Tuesday, November 3, 2015

Stirling's Approximation: Derivation and Corollaries


Lemma 1: \(\lim_{n \rightarrow \infty} \sqrt[n]{n!}/n=1/e\)

Way 1 (somewhat rigorous)

From elementary calculus, we have that: \[ \int_{0}^{1} \ln(x) dx =-1 \] Taking this as a Riemann sum, as done in introductory calculus, we have: \[ -1=\int_{0}^{1}\ln(x)dx=\lim_{N \rightarrow \infty} \sum_{k=1}^{N}\ln\left (\frac{k}{N} \right ) \cdot \frac{1}{N} \] \[ -1=\lim_{N \rightarrow \infty} -\ln(N)+\frac{1}{N} \sum_{k=1}^{N}\ln\left (k \right ) \] \[ -1=\lim_{N \rightarrow \infty} -\ln(N)+\frac{1}{N} \ln\left (N! \right ) \] Therefore, \[ \lim_{N \rightarrow \infty} \frac{\sqrt[N]{N!}}{N}=\frac{1}{e} \]

Way 2 (less rigorous)

\[ \lim_{n \rightarrow \infty} \frac{\sqrt[n]{n!}}{n}=x \] So, for n big, in a certain sense: \[ n! \approx (nx)^n \] \[ \frac{(n+1)!}{n!(n+1)}=1 \approx \frac{((n+1)x)^{n+1}}{(nx)^n (n+1)}=\left ( 1+ \frac{1}{n} \right )^n x \] Thus, in order to get equality in the limit, we must have: \[ x = \lim_{n \rightarrow \infty} \left ( 1+ \frac{1}{n} \right )^{-n}=\frac{1}{e} \]

Lemma 2: Wallis Product in Factorial Form

Recall from this article the following expression for pi: \[ \frac{\pi}{2}=\prod_{k=1}^{\infty}\frac{2k \cdot 2k}{(2k-1)(2k+1)}=\lim_{N \rightarrow \infty}\prod_{k=1}^{N}\frac{2k \cdot 2k}{(2k-1)(2k+1)}=\lim_{N \rightarrow \infty} \frac{\left ( 2^N \cdot N! \right )^4}{\left ( (2N)! \right )^2(2N+1)} \]

Lemma 3: An Inequality for the Natural Logarithm

Let \(x,y > 0\). Clearly \[ 0 \leq \frac{1}{y^2 (1+y)^2 (2y+1)^2} \] Therefore \[ 0 \leq \int_{x}^{\infty}\frac{dy}{y^2 (1+y)^2 (2y+1)^2}=\frac{1}{x}+\frac{1}{x+1}+\frac{4}{x+1/2}-6\ln \left ( 1+\frac{1}{x} \right ) \] \[ 6\ln \left ( 1+\frac{1}{x} \right ) -\frac{6}{x+1/2} \leq \frac{1}{x}+\frac{1}{x+1}-\frac{2}{x+1/2} \] \[ (x+\tfrac{1}{2})\ln \left ( 1+\frac{1}{x} \right ) -1 \leq \frac{(x+\tfrac{1}{2})}{6}\left (\frac{1}{x}+\frac{1}{x+1} \right )-\frac{1}{3}=\frac{1}{12x(x+1)} \] Also, clearly \[ 0 \leq \frac{16y^2+41y+24}{y(1+y)^2 (2+y)^2 (2y+1)^2} \] Therefore \[ 0 \leq \int_{x}^{\infty}\frac{16y^2+41y+24}{y(1+y)^2 (2+y)^2 (2y+1)^2} dy=6\left (\ln \left ( 1+\frac{1}{x} \right )-\frac{1}{x+\tfrac{1}{2}} \right)-\frac{1}{2(x+\tfrac{1}{2})(x+1)(x+2)} \] And so \[ \frac{1}{12(x+1)(x+2)} \leq (x+\tfrac{1}{2})\ln \left ( 1+\frac{1}{x} \right )-1 \]

