The Wallis product for pi, proved geometrically
Why Mathematics Stack Exchange recommends it
Mathematics Stack Exchange users link it as an accessible geometric take on the Wallis product and its connections to pi, and one asker cites it when questioning whether the proof is new and rigorous.
Recommended by 2 people across 2 different months, 2018 to 2023.
What it is
A 3Blue1Brown animated explainer that derives the Wallis product for pi using a geometric argument with roots of unity on the unit circle and products of distances. Grant Sanderson presents the proof and points to dominated convergence for a fully rigorous treatment.
Cited in 1 Wikipedia article, including Wallis product.
What people said
Strong rec Christian E. Ramirez · 2023-07 · Mathematics Stack Exchange · on “How to define $\pi$ from geometry to the limit of a sequence?”
The Basel problem does actually have somewhat of a geometric interpretation. 3Blue1Brown has an excellent video [LINK] on the subject, as well as a video [this video] on a geometric proof of the Wallis product.
Recommends Roman Odaisky · 2018-08 · Mathematics Stack Exchange · on “Is there a sufficiently reachable plausibility argument that $\pi$ is irrational?”
@PM2Ring 3Blue1Brown has a video on the Wallis product that might be accessible to a bright 12 year old. [this video]
Cites Zain Patel · 2018-04 · Mathematics Stack Exchange · on “Is this a genuinely new proof of Wallis' Product?”
This (reputable) Youtube video [this video] claims to have found a new proof for Wallis' Product $$\prod_{n=1}^{\infty} \left(\frac{2n}{2n-1}\cdot \frac{2n}{2n+1}\right) = \frac{\pi}{2}.$$ There is a supplementary blog post located here [LINK] that provides some detail and supplemental rigour. The general gist of the argument is that we consider the $n$th roots of unity, along with two points on the unit circle. We then take a product of their distances to the roots of unity and use Vieta's formula to get a closed form for the products. Is this a genuinely new proof for Wallis' Product? Is it completely rigorous? If not, do you know of any existing literature that outlines a similar proof? I pers