In Exercises 131 and 132, find the exact value of and then approximate using Stirlings

Chapter 1, Problem 132

(choose chapter or problem)

For large values of  \(\boldsymbol{n}\),  \(n !=1 \cdot 2 \cdot 3 \cdot 4 \cdot \cdots(n-1) \cdot n\)

can be approximated by Stirling's Formula,

\(n ! \approx\left(\frac{n}{e}\right)^{n} \sqrt{2 \pi n}\)

In Exercises 131 and 132, find the exact value of  \(n\) !, and then approximate  \(n\) ! using Stirling's Formula.

n = 15

Text Transcription:

n  

n! =1 cdot 2 cdot 3 cdot 4 cdot cdots(n - 1) cdot n

n! approx (n / e)^{n} sqrt{2 pi n}