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 ($$\frac{\mathrm{\Delta }{y}_0-\mathrm{\Delta }{y}_{-1}}{2}$$) + $$\frac{P^2}{2!}$$ $$\mathrm{\Delta }$$$${}^2$$y₋₁ + ($$\frac{p(p^2-1)}{3!}$$ ($$\frac{{\mathrm{\Delta }}^3{y}_{-1}+{\mathrm{\Delta }}^3{y}_{-2}}{2}$$) $$\frac{p^2(p^2-1)}{4!}$$) $$\mathrm{\Delta }$$$${}^4$$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$$\mathrm{\Delta }$$yₒ + $$\frac{p(p-1)}{2!}$$ $$\mathrm{\Delta }$$$${}^2$$y₋₁ + $$\frac{p(p+1)(p-1)}{3!}$$ $$\mathrm{\Delta }$$$${}^3$$y₋₁ $$\frac{p(p-1)(p+1)(p-2)}{4!}$$ + $$\mathrm{\Delta }$$$${}^4$$y₋₂+ …..

Gauss Backward Difference Formula is Given as:

yₚ = yₒ + p$$\mathrm{\Delta }$$y₋₁ + $$\frac{p(p-1)}{2!}$$ $$\mathrm{\Delta }$$$${}^2$$y₋₁ + $$\frac{p(p+1)(p-1)}{3!}$$ $$\mathrm{\Delta }$$$${}^3$$y₋₂ $$\frac{p(p-1)(p+1)(p-2)}{4!}$$ + $$\mathrm{\Delta }$$$${}^4$$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 ($$\frac{\mathrm{\Delta }{y}_0-\mathrm{\Delta }{y}_{-1}}{2}$$) + $$\frac{P^2}{2!}$$ $$\mathrm{\Delta }$$$${}^2$$y₋₁ + $$\frac{p(p^2-1)}{3!}$$ ($$\frac{{\mathrm{\Delta }}^3{y}_{-1}-{\mathrm{\Delta }}^3{y}_{-2}}{2}$$) $$\frac{p^2(p^2-1)}{4!}$$ $$\mathrm{\Delta }$$$${}^4$$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, $$\mathrm{\Delta }$$y₀ = -1310, $$\mathrm{\Delta }$$$${}^2$$y₋₁ = - 230  $$\mathrm{\Delta }$$$${}^3$$y₋₁ = - 80, $$\mathrm{\Delta }$$$${}^4$$y₋₁= - 21

Stirling formula is given as:

yₚ = yₒ + p ($$\frac{\mathrm{\Delta }{y}_0-\mathrm{\Delta }{y}_{-1}}{2}$$) + $$\frac{P^2}{2!}$$ $$\mathrm{\Delta }$$$${}^2$$y₋₁ + $$\frac{p(p^2-1)}{3!}$$ ($$\frac{{\mathrm{\Delta }}^3{y}_{-1}-{\mathrm{\Delta }}^3{y}_{-2}}{2}$$) $$\frac{p^2(p^2-1)}{4!}$$ $$\mathrm{\Delta }$$$${}^4$$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