Editing Stirling's approximation

{{short description|Approximation for factorials}}
[[File:Mplwp factorial gamma stirling.svg|thumb|right|upright=1.35|Comparison of Stirling's approximation with the factorial]]
In [[mathematics]], '''Stirling's approximation''' (or '''Stirling's formula''') is an [[Asymptotic analysis|asymptotic]] approximation for [[factorial]]s. It is a good approximation, leading to accurate results even for small values of <math>n</math>. It is named after [[James Stirling (mathematician)|James Stirling]], though a related but less precise result was first stated by [[Abraham de Moivre]].{{r|dutka|LeCam1986|Pearson1924}}

One way of stating the approximation involves the [[logarithm]] of the factorial:
<math display=block>\ln(n!) = n\ln n - n +O(\ln n),</math>
where the [[big O notation]] means that, for all sufficiently large values of <math>n</math>, the difference between <math>\ln(n!)</math> and <math>n\ln n-n</math> will be at most proportional to the logarithm of <math>n</math>. In computer science applications such as the [[Comparison sort#Number of comparisons required to sort a list|worst-case lower bound for comparison sorting]], it is convenient to instead use the [[binary logarithm]], giving the equivalent form
<math display=block>\log_2 (n!) = n\log_2 n - n\log_2 e +O(\log_2 n).</math> The error term in either base can be expressed more precisely as <math>\tfrac12\log_2(2\pi n)+O(\tfrac1n)</math>, corresponding to an approximate formula for the factorial itself,
<math display=block>n! \sim \sqrt{2 \pi n}\left(\frac{n}{e}\right)^n.</math>
Here the sign <math>\sim</math> means that the two quantities are asymptotic, that is, their ratio tends to 1 as <math>n</math> tends to infinity.

== Derivation ==
Roughly speaking, the simplest version of Stirling's formula can be quickly obtained by approximating the sum
<math display=block>\ln(n!) = \sum_{j=1}^n \ln j</math>
with an [[integral]]:
<math display=block>\sum_{j=1}^n \ln j \approx \int_1^n \ln x \,{\rm d}x = n\ln n - n + 1.</math>

The full formula, together with precise estimates of its error, can be derived as follows. Instead of approximating <math>n!</math>, one considers its [[natural logarithm]], as this is a [[slowly varying function]]:
<math display=block>\ln(n!) = \ln 1 + \ln 2 + \cdots + \ln n.</math>

The right-hand side of this equation minus
<math display=block>\tfrac{1}{2}(\ln 1 + \ln n) = \tfrac{1}{2}\ln n</math>
is the approximation by the [[trapezoid rule]] of the integral
<math display=block>\ln(n!) - \tfrac{1}{2}\ln n \approx \int_1^n \ln x\,{\rm d}x = n \ln n - n + 1,</math>

and the error in this approximation is given by the [[Euler–Maclaurin formula]]:
<math display=block>\begin{align}
\ln(n!) - \tfrac{1}{2}\ln n & = \tfrac{1}{2}\ln 1 + \ln 2 + \ln 3 + \cdots + \ln(n-1) + \tfrac{1}{2}\ln n\\
& = n \ln n - n + 1 + \sum_{k=2}^{m} \frac{(-1)^k B_k}{k(k-1)} \left( \frac{1}{n^{k-1}} - 1 \right) + R_{m,n},
\end{align}</math>

where <math>B_k</math> is a [[Bernoulli number]], and {{math|''R''<sub>''m'',''n''</sub>}} is the remainder term in the Euler–Maclaurin formula. Take limits to find that
<math display=block>\lim_{n \to \infty} \left( \ln(n!) - n \ln n + n - \tfrac{1}{2}\ln n \right) = 1 - \sum_{k=2}^{m} \frac{(-1)^k B_k}{k(k-1)} + \lim_{n \to \infty} R_{m,n}.</math>

