Prove Of Sterling Formulae

Sterling formulae n! = √2π nⁿ⁺¹/² e⁻ⁿ

logn! = log√2π × lognⁿ⁺¹/² × (-n)loge eq1

According to the law of logarithm which says whenever you have any thing like log10ʰ therefore is the same as hlog10.

= log(2π)¹/² × (n + 1/2)logn × -nloge eq2

= 1/2log2π × (n + 0.5)logn × -nloge  eq3

=  log[1/2(2π) × (n +1/2)n × -ne] eq4

By removing the logarithm

Remember this is proving of the formula, applying anti-logarithm is not required. But when n become any ℕ natural number. hence compute your solution with eq3.

 

Therefore 

n! = √2π nⁿ⁺¹/² e⁻ⁿ

Note: 1/2 is also sometimes square root √ ;

1/2 if devide is the same thing as 0.5

See example of sterling formulae of 11! here.

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