Stirling's approximation

Comparison of Stirling's approximation with the factorial

In mathematics, Stirling's approximation (or Stirling's formula) is an asymptotic approximation for factorials. It is a good approximation, leading to accurate results even for small values of <math>n</math>. It is named after James Stirling, though a related but less precise result was first stated by Abraham de Moivre.Template:R

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 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.

DerivationEdit

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 Template:Math 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 Template:Math 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 Template:Math, 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, 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 derivationsEdit

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 Template:Math, 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 ordersEdit

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.Template:NoteTag

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 versionEdit

A complex-analysis version of this methodTemplate:R is to consider <math>\frac{1}{n!}</math> as a 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 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 distributionEdit

An alternative version uses the fact that the Poisson distribution converges to a normal distribution by the Central Limit Theorem.<ref>Template:Cite book</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 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 estimatesEdit

File:Stirling series relative error.svg

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 Template:Math.

Stirling's formula is in fact the first approximation to the following series (now called the Stirling series):Template:R <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.Template:R Further terms are listed in the On-Line Encyclopedia of Integer Sequences as A001163 and A001164. The first graph in this section shows the relative error vs. <math>n</math>, for 1 through all 5 terms listed above. (Bender and Orszag<ref>Template:Cite book</ref> p. 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

The relative error in a truncated Stirling series vs. the number of terms used

As Template:Math, 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 Template:Math be the Stirling series to <math>t</math> terms evaluated at <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.Template:Citation needed

Other bounds, due to Robbins,Template:R 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 functionEdit

For all positive integers, <math display=block>n! = \Gamma(n + 1),</math> where Template:Math 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 Template:Math, 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 Template:Math, where Template:Mvar is positive, with an error term of Template:Math. 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 Template:Math.Template:R

A further application of this asymptotic expansion is for complex argument Template:Mvar with constant Template:Math. See for example the Stirling formula applied in Template:Math of the Riemann–Siegel theta function on the straight line Template:Math.

A convergent version of Stirling's formulaEdit

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.Template:R Obtaining a convergent version of Stirling's formula entails evaluating 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 factorials. 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 Template:Math 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 Template:Math. Stirling's formula may also be given in convergent form as<ref>Template:Cite book</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 calculatorsEdit

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 Template:Mvar 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.Template:R

Gergő Nemes proposed in 2007 an approximation which gives the same number of exact digits as the Windschitl approximation but is much simpler:Template:R <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>Template:Citation</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 Template:Math. The equivalent approximation for Template:Math has an asymptotic error of Template:Math 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 isTemplate:R <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>

HistoryEdit

The formula was first discovered by Abraham de Moivre Template:R 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>.Template:R