Show that? (P(N', n))/(P(N,n)) ~~ ((N')/N)^n (N', N, n) in ZZ^+ (N', N )">>" n P(x,y) = (x!)/((x-y)!) Stirling Approx allowed

1 Answer
Jul 18, 2018

Stirling's approximation says that
x! approx sqrt(2pi n) (n/e)^n = sqrt(2pi) * sqrt(n) * n^n * e^-n

In this way, we can approximate P(x, y):

(x!)/((x-y)!) approx sqrt(2pi)/sqrt(2pi) cdot sqrt(x/(x-y)) cdot (x^x/(x-y)^(x-y)) e^(-x)/e^(-(x-y))

Since the second condition says that y is far less than x for our purposes, so
sqrt(x) approx sqrt(x-y)
And since all of these we know that x and x-y are far greater than e, it is clear that the dominating term is
P(x, y) approx (x^x)/((x-y)^(x-y)) = (x/(x-y))^x (x-y)^y approx (x/x)^x (x-y)^y
P(x, y) approx x^y

Returning to the original question,
(P(N', n))/(P(N, n)) approx ((N')^n)/((N)^n) = ((N')/N)^n

completing the derivation.