Theorem: Stirling's Approximation

Let us define a function and sequence of coefficients as follows: \[ g(n)=\ln\left ( \frac{n!}{\left ( \tfrac{n}{e} \right )^n \sqrt{2\pi n}} \right )=\sum_{k=-\infty}^{\infty} A_k n^k \] We then have, from lemma 1, \[ \frac{1}{e}=\lim_{n \rightarrow \infty} \frac{\sqrt[n]{n!}}{n}=\lim_{n \rightarrow \infty} \frac{\sqrt[n]{\left (\tfrac{n}{e} \right )^n \sqrt{2\pi n} \cdot e^{g(n)}}}{n}=\frac{1}{e} \lim_{n \rightarrow \infty} \sqrt[2n]{2\pi n} \cdot e^{g(n)/n} \] Thus \[ 1=\lim_{n \rightarrow \infty} e^{g(n)/n}=\exp\left (\lim_{n \rightarrow \infty} \sum_{k=-\infty}^{\infty} A_k n^{k-1} \right )=\exp\left (\lim_{n \rightarrow \infty} \sum_{k=1}^{\infty} A_k n^{k-1} \right ) \] And therefore \(A_k=0\) for \(k \geq 1\). From lemma 2, \[ \frac{\pi}{2}=\lim_{n \rightarrow \infty} \frac{\left ( 2^n \cdot n! \right )^4}{\left ( (2n)! \right )^2(2n+1)}=\lim_{n \rightarrow \infty} \frac{\left ( 2^n \cdot \left (\tfrac{n}{e} \right )^n \sqrt{2\pi n} \cdot e^{g(n)} \right )^4}{\left ( \left (\tfrac{2n}{e} \right )^{2n} \sqrt{4\pi n} \cdot e^{g(2n)} \right )^2(2n+1)} \] \[ \frac{\pi}{2}=\lim_{n \rightarrow \infty} \frac{\left ( 2^n \cdot n! \right )^4}{\left ( (2n)! \right )^2(2n+1)}=\lim_{n \rightarrow \infty} \frac{\left ( 2^n \cdot \left (\tfrac{n}{e} \right )^n \sqrt{2\pi n} \cdot e^{g(n)} \right )^4}{\left ( \left (\tfrac{2n}{e} \right )^{2n} \sqrt{4\pi n} \cdot e^{g(2n)} \right )^2(2n+1)} \] \[ \frac{\pi}{2}=\lim_{n \rightarrow \infty} \frac{n \pi}{2n+1} \cdot e^{4g(n)-2g(2n)} \] \[ 0=\lim_{n \rightarrow \infty} 2g(n)-g(2n)=2A_0-A_0=A_0 \] Therefore, \(A_k=0\) for \(k \geq 0\), and thus \(\lim_{n \rightarrow \infty} g(n)=0\). Thus it follows that \[ \lim_{n \rightarrow \infty} \frac{n!}{\left ( \tfrac{n}{e} \right )^n \sqrt{2\pi n}}=1 \] This fact is known as Stirling's Approximation. Moreover, we have \[ g(n)-g(n+1)=\ln\left ( \frac{n!\left ( \tfrac{n+1}{e} \right )^{n+1} \sqrt{2\pi (n+1)}}{(n+1)!\left ( \tfrac{n}{e} \right )^n \sqrt{2\pi n}} \right )=\ln\left ( \frac{(n+1)^{n+\tfrac{1}{2}}}{e \cdot n^{n+\tfrac{1}{2}}} \right ) \] \[ g(n)-g(n+1)=(n+\tfrac{1}{2})\ln\left ( 1+\frac{1}{n} \right )-1 \] By lemma 3, we then have \[ \frac{1}{12(n+1)(n+2)} \leq g(n)-g(n+1) \leq \frac{1}{12n(n+1)} \] \[ \sum_{k=n}^{\infty} \frac{1}{12(k+1)(k+2)}=\frac{1}{12(n+1)} \leq \sum_{k=n}^{\infty} g(k)-g(k+1)=g(n)-g(\infty)=g(n) \leq \sum_{k=n}^{\infty} \frac{1}{12k(k+1)}=\frac{1}{12n} \] That is \(\tfrac{1}{12(n+1)} \leq g(n) \leq \tfrac{1}{12n}\). And therefore: \[ \left (\tfrac{n}{e} \right )^n \sqrt{2\pi n}\cdot e^{\tfrac{1}{12(n+1)}} \leq n! \leq \left (\tfrac{n}{e} \right )^n \sqrt{2\pi n} \cdot e^{\tfrac{1}{12n}} \] In fact, it is possible to obtain exact formulas for \(g(n)\). For example, by more advanced calculations, we can show that \[ g(n)=\int_{0}^{\infty}\frac{2 \tan^{-1}\left ( \tfrac{y}{n} \right )}{e^{2\pi y}-1}=\sum_{k=1}^{\infty} \frac{B_{2k}}{2k(2k-1)n^{2k-1}}=\frac{1}{12n}-\frac{1}{360n^3}+\frac{1}{1260n^5}- \cdots \] Where \(B_m\) is the mth Bernoulli number. These two expressions are, respectively Binet's second expression and Stirling's series.