Denote this limit as <math>y</math>.  Because the remainder {{math|''R''<sub>''m'',''n''</sub>}} in the Euler–Maclaurin formula satisfies
<math display=block>R_{m,n} = \lim_{n \to \infty} R_{m,n} + O \left( \frac{1}{n^m} \right),</math>

where [[big-O notation]] is used, combining the equations above yields the approximation formula in its logarithmic form:
<math display=block>\ln(n!) = n \ln \left( \frac{n}{e} \right) + \tfrac{1}{2}\ln n + y + \sum_{k=2}^{m} \frac{(-1)^k B_k}{k(k-1)n^{k-1}} + O \left( \frac{1}{n^m} \right).</math>

Taking the exponential of both sides and choosing any positive integer <math>m</math>, one obtains a formula involving an unknown quantity <math>e^y</math>. For {{math|''m'' {{=}} 1}}, the formula is
<math display=block>n! = e^y \sqrt{n} \left( \frac{n}{e} \right)^n \left( 1 + O \left( \frac{1}{n} \right) \right).</math>

The quantity <math>e^y</math> can be found by taking the limit on both sides as <math>n</math> tends to infinity and using [[Wallis product|Wallis' product]], which shows that <math>e^y=\sqrt{2\pi}</math>. Therefore, one obtains Stirling's formula:
<math display=block>n! = \sqrt{2 \pi n} \left( \frac{n}{e} \right)^n \left( 1 + O \left( \frac{1}{n} \right) \right).</math>

== Alternative derivations ==
An alternative formula for <math>n!</math> using the [[gamma function]] is
<math display=block> n! = \int_0^\infty x^n e^{-x}\,{\rm d}x.</math>
(as can be seen by repeated integration by parts). Rewriting and changing variables {{math|''x'' {{=}} ''ny''}}, one obtains
<math display=block> n! = \int_0^\infty e^{n\ln x-x}\,{\rm d}x = e^{n \ln n} n \int_0^\infty e^{n(\ln y -y)}\,{\rm d}y.</math>
Applying [[Laplace's method]] one has
<math display=block>\int_0^\infty e^{n(\ln y -y)}\,{\rm d}y \sim \sqrt{\frac{2\pi}{n}} e^{-n},</math>
which recovers Stirling's formula:
<math display=block>n! \sim e^{n \ln n} n \sqrt{\frac{2\pi}{n}} e^{-n}
= \sqrt{2\pi n}\left(\frac{n}{e}\right)^n.
</math>

=== Higher orders ===
In fact, further corrections can also be obtained using Laplace's method. From previous result, we know that <math>\Gamma(x) \sim x^x e^{-x}</math>, so we "peel off" this dominant term, then perform two changes of variables, to obtain:<math display="block">x^{-x}e^x\Gamma(x) = \int_\R e^{x(1+t-e^t)}dt</math>To verify this: <math>\int_\R e^{x(1+t-e^t)}dt \overset{t \mapsto \ln t}{=} e^x \int_0^\infty t^{x-1} e^{-xt} dt \overset{t \mapsto t/x}{=} x^{-x} e^x \int_0^\infty e^{-t} t^{x-1} dt = x^{-x} e^x \Gamma(x)</math>.

Now the function <math>t \mapsto 1+t - e^t</math> is unimodal, with maximum value zero. Locally around zero, it looks like <math>-t^2/2</math>, which is why we are able to perform Laplace's method. In order to extend Laplace's method to higher orders, we perform another change of variables by <math>1+t-e^t = -\tau^2/2</math>. This equation cannot be solved in closed form, but it can be solved by serial expansion, which gives us <math>t = \tau - \tau^2/6 + \tau^3/36 + a_4 \tau^4 + O(\tau^5) </math>. Now plug back to the equation to obtain<math display="block">x^{-x}e^x\Gamma(x) = \int_\R e^{-x\tau^2/2}(1-\tau/3 + \tau^2/12 + 4a_4 \tau^3 + O(\tau^4)) d\tau = \sqrt{2\pi}(x^{-1/2} + x^{-3/2}/12) + O(x^{-5/2})</math>notice how we don't need to actually find <math>a_4</math>, since it is cancelled out by the integral. Higher orders can be achieved by computing more terms in <math>t = \tau + \cdots</math>, which can be obtained programmatically.{{NoteTag|note=For example, a program in Mathematica: <syntaxhighlight lang="mathematica">
series = tau - tau^2/6 + tau^3/36 + tau^4*a + tau^5*b;
(*pick the right a,b to make the series equal 0 at higher orders*)
Series[tau^2/2 + 1 + t - Exp[t] /. t -> series, {tau, 0, 8}]

