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

Question

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.

score 3 · Accepted Answer · answered Sep 18 '16 at 13:48

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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?

answered Sep 18 '16 at 13:48

MvG

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Thanks for your opinion. To be honest, I am not sure the lower bound of 2^n is correct. Is there a lower bound that you can think of ? – xxx_yyyy Sep 18 '16 at 13:52
But I saw the bound of 2m choose m, which seems to have higher and lower bound of constant times 4^n. – xxx_yyyy Sep 18 '16 at 13:53
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Per Stirling formula, one can see that `binom{2n}{n}=Theta(4^n/sqrt(n))`. The square root factor makes the requested lower bound impossible. As you can test via WA http://www.wolframalpha.com/input/?i=plot+4^n+%2F+sqrt%28n%29+%2F+binomial%282*n,+n%29+for+n+from+0+to+1000 – Lutz Lehmann Sep 18 '16 at 15:08

score 3 · Answer 2 · answered Sep 19 '16 at 01:58

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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

answered Sep 19 '16 at 01:58

Paul Hankin

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score 2 · Answer 3 · answered Sep 18 '16 at 13:30

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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.

answered Sep 18 '16 at 13:30

Lutz Lehmann

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Thanks, but with Stirling's formula, I still can not prove the lower bound. Any suggestion? – xxx_yyyy Sep 18 '16 at 13:47
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@xxx_yyyy Note that this is *equality*. So `theta == 0` is a lower bound, and `theta == 1` is an upper bound. – Tavian Barnes Sep 18 '16 at 14:35
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You can not prove the lower bound since it is wrong. As you already found out, it has to be in terms of `2^n/sqrt(n)`. – Lutz Lehmann Sep 18 '16 at 15:11

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

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