Wallis's Product
Theorem
\(\ds \prod_{n \mathop = 1}^\infty \frac {2 n} {2 n - 1} \cdot \frac {2 n} {2 n + 1}\) | \(=\) | \(\ds \frac 2 1 \cdot \frac 2 3 \cdot \frac 4 3 \cdot \frac 4 5 \cdot \frac 6 5 \cdot \frac 6 7 \cdot \frac 8 7 \cdot \frac 8 9 \cdots\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac \pi 2\) |
Proof 1
Into Euler Formula for Sine Function:
\(\ds \dfrac {\sin x} x\) | \(=\) | \(\ds \paren {1 - \dfrac {x^2} {\pi^2} } \paren {1 - \dfrac {x^2} {4 \pi^2} } \paren {1 - \dfrac {x^2} {9 \pi^2} } \cdots\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \prod_{n \mathop = 1}^\infty \paren {1 - \dfrac {x^2} {n^2 \pi^2} }\) |
we substitute $x = \dfrac \pi 2$.
From Sine of Half-Integer Multiple of Pi:
- $\sin \dfrac \pi 2 = 1$
Hence:
\(\ds \frac 2 \pi\) | \(=\) | \(\ds \prod_{n \mathop = 1}^\infty \paren {1 - \frac 1 {4 n^2} }\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds \frac \pi 2\) | \(=\) | \(\ds \prod_{n \mathop = 1}^\infty \paren {\frac {4 n^2} {4 n^2 - 1} }\) | |||||||||||
\(\ds \) | \(=\) | \(\ds \prod_{n \mathop = 1}^\infty \frac {\paren {2 n} \paren {2 n} } {\paren {2 n - 1} \paren {2 n + 1} }\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac 2 1 \cdot \frac 2 3 \cdot \frac 4 3 \cdot \frac 4 5 \cdot \frac 6 5 \cdot \frac 6 7 \cdot \frac 8 7 \cdot \frac 8 9 \cdots\) |
$\blacksquare$
Wallis's Original Proof
Wallis, of course, had no recourse to Euler's techniques.
He did this job by comparing $\ds \int_0^\pi \sin^n x \rd x$ for even and odd values of $n$, and noting that for large $n$, increasing $n$ by $1$ makes little change.
From the Reduction Formula for Integral of Power of Sine, we have:
- $\ds (1): \quad \int \sin^n x \rd x = - \frac 1 n \sin^{n - 1} x \cos x + \frac {n - 1} n \int \sin^{n - 2} x \rd x$
Let $I_n$ be defined as:
- $\ds I_n = \int_0^{\pi / 2} \sin^n x \rd x$
As $\cos \dfrac \pi 2 = 0$ from Shape of Cosine Function, we have from $(1)$ that:
- $(2): \quad I_n = \dfrac {n-1} n I_{n - 2}$
To start the ball rolling, we note that:
- $\ds I_0 = \int_0^{\pi / 2} \rd x = \frac \pi 2$
- $\ds I_1 = \int_0^{\pi / 2} \sin x \rd x = \bigintlimits {-\cos x} 0 {\pi / 2} = 1$
We need to separate the cases where the subscripts are even and odd:
\(\ds I_{2 n}\) | \(=\) | \(\ds \frac {2 n - 1} {2 n} I_{2 n - 2}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac {2 n - 1} {2 n} \cdot \frac {2 n - 3} {2 n - 2} I_{2 n - 4}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \cdots\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac {2 n - 1} {2 n} \cdot \frac {2 n - 3} {2 n - 2} \cdot \frac {2 n - 5} {2 n - 4} \cdots \frac 3 4 \cdot \frac 1 2 I_0\) | ||||||||||||
\(\text {(A)}: \quad\) | \(\ds \) | \(=\) | \(\ds \frac {2 n - 1} {2 n} \cdot \frac {2 n - 3} {2 n - 2} \cdot \frac {2 n - 5} {2 n - 4} \cdots \frac 3 4 \cdot \frac 1 2 \cdot \frac \pi 2\) |
\(\ds I_{2 n+1}\) | \(=\) | \(\ds \frac {2 n} {2 n + 1} I_{2 n - 1}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac {2 n} {2 n + 1} \cdot \frac {2 n - 2} {2 n - 1} I_{2 n - 3}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \cdots\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac {2 n} {2 n + 1} \cdot \frac {2 n - 2} {2 n - 1} \cdot \frac {2 n - 4} {2 n - 3} \cdots \frac 4 5 \cdot \frac 2 3 I_1\) | ||||||||||||
