{"id":31167,"date":"2021-03-27T13:00:03","date_gmt":"2021-03-27T13:00:03","guid":{"rendered":"http:\/\/toposuranos.com\/material\/?p=31167"},"modified":"2025-01-07T03:57:44","modified_gmt":"2025-01-07T03:57:44","slug":"la-formule-de-stirling","status":"publish","type":"post","link":"http:\/\/toposuranos.com\/material\/fr\/la-formule-de-stirling\/","title":{"rendered":"La formule de Stirling"},"content":{"rendered":"<style>\n    p, ul, ol {\n        text-align: justify;\n    }\n    h1, h2 {\n        text-align: center;\n    }\n<\/style>\n<h1>La Formule de Stirling<\/h1>\n<p><em>La formule de Stirling est un outil essentiel pour simplifier les calculs de factorielles de grands nombres, offrant une approximation rapide et pratique.<\/p>\n<p>Ce r\u00e9sultat est particuli\u00e8rement utile dans des domaines tels que la thermodynamique, la probabilit\u00e9 et l&#8217;analyse asymptotique, o\u00f9 l&#8217;on travaille souvent avec des nombres extr\u00eamement grands. Comprendre son d\u00e9veloppement ne facilite pas seulement son application, mais permet aussi d&#8217;appr\u00e9cier sa pertinence dans le calcul efficace et la r\u00e9solution de probl\u00e8mes complexes.<br \/>\n<\/em><\/p>\n<p style=\"text-align:center;\"><strong>Objectifs d&#8217;apprentissage :<\/strong><br \/>\n\u00c0 la fin de ce cours, l&#8217;\u00e9tudiant sera capable de<\/p>\n<ol>\n<li><strong>Comprendre<\/strong> la d\u00e9duction de la formule de Stirling \u00e0 partir de la d\u00e9finition de la factorielle via la fonction Gamma.<\/li>\n<li><strong>Appliquer<\/strong> la formule de Stirling pour approximer les factorielles de tr\u00e8s grands nombres.<\/li>\n<li><strong>Calculer<\/strong> des approximations logarithmiques de factorielles en utilisant des outils de base en logarithmes et exponentielles.<\/li>\n<\/ol>\n<p style=\"text-align:center;\"><strong><u>TABLE DES MATI\u00c8RES<\/u> :<\/strong><br \/>\n<a href=\"#1\">D\u00e9monstration de la formule de Stirling<\/a><br \/>\n<a href=\"#2\">Approximation logarithmique de la factorielle<\/a><br \/>\n<a href=\"#3\">Exemple : Approximation de la Factorielle d&#8217;un Tr\u00e8s Grand Nombre<\/a>\n<\/p>\n<p><a name=\"1\"><\/a><\/p>\n<h2>D\u00e9monstration de la formule de Stirling<\/h2>\n<p><center><iframe class=\"lazyload\" width=\"560\" height=\"315\" data-src=\"https:\/\/www.youtube.com\/embed\/CcHCyRR1WrY?si=eTZsj4wBqQ2krELG\" title=\"Lecteur vid\u00e9o YouTube\" frameborder=\"0\" allow=\"accelerometer; autoplay; clipboard-write; encrypted-media; gyroscope; picture-in-picture; web-share\" referrerpolicy=\"strict-origin-when-cross-origin\" allowfullscreen><\/iframe><\/center><\/p>\n<p>Le d\u00e9veloppement de la formule de Stirling commence par la d\u00e9finition de la factorielle via la fonction Gamma, qui est :<\/p>\n<p style=\"text-align:center;\"><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">n! =\\Gamma(n+1) = \\displaystyle \\int_0^\\infty t^n e^{-t} \\, dt<\/span><\/span><\/p>\n<p>En utilisant cette expression, nous effectuons un changement de variable : <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">t = nx<\/span><\/span>. Cela implique que <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">x \\in [0, \\infty[<\/span><\/span> et <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">dt = n dx<\/span><\/span>. Avec ce changement, l&#8217;int\u00e9grale se transforme comme suit :<\/p>\n<p style=\"text-align:center;\"><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">n! = \\Gamma(n+1) = \\displaystyle \\int_0^\\infty (nx)^n e^{-nx} n \\, dx = n^{n+1} \\int_0^\\infty