Example Of Sterling Formular
First of all introduce the sterling formulae. n! = 11!
n! = √2π nⁿ⁺¹/² e⁻ⁿ
by applying logarithm we have
log√2π * lognⁿ⁺¹/² * loge⁻ⁿ
log2π¹/² * lognⁿ⁺¹/² * loge⁻ⁿ
¹/2log2π * (n+1/2)logn * (-n)loge
Subtitute n with 11 and e = 2.718
0.3991*11.9760 * -4.7767
= 2.9216 ×10⁻³
= now we take the antilog of 0.9216
= 8.3483
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