The Wallis product for pi, proved geometrically

Q: What are the two lemmas about the distance product, and how do they stay true for any number of lighthouses n?

Lemma 1: If the observer sits exactly halfway between two adjacent lighthouses on the unit circle, the distance product equals 2 for any n. Lemma 2: If the observer sits on top of one lighthouse, the full product is 0, but after removing that lighthouse’s contribution, the remaining distance product equals n. Both results come from expressing the product of distances using polynomial identities for the n-th roots of unity.

Q: How do roots of unity and the polynomial x^n − 1 connect to multiplying distances?

The n lighthouses are the roots of x^n − 1 = 0. Factoring x^n − 1 as ∏(x − ζ_k) lets an observer’s complex coordinate z be substituted into the polynomial. The magnitude of z^n − 1 becomes the product of distances from z to each root ζ_k (up to the geometric normalization of the unit circle). This turns a hard-looking distance multiplication into a simple algebraic expression.

Q: Why does the ratio “keeper distance product / sailor distance product” produce the specific rational factors 2/1, 2/3, 4/3, 4/5, …?

For each lighthouse, the ratio contributes a factor (distance to keeper)/(distance to sailor). As n becomes large, the keeper and sailor are separated by a tiny arc, so for a lighthouse k steps away, the angle from that lighthouse to the keeper is approximately a rational multiple of the angle to the sailor. In the limit, those distance ratios approach simple fractions, yielding the interleaved pattern 2/1, 2/3, 4/3, 4/5, 6/5, 6/7, etc.

Q: What role does π enter, and why does the limit become π/2?

On the unit circle, the distance between adjacent points is the chord corresponding to an arc of size 2π/n, so the separation between the keeper and sailor (halfway between adjacent points) approaches π/n. The earlier ratio formula gives n·(distance between observers)/2, which tends to n·(π/n)/2 = π/2. Since that ratio equals the Wallace product’s value in the limit, the product converges to π/2.

Q: What subtle issue arises when turning infinitely many factors into a limit, and how is it handled?

Interchanging “product of limits” with “limit of products” can fail for infinite collections of factors. The proof relies on a specific interleaving of positive and negative-index contributions (clockwise and counterclockwise lighthouses). Dominated convergence provides the condition under which the interchange is valid here, making the Wallace product’s limit rigorous.

TL;DR

The Wallace product is an interleaved infinite product (2/1)(2/3)(4/3)(4/5)(6/5)(6/7)… that converges to π/2.

Briefing Cornell Notes

Briefing

A carefully chosen infinite product of simple fractions— (2/1)·(2/3)·(4/3)·(4/5)·(6/5)·(6/7)·…—converges to π/2. The result, known as the Wallace product, matters because it turns a seemingly arithmetic expression into a geometric quantity tied to circles, and it does so with a proof that leans on complex numbers and the algebra of evenly spaced points.

The construction starts by looking at products built from even and odd integers. If all even numbers sit in the numerator and all odd numbers in the denominator, the partial products grow without bound. If the roles are reversed by shifting indices slightly, the partial products shrink toward 0. The Wallace product sits between these extremes: it interleaves factors so the partial products rise and fall in smaller and smaller steps, settling to a finite limit.

To connect this interleaved product to π, the argument switches to geometry. Imagine n “lighthouses” equally spaced on the unit circle in the complex plane, with one extra “observer” point. Instead of adding distances (as in the Basel-problem lighthouse story), the proof multiplies distances from the observer to each lighthouse, defining a “distance product.” Two key lemmas about this distance product drive everything:

1) If the observer is exactly halfway between two adjacent lighthouses, the distance product equals 2, regardless of n. 2) If the observer sits on top of one lighthouse, the full distance product is 0, but removing that lighthouse’s contribution leaves a distance product equal to n.

Complex numbers make these lemmas tractable. The n lighthouses are the n-th roots of unity, the solutions to x^n − 1 = 0. Plugging an observer’s complex coordinate into the polynomial factorization turns the product of distances into the magnitude of a simple expression involving z^n − 1. When the observer lies at a fraction f of the way around the circle between two lighthouses, the distance product becomes the chord length “cord of f,” and the special case f = 1/2 gives the constant value 2.

