{"id":34082,"date":"2021-03-27T13:00:56","date_gmt":"2021-03-27T13:00:56","guid":{"rendered":"https:\/\/toposuranos.com\/material\/?p=34082"},"modified":"2025-08-16T06:38:25","modified_gmt":"2025-08-16T06:38:25","slug":"die-stirling-formel","status":"publish","type":"post","link":"http:\/\/toposuranos.com\/material\/de\/die-stirling-formel\/","title":{"rendered":"Die Stirling-Formel"},"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>Die Stirling-Formel<\/h1>\n<p><em>Die Stirling-Formel ist ein wesentliches Werkzeug, um Berechnungen mit Fakult\u00e4ten gro\u00dfer Zahlen zu vereinfachen, da sie eine schnelle und praktische Ann\u00e4herung bietet.<\/p>\n<p>Dieses Resultat ist besonders n\u00fctzlich in Bereichen wie der Thermodynamik, der Wahrscheinlichkeitstheorie und der asymptotischen Analyse, wo der Umgang mit extrem gro\u00dfen Zahlen \u00fcblich ist. Das Verst\u00e4ndnis ihrer Herleitung erleichtert nicht nur ihre Anwendung, sondern erm\u00f6glicht auch, ihre Relevanz in der effizienten Berechnung und in der L\u00f6sung komplexer Probleme zu sch\u00e4tzen.<br \/>\n<\/em><\/p>\n<p style=\"text-align:center;\"><strong>Lernziele:<\/strong><br \/>\nAm Ende dieser Klasse wird der Student in der Lage sein,<\/p>\n<ol>\n<li><strong>Die Herleitung<\/strong> der Stirling-Formel aus der Definition der Fakult\u00e4t mittels der Gammafunktion zu verstehen.<\/li>\n<li><strong>Die Stirling-Formel anzuwenden<\/strong>, um Fakult\u00e4ten sehr gro\u00dfer Zahlen zu approximieren.<\/li>\n<li><strong>Logarithmische Approximationen<\/strong> von Fakult\u00e4ten mithilfe grundlegender Werkzeuge der Logarithmen und Exponenten zu berechnen.<\/li>\n<\/ol>\n<p style=\"text-align:center;\"><strong><u>INHALTSVERZEICHNIS<\/u>:<\/strong><br \/>\n<a href=\"#1\">Beweis der Stirling-Formel<\/a><br \/>\n<a href=\"#2\">Logarithmische Approximation der Fakult\u00e4t<\/a><br \/>\n<a href=\"#3\">Beispiel: Approximation der Fakult\u00e4t einer sehr gro\u00dfen Zahl<\/a>\n<\/p>\n<p><a name=\"1\"><\/a><\/p>\n<h2>Beweis der Stirling-Formel<\/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>Die Herleitung der Stirling-Formel beginnt mit der Definition der Fakult\u00e4t mittels der Gammafunktion, die lautet:<\/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>Mit diesem Ausdruck f\u00fchren wir eine Variablen\u00e4nderung durch: <span class=\"katex-eq\" data-katex-display=\"false\">t = nx<\/span>. Dies impliziert, dass <span class=\"katex-eq\" data-katex-display=\"false\">x \\in [0, \\infty[<\/span> und <span class=\"katex-eq\" data-katex-display=\"false\">dt = n dx<\/span>. Mit dieser \u00c4nderung transformiert sich das Integral wie folgt:<\/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>Als N\u00e4chstes f\u00fchren wir eine zweite Variablen\u00e4nderung durch: <span class=\"katex-eq\" data-katex-display=\"false\">x = 1 + \\dfrac{s}{\\sqrt{n}}<\/span>. Dies impliziert:<\/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>Mit dieser Variablen\u00e4nderung nimmt das Integral die folgende Form an:<\/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>Nun verwenden wir die Taylorreihenentwicklung f\u00fcr den nat\u00fcrlichen Logarithmus:<\/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>Wenden wir diese Entwicklung auf <span class=\"katex-eq\" data-katex-display=\"false\">\\ln\\left(1+\\dfrac{s}{\\sqrt{n}}\\right)<\/span> an, so entfalten wir den Ausdruck der Exponentialfunktion wie folgt:<\/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>Auf diese Weise k\u00f6nnen wir den vollst\u00e4ndigen Ausdruck wie folgt schreiben:<\/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>Dieses Ergebnis ist grundlegend f\u00fcr die Berechnung von Fakult\u00e4ten sehr gro\u00dfer Zahlen. Wenn <span class=\"katex-eq\" data-katex-display=\"false\">n<\/span> w\u00e4chst, tendieren die Terme in der Summe innerhalb der Exponentialfunktion gegen null, sodass nur der dominante Term \u00fcbrig bleibt. Dies vereinfacht das Integral, das als gau\u00dfsches Integral gel\u00f6st werden kann:<\/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>Dieses Ergebnis ist als die Stirling-Formel f\u00fcr die Fakult\u00e4t gro\u00dfer Zahlen bekannt:<\/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>Logarithmische Approximation der