{"id":30320,"date":"2021-03-27T13:00:51","date_gmt":"2021-03-27T13:00:51","guid":{"rendered":"http:\/\/toposuranos.com\/material\/?p=30320"},"modified":"2025-01-07T03:29:11","modified_gmt":"2025-01-07T03:29:11","slug":"la-formula-de-stirling","status":"publish","type":"post","link":"https:\/\/toposuranos.com\/material\/es\/la-formula-de-stirling\/","title":{"rendered":"La F\u00f3rmula de Stirling"},"content":{"rendered":"<style>\n\tp, ul, ol {\n\t\ttext-align: justify;\n\t}\n\th1, h2 {\n\ttext-align:center;\n\t}\n<\/style>\n<h1>La F\u00f3rmula de Stirling<\/h1>\n<p><em>La f\u00f3rmula de Stirling es una herramienta esencial para simplificar c\u00e1lculos con factoriales de n\u00fameros grandes, ofreciendo una aproximaci\u00f3n r\u00e1pida y pr\u00e1ctica.<\/p>\n<p>Este resultado es especialmente \u00fatil en \u00e1reas como la termodin\u00e1mica, la probabilidad y el an\u00e1lisis asint\u00f3tico, donde trabajar con n\u00fameros extremadamente grandes es com\u00fan. Comprender su desarrollo no solo facilita su aplicaci\u00f3n, sino que tambi\u00e9n permite apreciar su relevancia en el c\u00e1lculo eficiente y en la resoluci\u00f3n de problemas complejos.<br \/>\n<\/em><\/p>\n<p style=\"text-align:center;\"><strong>Objetivos de Aprendizaje:<\/strong><br \/>\nAl Finalizar esta clase el estudiante ser\u00e1 capaz de<\/p>\n<ol>\n<li><strong>Comprender<\/strong> la deducci\u00f3n de la f\u00f3rmula de Stirling desde la definici\u00f3n del factorial mediante la funci\u00f3n Gamma.<\/li>\n<li><strong>Aplicar<\/strong> la f\u00f3rmula de Stirling para aproximar factoriales de n\u00fameros muy grandes.<\/li>\n<li><strong>Calcular<\/strong> aproximaciones logar\u00edtmicas de factoriales mediante herramientas b\u00e1sicas de logaritmos y exponentes.<\/li>\n<\/ol>\n<p style=\"text-align:center;\"><strong><u>\u00cdNDICE DE CONTENIDOS<\/u>:<\/strong><br \/>\n<a href=\"#1\">Demostraci\u00f3n de la f\u00f3rmula de Stirling<\/a><br \/>\n<a href=\"#2\">Aproximaci\u00f3n logar\u00edtmica del factorial<\/a><br \/>\n<a href=\"#3\">Ejemplo: Aproximaci\u00f3n del Factorial de un N\u00famero Muy Grande<\/a>\n<\/p>\n<p><a name=\"1\"><\/a><\/p>\n<h2>Demostraci\u00f3n de la 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>El desarrollo de la f\u00f3rmula de Stirling comienza con la definici\u00f3n del factorial mediante la funci\u00f3n Gamma, que es:<\/p>\n<p style=\"text-align:center;\"><span class=\"katex-eq\" data-katex-display=\"false\">n! =\\Gamma(n+1) = \\displaystyle \\int_0^\\infty t^n e^{-t} \\, dt<\/span>\n<p>Usando esta expresi\u00f3n, realizamos un cambio de variable: <span class=\"katex-eq\" data-katex-display=\"false\">t = nx<\/span>. Esto implica que <span class=\"katex-eq\" data-katex-display=\"false\">x \\in [0, \\infty[<\/span> y <span class=\"katex-eq\" data-katex-display=\"false\">dt = n dx<\/span>. Con este cambio, la integral se transforma de la siguiente manera:<\/p>\n<p style=\"text-align:center;\"><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>\n<p>A continuaci\u00f3n, realizamos un segundo cambio de variable: <span class=\"katex-eq\" data-katex-display=\"false\">x = 1 + \\dfrac{s}{\\sqrt{n}}<\/span>. Esto 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>Con este cambio de variable, la integral toma la siguiente 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>Ahora utilizamos la expansi\u00f3n en series de Taylor para el logaritmo natural:<\/p>\n<p style=\"text-align:center;\"><span class=\"katex-eq\" data-katex-display=\"false\">\\ln(1+x) = \\displaystyle\\sum_{k=1}^{\\infty} \\dfrac{(-1)^{k+1}x^k}{k}   <\/span>\n<p>Al aplicar esta expansi\u00f3n en <span class=\"katex-eq\" data-katex-display=\"false\">\\ln\\left(1+\\dfrac{s}{\\sqrt{n}}\\right)<\/span>, desarrollamos la expresi\u00f3n de la exponencial como sigue:<\/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 este modo, podemos escribir la expresi\u00f3n completa como:<\/p>\n<p style=\"text-align:center;\"><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>\n<p>Este resultado es fundamental para calcular factoriales de n\u00fameros muy grandes. A medida que <span class=\"katex-eq\" data-katex-display=\"false\">n<\/span> crece, los t\u00e9rminos en la sumatoria dentro de la exponencial tienden a cero, dejando \u00fanicamente el t\u00e9rmino dominante. Esto simplifica la integral, que se puede resolver como una integral Gaussiana:<\/p>\n<p style=\"text-align:center;\"><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>\n<p><strong>Este resultado es conocido como la f\u00f3rmula de Stirling para el factorial de n\u00fameros grandes:<\/strong><\/p>\n<p style=\"text-align:center;\"><span class=\"katex-eq\" data-katex-display=\"false\">\\boxed{n! \\approx \\sqrt{2\\pi n}\\left(\\dfrac{n}{e}\\right)^{n}}<\/span>\n<p><a name=\"2\"><\/a><\/p>\n<h2>Aproximaci\u00f3n