How to prove binomial coefficient is asymptotic big theta of two to the power n?

狂风中的少年 提交于 2021-02-20 04:29:13

问题


I am stuck at this problem. I think it is equivalent to show 2m choose m is big theta of 4 to the power n, but still find difficult to prove it.

Thanks of @LutzL's suggestion. I thought of stirling's approximation before.


回答1:


The O-part should be easy. Choosing exactly n/2 elements out of n is a special case of choosing arbitrary combinations out of n elements, i.e. deciding for each of these n elements whether to choose it or not.

The Ω-part is harder. In fact, plotting 4n / binomial(2 n, n) for moderately large n I see no indication that this would flatten to stay below some constant. Speaking intuitively, the larger n is, the more special is the case when you take a random pick from n elements and coincidentially happen to choose exactly n/2 of them. That probability should tend to zero as n increases, meaning that 2n should always grow faster than n choose n/2. Are you certain you understood this part of your task correctly?




回答2:


You could use Stirlings formula for the factorials.

n! = sqrt(2*pi*(n+theta)) * (n/e)^n

where theta is between 0 and 1, with a strong tendency towards 0.




回答3:


It's not -- it's Theta(2^n/sqrt(n)), and in fact choose(n, n/2) ~ 2^n/sqrt(pi * (n/2)) as n->infinity). See https://en.wikipedia.org/wiki/Central_binomial_coefficient



来源:https://stackoverflow.com/questions/39558086/how-to-prove-binomial-coefficient-is-asymptotic-big-theta-of-two-to-the-power-n

易学教程内所有资源均来自网络或用户发布的内容,如有违反法律规定的内容欢迎反馈
该文章没有解决你所遇到的问题?点击提问,说说你的问题,让更多的人一起探讨吧!