Stirling's Approximation

1. Stirling’s
Approximation
Dr. Anjali Devi J S
Guest Faculty
School of Chemical Sciences
M G University

2. James Stirling and Stirling’s Approximation
James Stirling (Scottish Mathematician)
Born May 1692,
Garden, Stirlingshire
Died 5 December 1770 (aged 78)
Edinburgh, Scotland
Resting place Greyfriars Kirkyard
Nationality Scottish
Known for •Stirling's approximation
•Stirling numbers
Scientific career
Fields •Mathematics

3. Why Stirling’s Approximation?
 Factorials are HUGE!!
Find 10!
Ans: 3628
In Statistics, We have to deal with NA! Avogadro Number NA=
6.023 X1023
Find 1023!
To solve large, we use Stirling’s Approximation

4. Derivation of Stirling’s Approximation
N!= 1x2x3x…xN
HUGE NUMBER
Taking Natural Logarithm
ln (N!)= ln(1)+ln(2)+ ln(3).…+ln(N)
ln (N!)= 𝑖=1
𝑛
ln(𝑥)
If N is large (say 1023)
ln (N!)= 𝑖=1
𝑛
ln 𝑥 = 1
𝑁
𝑙𝑛𝑥 𝑑𝑥

5. Derivation of Stirling’s Approximation
ln (N!)= 1
𝑁
𝑙𝑛𝑥 𝑑𝑥
= 1
𝑁
1. 𝑙𝑛𝑥 𝑑𝑥
1st function
2nd function 𝐼𝑛𝑡𝑒𝑔𝑟𝑎𝑡𝑖𝑜𝑛 by parts
𝑢𝑣𝑑𝑥 = 𝑢 𝑣 𝑑𝑥 − 𝑢′( 𝑣 𝑑𝑥)𝑑𝑥
1
𝑁
ln(𝑥) 1 𝑑𝑥 = ln 𝑥
1
𝑁
1 𝑑𝑥 −
1
𝑁
(
1
𝑥
)(
1
𝑁
1 𝑑𝑥) 𝑑𝑥
= ln (N) N- ln(1) 1-N-1
=N ln(N)-N+1
=N ln (N) -N
ln (N!)≈𝑁 ln 𝑁 − 𝑁
This is Stirling’s Approximation

6. Accurate form of Stirling’s Approximation
A more accurate form of Stirling’s approximation is
𝑁! ≈ 2𝜋
1
2 𝑁𝑁+
1
2𝑒−𝑁

7. Question
ln (N!)≈𝑁 ln 𝑁 − 𝑁
Calculate the following using Stirling’s approximation:
(a) 10!
(b) 50!
N! Calculated
value
ln N!
(by
calculation)
ln N!
(by Stirling’s
Approximatio
n)
Error (%)
10! 3.63 x106 15.1 13.02 13.8
N! Calculated
value
ln N!
(by
calculation)
ln N!
(by Stirling’s
Approximatio
n)
Error (%)
50! 3.04 x1064 148.4 145.6 1.88

8. N! Calculated
value
ln N!
(by calculator)
ln N!
(by Stirling’s
Approximation)
Error
(%)
100! 9.33 x10157 363.7 360.5 0.88%
Question
ln (N!)≈𝑁 ln 𝑁 − 𝑁
Calculate the following using Stirling’s approximation:
(c) 100!
(d) 150! N! Calculated
value
ln N!
(by
calculation)
ln N!
(by Stirling’s
Approximation)
Error
(%)
150! 5.71 x10262 605.0 601.6 0.56%