{"id":31153,"date":"2021-03-27T13:00:16","date_gmt":"2021-03-27T13:00:16","guid":{"rendered":"http:\/\/toposuranos.com\/material\/?p=31153"},"modified":"2025-01-07T03:37:40","modified_gmt":"2025-01-07T03:37:40","slug":"a-formula-de-stirling","status":"publish","type":"post","link":"http:\/\/toposuranos.com\/material\/pt\/a-formula-de-stirling\/","title":{"rendered":"A F\u00f3rmula de Stirling"},"content":{"rendered":"<style>\n\tp, ul, ol {\n\t\ttext-align: justify;\n\t}\n\th1, h2 {\n\t\ttext-align: center;\n\t}\n<\/style>\n<h1>A F\u00f3rmula de Stirling<\/h1>\n<p><em>A f\u00f3rmula de Stirling \u00e9 uma ferramenta essencial para simplificar c\u00e1lculos com fatoriais de n\u00fameros grandes, oferecendo uma aproxima\u00e7\u00e3o r\u00e1pida e pr\u00e1tica.<\/p>\n<p>Este resultado \u00e9 especialmente \u00fatil em \u00e1reas como termodin\u00e2mica, probabilidade e an\u00e1lise assint\u00f3tica, onde trabalhar com n\u00fameros extremamente grandes \u00e9 comum. Compreender seu desenvolvimento n\u00e3o apenas facilita sua aplica\u00e7\u00e3o, mas tamb\u00e9m permite apreciar sua relev\u00e2ncia no c\u00e1lculo eficiente e na resolu\u00e7\u00e3o de problemas complexos.<br \/>\n<\/em><\/p>\n<p style=\"text-align:center;\"><strong>Objetivos de Aprendizagem:<\/strong><br \/>\nAo finalizar esta aula, o aluno ser\u00e1 capaz de:<\/p>\n<ol>\n<li><strong>Compreender<\/strong> a dedu\u00e7\u00e3o da f\u00f3rmula de Stirling a partir da defini\u00e7\u00e3o do fatorial atrav\u00e9s da fun\u00e7\u00e3o Gamma.<\/li>\n<li><strong>Aplicar<\/strong> a f\u00f3rmula de Stirling para aproximar fatoriais de n\u00fameros muito grandes.<\/li>\n<li><strong>Calcular<\/strong> aproxima\u00e7\u00f5es logar\u00edtmicas de fatoriais utilizando ferramentas b\u00e1sicas de logaritmos e expoentes.<\/li>\n<\/ol>\n<p style=\"text-align:center;\"><strong><u>\u00cdNDICE DE CONTE\u00daDO<\/u>:<\/strong><br \/>\n<a href=\"#1\">Demonstra\u00e7\u00e3o da f\u00f3rmula de Stirling<\/a><br \/>\n<a href=\"#2\">Aproxima\u00e7\u00e3o logar\u00edtmica do fatorial<\/a><br \/>\n<a href=\"#3\">Exemplo: Aproxima\u00e7\u00e3o do Fatorial de um N\u00famero Muito Grande<\/a>\n<\/p>\n<p><a name=\"1\"><\/a><\/p>\n<h2>Demonstra\u00e7\u00e3o da f\u00f3rmula de Stirling<\/h2>\n<p><center><iframe class=\"lazyload\" width=\"560\" height=\"315\" data-src=\"https:\/\/www.youtube.com\/embed\/CcHCyRR1WrY?si=eTZsj4wBqQ2krELG\" title=\"YouTube video player\" 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>O desenvolvimento da f\u00f3rmula de Stirling come\u00e7a com a defini\u00e7\u00e3o do fatorial atrav\u00e9s da fun\u00e7\u00e3o Gamma, que \u00e9:<\/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>Utilizando essa express\u00e3o, fazemos uma mudan\u00e7a de vari\u00e1vel: <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">t = nx<\/span><\/span>. Isso implica que <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">x \\in [0, \\infty[<\/span><\/span> e <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">dt = n dx<\/span><\/span>. Com essa mudan\u00e7a, a integral se transforma da seguinte forma:<\/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>Em seguida, realizamos uma segunda mudan\u00e7a de vari\u00e1vel: <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">x = 1 + \\dfrac{s}{\\sqrt{n}}<\/span><\/span>. Isso implica:<\/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>Com essa mudan\u00e7a de vari\u00e1vel, a integral assume a seguinte forma:<\/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>Agora utilizamos a expans\u00e3o em s\u00e9ries de Taylor para o logaritmo natural:<\/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>Aplicando essa expans\u00e3o em <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\ln\\left(1+\\dfrac{s}{\\sqrt{n}}\\right)<\/span><\/span>, desenvolvemos a express\u00e3o exponencial da seguinte maneira:<\/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>Dessa forma, podemos escrever a express\u00e3o completa como:<\/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>Esse resultado \u00e9 fundamental para calcular fatoriais de n\u00fameros muito grandes. \u00c0 medida que <span class=\"katex-eq\" data-katex-display=\"false\">n<\/span> cresce, os termos na soma dentro do expoente tendem a zero, deixando apenas o termo dominante. Isso simplifica a integral, que pode ser resolvida como uma integral Gaussiana:<\/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>Esse resultado \u00e9 conhecido como a f\u00f3rmula de Stirling para o fatorial de n\u00fameros grandes:<\/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>Aproxima\u00e7\u00e3o Logar\u00edtmica do Fatorial<\/h2>\n<p><center><iframe