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http://books.google.com/books/about/An_Introduction_to_the_Approximation_of.html?id=VTW2cmjC43YC

I'd be thankful if someone would explain how the inequality near the top of page 17 was gotten.

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- #1

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http://books.google.com/books/about/An_Introduction_to_the_Approximation_of.html?id=VTW2cmjC43YC

I'd be thankful if someone would explain how the inequality near the top of page 17 was gotten.

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Stephen Tashi

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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.

- #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".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.

The inequality is

n is even

[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]"Sharp form fo Stirling's inequality" is

Is what right? Do you mean that you used k = 4 or n = 4 ?Is it right? Tried with 4.

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I tried it with 4 and it seemed not to hold, the inequality. Tried with 2 and same problem.

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Stephen Tashi

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[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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