What is the Wallis Product?

π/2 = ∏ 4n²/(4n²-1)
π = 2 · (2/1) · (2/3) · (4/3) · (4/5) · (6/5) · (6/7) ⋯ Wallis, 1655.

The Wallis product writes π/2 as an infinite product of simple fractions: (2/1) × (2/3) × (4/3) × (4/5) × (6/5) × (6/7) × ⋯ Each even number appears twice, once larger and once smaller than its neighbours. Multiply enough terms and the product converges to π/2 ≈ 1.5708.

Wallis partial products approaching π/2
1.331.41.471.54π/2W(n)151014n

Wallis product: (2/1)(2/3)(4/3)(4/5)(6/5)(6/7)... The partial products converge to π/2 ≈ 1.5708 from below, oscillating around the limit.

John Wallis derived this formula in 1655 from the integral ∫₀^(π/2) sinⁿ(x) dx, comparing the cases of even and odd n. What makes it remarkable is that it derives π from pure multiplication of rational numbers, with no geometry involved. The same product emerges from the Gamma function identity: π = Γ(1/2)².

The Wallis product: alternating even fractions
π/2 = (2/1)·(2/3)·(4/3)·(4/5)·(6/5)·(6/7)·…
= Π_{n=1}^∞ (4n²)/(4n²−1)
Wallis derived this in 1655 by comparing integrals of powers of sin(x). It was the first product formula for π.

The Wallis product converges very slowly: after n pairs the error is of order 1/(4n). It has enormous theoretical importance as one of the first infinite products ever studied, opening the path to the analysis of sin(x) = x∏(1 - x²/n²π²) and the entire theory of infinite products in complex analysis.

Integrals of sin^n(x) from 0 to π/2: even/odd pattern produces Wallis
0.530.881.221.57∫₀^(π/2) sinⁿx dx0235n

Even n: I(n) = (π/2)·(1/2)·(3/4)·(5/6)…(n−1)/n. Odd n: I(n) = 1·(2/3)·(4/5)…(n−1)/n. The ratio of adjacent integrals I(2n)/I(2n+1) → 1, giving the Wallis product.

Related topics
Pi Fundamental Theorem Calculus Taylor Series
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