(*now do the integral*)
integral = Integrate[Exp[-x*tau^2/2] * D[series /. a -> 0 /. b -> 0, tau], {tau, -Infinity, Infinity}];
Simplify[integral/Sqrt[2*Pi]*Sqrt[x]]

</syntaxhighlight>|name=mathematica-program|content=content|text=text}}

Thus we get Stirling's formula to two orders:<math display="block"> n! = \sqrt{2\pi n}\left(\frac{n}{e}\right)^n \left(1 + \frac{1}{12 n}+O\left(\frac{1}{n^2}\right) \right).
</math>

=== Complex-analytic version ===
A complex-analysis version of this method{{r|flajolet-sedgewick}} is to consider <math>\frac{1}{n!}</math> as a [[Taylor series|Taylor coefficient]] of the exponential function <math>e^z = \sum_{n=0}^\infty \frac{z^n}{n!}</math>, computed by [[Cauchy's integral formula]] as
<math display="block">\frac{1}{n!} = \frac{1}{2\pi i} \oint\limits_{|z|=r} \frac{e^z}{z^{n+1}} \, \mathrm dz. </math>

This line integral can then be approximated using the [[Method of steepest descent|saddle-point method]] with an appropriate choice of contour radius <math>r = r_n</math>. The dominant portion of the integral near the saddle point is then approximated by a real integral and Laplace's method, while the remaining portion of the integral can be bounded above to give an error term.

=== Using the Central Limit Theorem and the Poisson distribution ===
An alternative version uses the fact that the [[Poisson distribution]] converges to a [[normal distribution]] by the [[Central limit theorem|Central Limit Theorem]].<ref>{{Cite book |last=MacKay |first=David J. C. |title=Information theory, inference, and learning algorithms |date=2019 |publisher=Cambridge University Press |isbn=978-0-521-64298-9 |edition=22nd printing |location=Cambridge}}</ref>

Since the Poisson distribution with parameter <math>\lambda</math> converges to a normal distribution with mean <math>\lambda</math> and variance <math>\lambda</math>, their [[Probability density function|density functions]] will be approximately the same:

<math>\frac{\exp(-\mu)\mu^x}{x!}\approx \frac{1}{\sqrt{2\pi\mu}}\exp(-\frac{1}{2}(\frac{x-\mu}{\sqrt{\mu}}))</math>

Evaluating this expression at the mean, at which the approximation is particularly accurate, simplifies this expression to:

<math>\frac{\exp(-\mu)\mu^\mu}{\mu!}\approx \frac{1}{\sqrt{2\pi\mu}}</math>

Taking logs then results in:

<math>-\mu+\mu\ln\mu-\ln\mu!\approx -\frac{1}{2}\ln 2\pi\mu</math>

which can easily be rearranged to give:

<math>\ln\mu!\approx  \mu\ln\mu - \mu + \frac{1}{2}\ln 2\pi\mu</math>

Evaluating at <math>\mu=n</math> gives the usual, more precise form of Stirling's approximation.

== Speed of convergence and error estimates ==
[[File:Stirling series relative error.svg|thumb|upright=1.8|The relative error in a truncated Stirling series vs. <math>n</math>, for 0 to 5 terms. The kinks in the curves represent points where the truncated series coincides with {{math|&Gamma;(''n'' + 1)}}.]]

