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Inequality from Stirling's formula

  1. Oct 27, 2011 #1
  2. jcsd
  3. Oct 29, 2011 #2

    Stephen Tashi

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    I don't see how to navigate to page 17 in that link. (I'm not enthusiastic about how the interface to Google books behaves on my browser and internet connection. It seems darn slow and tedious.) You'd best type out your question.
     
  4. Oct 31, 2011 #3
    Thank you for writing back.

    The inequality is

    n is even
    C(n,n/2)/2^(n+1) > 1/(2*sqrt(n)).

    "Sharp form fo Stirling's inequality" is

    sqrt(2*pi*k) * k^k * e^-k < k! < sqrt(2*pi*k) * k^k * e^-k * (1+1/(4*k))

    Is it right? Tried with 4.

    With Google books, by clicking on the book in the upper left hand corner, it will appear big on the screen and you can click on the big image and page up and down.
    Think I'm missing something here.
     
  5. Oct 31, 2011 #4

    Stephen Tashi

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    I'm missing all pages after page 15. It does say "some pages are omitted from the preview".


    [tex] \frac{ \binom{n}{n/2}}{2^{n+1}} > \frac{1}{2 \sqrt{n}} [/tex]

    [tex] (\sqrt{2 \pi k}) k^k e^{-k} < k! < (\sqrt{2 \pi k}) k^k e^{-k } (1+\frac{1}{ 4k} ) [/tex]

    Is what right? Do you mean that you used k = 4 or n = 4 ?
     
  6. Oct 31, 2011 #5
    Sorry, n.
    I tried it with 4 and it seemed not to hold, the inequality. Tried with 2 and same problem.
     
  7. Oct 31, 2011 #6

    Stephen Tashi

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    This inequality is not correct:
    [tex] \frac{ \binom{n}{n/2}}{2^{n+1}} > \frac{1}{2 \sqrt{n}} [/tex]


    Assuming the inequality
    [tex] (\sqrt{2 \pi k}) k^k e^{-k} < k! < (\sqrt{2 \pi k}) k^k e^{-k } (1+\frac{1}{ 4k} ) [/tex]

    the only similar inequality that I see is:

    [tex] \frac{ \binom{n}{n/2}}{(1 + \frac{1}{2n})^2} > \frac{\sqrt{2}}{\sqrt{\pi n}} > \frac{1}{\sqrt{2n}} [/tex]
     
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