x^n e^{-nx} dx<\/span><\/span><\/p>\n<p>Ensuite, nous effectuons un second changement de variable : <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">x = 1 + \\dfrac{s}{\\sqrt{n}}<\/span><\/span>. Cela implique :<\/p>\n<p style=\"text-align:center;\"><span class=\"katex-eq\" data-katex-display=\"false\">\\begin{array}{rl}\n\n &amp; s = (x-1)\\sqrt{n}, \\quad s \\in [-\\sqrt{n}, \\infty[ \\\\ \\\\\n\n &amp; dx = \\dfrac{ds}{\\sqrt{n}}\n\n\\end{array}<\/span>\n<p>Avec ce changement de variable, l&#8217;int\u00e9grale prend la forme suivante :<\/p>\n<p style=\"text-align:center;\"><span class=\"katex-eq\" data-katex-display=\"false\">\n\\begin{array}{rl}\n\nn! = \\Gamma(n+1) &amp;= \\displaystyle n^{n+1} \\int_{-\\sqrt{n}}^\\infty \\left( 1 + \\dfrac{s}{\\sqrt{n}} \\right)^n e^{-n\\left(1+\\dfrac{s}{\\sqrt{n}}\\right)} \\dfrac{ds}{\\sqrt{n}} \\\\ \\\\\n\n&amp;= \\displaystyle \\dfrac{n^{n+1}}{\\sqrt{n}} \\int_{-\\sqrt{n}}^\\infty e^{n\\ln\\left( 1 + \\dfrac{s}{\\sqrt{n}} \\right)} e^{-n - s\\sqrt{n}} ds \\\\ \\\\\n\n&amp;= \\displaystyle n^n e^{-n} \\sqrt{n} \\int_{-\\sqrt{n}}^\\infty e^{n\\ln\\left(1+\\dfrac{s}{\\sqrt{n}}\\right) - s\\sqrt{n}} ds\n\n\\end{array}\n\n<\/span>\n<p>Nous utilisons maintenant le d\u00e9veloppement en s\u00e9rie de Taylor pour le logarithme naturel :<\/p>\n<p style=\"text-align:center;\"><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\ln(1+x) = \\displaystyle\\sum_{k=1}^{\\infty} \\dfrac{(-1)^{k+1}x^k}{k}   <\/span><\/span><\/p>\n<p>En appliquant ce d\u00e9veloppement \u00e0 <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\ln\\left(1+\\dfrac{s}{\\sqrt{n}}\\right)<\/span><\/span>, nous d\u00e9veloppons l&#8217;expression exponentielle comme suit :<\/p>\n<p style=\"text-align:center;\"><span class=\"katex-eq\" data-katex-display=\"false\">\\begin{array}{rl}\n\nn\\ln\\left(1+\\dfrac{s}{\\sqrt{n}}\\right) - s\\sqrt{n} &amp; = \\displaystyle n \\left[\\sum_{k=1}^{\\infty} \\dfrac{(-1)^{k+1}\\left(\\dfrac{s}{\\sqrt{n}} \\right)^k}{k} \\right] - s\\sqrt{n} \\\\ \\\\\n\n&amp; = n \\left[ \\dfrac{s}{\\sqrt{n}} - \\dfrac{s^2}{2n} + \\dfrac{s^3}{3n\\sqrt{n}} - \\dfrac{s^4}{4n^2} + \\dfrac{s^5}{5n^2\\sqrt{n}} \\cdots \\right] - s\\sqrt{n} \\\\ \\\\\n\n&amp; = s\\sqrt{n} - \\dfrac{s^2}{2} + \\dfrac{s^3}{3\\sqrt{n}} - \\dfrac{s^4}{4n} + \\dfrac{s^5}{5n\\sqrt{n}} \\cdots - s\\sqrt{n} \\\\ \\\\\n\n&amp; = - \\dfrac{s^2}{2} + \\dfrac{s^3}{3\\sqrt{n}} - \\dfrac{s^4}{4n} + \\dfrac{s^5}{5n\\sqrt{n}} \\cdots \\\\ \\\\\n\n&amp; = - \\dfrac{s^2}{2} + \\displaystyle \\sum_{k=3}^\\infty \\dfrac{(-1)^{k+1}s^k}{k\\sqrt{n^{k-2}}}\n\n\\end{array}\n\n<\/span>\n<p>De cette mani\u00e8re, nous pouvons \u00e9crire l&#8217;expression compl\u00e8te comme :<\/p>\n<p style=\"text-align:center;\"><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">n! = \\Gamma(n+1) = \\displaystyle n^n e^{-n} \\sqrt{n} \\int_{-\\sqrt{n}}^\\infty e^{- \\dfrac{s^2}{2} + \\displaystyle \\sum_{k=3}^\\infty \\dfrac{(-1)^{k+1}s^k}{k\\sqrt{n^{k-2}}}} ds <\/span><\/span><\/p>\n<p>Ce r\u00e9sultat est fondamental pour calculer les factorielles de tr\u00e8s grands nombres. Lorsque <span class=\"katex-eq\" data-katex-display=\"false\">n<\/span> augmente, les termes de la somme dans l&#8217;exponentielle tendent vers z\u00e9ro, ne laissant que le terme dominant. Cela simplifie l&#8217;int\u00e9grale, qui peut \u00eatre r\u00e9solue comme une int\u00e9grale gaussienne :<\/p>\n<p style=\"text-align:center;\"><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">n! = \\Gamma(n+1) \\approx \\displaystyle n^n e^{-n} \\sqrt{n} \\int_{-\\infty}^\\infty