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I am aware of the following approximation:

$$\dbinom{2m}m \sim \dfrac{4^m}{\sqrt{m\pi}}$$

Which is equivalent to saying the follwing (as $m \to \infty$):

Given $m$ objects, the number of ways to pick half of those objects is roughly

$$\frac{2^{m+\frac{1}{2}}}{\sqrt{m\pi}}$$

Is there an intuitive/visual explanation as to why $\pi$ turns up here?

Note: I am not looking for a proof of the asymptotic approximation, but more of an intuitive or visual explanation as to why circles are 'involved' in choosing objects.

RobPratt
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Ultra
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2 Answers2

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This may not be fully satisfactory, but one way to interpret your quantity is probabilistically. Namely, you have $$ \frac{1}{4^m}\binom{2m}{m} =\mathbb{P}\Big\{S_{2m} = m\Big\} $$ where $S_{2m}$ is a Binomial random variable; for instance, it is the probability that you have $m$ heads in $2m$ independent tosses of a fair coin.

By the central limit theorem, we have the following Gaussian approximation, $$ \sqrt{8m}\frac{S_{2m} - m}{2m} \approx N(0, 1). $$ Hence you would expect that your probability is of the order $$ \mathbb{P}\{|S_{2m} - m| < 1/2\} \approx \mathbb{P}\{|N(0, 1)| < \sqrt{1/(2m)}\} = 2 \int_{0}^{\sqrt{1/(2m)}} \phi(t) \, dt, $$ where $\phi$ is the standard Gaussian density. Clearly, we have $$ 2 \int_{0}^{\sqrt{1/(2m)}} \phi(t) \, dt \sim 2 \sqrt{\frac{1}{2m}}\phi(0) = \frac{1}{\sqrt{m \pi}}, \quad \mbox{as}~m\to\infty. $$ Putting the pieces together you will obtain $$ \binom{2m}{m} \approx \frac{4^m}{\sqrt{m \pi}}. $$

Stated briefly: the factor $\pi$ comes in simply because it is arising from normalization in the Gaussian density.

Drew Brady
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I think one somewhat simple interpretation connecting it with actual circles would be as follows:

It is a standard calculus result that the $n$-th term of $\frac{1}{\sqrt{1-x}}$ is $\frac{1}{4^n}\binom{2n}{n}$. For any integer $k$, in the expansion $$ \frac{1}{(1-x)^{k+1}} = \sum\limits_{n=0}^\infty \binom{n+k}{k} x^n, $$ the coefficients near $x^n$ asymptotically grows as $\frac{(n+1)\dots(n+k)}{k!} \sim \frac{n^k}{k!}$. For $k=-\frac{1}{2}$ a similar result should stand:

$$ \frac{1}{4^n}\binom{2n}{n} \sim \frac{1}{(-\frac{1}{2})! \sqrt n} $$

Well, and it so happens that $(-\frac{1}{2})! = \sqrt{\pi}$. Why? The most standard, but not very intuitive explanation is the fact that it is $\Gamma(\frac{1}{2})$, the value of the Gamma function $\Gamma(z)$ in $z=\frac{1}{2}$.

A more intuitive geometric interpretation is shown e.g. in this video. To describe it very briefly, let $S_n(r)$ and $B_n(r)$ be the area and the volume of the $n$-dimensional sphere and ball correspondingly. Then, you can show

$$\begin{cases} S_{n-1}(r) &= \frac{n}{r} &B_n(r), \\ S_n(r) &= 2 \pi r & B_{n-1}(r) \end{cases} $$

Joined together, these formulas yield

$$ B_n(r) = \frac{2}{n} \pi r^2 B_{n-2}(r), $$

which expands into

$$ B_n(r) = \frac{2}{n} \frac{2}{n-2}\dots \frac{2}{n-2k} \pi^k r^{2k} B_{n-2k}(r). $$

Note that you essentially divide by $\frac{n}{2}(\frac{n}{2}-1)(\frac{n}{2}-2)\dots$, which simplifies for even $n$ into $$ \boxed{B_n(r) = \frac{\pi^{n/2}}{(\frac{n}{2})!} r^n} $$ Then, it is natural to want for this formula to also work for odd $n$, and the most natural way to do this is by defining $(\frac{1}{2})! = \frac{\sqrt{\pi}}{2}$, which will also imply the desired $(-\frac{1}{2})!=\sqrt \pi$.

Oleksandr Kulkov
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