A Geometric Proof of the Wallis Product for π

The Wallis product connects a sequence of fractions to the geometry of a circle. Unlike familiar series for π, this result multiplies terms. Sridhar, a member of the 3Blue1Brown team, independently found the geometric argument presented here; the presenters leave open whether a similar proof exists elsewhere.

The product (2/1)(4/3)(6/5)… grows without bound, while (2/3)(4/5)(6/7)… tends to zero. Interleave them one for one: (2/1)(2/3)(4/3)(4/5)… . Its alternating increases and decreases approach π/2.

Place evenly spaced lighthouses on a unit circle and choose an observer. Instead of adding inverse-square distances as in the earlier Basel-problem video, multiply the distances from the observer to the lighthouses. Call that the distance product.

Two facts drive the proof. An observer halfway between adjacent lighthouses has distance product 2, regardless of their number. An observer on one lighthouse, with that lighthouse omitted from the product, has a distance product equal to the original number of lighthouses.

Put the circle in the complex plane. Its n evenly spaced points are the nth roots of unity: raising each to the nth power gives 1. Therefore the monic polynomial with those points as roots factors as z^n − 1 = ∏(z − l_k).

At any observer position z, taking the magnitude of the product on the right gives the product of distances to all lighthouses. The simpler expression |z^n − 1| gives the same value without calculating every chord separately.

If the observer is a fraction f of the way from one lighthouse to the next, z^n lands f of a full turn around the circle. The distance product is the corresponding chord length, 2 sin(πf) for the fraction considered. At f = 1/2 this is 2, proving the first fact.

For the second fact, place the observer at 1 and omit that lighthouse. Divide z^n − 1 by z − 1 to remove its factor. The quotient is 1 + z + … + z^(n−1); at z = 1 every term is 1, so the product of the remaining distances is n.

Now use two observers: the keeper on a lighthouse and the sailor halfway to its neighbor. Remove the keeper’s lighthouse from both products. The keeper’s product is n; the sailor’s original product was 2, and removing that lighthouse divides it by the distance d between the observers. Their ratio is therefore nd/2.

Compute that same ratio one lighthouse at a time. For a lighthouse on one side, the ratio of its distance to the keeper to its distance to the sailor approaches 2/1, then 4/3, then 6/5 as n grows. On the other side the corresponding factors approach 2/3, 4/5, 6/7 and so on.

Interleaving the contributions from the two sides yields the Wallis product. Meanwhile the observers are separated by half of one of n equal arcs; as n grows, their chord distance is asymptotic to π/n. Thus nd/2 approaches π/2.

There is a subtle gap in the informal argument: taking a limit of each factor and multiplying infinitely many results need not equal taking the limit of the finite products. These two operations cannot be exchanged automatically.

The video gives a counterexample with rows containing one 7 and otherwise 1s. Every row product is 7, but if the 7 moves farther along each row, the limit of each fixed column is 1; multiplying those column limits gives 1.

Dominated convergence is the tool the presenters identify for justifying the interchange in this case. They do not supply the full rigorous argument in the video; the description notes that their former supplementary blog post was lost during a website transition.

The ordering of the infinite product also matters. The factors from one side diverge upward while those from the other approach zero. Pairing them one for one yields π/2; a different interleaving can produce a different result.

The same geometry generalizes to a sailor at any fraction f between adjacent lighthouses. The sailor’s distance product becomes the chord length 2 sin(πf), and the keeper-to-sailor ratio tends to fπ/sin(πf).

Individual lighthouses contribute factors k/(k−f), with positive and negative k bundled carefully. Rearranging the resulting expression gives an infinite product for sine. This links the geometric Wallis argument to Euler’s product for sine and, in turn, to the Basel problem.