e^{- \\frac{s^2}{2}} ds = n^n e^{-n} \\sqrt{n} \\sqrt{2\\pi} <\/span><\/span><\/p>\n<p><strong>Ce r\u00e9sultat est connu sous le nom de formule de Stirling pour la factorielle des grands nombres :<\/strong><\/p>\n<p style=\"text-align:center;\"><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\boxed{n! \\approx \\sqrt{2\\pi n}\\left(\\dfrac{n}{e}\\right)^{n}}<\/span><\/span><\/p>\n<p><a name=\"2\"><\/a><\/p>\n<h2>Approximation logarithmique de la factorielle<\/h2>\n<p><center><iframe class=\"lazyload\" width=\"560\" height=\"315\" data-src=\"https:\/\/www.youtube.com\/embed\/ASifSl6YgTk?si=8rHCOoUbWoiH962o\" title=\"Lecteur vid\u00e9o YouTube\" frameborder=\"0\" allow=\"accelerometer; autoplay; clipboard-write; encrypted-media; gyroscope; picture-in-picture; web-share\" referrerpolicy=\"strict-origin-when-cross-origin\" allowfullscreen><\/iframe><\/center><\/p>\n<p>Un r\u00e9sultat direct de la formule de Stirling est l&#8217;approximation logarithmique de la factorielle. En prenant le logarithme naturel de la formule de Stirling, nous obtenons :<\/p>\n<p style=\"text-align:center;\"><span class=\"katex-eq\" data-katex-display=\"false\">\\begin{array}{rcl}\n\n\\ln(n!) \\approx \\ln\\left( \\sqrt{2n\\pi}\\left(\\dfrac{n}{e}\\right)^{n} \\right) &amp;=&amp; \\dfrac{1}{2}\\ln(2n\\pi) + n\\ln\\left(\\dfrac{n}{e}\\right) \\\\ \\\\\n\n&amp;=&amp;  \\dfrac{1}{2}\\ln(2n\\pi) + n\\ln(n) - n \\\\ \\\\\n\n&amp;\\approx &amp; n\\ln(n) - n\n\n\\end{array}<\/span>\n<p>Dans la derni\u00e8re \u00e9tape, une approximation suppl\u00e9mentaire est r\u00e9alis\u00e9e en n\u00e9gligeant le terme <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\dfrac{1}{2}\\ln(2n\\pi)<\/span><\/span>. Ce terme devient insignifiant par rapport \u00e0 <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">n\\ln(n) - n<\/span><\/span> pour de grandes valeurs de <span class=\"katex-eq\" data-katex-display=\"false\">n<\/span>.<\/p>\n<p>La validit\u00e9 de cette approximation est justifi\u00e9e en calculant l&#8217;erreur relative entre les deux expressions :<\/p>\n<p style=\"text-align:center;\"><span class=\"katex-eq\" data-katex-display=\"false\">\\begin{array}{rcl}\n\n\\text{Approximation Initiale} &amp; = &amp; \\dfrac{1}{2}\\ln(2n\\pi) + n\\ln(n) - n \\\\ \\\\\n\n\\text{Approximation Finale} &amp; = &amp; n\\ln(n) - n \\\\ \\\\\n\n\\text{Erreur Relative} &amp;=&amp; \\dfrac{\\text{Approximation Finale} - \\text{Approximation Initiale}}{\\text{Approximation Initiale}} \\\\ \\\\\n\n&amp;=&amp; \\dfrac{-\\dfrac{1}{2}\\ln(2n\\pi)}{\\dfrac{1}{2}\\ln(2n\\pi) + n\\ln(n) - n}\n\n\\end{array}<\/span>\n<p>En calculant la limite lorsque <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">n \\to \\infty<\/span><\/span> :<\/p>\n<p style=\"text-align:center;\"><span class=\"katex-eq\" data-katex-display=\"false\">\\begin{array}{rl}\n\n\\displaystyle \\lim_{n\\to\\infty} \\text{Erreur Relative} &amp; = \\displaystyle \\lim_{n\\to\\infty} \\dfrac{-\\dfrac{1}{2}\\ln(2n\\pi)}{\\dfrac{1}{2}\\ln(2n\\pi) + n\\ln(n) - n} \\\\ \\\\\n\n&amp; = \\displaystyle \\lim_{n\\to\\infty} \\dfrac{-\\dfrac{1}{2n}}{\\dfrac{1}{2n} + \\ln(n) + 1 - 1} = 0\n\n\\end{array}<\/span>\n<p>Par cons\u00e9quent, puisque l&#8217;erreur tend vers z\u00e9ro pour de grandes valeurs de <span class=\"katex-eq\" data-katex-display=\"false\">n<\/span>, nous pouvons utiliser l&#8217;approximation logarithmique suivante avec confiance :<\/p>\n<p style=\"text-align:center;\"><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\boxed{\\ln(n!) \\approx n\\ln(n) - n}<\/span><\/span><\/p>\n<p><a