristique est:$$P(x)=-x^2+x-1.$$Ses racines sont alors \\(\\alpha=\\text{e}^{\\text{i}\\frac{\\pi}{3}}\\) et \\(\\beta=\\text{e}^{-\\text{i}\\frac{\\pi}{3}}\\). On pose alors:$$v_n=\\frac{u_n-\\alpha}{u_n-\\beta}.$$La suite \\((v_n)\\) est alors g\u00e9om\u00e9trique de raison \\(q=\\text{e}^{-2\\text{i}\\frac{\\pi}{3}}\\). On a alors \\(v_0=\\frac{1-\\alpha}{1-\\beta}\\) et:$$u_n=\\frac{ \\beta(u_0-\\alpha)q^n &#8211; \\alpha(u_0-\\beta)}{(u_0-\\beta)q^n-(u_0-\\beta)}.$$Le calcul des termes successifs donne avec \\(u_0=\\frac{1}{2}\\): \\(u_1=-1\\), \\(u_2=2\\), \\(u_3=\\frac{1}{2}=u_0\\). La suite est cyclique.<\/p>\n<\/blockquote>\n\n\n\n<p class=\"wp-block-paragraph\">Les suites homographiques dont le polyn\u00f4me caract\u00e9ristique admet deux racines complexes sont cycliques. La preuve est assez simple \u00e0 faire mais comme j&rsquo;ai mis toute mon \u00e2me dans cet article, et que je suis \u00e9puis\u00e9, je laisse le soin au lecteur (ou \u00e0 la lectrice) de s&#8217;emparer de cette preuve.<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Si les suites vous passionnent, n&rsquo;h\u00e9sitez pas \u00e0 vous procurer le magnifique livre \u00ab\u00a0Ainsi de suite\u00a0\u00bb:<\/p>\n\n\n\n<figure class=\"wp-block-embed is-type-wp-embed is-provider-livres-de-maths-pdf wp-block-embed-livres-de-maths-pdf\"><div class=\"wp-block-embed__wrapper\">\n<blockquote class=\"wp-embedded-content\" data-secret=\"RB450yNsUn\"><a href=\"https:\/\/livres-maths-pdf.mathweb.fr\/produit\/ainsi-de-suite-2025\/\">Ainsi de suite 2025<\/a><\/blockquote><iframe loading=\"lazy\" class=\"wp-embedded-content\" sandbox=\"allow-scripts\" security=\"restricted\" style=\"position: absolute; visibility: hidden;\" title=\"\u00ab\u00a0Ainsi de suite 2025\u00a0\u00bb &#8212; Livres de maths PDF\" src=\"https:\/\/livres-maths-pdf.mathweb.fr\/produit\/ainsi-de-suite-2025\/embed\/#?secret=dqdI7aIjjH#?secret=RB450yNsUn\" data-secret=\"RB450yNsUn\" width=\"600\" height=\"338\" frameborder=\"0\" marginwidth=\"0\" marginheight=\"0\" scrolling=\"no\"><\/iframe>\n<\/div><\/figure>\n","protected":false},"excerpt":{"rendered":"<p>Suites homographiques: quels sont leurs secrets ? Comment les \u00e9tudier ? Nous allons voir dans cet article comment trouver une suite auxiliaire qui pourra nous permettre de trouver son terme g\u00e9n\u00e9ral.<\/p>\n","protected":false},"author":1,"featured_media":11509,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[6],"tags":[],"class_list":["post-11482","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-mathematiques"],"yoast_head":"<!-- This site is optimized with the Yoast SEO plugin v28.5 - https:\/\/yoast.com\/product\/yoast-seo-wordpress\/ -->\n<title>Suites homographiques: comment les \u00e9tudier ? - Mathweb.fr<\/title>\n<meta name=\"description\" content=\"Suites homographiques: quels sont leurs secrets? Comment les \u00e9tudier ? 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