\(\text {(B)}: \quad\) | \(\ds \) | \(=\) | \(\ds \frac {2 n} {2 n + 1} \cdot \frac {2 n - 2} {2 n - 1} \cdot \frac {2 n - 4} {2 n - 3} \cdots \frac 4 5 \cdot \frac 2 3\) |
By Shape of Sine Function, we have that on $0 \le x \le \dfrac \pi 2$:
- $0 \le \sin x \le 1$
Therefore:
- $0 \le \sin^{2 n + 2} x \le \sin^{2 n +1} x \le \sin^{2 n} x$
It follows from Relative Sizes of Definite Integrals that:
- $\ds 0 < \int_0^{\pi / 2} \sin^{2 n + 2} x \rd x \le \int_0^{\pi / 2} \sin^{2 n + 1} x \rd x \le \int_0^{\pi / 2} \sin^{2 n} x \rd x$
That is:
- $(3): \quad 0 < I_{2 n + 2} \le I_{2 n + 1} \le I_{2 n}$
By $(2)$ we have:
- $\dfrac {I_{2 n + 2} } {I_{2 n} } = \dfrac {2 n + 1} {2 n + 2}$
Dividing $(3)$ through by $I_{2n}$ then, we have:
- $\dfrac {2 n + 1} {2 n + 2} \le \dfrac {I_{2 n + 1}} {I_{2 n}} \le 1$
By Squeeze Theorem, it follows that:
- $\dfrac {I_{2 n + 1} } {I_{2 n} } \to 1$ as $n \to \infty$
which is equivalent to:
- $\dfrac {I_{2 n} } {I_{2 n + 1} } \to 1$ as $n \to \infty$
Now we take $(B)$ and divide it by $(A)$ to get:
- $\dfrac {I_{2 n + 1} } {I_{2 n} } = \dfrac 2 1 \cdot \dfrac 2 3 \cdot \dfrac 4 3 \cdot \dfrac 4 5 \cdots \dfrac {2 n} {2 n - 1} \cdot \dfrac {2 n} {2 n + 1} \cdot \dfrac 2 \pi$
So:
- $\dfrac \pi 2 = \dfrac 2 1 \cdot \dfrac 2 3 \cdot \dfrac 4 3 \cdot \dfrac 4 5 \cdots \dfrac {2 n} {2 n - 1} \cdot \dfrac {2 n} {2 n + 1} \cdot \paren {\dfrac {I_{2 n} } {I_{2 n + 1} } }$
Taking the limit as $n \to \infty$ gives the result.
$\blacksquare$
Also presented as
This result can also be seen presented as:
- $\ds \prod_{n \mathop = 1}^\infty \frac n {n - \frac 1 2} \cdot \frac n {n + \frac 1 2} = \frac \pi 2$
Source of Name
This entry was named for John Wallis.
Historical Note
Wallis's Product was discovered by John Wallis in $1656$.
Sources
- 1968: Murray R. Spiegel: Mathematical Handbook of Formulas and Tables ... (previous) ... (next): $\S 38$: Infinite Products: $38.9$: Wallis' Product
- 1986: David Wells: Curious and Interesting Numbers ... (previous) ... (next): $3 \cdotp 14159 \, 26535 \, 89793 \, 23846 \, 26433 \, 83279 \, 50288 \, 41972 \ldots$
- 1989: Ephraim J. Borowski and Jonathan M. Borwein: Dictionary of Mathematics ... (previous) ... (next): Wallis's product
- 1997: Donald E. Knuth: The Art of Computer Programming: Volume 1: Fundamental Algorithms (3rd ed.) ... (previous) ... (next): $\S 1.2.5$: Permutations and Factorials: Exercise $18$
- 1997: David Wells: Curious and Interesting Numbers (2nd ed.) ... (previous) ... (next): $3 \cdotp 14159 \, 26535 \, 89793 \, 23846 \, 26433 \, 83279 \, 50288 \, 41971 \ldots$
- 2008: David Nelson: The Penguin Dictionary of Mathematics (4th ed.) ... (previous) ... (next): Wallis's product
- 2014: Christopher Clapham and James Nicholson: The Concise Oxford Dictionary of Mathematics (5th ed.) ... (previous) ... (next): Wallis's Product
- 2021: Richard Earl and James Nicholson: The Concise Oxford Dictionary of Mathematics (6th ed.) ... (previous) ... (next): Wallis's Product