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Stirling's approximation proof?

  1. Nov 23, 2009 #1
    I read this in a book (it was stats and about poisson approx to normal)
    Given was this:

    [tex]n(n-1)(n-2) \cdots (n-r+1) = \frac{n!}{(n-r)!} \approx n^r[/tex]
    Stating that "Stirling's approximation" had been used.
    So I looked the up and found:

    [tex]\ln n! \approx n\ln n - n\ [/tex]


    In the poisson distribution n is very large and [tex]r[/tex] is very small compared to [tex]n[/tex] so all the terms in the given equation approximate to [tex]n[/tex]... This gives me my [tex]\approx n^r[/tex]

    But I just wondered where the Stirling equation comes in to it...

    [tex]\ln (\frac{n!}{(n-r)!}) = \ln(n!) - \ln((n-r)!) \ [/tex]
    [tex]\approx n\ln n - n - \left[ (n-r)\ln((n-r)) - (n-r) \right]\ [/tex]
    [tex]\approx n\ln n - n - (n-r)\ln((n-r)) + n - r \ [/tex]
    [tex]\approx n\ln n - (n-r)\ln((n-r)) -r \ [/tex]
    ...
    That's as far as I got...

    [tex]\approx \ln (n^n) - \ln((n-r)^{(r-n)}) -r \ [/tex]

    Unless taking logs, instead of to base e, to base n...

    [tex]\approx Log_n (n^n) - Log_n ((n-r)^{(r-n)}) -r \ [/tex]
    Then...
    [tex]Log_n (n^n) - Log_n ((n-r)^{(r-n)}) = r\ [/tex]
    [tex]n^n - (n-r)^{(n-r)} = n^r\ [/tex]

    ^ not sure if that's correct though

    Can anyone help?
     
  2. jcsd
  3. Nov 23, 2009 #2

    HallsofIvy

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    This is "number theory", not "algebra and linear algebra" so I am moving it.
     
  4. Nov 23, 2009 #3
    [tex]\ln(n-r) \approx ln(n) - r/n[/tex]

    [tex]
    n\ln n - (n-r)\ln((n-r)) -r \approx n \ln n - n \ln n + n r/n + r (\ln n - r/n) - r\approx r \ln n - r^2/n
    [/tex]
     
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