Stirling's formula is in fact the first approximation to the following series (now called the '''Stirling series'''):{{r|nist}}
<math display=block>
n! \sim \sqrt{2\pi n}\left(\frac{n}{e}\right)^n \left(1 +\frac{1}{12n}+\frac{1}{288n^2} - \frac{139}{51840n^3} -\frac{571}{2488320n^4}+ \cdots \right).</math>

An explicit formula for the coefficients in this series was given by G. Nemes.{{r|Nemes2010-2}} Further terms are listed in the [[On-Line Encyclopedia of Integer Sequences]] as {{OEIS link|A001163}} and {{OEIS link|A001164}}. The first graph in this section shows the [[Approximation error|relative error]] vs. <math>n</math>, for 1 through all 5 terms listed above. (Bender and Orszag<ref>{{Cite book |last1=Bender |first1=Carl M. |title=Advanced mathematical methods for scientists and engineers. 1: Asymptotic methods and perturbation theory |last2=Orszag |first2=Steven A. |date=2009 |publisher=Springer |isbn=978-0-387-98931-0 |edition=Nachdr. |location=New York, NY}}</ref> p.&nbsp;218) gives the asymptotic formula for the coefficients:<math display="block">A_{2 j+1} \sim(-1)^j 2(2 j) ! /(2 \pi)^{2(j+1)}</math>which shows that it grows superexponentially, and that by the [[ratio test]] the [[radius of convergence]] is zero.

[[File:Stirling error vs number of terms.svg|thumb|upright=1.8|The relative error in a truncated Stirling series vs. the number of terms used]]

As {{math|''n'' → ∞}}, the error in the truncated series is asymptotically equal to the first omitted term. This is an example of an [[asymptotic expansion]]. It is not a [[convergent series]]; for any ''particular'' value of <math>n</math> there are only so many terms of the series that improve accuracy, after which accuracy worsens.  This is shown in the next graph, which shows the relative error versus the number of terms in the series, for larger numbers of terms. More precisely, let {{math|''S''(''n'', ''t'')}} be the Stirling series to <math>t</math> terms evaluated at&nbsp;<math>n</math>.  The graphs show 
<math display=block>\left | \ln \left (\frac{S(n, t)}{n!} \right) \right |, </math>
which, when small, is essentially the relative error.

Writing Stirling's series in the form
<math display=block>\ln(n!) \sim n\ln n - n + \tfrac12\ln(2\pi n) +\frac{1}{12n} - \frac{1}{360n^3} + \frac{1}{1260n^5} - \frac{1}{1680n^7} + \cdots,</math>
it is known that the error in truncating the series is always of the opposite sign and at most the same magnitude as the first omitted term.{{Citation needed|date=December 2024}}

Other bounds, due to Robbins,{{r|Robbins1955}} valid for all positive integers <math>n</math> are
<math display=block>\sqrt{2\pi n}\left(\frac{n}{e}\right)^n e^{\frac{1}{12n + 1}} < n! < \sqrt{2\pi n}\left(\frac{n}{e}\right)^n e^{\frac{1}{12n}}. </math>
This upper bound corresponds to stopping the above series for <math>\ln(n!)</math> after the <math>\frac{1}{n}</math> term. The lower bound is weaker than that obtained by stopping the series after the <math>\frac{1}{n^3}</math> term. A looser version of this bound is that <math>\frac{n! e^n}{n^{n+\frac12}} \in (\sqrt{2 \pi}, e]</math> for all <math>n \ge 1</math>.

==Stirling's formula for the gamma function==
For all positive integers,
<math display=block>n! = \Gamma(n + 1),</math>
where {{math|Γ}} denotes the [[gamma function]].

However, the gamma function, unlike the factorial, is more broadly defined for all complex numbers other than non-positive integers; nevertheless, Stirling's formula may still be applied. If {{math|Re(''z'') > 0}}, then
<math display=block>\ln\Gamma (z) = z\ln z - z + \tfrac12\ln\frac{2\pi}{z} + \int_0^\infty\frac{2\arctan\left(\frac{t}{z}\right)}{e^{2\pi t}-1}\,{\rm d}t.</math>

Repeated integration by parts gives
<math display=block>\begin{align}
\ln\Gamma(z) \sim z\ln z - z + \tfrac12\ln\frac{2\pi}{z}
+ \sum_{n=1}^{N-1} \frac{B_{2n}}{2n(2n-1)z^{2n-1}} \\
= z\ln z - z + \tfrac12\ln\frac{2\pi}{z}
+\frac{1}{12z}
-\frac{1}{360z^3}
+\frac{1}{1260z^5}+\dots
,\end{align}</math>