logar\u00edtmica del factorial<\/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>Un resultado directo de la f\u00f3rmula de Stirling es la aproximaci\u00f3n logar\u00edtmica del factorial. Al tomar el logaritmo natural de la f\u00f3rmula de Stirling, obtenemos:<\/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>En el \u00faltimo paso, se realiza una aproximaci\u00f3n adicional al despreciar el t\u00e9rmino <span class=\"katex-eq\" data-katex-display=\"false\">\\dfrac{1}{2}\\ln(2n\\pi)<\/span>. Este t\u00e9rmino se vuelve insignificante en comparaci\u00f3n con <span class=\"katex-eq\" data-katex-display=\"false\">n\\ln(n) - n<\/span> para valores grandes de <span class=\"katex-eq\" data-katex-display=\"false\">n<\/span>.<\/p>\n<p>La validez de esta aproximaci\u00f3n se justifica calculando el error relativo entre ambas expresiones:<\/p>\n<p style=\"text-align:center;\"><span class=\"katex-eq\" data-katex-display=\"false\">\\begin{array}{rcl}\n\n\\text{Aproximaci\u00f3n Inicial} &amp; = &amp; \\dfrac{1}{2}\\ln(2n\\pi) + n\\ln(n) - n \\\\ \\\\\n\n\\text{Aproximaci\u00f3n Final} &amp; = &amp; n\\ln(n) - n \\\\ \\\\\n\n\\text{Error Relativo} &amp;=&amp; \\dfrac{\\text{Aproximaci\u00f3n Final} - \\text{Aproximaci\u00f3n Inicial}}{\\text{Aproximaci\u00f3n 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>Calculando el l\u00edmite cuando <span class=\"katex-eq\" data-katex-display=\"false\">n \\to \\infty<\/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{Error 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>Por lo tanto, dado que el error tiende a cero  para valores grandes de <span class=\"katex-eq\" data-katex-display=\"false\">n<\/span>, podemos usar la siguiente aproximaci\u00f3n logar\u00edtmica con confianza:<\/p>\n<p style=\"text-align:center;\"><span class=\"katex-eq\" data-katex-display=\"false\">\\boxed{\\ln(n!) \\approx n\\ln(n) - n}<\/span>\n<p><a name=\"3\"><\/a><\/p>\n<h2>Ejemplo: Aproximaci\u00f3n del Factorial de un N\u00famero Muy 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 el factorial de n\u00fameros extremadamente grandes, como <em>10.000!<\/em>, es pr\u00e1cticamente imposible con herramientas convencionales debido al tama\u00f1o del resultado. Sin embargo, utilizando la aproximaci\u00f3n logar\u00edtmica del factorial derivada de la f\u00f3rmula de Stirling, podemos hacerlo manejable incluso con calculadoras b\u00e1sicas.<\/p>\n<p>La f\u00f3rmula logar\u00edtmica del factorial nos dice:<\/p>\n<p style=\"text-align:center;\"><span class=\"katex-eq\" data-katex-display=\"false\">\\ln(10.000!) \\approx 10.000 \\ln(10.000) - 10.000<\/span>\n<p>Para convertir de logaritmos naturales (<span class=\"katex-eq\" data-katex-display=\"false\">\\ln<\/span>) a logaritmos base 10 (<span class=\"katex-eq\" data-katex-display=\"false\">\\log<\/span>), usamos la relaci\u00f3n:<\/p>\n<p style=\"text-align:center;\"><span class=\"katex-eq\" data-katex-display=\"false\">\\ln(10.000!) = \\dfrac{\\log(10.000!)}{\\log(e)}<\/span>\n<p>Esto implica que:<\/p>\n<p style=\"text-align:center;\"><span class=\"katex-eq\" data-katex-display=\"false\">\\log(10.000!) \\approx \\log(e) \\cdot (10.000 \\ln(10.000) - 10.000)<\/span>\n<p>Por lo tanto:<\/p>\n<p style=\"text-align:center;\"><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>\n<p>Aqu\u00ed es donde notamos que la expresi\u00f3n en el exponente resulta manejable para la mayor\u00eda de las calculadoras. De este modo, aunque no podamos visualizar el n\u00famero debido a su inmenso tama\u00f1o, sabemos que cuenta con aproximadamente 35,657 d\u00edgitos. Este enfoque transforma un c\u00e1lculo aparentemente inalcanzable en algo realizable.<\/p>\n","protected":false},"excerpt":{"rendered":"<p>La F\u00f3rmula de Stirling La f\u00f3rmula de Stirling es una herramienta esencial para simplificar c\u00e1lculos con factoriales de n\u00fameros grandes, ofreciendo una aproximaci\u00f3n r\u00e1pida y pr\u00e1ctica. Este resultado es especialmente \u00fatil en \u00e1reas como la termodin\u00e1mica, la probabilidad y el an\u00e1lisis asint\u00f3tico, donde trabajar con n\u00fameros extremadamente grandes es com\u00fan. Comprender su desarrollo no solo [&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":120,"footnotes":""},"categories":[563,668,917],"tags":[],"class_list":["post-30320","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-matematica","category-probabilidades-y-estadistica","category-termodinamica"],"yoast_head":"<!-- This site is optimized with the Yoast SEO plugin v27.4 - https:\/\/yoast.com\/product\/yoast-seo-wordpress\/ -->\n<title>La F\u00f3rmula de Stirling - toposuranos.com\/material<\/title>\n<meta name=\"description\" content=\"La F\u00f3rmula de Stirling: n! \u2248 \u221a(2\u03c0n) * (n\/e)^n. 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