Wallis' product exploration

[pinkman]

(02-08-2020 08:58 PM)Allen Wrote:  

(02-08-2020 07:53 PM)Albert Chan Wrote:  That is because the formula had this denominator (squared!)
\(\Large 2\prod _{k=1} ^{k=n} {4k^2 \over 4k^2-1} =
{2^{4n+2} \over 4n+2} ÷ \binom{2n}{n}^2 \)

Interesting!!

Since the central binomial coefficients are related to Catalan numbers, Wallis' product \( \pi \) approximation for each \( n \) is inversely proportional to the number of ways a regular \( n \)-gon can be divided into \( n-2 \) triangles.

As \( n \rightarrow \infty \) then the \(n\)-gon shape approaches a circle.

Wait, \( \pi \) is related to circles because of triangles?

Really interesting. And funny
So, this apparently abstract formula of Wallis is the same approach to find PI than any other method: converge to a circle. Well, what did I expect? No PI without circles!
It also leaded me to wake up my memory about binomial coefficients, that I had a bit (completely) forgotten...