where <math>B_n</math> is the <math>n</math>th [[Bernoulli number]] (note that the limit of the sum as <math>N \to \infty</math> is not convergent, so this formula is just an [[asymptotic expansion]]). The formula is valid for <math>z</math> large enough in absolute value, when {{math|{{abs|arg(''z'')}} < π − ''ε''}}, where {{mvar|ε}} is positive, with an error term of {{math|''O''(''z''<sup>−2''N''+ 1</sup>)}}. The corresponding approximation may now be written:
<math display=block>\Gamma(z) = \sqrt{\frac{2\pi}{z}}\,{\left(\frac{z}{e}\right)}^z \left(1 + O\left(\frac{1}{z}\right)\right).</math>

where the expansion is identical to that of Stirling's series above for <math>n!</math>, except that <math>n</math> is replaced with {{math|''z'' − 1}}.{{r|spiegel}}

A further application of this asymptotic expansion is for complex argument {{mvar|z}} with constant {{math|Re(''z'')}}. See for example the Stirling formula applied in {{math|Im(''z'') {{=}} ''t''}} of the [[Riemann–Siegel theta function]] on the straight line {{math|{{sfrac|1|4}} + ''it''}}.

==A convergent version of Stirling's formula==
[[Thomas Bayes]] showed, in a letter to [[John Canton]] published by the [[Royal Society]] in 1763, that Stirling's formula did not give a [[convergent series]].{{r|bayes-canton}}  Obtaining a convergent version of Stirling's formula entails evaluating [[Gamma function#Raabe's formula|Binet's formula]]:
<math display=block>\int_0^\infty \frac{2\arctan\left(\frac{t}{x}\right)}{e^{2\pi t}-1}\,{\rm d}t = \ln\Gamma(x) - x\ln x + x - \tfrac12\ln\frac{2\pi}{x}.</math>

One way to do this is by means of a convergent series of inverted [[rising factorial]]s. If
<math display=block>z^{\bar n} = z(z + 1) \cdots (z + n - 1),</math>
then
<math display=block>\int_0^\infty \frac{2\arctan\left(\frac{t}{x}\right)}{e^{2\pi t} - 1}\,{\rm d}t = \sum_{n=1}^\infty \frac{c_n}{(x + 1)^{\bar n}},</math>
where
<math display=block>c_n = \frac{1}{n} \int_0^1 x^{\bar n} \left(x - \tfrac{1}{2}\right)\,{\rm d}x = \frac{1}{2n}\sum_{k=1}^n \frac{k|s(n, k)|}{(k + 1)(k + 2)},</math>
where {{math|''s''(''n'',&nbsp;''k'')}} denotes the [[Stirling numbers of the first kind]]. From this one obtains a version of Stirling's series
<math display=block>\begin{align}
\ln\Gamma(x) &= x\ln x - x + \tfrac12\ln\frac{2\pi}{x} + \frac{1}{12(x+1)} + \frac{1}{12(x+1)(x+2)} + \\
  &\quad + \frac{59}{360(x+1)(x+2)(x+3)} + \frac{29}{60(x+1)(x+2)(x+3)(x+4)} + \cdots,
\end{align}</math>
which converges when {{math|Re(''x'') > 0}}. 
Stirling's formula may also be given in convergent form as<ref>{{cite book |last1=Artin |first1=Emil |title=The Gamma Function |date=2015 |page = 24|publisher=Dover }}</ref>
<math display=block>
\Gamma(x)=\sqrt{2\pi}x^{x-\frac{1}{2}}e^{-x+\mu(x)}
</math>
where 
<math display=block>
\mu\left(x\right)=\sum_{n=0}^{\infty}\left(\left(x+n+\frac{1}{2}\right)\ln\left(1+\frac{1}{x+n}\right)-1\right).
</math>

==Versions suitable for calculators==
The approximation
<math display=block>\Gamma(z) \approx \sqrt{\frac{2 \pi}{z}} \left(\frac{z}{e} \sqrt{z \sinh\frac{1}{z} + \frac{1}{810z^6} } \right)^z</math>
and its equivalent form
<math display=block>2\ln\Gamma(z) \approx \ln(2\pi) - \ln z + z \left(2\ln z + \ln\left(z\sinh\frac{1}{z} + \frac{1}{810z^6}\right) - 2\right)</math>
can be obtained by rearranging Stirling's extended formula and observing a coincidence between the resultant [[power series]] and the [[Taylor series]] expansion of the [[hyperbolic sine]] function. This approximation is good to more than 8 decimal digits for {{mvar|z}} with a real part greater than 8. Robert H. Windschitl suggested it in 2002 for computing the gamma function with fair accuracy on calculators with limited program or register memory.{{r|toth}}

