5

$$\log_2\binom{N}{\frac{N}{2}}\approx N\log_2N - 2(N-\frac{N}{2})\log_2(N-\frac{N}{2})=N\log_2N - 2\frac{N}{2}\log_2(\frac{N}{2})$$ $$=N\log_2N - {N}{}\log_2({N}) + {N}{}=N$$ $$\implies\binom{N}{\frac{N}{2}}\approx 2^{N}$$

Is my approximation correct?

Or the following one correct?

$$\begin{align} \log {n\choose m} & \approx (n+\tfrac{1}{2})\log n - (m+\tfrac{1}{2})\log m - (n-m+\tfrac{1}{2})\log (n-m) - \tfrac{1}{2}\log 2\pi \end{align}$$ $$\begin{align} \log {n\choose \frac{n}{2}} & \approx (n+\tfrac{1}{2})\log n - (\frac{n}{2}+\tfrac{1}{2})\log \frac{n}{2} - (n-\frac{n}{2}+\tfrac{1}{2})\log (n-\frac{n}{2}) - \tfrac{1}{2}\log 2\pi\end{align}$$ $$\begin{align}\approx(n+\tfrac{1}{2})\log n - (\frac{n}{2}+\tfrac{1}{2})\log \frac{n}{2} - (\frac{n}{2}+\tfrac{1}{2})\log (\frac{n}{2}) - \tfrac{1}{2}\log 2\pi\end{align}$$ $$\begin{align}=(n+\tfrac{1}{2})\log n - (n+1)\log \frac{n}{2}=(n+\tfrac{1}{2})\log n - (n+1)\log {n} + (n+1) - \tfrac{1}{2}\log 2\pi \end{align}$$ $$\begin{align}=-\tfrac{1}{2}\log n + (n+1) - \tfrac{1}{2}\log 2\pi \end{align}$$ $$\implies{n\choose \frac{n}{2}}\approx\frac{2^{n+1}}{\sqrt{\pi n}}$$

What is a reasonable approximation?

Turbo
  • 6,319

1 Answers1

8

Wikipedia gives the approximation $$\frac{2^{n+\frac12}}{\sqrt{\pi n}}$$ This comes from Stirling's formula, and the quotient of this by the binomial coefficient approaches 1 as $n\to\infty$.

Here is the article.

Matt Samuel
  • 59,287
  • Always good to include a link when referencing Wikipedia. – Thomas Andrews Dec 24 '14 at 19:15
  • @ThomasAndrews added. I went to the link on my computer and I'm writing the answer on my phone, which is pretty hard. Only now did I realize that I could've gone to the site on my phone and copied and pasted the link. Live and learn. – Matt Samuel Dec 24 '14 at 19:20