name=\"3\"><\/a><\/p>\n<h2>Exemple : Approximation de la factorielle d&#8217;un tr\u00e8s grand nombre<\/h2>\n<p><center><iframe class=\"lazyload\" width=\"560\" height=\"315\" data-src=\"https:\/\/www.youtube.com\/embed\/kja9niRWjpg?si=eHkcZYaq0Fgntc3G\" title=\"Lecteur vid\u00e9o YouTube\" frameborder=\"0\" allow=\"accelerometer; autoplay; clipboard-write; encrypted-media; gyroscope; picture-in-picture; web-share\" referrerpolicy=\"strict-origin-when-cross-origin\" allowfullscreen><\/iframe><\/center><\/p>\n<p>Calculer la factorielle de nombres extr\u00eamement grands, comme <em>10.000!<\/em>, est pratiquement impossible avec des outils conventionnels en raison de la taille du r\u00e9sultat. Cependant, en utilisant l&#8217;approximation logarithmique de la factorielle d\u00e9riv\u00e9e de la formule de Stirling, nous pouvons rendre cela faisable m\u00eame avec des calculatrices basiques.<\/p>\n<p>La formule logarithmique de la factorielle nous indique :<\/p>\n<p style=\"text-align:center;\"><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\ln(10.000!) \\approx 10.000 \\ln(10.000) - 10.000<\/span><\/span><\/p>\n<p>Pour convertir des logarithmes naturels (<span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\ln<\/span><\/span>) en logarithmes de base 10 (<span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\log<\/span><\/span>), nous utilisons la relation :<\/p>\n<p style=\"text-align:center;\"><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\ln(10.000!) = \\dfrac{\\log(10.000!)}{\\log(e)}<\/span><\/span><\/p>\n<p>Cela implique que :<\/p>\n<p style=\"text-align:center;\"><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\log(10.000!) \\approx \\log(e) \\cdot (10.000 \\ln(10.000) - 10.000)<\/span><\/span><\/p>\n<p>Par cons\u00e9quent :<\/p>\n<p style=\"text-align:center;\"><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">10.000! \\approx 10^{\\log(e) \\cdot (10.000 \\ln(10.000) - 10.000)} \\approx 10^{35.657,06}<\/span><\/span><\/p>\n<p>Voici o\u00f9 nous remarquons que l&#8217;expression dans l&#8217;exposant devient g\u00e9rable pour la plupart des calculatrices. Ainsi, bien que nous ne puissions pas visualiser le nombre en raison de sa taille immense, nous savons qu&#8217;il compte environ 35.657 chiffres. Cette approche transforme un calcul apparemment inaccessible en quelque chose de r\u00e9alisable.<\/p>\n","protected":false},"excerpt":{"rendered":"<p>La Formule de Stirling La formule de Stirling est un outil essentiel pour simplifier les calculs de factorielles de grands nombres, offrant une approximation rapide et pratique. Ce r\u00e9sultat est particuli\u00e8rement utile dans des domaines tels que la thermodynamique, la probabilit\u00e9 et l&#8217;analyse asymptotique, o\u00f9 l&#8217;on travaille souvent avec des nombres extr\u00eamement grands. Comprendre son [&hellip;]<\/p>\n","protected":false},"author":1,"featured_media":30374,"comment_status":"open","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"iawp_total_views":5,"footnotes":""},"categories":[569,682,931],"tags":[],"class_list":["post-31167","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-mathematiques","category-probabilites-et-statistiques","category-thermodynamique"],"yoast_head":"<!-- This site is optimized with the Yoast SEO plugin v26.7 - https:\/\/yoast.com\/wordpress\/plugins\/seo\/ -->\n<title>La formule de Stirling - toposuranos.com\/material<\/title>\n<meta name=\"description\" content=\"La formule de Stirling : n! \u2248 \u221a(2\u03c0n) * (n\/e)^n. 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