Gergő Nemes proposed in 2007 an approximation which gives the same number of exact digits as the Windschitl approximation but is much simpler:{{r|Nemes2010}}
<math display=block>\Gamma(z) \approx \sqrt{\frac{2\pi}{z} } \left(\frac{1}{e} \left(z + \frac{1}{12z - \frac{1}{10z}}\right)\right)^z,</math>
or equivalently,
<math display=block> \ln\Gamma(z) \approx \tfrac{1}{2} \left(\ln(2\pi) - \ln z\right) + z\left(\ln\left(z + \frac{1}{12z - \frac{1}{10z}}\right) - 1\right). </math>

An alternative approximation for the gamma function stated by [[Srinivasa Ramanujan]] in [[Ramanujan's lost notebook]]<ref>{{citation|url=https://archive.org/details/lost-notebook/page/n337/|title=Lost Notebook and Other Unpublished Papers|first=Srinivasa|last=Ramanujan|date=14 August 1920 |author-link=Srinivasa Ramanujan|via=Internet Archive|page=339}}</ref> is
<math display=block>\Gamma(1+x) \approx \sqrt{\pi} \left(\frac{x}{e}\right)^x  \left( 8x^3 + 4x^2 + x + \frac{1}{30} \right)^{\frac{1}{6}}</math>
for {{math|''x'' ≥ 0}}. The equivalent approximation for {{math|ln ''n''!}} has an asymptotic error of {{math|{{sfrac|1|1400''n''<sup>3</sup>}}}} and is given by
<math display=block>\ln n! \approx n\ln n - n + \tfrac{1}{6}\ln(8n^3 + 4n^2 + n + \tfrac{1}{30}) + \tfrac{1}{2}\ln\pi .</math>

The approximation may be made precise by giving paired upper and lower bounds; one such inequality is{{r|E.A.Karatsuba|Mortici2011-1|Mortici2011-2|Mortici2011-3}}
<math display=block> \sqrt{\pi} \left(\frac{x}{e}\right)^x \left( 8x^3 + 4x^2 + x + \frac{1}{100} \right)^{1/6} < \Gamma(1+x) <  \sqrt{\pi} \left(\frac{x}{e}\right)^x \left( 8x^3 + 4x^2 + x + \frac{1}{30} \right)^{1/6}.</math>

== History ==
The formula was first discovered by [[Abraham de Moivre]]{{r|LeCam1986}} in the form
<math display=block>n! \sim [{\rm constant}] \cdot n^{n+\frac12} e^{-n}.</math>

De Moivre gave an approximate rational-number expression for the natural logarithm of the constant. Stirling's contribution consisted of showing that the constant is precisely <math>\sqrt{2\pi} </math>.{{r|Pearson1924}}

==See also==
* [[Lanczos approximation]]
* [[Spouge's approximation]]

==References==
{{Reflist|refs=

<ref name=bayes-canton>{{citation |url=http://www.york.ac.uk/depts/maths/histstat/letter.pdf |title=A letter from the late Reverend Mr. Thomas Bayes, F. R. S. to John Canton, M. A. and F. R. S.|date=24 November 1763 |bibcode=1763RSPT...53..269B |access-date=2012-03-01 |url-status=live |archive-url=https://web.archive.org/web/20120128050439/http://www.york.ac.uk/depts/maths/histstat/letter.pdf |archive-date=2012-01-28 |last1=Bayes |first1=Thomas |journal= Philosophical Transactions |volume=53 |page=269 }}</ref>

<ref name=dutka>{{citation 
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<ref name=E.A.Karatsuba>{{citation
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<ref name=LeCam1986>{{citation
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 }}; see p. 81, "The result, obtained using a formula originally proved by de Moivre but now called Stirling's formula, occurs in his 'Doctrine of Chances' of 1733."</ref>

