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

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