With the lemmas in place, two observers are introduced: a “keeper” placed on a lighthouse and a “sailor” placed halfway between that lighthouse and the next. Comparing the keeper’s distance product (with the zeroed lighthouse removed) to the sailor’s distance product yields a ratio that equals n times the distance between the observers divided by 2. The same ratio can also be computed lighthouse-by-lighthouse: as n grows, each lighthouse’s contribution approaches a rational factor like 2/1, 2/3, 4/3, 4/5, 6/5, 6/7, and so on. Meanwhile, the distance between adjacent points on the unit circle is about π/n, so the ratio tends to π/2. That limit is exactly the Wallace product.

The proof also flags a subtlety: exchanging limits and infinite products is not automatically valid. Here, dominated convergence supplies the needed justification for the specific one-for-one interleaving of factors. Finally, the same framework generalizes: moving the sailor to a fraction f between lighthouses produces an infinite product formula for sin(πf), linking the Wallace-style circle geometry back to Euler’s product for sine and, indirectly, to the Basel problem’s heritage.

Cornell Notes

The Wallace product is an interleaved infinite product of fractions built from even and odd integers: (2/1)(2/3)(4/3)(4/5)(6/5)(6/7)… . By placing n equally spaced points (the n-th roots of unity) on the unit circle and defining a “distance product” as the product of distances from an observer to all points, two facts emerge: halfway between adjacent points the distance product is always 2, and when the observer sits on a point, removing that point’s zero contribution leaves distance product n. Comparing a keeper (on a point) to a sailor (halfway between points) gives a ratio that also factors into the rational terms of the Wallace product. As n→∞, the ratio approaches π/2, so the product converges to π/2. A dominated-convergence argument justifies the limit/product interchange for this specific interleaving.

Why do the “all-even over all-odd” and “shifted” products behave so differently, and what does the Wallace product change?

With all even numerators over all odd denominators, every new factor is >1, so partial products increase without bound (diverge to infinity). With a slight shift—odds in the numerator and evens in the denominator—every factor is <1, so partial products decrease to 0. The Wallace product interleaves the two behaviors: factors alternate so the partial products rise and fall in smaller increments, producing a finite limit instead of divergence.

What are the two lemmas about the distance product, and how do they stay true for any number of lighthouses n?

Lemma 1: If the observer sits exactly halfway between two adjacent lighthouses on the unit circle, the distance product equals 2 for any n. Lemma 2: If the observer sits on top of one lighthouse, the full product is 0, but after removing that lighthouse’s contribution, the remaining distance product equals n. Both results come from expressing the product of distances using polynomial identities for the n-th roots of unity.

How do roots of unity and the polynomial x^n − 1 connect to multiplying distances?

The n lighthouses are the roots of x^n − 1 = 0. Factoring x^n − 1 as ∏(x − ζ_k) lets an observer’s complex coordinate z be substituted into the polynomial. The magnitude of z^n − 1 becomes the product of distances from z to each root ζ_k (up to the geometric normalization of the unit circle). This turns a hard-looking distance multiplication into a simple algebraic expression.

Why does the ratio “keeper distance product / sailor distance product” produce the specific rational factors 2/1, 2/3, 4/3, 4/5, …?

For each lighthouse, the ratio contributes a factor (distance to keeper)/(distance to sailor). As n becomes large, the keeper and sailor are separated by a tiny arc, so for a lighthouse k steps away, the angle from that lighthouse to the keeper is approximately a rational multiple of the angle to the sailor. In the limit, those distance ratios approach simple fractions, yielding the interleaved pattern 2/1, 2/3, 4/3, 4/5, 6/5, 6/7, etc.

What role does π enter, and why does the limit become π/2?

On the unit circle, the distance between adjacent points is the chord corresponding to an arc of size 2π/n, so the separation between the keeper and sailor (halfway between adjacent points) approaches π/n. The earlier ratio formula gives n·(distance between observers)/2, which tends to n·(π/n)/2 = π/2. Since that ratio equals the Wallace product’s value in the limit, the product converges to π/2.

What subtle issue arises when turning infinitely many factors into a limit, and how is it handled?