Stirling's Formula

Stirling formula or Stirling approximation, named after Scottish Mathematician James Stirling, is a formula used to find the approximate value of large factorials (written n!; eg; 3! = 3 x 2 x 1) that make use of Mathematical constant e (the base of the natural logarithm) and π.

The Stirling formula offers the easiest way to calculate the factorial n! especially for a large positive integer n which is a very tedious task to calculate.

Stirling Approximation Formula

The factorial function n! Is approximated by the formula:

n! ~ \[\sqrt{2 \pi n (\frac{n}{e})^{n}}\]

Furthermore, for any positive integer, we have the limits

\[\sqrt{2 \pi}\] nⁿ⁺ \[\frac{1}{2}\] e⁻ⁿ ≤  n!  ≤ nⁿ⁺ \[\frac{1}{2}\] e⁻ⁿ

Stirling Interpolation Formula 

Statement:

 If...x₋₂, x₋₁, x₀, x₁, x₂ are given as a set observation with a common difference n and let y₋₂, y₋₁, y₀, y₁, y₂ are their corresponding values where y = f(x) be the given function, then 

yₚ = yₒ + p (\[\frac{\Delta y_{0} - \Delta y_{-1}}{2}\]) + \[\frac{P^{2}}{2!}\] \[\Delta\]\[^{2}\]y₋₁ + (\[\frac{p(p^{2} - 1)}{3!}\] (\[\frac{\Delta^{3} y_{-1} + \Delta^{3}y_{-2}}{2}\]) \[\frac{p^{2}(p^{2} - 1)}{4!}\]) \[\Delta\]\[^{4}\]y₋₂   

Here, p = \[\frac{x - x_{0}}{h}\]

Proof: 

Stirling interpolation formula will be obtained by taking the average of Gauss Forward difference formula and Gauss Backward difference formula.

Gauss Forward Difference Formula is Given as:

yₚ = yₒ + p\[\Delta\]yₒ + \[\frac{p(p - 1)}{2!}\] \[\Delta\]\[^{2}\]y₋₁ + \[\frac{p(p + 1)(p - 1)}{3!}\] \[\Delta\]\[^{3}\]y₋₁ \[\frac{p(p - 1)(p + 1)(p - 2)}{4!}\] + \[\Delta\]\[^{4}\]y₋₂+ …..

Gauss Backward Difference Formula is Given as:

yₚ = yₒ + p\[\Delta\]y₋₁ + \[\frac{p(p - 1)}{2!}\] \[\Delta\]\[^{2}\]y₋₁ + \[\frac{p(p + 1)(p - 1)}{3!}\] \[\Delta\]\[^{3}\]y₋₂ \[\frac{p(p - 1)(p + 1)(p - 2)}{4!}\] + \[\Delta\]\[^{4}\]y₋₂+ …..

Now, stirling formulas is given as 

\[\frac{\text{Gauss Forward Difference Formula + Gauss Backward Difference Formula}}{2}\]

yₚ = yₒ + p

yₚ = yₒ + p (\[\frac{\Delta y_{0} - \Delta y_{-1}}{2}\]) + \[\frac{P^{2}}{2!}\] \[\Delta\]\[^{2}\]y₋₁ + \[\frac{p(p^{2} - 1)}{3!}\] (\[\frac{\Delta^{3} y_{-1} - \Delta^{3} y_{-2}}{2}\]) \[\frac{p^{2}(p^{2} - 1)}{4!}\] \[\Delta\]\[^{4}\]y₋₂

Stirling Formula Example

1. Find the Factorial 11 Using the Stirling  Formula.

Solution:

Using the Stirling Formula  - \[\sqrt{2 \pi}\] nⁿ⁺ \[\frac{1}{2}\] e⁻ⁿ 

We have:

11! = \[\sqrt{2 \pi}\] 11¹¹⁺ \[\frac{1}{2}\] x e⁻¹¹ = 39615625.05

Therefore, the value of factorial 11 = 39615625.05

2. Calculate the Solution Using the Stirling Formula

x = 28

Solution:

The table for the values of x and y

Stirling methods to determine the solution.

h = 25 - 20 = 5

Considering x₀ = 30 and p = \[\frac{x - x_{0}}{h}\] = \[\frac{x - 30}{5}\] 

The difference table is given below:

x = 28

p =  \[\frac{x - x_{0}}{h}\] =  \[\frac{28 - 30}{5}\] = 0.4

y₀ = 47236, \[\Delta\]y₀ = -1310, \[\Delta\]\[^{2}\]y₋₁ = - 230  \[\Delta\]\[^{3}\]y₋₁ = - 80, \[\Delta\]\[^{4}\]y₋₁= - 21

Stirling formula is given as:

yₚ = yₒ + p (\[\frac{\Delta y_{0} - \Delta y_{-1}}{2}\]) + \[\frac{P^{2}}{2!}\] \[\Delta\]\[^{2}\]y₋₁ + \[\frac{p(p^{2} - 1)}{3!}\] (\[\frac{\Delta^{3} y_{-1} - \Delta^{3} y_{-2}}{2}\]) \[\frac{p^{2}(p^{2} - 1)}{4!}\] \[\Delta\]\[^{4}\]y₋₂

y\[_{-0.4}\] =  47236 + (-0.4) . \[\frac{(-1310 - 1080)}{2}\] + \[\frac{(0.16)}{2}\] . (-230) + \[\frac{(-0.4)(0.16 - 1)}{6}\] + \[\frac{(-80 + 59)}{2}\] + \[\frac{(-0.16)(0.16 - 1)}{24}\] . (-21)

y\[_{-0.4}\] = 47236 + 478 - 18.4 - 3.892 + 0.1176

y\[_{-0.4}\] = 47691.8256

Therefore, the solution of Stirling interpolation is 

y(28) = 47691.8256