{"id":31145,"date":"2021-03-27T13:00:23","date_gmt":"2021-03-27T13:00:23","guid":{"rendered":"http:\/\/toposuranos.com\/material\/?p=31145"},"modified":"2025-01-07T03:32:53","modified_gmt":"2025-01-07T03:32:53","slug":"the-stirling-formula","status":"publish","type":"post","link":"https:\/\/toposuranos.com\/material\/en\/the-stirling-formula\/","title":{"rendered":"The Stirling Formula"},"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>The Stirling Formula<\/h1>\n<p><em>The Stirling formula is an essential tool for simplifying calculations with factorials of large numbers, offering a quick and practical approximation.<\/p>\n<p>This result is especially useful in areas such as thermodynamics, probability, and asymptotic analysis, where working with extremely large numbers is common. Understanding its derivation not only facilitates its application but also allows appreciation of its relevance in efficient calculation and solving complex problems.<br \/>\n<\/em><\/p>\n<p style=\"text-align:center;\"><strong>Learning Objectives:<\/strong><br \/>\nAt the end of this class, the student will be able to:<\/p>\n<ol>\n<li><strong>Understand<\/strong> the derivation of Stirling&#8217;s formula from the definition of the factorial using the Gamma function.<\/li>\n<li><strong>Apply<\/strong> Stirling&#8217;s formula to approximate factorials of very large numbers.<\/li>\n<li><strong>Calculate<\/strong> logarithmic approximations of factorials using basic logarithmic and exponential tools.<\/li>\n<\/ol>\n<p style=\"text-align:center;\"><strong><u>TABLE OF CONTENTS<\/u>:<\/strong><br \/>\n<a href=\"#1\">Derivation of Stirling&#8217;s Formula<\/a><br \/>\n<a href=\"#2\">Logarithmic Approximation of Factorials<\/a><br \/>\n<a href=\"#3\">Example: Approximation of the Factorial of a Very Large Number<\/a>\n<\/p>\n<p><a name=\"1\"><\/a><\/p>\n<h2>Derivation of Stirling&#8217;s Formula<\/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>The derivation of Stirling&#8217;s formula begins with the definition of the factorial using the Gamma function, which is:<\/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>Using this expression, we perform a variable change: <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">t = nx<\/span><\/span>. This implies that <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">x \\in [0, \\infty[<\/span><\/span> and <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">dt = n dx<\/span><\/span>. With this change, the integral transforms as follows:<\/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>Next, we perform a second variable change: <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">x = 1 + \\dfrac{s}{\\sqrt{n}}<\/span><\/span>. This implies:<\/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>With this variable change, the integral takes the following form:<\/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>Now we use the Taylor series expansion for the natural logarithm:<\/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>Applying this expansion to <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\ln\\left(1+\\dfrac{s}{\\sqrt{n}}\\right)<\/span><\/span>, we expand the exponential expression as follows:<\/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>Thus, we can write the full expression as:<\/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>This result is fundamental for calculating factorials of very large numbers. As <span class=\"katex-eq\" data-katex-display=\"false\">n<\/span> grows, the terms in the summation inside the exponential tend to zero, leaving only the dominant term. This simplifies the integral, which can be solved as a Gaussian integral:<\/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>This result is known as Stirling&#8217;s formula for the factorial of large numbers:<\/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>Logarithmic Approximation of Factorials<\/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>A direct result of Stirling&#8217;s formula is the logarithmic approximation of the factorial. Taking the natural logarithm of Stirling&#8217;s formula, we obtain:<\/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>In the last step, an additional approximation is made by neglecting the term <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\dfrac{1}{2}\\ln(2n\\pi)<\/span><\/span>. This term becomes insignificant compared to <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">n\\ln(n) - n<\/span><\/span> for large values of <span class=\"katex-eq\" data-katex-display=\"false\">n<\/span>.<\/p>\n<p>The validity of this approximation is justified by calculating the relative error between the two expressions:<\/p>\n<p style=\"text-align:center;\"><span class=\"katex-eq\" data-katex-display=\"false\">\\begin{array}{rcl}\n\n\\text{Initial Approximation} &amp; = &amp; \\dfrac{1}{2}\\ln(2n\\pi) + n\\ln(n) - n \\\\ \\\\\n\n\\text{Final Approximation} &amp; = &amp; n\\ln(n) - n \\\\ \\\\\n\n\\text{Relative Error} &amp;=&amp; \\dfrac{\\text{Final Approximation} - \\text{Initial Approximation}}{\\text{Initial 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>Now consider the limit as <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{Relative Error} &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>Thus, since the error tends to zero for large values of <span class=\"katex-eq\" data-katex-display=\"false\">n<\/span>, we can confidently use the following logarithmic approximation:<\/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>Example: Approximation of the Factorial of a Very Large Number<\/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>Calculating the factorial of extremely large numbers, such as <em>10,000!<\/em>, is practically impossible with conventional tools due to the size of the result. However, using the logarithmic approximation of the factorial derived from Stirling&#8217;s formula, we can make it manageable even with basic calculators.<\/p>\n<p>The logarithmic formula of the factorial states:<\/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>To convert from natural logarithms (<span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\ln<\/span><\/span>) to base-10 logarithms (<span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\log<\/span><\/span>), we use the relationship:<\/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>This implies that:<\/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>Thus:<\/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>Here, we observe that the exponent becomes manageable for most calculators. Thus, while we may not be able to visualize the number due to its immense size, we know that it contains approximately 35,657 digits. This approach turns an otherwise unattainable calculation into something feasible.<\/p>\n","protected":false},"excerpt":{"rendered":"<p>The Stirling Formula The Stirling formula is an essential tool for simplifying calculations with factorials of large numbers, offering a quick and practical approximation. This result is especially useful in areas such as thermodynamics, probability, and asymptotic analysis, where working with extremely large numbers is common. Understanding its derivation not only facilitates its application but [&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":25,"footnotes":""},"categories":[567,670,919],"tags":[],"class_list":["post-31145","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-mathematics","category-probabilities-and-statistics","category-thermodynamics"],"yoast_head":"<!-- This site is optimized with the Yoast SEO plugin v27.4 - https:\/\/yoast.com\/product\/yoast-seo-wordpress\/ -->\n<title>The Stirling Formula - toposuranos.com\/material<\/title>\n<meta name=\"description\" content=\"The Stirling Formula: n! \u2248 \u221a(2\u03c0n) * (n\/e)^n. 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