[英]the asymptotic growth of n choose floor(n/2)
How can I find the asymptotic growth of n choose floor(n/2) ?如何找到 n choose floor(n/2) 的渐近增长? I tried to use the expansion and got that it is equal to
我尝试使用扩展并得到它等于
[n*(n-1)*........*(floor(n/2)+1)] / (n-floor(n/2))!
Any idea how can i go from there?知道我怎么去那里吗? Any help is appreciated, prefer hints over answers
任何帮助表示赞赏,更喜欢提示而不是答案
I agree with the answer above but would like to provide more depth.我同意上面的答案,但想提供更多深度。 Assuming
n
is even, we have:假设
n
是偶数,我们有:
To upper bound this, we use the upper bound of Stirling's Approximation in the numerator and the lower bound in the denominator (eg we want the largest numerator and smallest denominator).为了设定上限,我们在分子中使用斯特林近似的上限,在分母中使用下限(例如,我们想要最大的分子和最小的分母)。 This will give us the upper bound:
这将为我们提供上限:
We then distribute the exponent in the denominator to get:然后我们将指数分布在分母中得到:
Cancel out抵消
, move
, 移动
from denominator to numerator and simplify;
从分母到分子并化简; we get:
我们得到:
Follow the same process with the lower bound, put Stirling's approx upper bound in the denominator, and lower bound in the numerator.对下界执行相同的过程,将斯特林近似上界放在分母中,将下界放在分子中。 This will yield:
这将产生:
We then know it's lower bounded by some constant times然后我们知道它的下限是一些常数时间
and it's upper bounded by another constant times
它的上限是另一个常数时间
.
.
Thus, we conclude its asymptotic growth is因此,我们得出结论它的渐近增长是
.
.
Using Stirling's approximation , you get使用斯特林近似,你得到
n! = \sqrt{2n\pi}(n/e)^n
If you substitute it into $\\choose{n}{n/2}$, you should eventually end up with如果你把它代入 $\\choose{n}{n/2}$,你最终应该得到
2^{n+1/2}/\sqrt{n\pi}
PS.附注。 you might want to check my math before you actually use the answer :-)
在实际使用答案之前,您可能想检查一下我的数学:-)
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