Corollary: Product of a Rational Function

Firstly, since \[ \prod_{k=1}^N \left(ak+b\right)=a^N\prod_{k=1}^N \left(k+\frac{b}{a}\right) \] We will just evaluate \[ \prod_{k=1}^N \left(k+b\right)=\frac{(N+b)!}{b!} \approx \left(\frac{N+b}{e}\right)^{N+b}\frac{\sqrt{2\pi(N+b)}}{b!}=N^{N+b+\tfrac{1}{2}}e^{-N}\frac{\sqrt{2\pi}}{b!}e^{-b}\left(1+\frac{b}{N}\right)^{N+b+\tfrac{1}{2}} \] \[ \prod_{k=1}^N \left(k+b\right)=\frac{(N+b)!}{b!} \approx N^{N+b+\tfrac{1}{2}}e^{-N}\frac{\sqrt{2\pi}}{b!} \] More generally, given the above, it is not difficult to demonstrate the following generalization. Let \(m,n > 0\). Let \(a_1,a_2,...,a_m\) and \(b_1,b_2,...,b_n\) and \(r_1,r_2,...,r_m\) and \(s_1,s_2,...,s_n\) be sequences of numbers, such that \[ \sum_{k=1}^m r_k=\sum_{k=1}^n s_k \] and \[ \sum_{k=1}^m a_k r_k=\sum_{k=1}^n b_k s_k \] Then \[ \prod_{k=1}^\infty\frac{(k+a_1)^{r_1}(k+a_2)^{r_2}\cdots (k+a_m)^{r_m}}{(k+b_1)^{s_1}(k+b_2)^{s_2}\cdots (k+b_n)^{s_n}}=\frac{\prod_{j=1}^n (b_j!)^{s_j}}{\prod_{j=1}^m (a_j!)^{r_j}} \] In cases where the coefficients are non-integral, we use the Gamma function (an extension of the factorial to non-integers), instead of factorials: \[ \prod_{k=1}^\infty\frac{(k+a_1)^{r_1}(k+a_2)^{r_2}\cdots (k+a_m)^{r_m}}{(k+b_1)^{s_1}(k+b_2)^{s_2}\cdots (k+b_n)^{s_n}}=\frac{\prod_{j=1}^n (\Gamma (b_j+1))^{s_j}}{\prod_{j=1}^m (\Gamma (a_j+1))^{r_j}} \] For instance \[ \prod_{k=0}^\infty \frac{(k+1)(k+a+b)}{(k+a)(k+b)}=\frac{\Gamma(a)\Gamma(b)}{\Gamma(a+b)}=B(a,b) \] \[ \frac{\sin(\pi x)}{\pi x}=\prod_{k=1}^\infty\frac{(k-x)(k+x)}{k^2}=\frac{\Gamma(1)^2}{\Gamma(1-x) \Gamma(1+x)}=\frac{1}{\Gamma(1-x)x \Gamma(x)} \] \[ \prod_{k=1}^\infty\frac{(1+\tfrac{1}{k})^x}{1+\tfrac{x}{k}}=\prod_{k=1}^\infty\frac{(k+1)^x k}{k^x (k+x)}=\frac{\Gamma(1)^x \Gamma (1+x)}{\Gamma(2)^x \Gamma(1)}=\Gamma(1+x) \]

