# Stirlings Formula

• Jul 26th 2011, 01:55 PM
Arron
Stirlings Formula
I need to use Stirlings Formula to determine a number ¬ such that

(4n)!/((2n)!)^2 ~ ¬ 2^4n/Route n as n ---> infinity

I have

By Stirling's Formula, and the Product and Quotient Rules for ~, we obtain

(4n)!/((2n)!)^2 ~ Route 8pi n(4n/e)^4n / (Route 4 pi n (2n/e)^2n)^2

= Route 8 pi n 2^4n / (route 4 pi n)^2

¬ = ? or this

Can anyone help?
• Jul 27th 2011, 04:30 AM
chisigma
Re: Stirlings Formula
Quote:

Originally Posted by Arron
I need to use Stirlings Formula to determine a number ¬ such that

(4n)!/((2n)!)^2 ~ ¬ 2^4n/Route n as n ---> infinity

I have

By Stirling's Formula, and the Product and Quotient Rules for ~, we obtain

(4n)!/((2n)!)^2 ~ Route 8pi n(4n/e)^4n / (Route 4 pi n (2n/e)^2n)^2

= Route 8 pi n 2^4n / (route 4 pi n)^2

¬ = ? or this

Can anyone help?

A 'non very popular' version of the Stirling's formula is the following...

$\displaystyle n! \sim \sqrt{2\ \pi n}\ (\frac{n}{e})^{n}\ e^{\frac{1}{12\ n}}$ (1)

... so that from (1)...

$\displaystyle (4 n)! \sim 2\ \sqrt{2\ \pi\ n}\ 2^{8 n} (\frac{n}{e})^{4 n}\ e^{\frac{1}{48 n}}$ (2)

... and...

$\displaystyle (2 n)! \sim \sqrt{2}\ \sqrt{2\ \pi\ n}\ 2^{2 n} (\frac{n}{e})^{2 n}\ e^{\frac{1}{24 n}}$ (3)

Now from (2) and (3)...

$\displaystyle \frac{(4 n)!}{\{(2n)!\}^{2}} \sim \frac{2^{4 n}}{\sqrt{2\ \pi\ n}}$ (4)

... and the constant You are looking for is $\displaystyle \gamma= \frac{1}{\sqrt{2\ \pi}}$ ...

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$