Wallis's Product/Proof 1
Jump to navigation
Jump to search
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
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:
\(\ds \frac 2 \pi\) | \(=\) | \(\ds \prod_{n \mathop = 1}^\infty \paren {1 - \frac 1 {4 n^2} }\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds \frac \pi 2\) | \(=\) | \(\ds \prod_{n \mathop = 1}^\infty \paren {\frac {4 n^2} {4 n^2 - 1} }\) | |||||||||||
\(\ds \) | \(=\) | \(\ds \prod_{n \mathop = 1}^\infty \frac {\paren {2 n} \paren {2 n} } {\paren {2 n - 1} \paren {2 n + 1} }\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac 2 1 \cdot \frac 2 3 \cdot \frac 4 3 \cdot \frac 4 5 \cdot \frac 6 5 \cdot \frac 6 7 \cdot \frac 8 7 \cdot \frac 8 9 \cdots\) |
$\blacksquare$
Source of Name
This entry was named for John Wallis.
Sources
- 1992: George F. Simmons: Calculus Gems ... (previous) ... (next): Chapter $\text {A}.21$: Euler ($\text {1707}$ – $\text {1783}$)
- 1992: George F. Simmons: Calculus Gems ... (previous) ... (next): Chapter $\text {B}.12$: Wallis's Product
- 1992: George F. Simmons: Calculus Gems ... (previous) ... (next): Chapter $\text {B}.14$: Euler's Discovery of the Formula $\ds \sum_1^\infty \frac 1 {n^2} = \frac {\pi^2} 6$