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I encountered the following claim

$$\frac{1}{n+1}2^{nH_2(k/n)} \le \binom{n}{k} \le 2^{nH_2(k/n)}$$

where $H_2$ is the binary entropy function. The upper bound is rather well known but how does one show the lower bound?

zyx
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user782220
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    Related: http://math.stackexchange.com/questions/235962/asymptotics-of-binomial-coefficients-and-the-entropy-function – leonbloy Jun 19 '13 at 12:32
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    Notice you can strengthen that to $\frac1{\sqrt{2n}}$ or even $\frac1{\sqrt{8k(1-k/n)}}$ by considering stirling approximations. – Thomas Ahle Dec 14 '15 at 12:06

1 Answers1

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The lower bound is a rewriting of $\int_0^1 x^k (1-x)^{n-k} \leq 2^{-nH_2(k/n)}$, which is estimation of the integral by (maximum value of function integrated, which occurs at $x=\frac{k}{n}$) x (length of interval).

zyx
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