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Wallis Formula — Definition, Formula & Examples

The Wallis Formula expresses π/2 as an infinite product of rational factors. It states that π/2 equals the product of (2n/(2n−1)) · (2n/(2n+1)) for n = 1, 2, 3, …, giving a remarkable way to approximate π using only fractions.

The Wallis product is the identity π2=∏n=1∞4n24n2−1\displaystyle\frac{\pi}{2} = \prod_{n=1}^{\infty}\frac{4n^2}{4n^2-1}, which can equivalently be written as π2=21⋅23⋅43⋅45⋅65⋅67⋯\frac{\pi}{2} = \frac{2}{1}\cdot\frac{2}{3}\cdot\frac{4}{3}\cdot\frac{4}{5}\cdot\frac{6}{5}\cdot\frac{6}{7}\cdots. It is derived from the recursive evaluation of the integral ∫0π/2sin⁡nx dx\int_0^{\pi/2}\sin^n x\,dx by comparing even and odd cases and taking a limit.

Key Formula

π2=∏n=1∞4n24n2−1=∏n=1∞(2n)(2n)(2n−1)(2n+1)\frac{\pi}{2} = \prod_{n=1}^{\infty}\frac{4n^2}{4n^2-1} = \prod_{n=1}^{\infty}\frac{(2n)(2n)}{(2n-1)(2n+1)}
Where:
  • nn = Positive integer index of the product, running from 1 to ∞

How It Works

Wallis's formula arises from computing the ratio of the integrals In=∫0π/2sin⁡nx dxI_n = \int_0^{\pi/2}\sin^n x\,dx for consecutive even and odd nn. Using the reduction formula In=n−1n In−2I_n = \frac{n-1}{n}\,I_{n-2}, you can write closed-form expressions for I2kI_{2k} and I2k+1I_{2k+1}. Because sin⁡2k+1x≤sin⁡2kx≤sin⁡2k−1x\sin^{2k+1}x \le \sin^{2k}x \le \sin^{2k-1}x on [0,π/2][0,\pi/2], sandwiching gives I2k+1≤I2k≤I2k−1I_{2k+1}\le I_{2k}\le I_{2k-1}, and the ratio I2k/I2k+1→1I_{2k}/I_{2k+1}\to 1. Rearranging produces the infinite product for π/2\pi/2. To approximate π numerically, compute the partial product up to some finite NN and multiply by 2.

Worked Example

Problem: Use the first 4 factors of the Wallis product to approximate π.
Step 1: Write each factor for n = 1, 2, 3, 4:
4(1)24(1)2−1⋅4(2)24(2)2−1⋅4(3)24(3)2−1⋅4(4)24(4)2−1=43⋅1615⋅3635⋅6463\frac{4(1)^2}{4(1)^2-1}\cdot\frac{4(2)^2}{4(2)^2-1}\cdot\frac{4(3)^2}{4(3)^2-1}\cdot\frac{4(4)^2}{4(4)^2-1} = \frac{4}{3}\cdot\frac{16}{15}\cdot\frac{36}{35}\cdot\frac{64}{63}
Step 2: Multiply the four fractions:
4⋅16⋅36⋅643⋅15⋅35⋅63=14745699225≈1.48608\frac{4 \cdot 16 \cdot 36 \cdot 64}{3 \cdot 15 \cdot 35 \cdot 63} = \frac{147456}{99225} \approx 1.48608
Step 3: Multiply by 2 to approximate π:
π≈2×1.48608=2.9722\pi \approx 2 \times 1.48608 = 2.9722
Answer: With only 4 factors, the Wallis product gives π ≈ 2.972, illustrating that convergence is slow. The true value is π ≈ 3.14159.

Why It Matters

The Wallis formula is historically one of the first infinite-product representations of π, predating Euler's work. It appears in integral calculus courses when studying reduction formulas for ∫sin⁡nx dx\int \sin^n x\,dx and connects to the Gamma function identity Γ(1/2)=π\Gamma(1/2)=\sqrt{\pi}. Understanding it strengthens intuition about how products and integrals relate to series convergence.

Common Mistakes

Mistake: Confusing whether the product equals π or π/2.
Correction: The Wallis product converges to π/2, not π. You must multiply the partial product by 2 to get an approximation of π.