1. 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

2. 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) =

3. … bound by integral method Integral Method ln (x) … ln (x+1) ln ln n2 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

4. Integral Method
Reminder:

5. Integral Method
Bounds on ln(n!)

6. In-Class Problem
Problem 1

7. Stirling’s Formula So

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

9. Asymptotic Equivalence
f(n) ~ g(n)

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

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

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

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

14. 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)

15. f(x) = O(g(x)) g(x) c· f(x) Asymptotic Notation ln c
blue stays below red 
log
scale 
f(x)

16. 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

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

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

19. 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))

20. In-Class Problem
Problems 2--4

21. Big Oh Mistakes
False Lemma:
Of course really

22. = 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).

23. In-Class Problem
Problem 5