class=\"lazyload\" width=\"560\" height=\"315\" data-src=\"https:\/\/www.youtube.com\/embed\/ASifSl6YgTk?si=8rHCOoUbWoiH962o\" title=\"YouTube video player\" 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>Um resultado direto da f\u00f3rmula de Stirling \u00e9 a aproxima\u00e7\u00e3o logar\u00edtmica do fatorial. Ao tomar o logaritmo natural da f\u00f3rmula de Stirling, obtemos:<\/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>No \u00faltimo passo, \u00e9 feita uma aproxima\u00e7\u00e3o adicional ao desprezar o termo <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\dfrac{1}{2}\\ln(2n\\pi)<\/span><\/span>. Esse termo torna-se insignificante em compara\u00e7\u00e3o com <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">n\\ln(n) - n<\/span><\/span> para valores grandes de <span class=\"katex-eq\" data-katex-display=\"false\">n<\/span>.<\/p>\n<p>A validade dessa aproxima\u00e7\u00e3o \u00e9 justificada calculando o erro relativo entre as duas express\u00f5es:<\/p>\n<p style=\"text-align:center;\"><span class=\"katex-eq\" data-katex-display=\"false\">\\begin{array}{rcl}\n\n\\text{Aproxima\u00e7\u00e3o Inicial} &amp; = &amp; \\dfrac{1}{2}\\ln(2n\\pi) + n\\ln(n) - n \\\\ \\\\\n\n\\text{Aproxima\u00e7\u00e3o Final} &amp; = &amp; n\\ln(n) - n \\\\ \\\\\n\n\\text{Erro Relativo} &amp;=&amp; \\dfrac{\\text{Aproxima\u00e7\u00e3o Final} - \\text{Aproxima\u00e7\u00e3o Inicial}}{\\text{Aproxima\u00e7\u00e3o Inicial}} \\\\ \\\\\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>Agora, consideramos o limite quando <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{Erro Relativo} &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>Assim, como o erro tende a zero para valores grandes de <span class=\"katex-eq\" data-katex-display=\"false\">n<\/span>, podemos usar a seguinte aproxima\u00e7\u00e3o logar\u00edtmica com confian\u00e7a:<\/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>Exemplo: Aproxima\u00e7\u00e3o do Fatorial de um N\u00famero Muito Grande<\/h2>\n<p><center><iframe class=\"lazyload\" width=\"560\" height=\"315\" data-src=\"https:\/\/www.youtube.com\/embed\/kja9niRWjpg?si=eHkcZYaq0Fgntc3G\" title=\"YouTube video player\" 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>Calcular o fatorial de n\u00fameros extremamente grandes, como <em>10.000!<\/em>, \u00e9 praticamente imposs\u00edvel com ferramentas convencionais devido ao tamanho do resultado. No entanto, usando a aproxima\u00e7\u00e3o logar\u00edtmica do fatorial derivada da f\u00f3rmula de Stirling, podemos torn\u00e1-lo vi\u00e1vel mesmo com calculadoras b\u00e1sicas.<\/p>\n<p>A f\u00f3rmula logar\u00edtmica do fatorial nos informa:<\/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>Para converter de logaritmos naturais (<span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\ln<\/span><\/span>) para logaritmos de base 10 (<span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\log<\/span><\/span>), usamos a rela\u00e7\u00e3o:<\/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>Isso implica 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>Portanto:<\/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>Aqui, notamos que a express\u00e3o no expoente torna-se gerenci\u00e1vel para a maioria das calculadoras. Assim, embora n\u00e3o possamos visualizar o n\u00famero devido ao seu tamanho imenso, sabemos que ele possui aproximadamente 35.657 d\u00edgitos. Essa abordagem transforma um c\u00e1lculo aparentemente inating\u00edvel em algo fact\u00edvel.<\/p>\n","protected":false},"excerpt":{"rendered":"<p>A F\u00f3rmula de Stirling A f\u00f3rmula de Stirling \u00e9 uma ferramenta essencial para simplificar c\u00e1lculos com fatoriais de n\u00fameros grandes, oferecendo uma aproxima\u00e7\u00e3o r\u00e1pida e pr\u00e1tica. Este resultado \u00e9 especialmente \u00fatil em \u00e1reas como termodin\u00e2mica, probabilidade e an\u00e1lise assint\u00f3tica, onde trabalhar com n\u00fameros extremamente grandes \u00e9 comum. Compreender seu desenvolvimento n\u00e3o apenas facilita sua aplica\u00e7\u00e3o, [&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":23,"footnotes":""},"categories":[571,672,921],"tags":[],"class_list":["post-31153","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-matematica-pt","category-probabilidades-e-estatistica","category-termodinamica-pt"],"yoast_head":"<!-- This site is optimized with the Yoast SEO plugin v26.7 - https:\/\/yoast.com\/wordpress\/plugins\/seo\/ -->\n<title>A F\u00f3rmula de Stirling - toposuranos.com\/material<\/title>\n<meta name=\"description\" content=\"A F\u00f3rmula de Stirling: n! \u2248 \u221a(2\u03c0n) * (n\/e)^n. 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