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Math Help - Stirlings Formula

  1. #1
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    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

    = ? not sure about this

    = ? or this

    Can anyone help?
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  2. #2
    MHF Contributor chisigma's Avatar
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    Re: Stirlings Formula

    Quote Originally Posted by Arron View Post
    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

    = ? not sure about this

    = ? or this

    Can anyone help?
    A 'non very popular' version of the Stirling's formula is the following...

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

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

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

    ... and...

    (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)...

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

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

    Kind regards

    \chi \sigma
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