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Help with Stirlings formula please

  1. Sep 2, 2009 #1
    Help with Stirlings formula please!!

    Show that

    4n is equivalent to lambda/n^1/2(16)^n as n-> infinity
    2n

    Now using


    p is eqquivalent to p!/q!(p-q)!
    q

    and n = (2*pi*n)^1/2(n/e)^n

    so 4n = (8*pi*n)^1/2(4n/e)^4n

    and 2n = (4*pi*n)^1/2(2n/e)^2n

    now try as I might I can't get it to work out - thus being able to find lambda??
    perhaps my algebra is not what is used to be!!

    Can anyone give me some gentle prods please......

    Many thanks

    James
     
  2. jcsd
  3. Sep 2, 2009 #2

    mathman

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    Re: Help with Stirlings formula please!!

    4n! ~ (8*pi*n)1/2(4n/e)4n
    (2n!)2 ~ (4*pi*n)(2n/e))4n


    4n!/(2n!)2 ~ [(2*pi*n)-1/2]24n = [(2*pi*n)-1/2]16n
     
  4. Sep 3, 2009 #3
    Re: Help with Stirlings formula please!!

    Hi thanks for the reply, but can you explain/expand your answer please!!

    I can see 4n! ~ (8*pi*n)1/2(4n/e)4n

    but where does (2n!)^2 ~ (4*pi*n)(2n/e))4n come from?

    This maybe a really silly question but here goes.....

    Why is (2n!)^2?

    Many Thanks

    James
     
  5. Sep 3, 2009 #4

    mathman

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    Re: Help with Stirlings formula please!!

    Your question is for [4n,2n] (the combination symbol) which is 4n!/{(2n!)(2n!)}

    (2n!)^2 is just (2n!)(2n!), where the second 2n is simply 4n-2n.
     
  6. Sep 4, 2009 #5
    Re: Help with Stirlings formula please!!

    Thanks Mathman, after a bit of chewing on last night I 'got it'


    Thanks for your help

    James
     
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