Wallis's Product

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Theorem

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


This is called Wallis's product.


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

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.

$\blacksquare$


Proof 3

\(\ds \prod_{n \mathop = 1}^\infty \frac {2 n} {2 n - 1} \cdot \frac {2 n} {2 n + 1}\) \(=\) \(\ds \prod_{n \mathop = 1}^\infty \frac {\paren {2 n - 0} \paren {2 n - 0} \times \dfrac 1 2 \times \dfrac 1 2} {\paren {2 n - 1} \paren {2 n + 1} \times \dfrac 1 2 \times \dfrac 1 2}\) multiplying top and bottom by $\dfrac 1 2 \times \dfrac 1 2$
\(\ds \) \(=\) \(\ds \prod_{n \mathop = 1}^\infty \frac {\paren {n - 0} \paren {n - 0} } {\paren {n - \dfrac 1 2} \paren {n + \dfrac 1 2} }\)
\(\ds \) \(=\) \(\ds \frac { \map \Gamma {\dfrac 1 2} \map \Gamma {\dfrac 3 2} } {\map \Gamma {1} \map \Gamma {1} }\) Infinite Product of Product of Sequence of n plus alpha over Sequence of n plus beta
\(\ds \) \(=\) \(\ds \frac { \sqrt \pi \times \dfrac {\sqrt \pi} 2 } {0! \times 0! }\) Gamma Function of One Half, Gamma Difference Equation, Gamma Function Extends Factorial
\(\ds \) \(=\) \(\ds \frac \pi 2\) Definition of Factorial

$\blacksquare$


Also presented as

Wallis's Product can also be seen presented as:

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

Some sources present it as:

$\ds \prod_{n \mathop = 1}^\infty \frac {4 n^2} {4 n^2 - 1} = \frac \pi 2$

which follows from the given form by an application of Difference of Two Squares on the denominator of the product.


Source of Name

This entry was named for John Wallis.


Historical Note

Wallis's product was published by John Wallis in $1656$, in his Arithmetica Infinitorum.

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.


Sources

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