Fakult\u00e4t<\/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>Ein direktes Resultat der Stirling-Formel ist die logarithmische Approximation der Fakult\u00e4t. Wenn wir den nat\u00fcrlichen Logarithmus der Stirling-Formel nehmen, erhalten wir:<\/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>Im letzten Schritt wird eine zus\u00e4tzliche Approximation vorgenommen, indem der Term <span class=\"katex-eq\" data-katex-display=\"false\">\\dfrac{1}{2}\\ln(2n\\pi)<\/span> vernachl\u00e4ssigt wird. Dieser Term wird im Vergleich zu <span class=\"katex-eq\" data-katex-display=\"false\">n\\ln(n) - n<\/span> f\u00fcr gro\u00dfe Werte von <span class=\"katex-eq\" data-katex-display=\"false\">n<\/span> unbedeutend.<\/p>\n<p>Die G\u00fcltigkeit dieser Approximation wird durch die Berechnung des relativen Fehlers zwischen beiden Ausdr\u00fccken gerechtfertigt:<\/p>\n<p style=\"text-align:center;\"><span class=\"katex-eq\" data-katex-display=\"false\">\\begin{array}{rcl}\n\n\\text{Anf\u00e4ngliche Approximation} &amp; = &amp; \\dfrac{1}{2}\\ln(2n\\pi) + n\\ln(n) - n \\\\ \\\\\n\n\\text{Endg\u00fcltige Approximation} &amp; = &amp; n\\ln(n) - n \\\\ \\\\\n\n\\text{Relativer Fehler} &amp;=&amp; \\dfrac{\\text{Endg\u00fcltige Approximation} - \\text{Anf\u00e4ngliche Approximation}}{\\text{Anf\u00e4ngliche Approximation}} \\\\ \\\\\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>Berechnen wir das Limes, wenn <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{Relativer Fehler} &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>Da also der Fehler f\u00fcr gro\u00dfe Werte von <span class=\"katex-eq\" data-katex-display=\"false\">n<\/span> gegen null geht, k\u00f6nnen wir die folgende logarithmische Approximation mit Vertrauen verwenden:<\/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>Beispiel: Approximation der Fakult\u00e4t einer sehr gro\u00dfen Zahl<\/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>Die Berechnung der Fakult\u00e4t extrem gro\u00dfer Zahlen, wie <em>10.000!<\/em>, ist mit herk\u00f6mmlichen Werkzeugen aufgrund der Gr\u00f6\u00dfe des Ergebnisses praktisch unm\u00f6glich. Mit der logarithmischen Approximation der Fakult\u00e4t, die aus der Stirling-Formel abgeleitet wurde, k\u00f6nnen wir dies jedoch selbst mit einfachen Taschenrechnern handhabbar machen.<\/p>\n<p>Die logarithmische Fakult\u00e4tsformel lautet:<\/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>Um von nat\u00fcrlichen Logarithmen (<span class=\"katex-eq\" data-katex-display=\"false\">\\ln<\/span>) zu Logarithmen zur Basis 10 (<span class=\"katex-eq\" data-katex-display=\"false\">\\log<\/span>) zu konvertieren, verwenden wir die Beziehung:<\/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>Dies impliziert:<\/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>Daraus folgt:<\/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>Hier stellen wir fest, dass der Ausdruck im Exponenten f\u00fcr die meisten Taschenrechner handhabbar ist. Auch wenn wir die Zahl aufgrund ihrer immensen Gr\u00f6\u00dfe nicht darstellen k\u00f6nnen, wissen wir, dass sie ungef\u00e4hr 35.657 Ziffern hat. Dieser Ansatz verwandelt eine scheinbar unerreichbare Berechnung in etwas Realisierbares.<\/p>\n","protected":false},"excerpt":{"rendered":"<p>Die Stirling-Formel Die Stirling-Formel ist ein wesentliches Werkzeug, um Berechnungen mit Fakult\u00e4ten gro\u00dfer Zahlen zu vereinfachen, da sie eine schnelle und praktische Ann\u00e4herung bietet. Dieses Resultat ist besonders n\u00fctzlich in Bereichen wie der Thermodynamik, der Wahrscheinlichkeitstheorie und der asymptotischen Analyse, wo der Umgang mit extrem gro\u00dfen Zahlen \u00fcblich ist. Das Verst\u00e4ndnis ihrer Herleitung erleichtert nicht [&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":[1302,1294,1366],"tags":[],"class_list":["post-34082","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-mathematik","category-thermodynamik","category-wahrscheinlichkeiten-und-statistik"],"yoast_head":"<!-- This site is optimized with the Yoast SEO plugin v26.7 - https:\/\/yoast.com\/wordpress\/plugins\/seo\/ -->\n<title>Die Stirling-Formel - toposuranos.com\/material<\/title>\n<meta name=\"description\" content=\"Die Stirling-Formel: n! \u2248 \u221a(2\u03c0n) * (n\/e)^n. 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