Asymptotics & Stirling’s Approximation Mathematics for Computer Science MIT 6.042J/18.062J Asymptotics & Stirling’s Approximation Copyright © Albert Meyer, 2002. Prof. Albert Meyer & Dr. Radhika Nagpal

Factorial defines a product: Integral Method Factorial defines a product: Turn product into a sum taking logs: ln(n!) = ln(1 · 2 · 3 · · · (n – 1) · n) = ln 1 + ln 2 +· · · + ln(n – 1) + ln(n) =

… bound by integral method Integral Method ln (x) … ln (x+1) ln ln n 2 3 1 4 5 n–2 n–1 n … ln (x+1) ln (x) … ln 2 ln 3 ln 4 ln 5 ln n-1 ln n

Integral Method Reminder:

Integral Method Bounds on ln(n!)

In-Class Problem Problem 1

Stirling’s Formula So

Stirling’s Formula A precise approximation: Stirling’s Formula

Asymptotic Equivalence f(n) ~ g(n)

Little Oh Asymptotically smaller: f(n) = o(g(n))

Big Oh Asymptotic Order of Growth: f(n) = O(g(n))

The Oh’s If f = o(g) or f ~ g then f = O(g) lim = 0 lim = 1 lim < 

The Oh’s If f = o(g), then g  O(f) lim =  lim = 0

c,n0  0 n  n0 |f(n)|  c·g(n) Big Oh Equivalently, f(n) = O(g(n)) c,n0  0 n  n0 |f(n)|  c·g(n)

Asymptotic Notation f(x) = O(g(x)) +c c g(x) blue stays below red f(x)

Lemma: xa = o(xb) for a < b Little Oh Lemma: xa = o(xb) for a < b Proof: and b - a > 0. So as x  , xb-a   and .

Lemma: ln x = o(x) for  > 0. Little Oh Lemma: ln x = o(x) for  > 0. Proof: for z 1.

Lemma: ln x = o(x) for  > 0. Little Oh Lemma: ln x = o(x) for  > 0. Proof: Let for  > .

f(n) = O(g(n)) and g(n) = O(f(n)) Theta Same Order of Growth: f(n) = (g(n)) f(n) = O(g(n)) and g(n) = O(f(n))

In-Class Problem Problems 2--4

Big Oh Mistakes False Lemma: Of course really

= n· O(1) = O(n). 0 = O(1), 1 = O(1), 2 = O(1),… So each i = O(1). So Big Oh Mistakes False Lemma: False Proof: 0 = O(1), 1 = O(1), 2 = O(1),… So each i = O(1). So = n· O(1) = O(n).

In-Class Problem Problem 5