{"id":2049,"date":"2020-02-25T16:52:54","date_gmt":"2020-02-25T15:52:54","guid":{"rendered":"https:\/\/www.mathweb.fr\/euclide\/?p=2049"},"modified":"2020-02-29T15:18:57","modified_gmt":"2020-02-29T14:18:57","slug":"la-suite-de-fibonacci","status":"publish","type":"post","link":"https:\/\/www.mathweb.fr\/euclide\/2020\/02\/25\/la-suite-de-fibonacci\/","title":{"rendered":"La suite de Fibonacci"},"content":{"rendered":"\n<p class=\"wp-block-paragraph\">La suite de Fibonacci est la suite d\u00e9finie par ses deux premiers termes \\(F_0=F_1=1\\) et par la relation de r\u00e9currence suivante:$$\\forall n\\in\\mathbb{N},\\ F_{n+2}=F_{n+1}+F_{n}.$$ Nous allons nous pencher sur cette suite afin de d\u00e9terminer une expression de son terme g\u00e9n\u00e9ral en fonction de son rang.<\/p>\n\n\n\n<div class=\"wp-block-image\"><figure class=\"aligncenter size-medium\"><a href=\"https:\/\/www.mathweb.fr\/euclide\/wp-content\/uploads\/2020\/02\/fibonacci.jpg\" data-fancybox=\"gallery\"><img loading=\"lazy\" decoding=\"async\" width=\"262\" height=\"300\" src=\"https:\/\/www.mathweb.fr\/euclide\/wp-content\/uploads\/2020\/02\/fibonacci-262x300.jpg\" alt=\"\" class=\"wp-image-2059\" srcset=\"https:\/\/www.mathweb.fr\/euclide\/wp-content\/uploads\/2020\/02\/fibonacci-262x300.jpg 262w, https:\/\/www.mathweb.fr\/euclide\/wp-content\/uploads\/2020\/02\/fibonacci-300x343.jpg 300w, https:\/\/www.mathweb.fr\/euclide\/wp-content\/uploads\/2020\/02\/fibonacci-600x686.jpg 600w, https:\/\/www.mathweb.fr\/euclide\/wp-content\/uploads\/2020\/02\/fibonacci-768x878.jpg 768w, https:\/\/www.mathweb.fr\/euclide\/wp-content\/uploads\/2020\/02\/fibonacci.jpg 787w\" sizes=\"auto, (max-width: 262px) 100vw, 262px\" \/><\/a><figcaption>Leonardo Bonacci, dit <em>Fibonacci<\/em><\/figcaption><\/figure><\/div>\n\n\n\n<!--more-->\n\n\n\n<p class=\"wp-block-paragraph\">La premi\u00e8re chose que j&rsquo;ai envie d&rsquo;\u00e9crire, c&rsquo;est:$$\\forall n\\in\\mathbb{N},\\ F_{n+2}-F_{n+1}-F_n=0.$$Ensuite, je me dis que \u00e7a serait cool si cette suite \u00e9tait g\u00e9om\u00e9trique&#8230; Bon, elle ne l&rsquo;est pas, mais j&rsquo;ai envie de voir un truc&#8230; Supposons alors que \\(F_n=q^n\\), o\u00f9 \\(q \\neq 0\\). Alors, la relation pr\u00e9c\u00e9dente devient:$$q^{n+2}-q^{n+1}-q^n=0$$ soit:$$q^n(q^2-q-1)=0.$$Comme \\(q\\) n&rsquo;est pas nul, cela signifie que \\(q^2-q-1=0\\), c&rsquo;est-\u00e0-dire, apr\u00e8s calcul du discriminant, je trouve deux valeurs possibles pour \\(q\\):$$q_1=\\frac{1-\\sqrt5}{2}\\text{ ou }q_2=\\frac{1+\\sqrt5}{2}.$$Mais bon&#8230; je ne suis pas si stupide que \u00e7a: je vois bien que ni \\((q_1^n)\\) ni \\((q_2^2)\\) ne convient car les deuxi\u00e8mes termes de ces deux suites ne co\u00efncident pas avec le deuxi\u00e8me terme de la suite de Fibonacci.<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">C&rsquo;est l\u00e0 que j&rsquo;ai une id\u00e9e : pourquoi ne pas consid\u00e9rer une combinaison lin\u00e9aire de ces deux suites ? Allez ! Je me lance ! Je pose pour tout entier naturel <em>n<\/em>:$$u_n=\\alpha q_1^n + \\beta q_2^n.$$Il est assez facile de constater que:$$\\begin{align}u_{n+2}-u_{n+1}-u_n &amp; = \\alpha q_1^n(q_1^2-q_1-1) + \\beta q_2^n(q_2^2-q_2-1)\\\\&amp; = 0\\end{align}$$car \\( q_1^2-q_1-1  = 0\\) et \\( q_2^2-q_2-1 = 0\\). Ainsi, la suite de Fibonacci fait partie des suites \\((u_n)\\). Il ne reste plus qu&rsquo;\u00e0 trouver les valeurs de \\(\\alpha\\) et \\(\\beta\\). Pour cela, on va consid\u00e9rer que:$$\\begin{cases}F_0 = \\alpha + \\beta &amp; = 1\\\\F_1=\\alpha q_1 + \\beta q_2 &amp; = 1\\end{cases}$$On arrive alors \u00e0:$$\\alpha=\\frac{5-\\sqrt5}{10}\\text{ et }\\beta=\\frac{5+\\sqrt5}{10}.$$Ainsi, la suite de Fibonacci peut s&rsquo;exprimer de la mani\u00e8re suivante:$$F_n=\\left( \\frac{5-\\sqrt5}{10}  \\right)\\left( \\frac{1-\\sqrt5}{2} \\right)^n +  \\left( \\frac{5+\\sqrt5}{10}  \\right)\\left( \\frac{1+\\sqrt5}{2} \\right)^n.$$<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Le nombre \\(\\displaystyle\\frac{1+\\sqrt5}{2}\\) qui appara\u00eet dans la formule est appel\u00e9 le <em>nombre d&rsquo;or<\/em>; on le note souvent \\(\\varphi\\) ou \\(\\phi\\) (\u00ab\u00a0phi\u00a0\u00bb). Ce qu&rsquo;il y a d&rsquo;int\u00e9ressant, c&rsquo;est que si on calcule les quotients successifs \\(\\displaystyle\\frac{F_{n+1}}{F_n}\\), on s&rsquo;aper\u00e7oit qu&rsquo;ils se rapprochent de plus en plus du nombre d&rsquo;or (voir <a href=\"https:\/\/www.mathweb.fr\/euclide\/2019\/03\/15\/python-et-le-nombre-dor\/\" target=\"_blank\" rel=\"noreferrer noopener\" aria-label=\"cet article (s\u2019ouvre dans un nouvel onglet)\">cet article<\/a>).<\/p>\n","protected":false},"excerpt":{"rendered":"<p>La suite de Fibonacci est la suite d\u00e9finie par ses deux premiers termes \\(F_0=F_1=1\\) et par la relation de r\u00e9currence suivante:$$\\forall n\\in\\mathbb{N},\\ F_{n+2}=F_{n+1}+F_{n}.$$ Nous allons nous pencher sur cette suite afin de d\u00e9terminer une expression de son terme g\u00e9n\u00e9ral en fonction de son rang.<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[21,6],"tags":[96,175],"class_list":["post-2049","post","type-post","status-publish","format-standard","hentry","category-enseignement","category-mathematiques","tag-fibonacci","tag-golden-number"],"yoast_head":"<!-- This site is optimized with the Yoast SEO plugin v27.8 - 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