Corollary: Asymptotic Behavior of Bernoulli Numbers

In this article, we found that \[ \zeta(2n)=\frac{1}{2}\frac{(2\pi)^{2n}}{(2n)!}\left | B_{2n} \right | \] Combining this with Stirling's approximation, we find that \[ \left | B_{2n} \right |=2\zeta(2n)\frac{(2n)!}{(2\pi)^{2n}} \approx 4\left ( \frac{n}{\pi e} \right )^{2n} \sqrt{n\pi} \cdot e^{1/24n} \]

Corollary: Approximation for Binomial Coefficients

\[ \binom{a}{b}=\frac{a!}{b!(a-b)!} \approx \frac{\left (\tfrac{a}{e} \right )^a \sqrt{2\pi a}} {\left (\tfrac{b}{e} \right )^b \sqrt{2\pi b}\left (\tfrac{a-b}{e} \right )^{a-b} \sqrt{2\pi (a-b)}}=\frac{1}{\sqrt{2\pi}}\sqrt{\frac{a}{b(a-b)}} \frac{a^a}{b^b (a-b)^{a-b}} \]

Corollary: Normal from Binomial

Let \(0 < p < 1\) and \(p+q=1\). Let \[ F_n(x)=\binom{n}{x}p^x q^{n-x} \] \[ f_n(x)=\sqrt{npq}F_n(np+x\sqrt{npq}) \] \[ \phi_n(x)=\ln(f_n(x)) \\ \\ \phi_n(x)=\ln(n!)-\ln((np+x\sqrt{npq})!)-\ln((nq-x\sqrt{npq})!)+(np+x\sqrt{npq})\ln(p)+(nq-x\sqrt{npq})\ln(q) \] Using Stirling's Approximation and some algebra \[ \phi_n(x) = -\tfrac{1}{2}\ln(2\pi)-\left (\tfrac{1}{2}+ np+x\sqrt{npq} \right )\ln\left ( 1+x\sqrt{\frac{q}{np}} \right)-\left (\tfrac{1}{2}+ nq-x\sqrt{npq} \right )\ln\left ( 1-x\sqrt{\frac{p}{nq}} \right)+O(\tfrac{1}{n}) \] Using the series expansion \(\ln(1+x)=x-\tfrac{1}{2}x^2+O(x^3) \) \[ \phi_n(x) = -\tfrac{1}{2}\ln(2\pi)-\tfrac{1}{2}x^2+O(\tfrac{1}{\sqrt{n}}) \] Thus, as \(n\) goes to infinity \[ \phi_\infty(x) = -\tfrac{1}{2}\ln(2\pi)-\tfrac{1}{2}x^2 \] \[ f_\infty(x) = \frac{e^{-x^2/2}}{\sqrt{2\pi}} \] Thus, in the limit, scaling for the changing means and variances, the binomial distribution tends to the normal distribution. Moreover, since the binomial distribution is normalized, we find that \[ \int_{-\infty}^{\infty}\frac{e^{-x^2/2}}{\sqrt{2\pi}}dx=1 \]