Stirling's formula: Difference between revisions
From Polymath Wiki
Jump to navigationJump to search
No edit summary |
No edit summary |
||
Line 14: | Line 14: | ||
:<math>\binom{n}{n/2 + k} = (\sqrt{\frac{2}{\pi}} + o(1)) \frac{1}{\sqrt{n}} 2^n</math> | :<math>\binom{n}{n/2 + k} = (\sqrt{\frac{2}{\pi}} + o(1)) \frac{1}{\sqrt{n}} 2^n</math> | ||
:<math>\binom{n}{n/3 + k} = (\sqrt{\frac{9}{4\pi}} + o(1)) \frac{1}{\sqrt{n}} 2 | :<math>\binom{n}{n/3 + k} = (\sqrt{\frac{9}{4\pi}} + o(1)) \frac{1}{\sqrt{n}} 3^n 2^{-2n/3 + k}</math> | ||
etc. | etc. | ||
Here is the [http://en.wikipedia.org/wiki/Stirling%27s_approximation Wikipedia article for the Stirling's formula]. | Here is the [http://en.wikipedia.org/wiki/Stirling%27s_approximation Wikipedia article for the Stirling's formula]. |
Revision as of 09:19, 16 February 2009
Stirling's formula: [math]\displaystyle{ n! = (1+o(1)) (2\pi n)^{1/2} n^n e^{-n} }[/math].
As a consequence, one has
- [math]\displaystyle{ \binom{n}{\alpha n + k} = (c(\alpha)+o(1)) \theta(\alpha)^k 2^{h(\alpha) n} \frac{1}{\sqrt{n}} }[/math]
for fixed [math]\displaystyle{ 0 \lt \alpha \lt 1 }[/math] and bounded k, where
- [math]\displaystyle{ c(\alpha) = \frac{1}{\sqrt{2\pi \alpha(1-\alpha)}} }[/math]
- [math]\displaystyle{ \theta(\alpha) = \frac{1-\alpha}{\alpha} }[/math]
- [math]\displaystyle{ h(\alpha) = \alpha \log_2 \frac{1}{\alpha} + (1-\alpha) \log_2 \frac{1}{1-\alpha}. }[/math]
Thus for instance
- [math]\displaystyle{ \binom{n}{n/2 + k} = (\sqrt{\frac{2}{\pi}} + o(1)) \frac{1}{\sqrt{n}} 2^n }[/math]
- [math]\displaystyle{ \binom{n}{n/3 + k} = (\sqrt{\frac{9}{4\pi}} + o(1)) \frac{1}{\sqrt{n}} 3^n 2^{-2n/3 + k} }[/math]
etc.
Here is the Wikipedia article for the Stirling's formula.