{"id":284,"date":"2019-02-02T19:13:52","date_gmt":"2019-02-02T18:13:52","guid":{"rendered":"https:\/\/www.leotheodon.com\/V4\/?p=284"},"modified":"2019-02-03T01:17:41","modified_gmt":"2019-02-03T00:17:41","slug":"formule-de-stirling","status":"publish","type":"post","link":"https:\/\/www.leotheodon.com\/V4\/2019\/02\/02\/formule-de-stirling\/","title":{"rendered":"Formule de Stirling"},"content":{"rendered":"\n

La formule de Stirling est c\u00e9l\u00e8bre pour donner une tr\u00e8s bonne approximation de la factorielle d’un nombre. En effet, elle se r\u00e9v\u00e8le extr\u00eamement efficace, avec une erreur inf\u00e9rieur \u00e0 2% d\u00e8s le cinqui\u00e8me rang. On rappelle la formule ci-dessous:<\/p>\n\n\n\n

  <\/span>   <\/span>\"\[ n! \sim \sqrt{2\pi n}\left(\dfrac{n}{e}\right)^n \]\"<\/p><\/p>\n\n\n\n

Nous allons donner l’id\u00e9e d’une d\u00e9monstration utilisant la m\u00e9thode du col. <\/strong><\/p>\n\n\n

D\u00e9monstration:<\/strong>
Exprimons tout d’abord la factorielle \u00e0 l’aide de la fonction Gamma d’Euler:<\/p>\n\n\n

  <\/span>   <\/span>\"\begin{align*}n! = \Gamma(n+1) & = \int_0^{+\infty}e^{-t}t^n dt\\& = \int_0^{+\infty}e^{-(t-n\ln(t))} dt\label{G1}\end{align*}\"<\/p><\/p>\n\n\n\n

On note \"f_n(t) = t-n\ln(t)\". On peut remarquer que cette fonction admet un minimum en calculant sa d\u00e9riv\u00e9e.<\/p>\n\n\n\n

<\/a>

(1) <\/span>   <\/span>\"\begin{equation*} f'_n(t) = 1-\dfrac{n}{t} \end{equation*}\"<\/p><\/p>\n\n\n\n

On remarque que l’on a \"f'_n(t)= 0 \Leftrightarrow t=n\". Ainsi, la fonction \"f\" admet un minimum en \"t_n=n\".
On peut alors \u00e9crire le d\u00e9veloppement de Taylor de \"f\" au voisinage de \"t_n\" sous la forme suivante:<\/p>\n\n\n\n

  <\/span>   <\/span>\"\begin{align*}f_n(t) & = f_n(t_n) + \underbrace{f_n'(t_n)}_{=0}(t-t_n) + \dfrac{1}{2} f_n''(t_n)(t-t_n)^2 + o((t-t_n)^2)\\& = n-n\ln(n) + \dfrac{1}{2n}(t-n)^2 + o((t-n)^2)t\label{G1}\end{align*}\"<\/p>
Ainsi, on peut \u00e9crire:<\/p>\n\n\n\n

  <\/span>   <\/span>\"\begin{align*}n! & \approx \int_0^{+\infty}e^{-(n-n\ln(n)+ \frac{1}{2n}(t-n)^2)} dt\\& \approx \left(\dfrac{n}{e}\right)^n \int_0^{+\infty}e^{-\frac{(t-n)^2}{2n}} dt\label{approx}\end{align*}\"<\/p><\/p>\n\n\n\n

En effectuant le changement de variable \"u=\dfrac{(t-n)}{\sqrt{2n}}\" et \"du = \dfrac{dt}{\sqrt{2n}}\", on obtient:<\/p>\n\n\n\n

<\/a>

(2) <\/span>   <\/span>\"\begin{equation*} n! \approx \left(\dfrac{n}{e}\right)^n \sqrt{2n}\int_{-\sqrt{\frac{n}{2}}}^{+\infty}e^{-u^2} du \end{equation*}\"<\/p><\/p>\n\n\n\n

De plus, pour \"n\" suffisamment grand, on a:<\/p>\n\n\n\n

<\/a>

(3) <\/span>   <\/span>\"\begin{equation*} \int_{-\sqrt{\frac{n}{2}}}^{+\infty}e^{-u^2} du \approx \int_{-\infty}^{+\infty}e^{-u^2}du = \sqrt{\pi} \end{equation*}\"<\/p><\/p>\n\n\n\n

Ainsi, on en d\u00e9duit la formule de Stirling:<\/p>\n\n\n\n

<\/a>

(4) <\/span>   <\/span>\"\begin{equation*}n! \approx \sqrt{2\pi n}\left(\dfrac{n}{e}\right)^n\end{equation*}\"<\/p><\/p>\n\n\n\n

Remarque:<\/strong>
On pourra noter que la d\u00e9monstration ci-dessus n’est pas compl\u00e8tement rigoureuse. Les notations sont lacunaires (il faudrait pr\u00e9ciser les passages \u00e0 la limite et les \u00e9quivalents \u00e0 l’infini), et on passe rapidement sur l’utilisation du d\u00e9veloppement de Taylor. Pour s’en convaincre, on peut regarder l’allure de la courbe associ\u00e9e \u00e0 la fonction \"f_n\".<\/p>\n\n\n\n

T\u00e9l\u00e9charger au format pdf :<\/a>T\u00e9l\u00e9charger<\/a><\/div>\n","protected":false},"excerpt":{"rendered":"

La formule de Stirling est c\u00e9l\u00e8bre pour donner une tr\u00e8s bonne approximation de la factorielle d’un nombre. En effet, elle se r\u00e9v\u00e8le extr\u00eamement efficace, avec une erreur inf\u00e9rieur \u00e0 2% d\u00e8s le cinqui\u00e8me rang....<\/p>\n","protected":false},"author":1,"featured_media":329,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"rop_custom_images_group":[],"rop_custom_messages_group":[],"rop_publish_now":"initial","rop_publish_now_accounts":[],"rop_publish_now_history":[],"rop_publish_now_status":"pending","_exactmetrics_skip_tracking":false,"_exactmetrics_sitenote_active":false,"_exactmetrics_sitenote_note":"","_exactmetrics_sitenote_category":0,"footnotes":""},"categories":[3,5],"tags":[50,26,49],"class_list":["post-284","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-education","category-math","tag-demonstration","tag-maths","tag-stirling"],"_links":{"self":[{"href":"https:\/\/www.leotheodon.com\/V4\/wp-json\/wp\/v2\/posts\/284","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.leotheodon.com\/V4\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.leotheodon.com\/V4\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.leotheodon.com\/V4\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.leotheodon.com\/V4\/wp-json\/wp\/v2\/comments?post=284"}],"version-history":[{"count":42,"href":"https:\/\/www.leotheodon.com\/V4\/wp-json\/wp\/v2\/posts\/284\/revisions"}],"predecessor-version":[{"id":477,"href":"https:\/\/www.leotheodon.com\/V4\/wp-json\/wp\/v2\/posts\/284\/revisions\/477"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/www.leotheodon.com\/V4\/wp-json\/wp\/v2\/media\/329"}],"wp:attachment":[{"href":"https:\/\/www.leotheodon.com\/V4\/wp-json\/wp\/v2\/media?parent=284"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.leotheodon.com\/V4\/wp-json\/wp\/v2\/categories?post=284"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.leotheodon.com\/V4\/wp-json\/wp\/v2\/tags?post=284"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}