Abrazolica

A Derivation of Wallis's Formula for Pi

For $\pi$ day plus one, here is a derivation of Wallis's formula for $\pi$ that is based on probability distributions. Start with the binomial distribution for a fair coin toss. The probability of getting $k$ heads on $n$ tosses is

$\binom{n}{k}\frac{1}{2^{n}}$

Assume an even number of tosses, $n=2m$. Then the distribution has a maximum at $k=m$ of

$\binom{2m}{m}\frac{1}{2^{2m}}$

For large $n$, the binomial distribution can be closely approximated by a Normal probability density function give by

$p(x)=\frac{1}{\sigma\sqrt{2\pi}}\exp{-\frac{(x-\mu)^{2}}{2\sigma^{2}}}$

where $\mu=n/2=m$ and $\sigma^2=n/4=m/2$. The maximum at $x=\mu$ is

$p(\mu)=\frac{1}{\sqrt{m\pi}}$

Comparing this with the maximum for the binomial distribution and taking the limit as $m\to\infty$, you get

$\frac{1}{\sqrt{\pi}}=\lim_{m\to\infty}\binom{2m}{m}\frac{\sqrt{m}}{2^{2m}}$

Now we just need the following double factorial identities

$(2m)!=(2m-1)!!2^{m}m!$

$2^{m}m!=(2m)!!$

Then we can write

$\binom{2m}{m}\frac{\sqrt{m}}{2^{2m}}=\frac{(2m-1)!!}{(2m)!!}\sqrt{m}$

and $\pi$ can be expressed as

$\pi=\lim_{m\to\infty}\frac{1}{m}\left(\frac{(2m)!!}{(2m-1)!!}\right)^{2}$

or in product form as

$\pi=\lim_{m\to\infty}\frac{1}{m}\prod_{k=1}^{m}\left(\frac{2k}{2k-1}\right)^{2}$

This is almost Wallis's formula. To get there, note that

$\prod_{k=1}^{m}(2k-1)^{2}=\frac{1}{2m+1}\prod_{k=1}^{m}(2k-1)(2k+1)$

Substitute this into the above formula, take the limit and you get Wallis's formula

$\pi=2\prod_{k=1}^{\infty}\frac{(2k)^{2}}{(2k-1)(2k+1)}$

Is that cool or what?

© 2010-2016 Stefan Hollos and Richard Hollos