Stirling ApproximationExample b(n, 2n, p)≈(4pq)πnb(n, ~2n, ~p) \approx {(4pq) \over \sqrt{\pi n}}b(n, 2n, p)≈πn(4pq) where nnn is the number of heads, 2n2n2n is the number of trials, and ppp is the probability of success. b(n, 2n, p)=(2nn)pn(1−p)2n−nb(n, ~2n, ~p) = {2n \choose n} p^n (1-p)^{2n-n}b(n, 2n, p)=(n2n)pn(1−p)2n−n =(2n)!n!n!pnqn= {(2n)! \over {n!n!}} p^n q^n=n!n!(2n)!pnqn By Stirling's approximation, ≈2π 2n 2n2n e−2n pn qn2πn nn e−n\approx {{\sqrt{2\pi ~ 2n} ~ {2n}^{2n} ~ e^{-2n} ~ p^n ~ q^n} \over {\sqrt{2 \pi n} ~ n^n ~ e^{-n}}}≈2πn nn e−n2π 2n 2n2n e−2n pn qn Cleaning up, =(4pq)πn ■= {(4pq) \over \sqrt{\pi n}} ~~~~~~~~~~ \blacksquare=πn(4pq) ■Backlinks (1)230126Comments (0)