= Z ¥ 0 xne xdx (8) This integral is the starting point for Stirling’s approximation. $\endgroup$ – Giuseppe Negro Sep 30 '15 at 18:21 I want a result which is the other way around - a combinatorial\probabilistic proof for Stirling's approximation. is a product N(N-1)(N-2)..(2)(1). Stirling's approximation for approximating factorials is given by the following equation. STIRLING’S APPROXIMATION FOR LARGE FACTORIALS 2 n! Proof of the Stirling's Formula. In confronting statistical problems we often encounter factorials of very large numbers. I've just scanned the link posted by jspecter and it looks good and reasonably elementary. … µ N e ¶N =) lnN! The inte-grand is a bell-shaped curve which a precise shape that depends on n. The maximum value of the integrand is found from d dx xne x = nxn 1e x xne x =0 (9) x max = n (10) xne x max = nne n (11) The result is applied often in combinatorics and probability, especially in the study of random walks. There’s something annoying about the proof – it uses a priori knowledge about . The factorial N! In mathematics, Stirling's approximation (or Stirling's formula) is an approximation for factorials. This completes our proof. 2 π n n e + − + θ1/2 /12 n n n <θ<0 1 Applications of Stirling’s formula can be found in di erent parts of Probability theory. \[ \ln(N! Stirling S Approximation To N Derivation For Info. to get Since the log function is increasing on the interval , we get for . It is named after James Stirling , though it was first stated by Abraham de Moivre . First take the log of n! $\begingroup$ Stirling's formula is a pretty hefty result, so the tools involved are going to go beyond things like routine application of L'Hopital's rule, although I am sure there is a way of doing it that involves L'Hopital's rule as a step. Stirling's approximation for approximating factorials is given by the following equation. (Set-up) Let . )\sim N\ln N - N + \frac{1}{2}\ln(2\pi N) \] I've seen lots of "derivations" of this, but most make a hand-wavy argument to get you to the first two terms, but only the full-blown derivation I'm going to work through will offer that third term, and also provides a means of getting additional terms. I'm not sure if this is possible, but to convince … dN … lnN: (1) The easy-to-remember proof is in the following intuitive steps: lnN! Stirling’s Approximation Last updated; Save as PDF Page ID 2013; References; Contributors and Attributions; Stirling's approximation is named after the Scottish mathematician James Stirling (1692-1770). Introduction of Formula In the early 18th century James Stirling proved the following formula: For some = ! … N lnN ¡N =) dlnN! The full approximation states that , and after the proof I challenge you to bound it from above by . It begins by approximating the ratio , so we had to know Stirling’s approximation beforehand to even think about this ratio. (because C 0). The Stirling formula gives an approximation to the factorial of a large number, N À 1. It is a good approximation, leading to accurate results even for small values of n . \[ \ln(n! Any application? By Stirling's theorem your approximation is off by a factor of $\sqrt{n}$, (which later cancels in the fraction expressing the binomial coefficients). Add the above inequalities, with , we get Though the first integral is improper, it is easy to show that in fact it is convergent. For example, it is used in the proof of thede Moivre-Laplace theorem, which states that thenormal distributionmay be used as an approximation to thebinomial distributionunder certain conditions. In its simple form it is, N!