Wallis's Product

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Theorem

\(\displaystyle \prod_{n \mathop = 1}^\infty \frac {2 n} {2 n - 1} \cdot \frac {2 n} {2 n + 1}\) \(=\) \(\displaystyle \frac 2 1 \cdot \frac 2 3 \cdot \frac 4 3 \cdot \frac 4 5 \cdot \frac 6 5 \cdot \frac 6 7 \cdot \frac 8 7 \cdot \frac 8 9 \cdots\)
\(\displaystyle \) \(=\) \(\displaystyle \frac \pi 2\)


Proof 1

Into Euler Formula for Sine Function:

\(\displaystyle \dfrac {\sin x} x\) \(=\) \(\displaystyle \paren {1 - \dfrac {x^2} {\pi^2} } \paren {1 - \dfrac {x^2} {4 \pi^2} } \paren {1 - \dfrac {x^2} {9 \pi^2} } \cdots\)
\(\displaystyle \) \(=\) \(\displaystyle \prod_{n \mathop = 1}^\infty \paren {1 - \dfrac {x^2} {n^2 \pi^2} }\)

we substitute $x = \dfrac \pi 2$.


From Sine of Half-Integer Multiple of Pi:

$\sin \dfrac \pi 2 = 1$

Hence:

\(\displaystyle \frac 2 \pi\) \(=\) \(\displaystyle \prod_{n \mathop = 1}^\infty \paren {1 - \frac 1 {4 n^2} }\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle \frac \pi 2\) \(=\) \(\displaystyle \prod_{n \mathop = 1}^\infty \paren {\frac {4 n^2} {4 n^2 - 1} }\)
\(\displaystyle \) \(=\) \(\displaystyle \prod_{n \mathop = 1}^\infty \frac {\paren {2 n} \paren {2 n} } {\paren {2 n - 1} \paren {2 n + 1} }\)
\(\displaystyle \) \(=\) \(\displaystyle \frac 2 1 \cdot \frac 2 3 \cdot \frac 4 3 \cdot \frac 4 5 \cdot \frac 6 5 \cdot \frac 6 7 \cdot \frac 8 7 \cdot \frac 8 9 \cdots\)


$\blacksquare$


Wallis's Original Proof

Wallis, of course, had no recourse to Euler's techniques.

He did this job by comparing $\displaystyle \int_0^\pi \sin^n x \rd x$ 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:

$\displaystyle (1): \quad \int \sin^n x \rd x = - \frac 1 n \sin^{n - 1} \cos x + \frac {n - 1} n \int \sin^{n - 2} x \rd x$


Let $I_n$ be defined as:

$\displaystyle I_n = \int_0^{\pi / 2} \sin^n x \rd x$

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:

$\displaystyle I_0 = \int_0^{\pi / 2} \rd x = \frac \pi 2 \qquad \qquad I_1 = \int_0^{\pi / 2} \sin x \rd x = \left[{- \cos x}\right]_0^{\pi / 2} = 1$

We need to separate the cases where the subscripts are even and odd:

\(\displaystyle I_{2 n}\) \(=\) \(\displaystyle \frac {2 n - 1} {2 n} I_{2 n - 2}\)
\(\displaystyle \) \(=\) \(\displaystyle \frac {2 n - 1} {2 n} \cdot \frac {2 n - 3} {2 n - 2} I_{2 n - 4}\)
\(\displaystyle \) \(=\) \(\displaystyle \cdots\)
\(\displaystyle \) \(=\) \(\displaystyle \frac {2 n - 1} {2 n} \cdot \frac {2 n - 3} {2 n - 2} \cdot \frac {2 n - 5} {2 n - 4} \cdots \frac 3 4 \cdot \frac 1 2 I_0\)
\((A):\quad\) \(\displaystyle \) \(=\) \(\displaystyle \frac {2 n - 1} {2 n} \cdot \frac {2 n - 3} {2 n - 2} \cdot \frac {2 n - 5} {2 n - 4} \cdots \frac 3 4 \cdot \frac 1 2 \cdot \frac \pi 2\)


\(\displaystyle I_{2 n+1}\) \(=\) \(\displaystyle \frac {2 n} {2 n + 1} I_{2 n - 1}\)
\(\displaystyle \) \(=\) \(\displaystyle \frac {2 n} {2 n + 1} \cdot \frac {2 n - 2} {2 n - 1} I_{2 n - 3}\)
\(\displaystyle \) \(=\) \(\displaystyle \cdots\)
\(\displaystyle \) \(=\) \(\displaystyle \frac {2 n} {2 n + 1} \cdot \frac {2 n - 2} {2 n - 1} \cdot \frac {2 n - 4} {2 n - 3} \cdots \frac 4 5 \cdot \frac 2 3 I_1\)
\((B):\quad\) \(\displaystyle \) \(=\) \(\displaystyle \frac {2 n} {2 n + 1} \cdot \frac {2 n - 2} {2 n - 1} \cdot \frac {2 n - 4} {2 n - 3} \cdots \frac 4 5 \cdot \frac 2 3\)


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:

$\displaystyle 0 < \int_0^{\pi / 2} \sin^{2 n + 2} x \rd x \le \int_0^{\pi / 2} \sin^{2 n + 1} x \rd x \le \int_0^{\pi / 2} \sin^{2 n} x \rd x$

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:

$\dfrac \pi 2 = \dfrac 2 1 \cdot \dfrac 2 3 \cdot \dfrac 4 3 \cdot \dfrac 4 5 \cdots \dfrac {2 n} {2 n - 1} \cdot \dfrac {2 n} {2 n + 1} \cdot \left({\dfrac {I_{2 n} } {I_{2 n + 1} } }\right)$

Taking the limit as $n \to \infty$ gives the result.

$\blacksquare$


Also presented as

This result can also be seen presented as:

$\displaystyle \prod_{n \mathop = 1}^\infty \frac n {n - \frac 1 2} \cdot \frac n {n + \frac 1 2} = \frac \pi 2$


Source of Name

This entry was named for John Wallis.


Historical Note

Wallis's Product was discovered by John Wallis in $1656$.


Sources