<ref name=flajolet-sedgewick>{{citation
 | last1 = Flajolet | first1 = Philippe
 | last2 = Sedgewick | first2 = Robert
 | doi = 10.1017/CBO9780511801655
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<ref name=Mortici2011-1>{{Citation |last=Mortici |first=Cristinel |year=2011 |title=Ramanujan's estimate for the gamma function via monotonicity arguments |journal=Ramanujan J. |volume=25 |issue=2 |pages=149–154|doi=10.1007/s11139-010-9265-y |s2cid=119530041 }}</ref>

<ref name=Mortici2011-2>{{Citation |last=Mortici |first=Cristinel |year=2011 |title=Improved asymptotic formulas for the gamma function |journal=Comput. Math. Appl. |volume=61 |issue=11 |pages=3364–3369|doi=10.1016/j.camwa.2011.04.036 }}.</ref>

<ref name=Mortici2011-3>{{Citation |last=Mortici |first=Cristinel |year=2011 |title=On Ramanujan's large argument formula for the gamma function |journal=Ramanujan J. |volume=26 |issue=2 |pages=185–192|doi=10.1007/s11139-010-9281-y |s2cid=120371952 }}.</ref>

<ref name=Nemes2010>{{citation
 | last = Nemes | first = Gergő
 | doi = 10.1007/s00013-010-0146-9
 | title = New asymptotic expansion for the Gamma function
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| s2cid = 121820640
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<ref name=Nemes2010-2>{{Citation|last=Nemes|first=Gergő|year=2010|title=On the coefficients of the asymptotic expansion of <math>n!</math>|journal=Journal of Integer Sequences|volume=13|issue=6|pages=5}}</ref>

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}}

==Further reading==
*{{citation |last1=Abramowitz |first1=M. |last2=Stegun |first2=I. |name-list-style=amp |title=Handbook of Mathematical Functions |year=2002 |title-link=Abramowitz and Stegun }}
*{{citation |last1=Paris |first1=R. B. |last2=Kaminski |first2=D. |name-list-style=amp |title=Asymptotics and Mellin–Barnes Integrals |year=2001 |publisher=Cambridge University Press |location=New York |isbn=978-0-521-79001-7 |url-access=registration |url=https://archive.org/details/asymptoticsmelli0000pari }}
*{{citation |last1=Whittaker |first1=E. T. |last2=Watson |first2=G. N. |name-list-style=amp |title=A Course in Modern Analysis |year=1996 |edition=4th |publisher=Cambridge University Press |location=New York |isbn=978-0-521-58807-2 }}
*{{citation
 | last = Romik | first = Dan
 | doi = 10.2307/2589351
 | issue = 6
 | journal = [[The American Mathematical Monthly]]
 | mr = 1767064
 | pages = 556–557
 | title = Stirling's approximation for <math>n!</math>: the ultimate short proof?
 | volume = 107
 | year = 2000| jstor = 2589351
 }}
*{{citation
 | last = Li | first = Yuan-Chuan
 | date = July 2006
 | issue = 1
 | journal = Real Analysis Exchange
 | mr = 2329236
 | pages = 267–271
 | title = A note on an identity of the gamma function and Stirling's formula
 | url = https://projecteuclid.org/euclid.rae/1184700051
 | volume = 32}}

<references group="note" />

{{Notelist}}

==External links==
{{Commons category}}

* {{SpringerEOM | title=Stirling_formula | id=Stirling_formula&oldid=44695}}
* [http://www.luschny.de/math/factorial/approx/SimpleCases.html Peter Luschny, ''Approximation formulas for the factorial function n!'']
* {{MathWorld | urlname=StirlingsApproximation | title=Stirling's Approximation | mode=cs2}}
* {{PlanetMath | urlname=StirlingsApproximation | title=Stirling's approximation}}

{{Calculus topics}}

[[Category:Approximations]]
[[Category:Asymptotic analysis]]
[[Category:Analytic number theory]]
[[Category:Gamma and related functions]]
[[Category:Theorems in mathematical analysis]]