Stirling’s Approximation Formula

Stirling’s Approximation Formula: A factorial, in mathematics, is defined for all positive integers as the product of all the integers preceding it and the integer itself. For example, n! called n factorial is calculated as n × (n-1) × (n-2) × (n-3) × …. 3 × 2 × 1. The above calculation gets tedious as the magnitude of the integers increases. This is when the Stirling formula is used.

Stirling’s approximation provides a formula for approximating the natural logarithm of a factorial, expressed as ln⁡(N!) = N ln⁡(N) – N. This approximation improves in accuracy as the number N increases.

What is Stirling’s Approximation Formula?

This formula was given by James Stirling. It is used to find the approximate value of the factorial of a given non-negative integer. It is to be noted that this formula yields the factorial value which is quite close to that of the real value of the factorial of the given integer. 

Stirling’s Approximation Formula

[Tex]n!≈\sqrt{2πn}[\frac{n}{e}]^n [/Tex]

where n is the given non- negative integer.

Solved Examples on Stirling’s Approximation Formula

Problem 1: Use the Stirling Formula to find the value of 6! and then calculate the error difference.

Solution:

As per Stirling formula, [Tex]n!≈\sqrt{2πn}[\frac{n}{e}]^n [/Tex]

Here n = 6.

⇒ [Tex]6!=\sqrt{2\timesπ\times6}[\frac{6}{e}]^6 [/Tex]

= 719.19

As 6! = 720, the Stirling formula gives an error of 0.11% .

Problem 2: Find the value of 11! using the Stirling formula and then calculating the error difference.

Solution:

As per Stirling formula, [Tex]n!≈\sqrt{2πn}[\frac{n}{e}]^n [/Tex]

Here n = 11.

⇒ [Tex]11!=\sqrt{2\timesπ\times11}[\frac{11}{e}]^{11} [/Tex]

= 39615625.05

As 11! = 39916800, the Stirling formula gives an error of 0.75% .

Problem 3: Find 13! using Stirling Formula and then calculate the error difference.

Solution:

As per Stirling formula, [Tex]n!≈\sqrt{2πn}[\frac{n}{e}]^n [/Tex]

Here n = 13.

⇒ [Tex]13!=\sqrt{2\timesπ\times13}[\frac{13}{e}]^{13} [/Tex]

= 6187239475.19

As 13! = 6227020800, the Stirling formula gives an error of 0.63% .

Problem 4: Find 5! using Stirling Formula and then calculate the error difference.

Solution:

As per Stirling formula, [Tex]n!≈\sqrt{2πn}[\frac{n}{e}]^n [/Tex]

Here n = 5.

⇒ [Tex]5!=\sqrt{2\timesπ\times5}[\frac{5}{e}]^{5} [/Tex]

= 118.96

As 5! = 120, the Stirling formula gives an error of 0.86% .

Problem 5: Find 7! using Stirling Formula and then calculate the error difference.

Solution:

As per Stirling formula, [Tex]n!≈\sqrt{2πn}[\frac{n}{e}]^n [/Tex]

Here n = 7.

⇒ [Tex]7!=\sqrt{2\timesπ\times7}[\frac{7}{e}]^{7} [/Tex]

= 4980.39

As 7! = 5040, the Stirling formula gives an error of 1.18% .

Problem 6: Find 9! using Stirling Formula and then calculate the error difference.

Solution:

As per Stirling formula, [Tex]n!≈\sqrt{2πn}[\frac{n}{e}]^n [/Tex]

Here n = 9.

⇒ 9!=\sqrt{2\timesπ\times9}[\frac{9}{e}]^{9}

= 359536.87

As 9! = 362880, the Stirling formula gives an error of 0.92% .

Problem 7: Find 3! using Stirling Formula and then calculate the error difference.

Solution:

As per Stirling formula, n!≈\sqrt{2πn}[\frac{n}{e}]^n

Here n = 3.

⇒ 3!=\sqrt{2\timesπ\times3}[\frac{3}{e}]^{3}

= 5.84

As 3! = 6, the Stirling formula gives an error of 2.67% .