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Math Help - Stirling's Formula

  1. #1
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    Stirling's Formula

    I need to use Stirling's Formula to find a value λ so that

    (4n)!/ ((2n)!)^2 ~ (λ/√n)(16)^n as n → ∞

    So far I have calculated

    (√(8πn)(4n/e)^4n)/ (√(4πn)(2n/e)^2n)^2

    = (√8πn)/4πn

    At which point I get stuck.
    Can anyone tell me whether I'm on the right track or not and where to go from here. Much appreciated.
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  2. #2
    Member Mauritzvdworm's Avatar
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    n!~ \sqrt{2\pi n}\left(\frac{n}{e}\right)^{n}

    hence we can write your expression in the following way

    \frac{(4n)!}{((2n)!)^{2}} ~  \frac{\sqrt{2\pi(4n)}\left(\frac{4n}{e}\right)^{4n  }}{\left[\sqrt{2\pi(2n)}\left(\frac{2n}{e}\right)^{2n}\righ  t]^{2}}

    =\frac{2\sqrt{2\pi n}2^{4n}\left(\frac{2n}{e}\right)^{4n}}{2(2\pi n)\left(\frac{2n}{e}\right)^{4n}}

    =\frac{16^{n}}{\sqrt{2\pi n}}

    now, if we let n\rightarrow n then the 2\pi tern doesnt realy make any contributions, hence we can conclude that

    \frac{(4n)!}{((2n)!)^{2}}~ \frac{16^{n}}{\sqrt{n}}

    or we can find the particular \lambda which will then just be \frac{1}{\sqrt{2\pi}}
    Last edited by Mauritzvdworm; August 24th 2009 at 12:09 PM. Reason: typo
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  3. #3
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    I totally understand how you've calculated ((2n)!) but I'm still having difficulty with the top part of the calculation.

    Why have you used x and 3n in the first step?
    Also, how does (4n/e)^4n become 2^4n(2n/e)^4n?
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  4. #4
    Member Mauritzvdworm's Avatar
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    I made a typo, fixed it though

    \left(\frac{2\cdot2n}{e}\right)^{4n}=2^{4n}\left(\  frac{2n}{e}\right)^{4n}=\left(2^{4}\right)^{n}\lef  t(\frac{2n}{e}\right)^{4n}=\left(16\right)^{n}\lef  t(\frac{2n}{e}\right)^{4n}
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  5. #5
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    Got it!
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