Steven R. Dunbar
Department of Mathematics
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Topics in
Probability Theory and Stochastic Processes
Steven R. Dunbar

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Wallis’ Formula

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_______________________________________________________________________________________________ ### Rating

Mathematically Mature: may contain mathematics beyond calculus with proofs.

_______________________________________________________________________________________________ ### Section Starter Question

Can you think of a sequence or a process that approximates $\mathrm{\pi }$? What is the intuition or reasoning behind that sequence?

_______________________________________________________________________________________________ ### Key Concepts

1. Wallis’ Formula is the amazing limit
$\underset{n\to \infty }{lim}\left(\frac{2\cdot 2\cdot 4\cdot 4\cdot 6\cdot 6\dots \left(2n\right)\cdot \left(2n\right)}{1\cdot 3\cdot 3\cdot 5\cdot 5\dots \left(2n-1\right)\cdot \left(2n-1\right)\cdot \left(2n+1\right)}\right)=\frac{\mathrm{\pi }}{2}.$

2. One proof of Wallis’ formula uses a recursion formula developed from integration of trigonometric functions.
3. Another proof uses only basic algebra, the Pythagorean Theorem, and the formula $\mathrm{\pi }{r}^{2}$ for the area of a circle of radius $r$.
4. Yet another proof uses Euler’s inﬁnite product representation for the sine function.

__________________________________________________________________________ ### Vocabulary

1. Wallis’ Formula is the amazing limit
$\underset{n\to \infty }{lim}\left(\frac{2\cdot 2\cdot 4\cdot 4\cdot 6\cdot 6\dots \left(2n\right)\cdot \left(2n\right)}{1\cdot 3\cdot 3\cdot 5\cdot 5\dots \left(2n-1\right)\cdot \left(2n-1\right)\cdot \left(2n+1\right)}\right)=\frac{\mathrm{\pi }}{2}.$

__________________________________________________________________________ ### Mathematical Ideas

#### Introduction

Wallis’ Formula is the amazing limit

 $\underset{n\to \infty }{lim}\left(\frac{2\cdot 2\cdot 4\cdot 4\cdot 6\cdot 6\dots \left(2n\right)\cdot \left(2n\right)}{1\cdot 3\cdot 3\cdot 5\cdot 5\dots \left(2n-1\right)\cdot \left(2n-1\right)\cdot \left(2n+1\right)}\right)=\frac{\mathrm{\pi }}{2}.$ (1)

Another way to write this is

$\frac{\mathrm{\pi }}{2}=\prod _{j=1}^{\infty }\frac{{\left(2j\right)}^{2}}{\left(2j-1\right)\left(2j+1\right)}.$

A “closed form” expression for the product in Wallis formula is

$\underset{n\to \infty }{lim}\frac{{2}^{4n}\cdot {\left(n!\right)}^{4}}{{\left(\left(2n\right)!\right)}^{2}\left(2n+1\right)}=\frac{\mathrm{\pi }}{2}$

or equivalently

$\underset{n\to \infty }{lim}\frac{{2}^{4n}}{{\left(\genfrac{}{}{0.0pt}{}{2n}{n}\right)}^{2}\left(2n+1\right)}=\frac{\mathrm{\pi }}{2}.$

Note that Wallis Formula is equivalent to saying that the “central binomial term” has the asymptotic expression

$\frac{1}{{2}^{2n}}\left(\genfrac{}{}{0.0pt}{}{2n}{n}\right)\sim \sqrt{\frac{2}{\left(2n+1\right)\mathrm{\pi }}}.$

See the subsection Central Binomial below for a proof of an equivalent inequality.

In the form

 ${w}_{n}=\prod _{j=1}^{n}\frac{{\left(2j\right)}^{2}}{\left(2j-1\right)\left(2j+1\right)}=\frac{{2}^{4n}}{{\left(\genfrac{}{}{0.0pt}{}{2n}{n}\right)}^{2}\left(2n+1\right)}$ (2)

it is easy to see that the sequence ${w}_{n}$ is increasing since $4{n}^{2}∕\left(4{n}^{2}-1\right)>1$. Figure 1 illustrates the increasing sequence. Figure 1: Convergence of the Wallis formula to $\mathrm{\pi }∕2$.

Doing a numerical linear regression of $log\left(\mathrm{\pi }∕2-{w}_{n}\right)$ versus $logn$ on the domain $n=1$ to $n=30$ indicates that ${w}_{n}$ approaches $\mathrm{\pi }∕2$ at a rate which is $O\left(1∕n\right)$.

#### A proof using integration and recursion

Theorem 1 (Wallis’ Formula).

 $\underset{n\to \infty }{lim}\left(\frac{2\cdot 2\cdot 4\cdot 4\cdot 6\cdot 6\dots \left(2n\right)\cdot \left(2n\right)}{1\cdot 3\cdot 3\cdot 5\cdot 5\dots \left(2n-1\right)\cdot \left(2n+1\right)}\right)=\frac{\mathrm{\pi }}{2}.$ (3)

Proof. Consider ${J}_{n}={\int }_{0}^{\mathrm{\pi }∕2}{cos}^{n}\left(x\right)\phantom{\rule{0.3em}{0ex}}\phantom{\rule{0.3em}{0ex}}dx$. Integrating by parts with $u={cos}^{n-1}\left(x\right)$ and $\phantom{\rule{0.3em}{0ex}}dv=cos\left(x\right)$ shows

$\begin{array}{llll}\hfill {\int }_{0}^{\mathrm{\pi }∕2}{cos}^{n}\left(x\right)\phantom{\rule{0.3em}{0ex}}\phantom{\rule{0.3em}{0ex}}dx& =\left(n-1\right){\int }_{0}^{\mathrm{\pi }∕2}{cos}^{n-2}\left(x\right){sin}^{2}\left(x\right)\phantom{\rule{0.3em}{0ex}}\phantom{\rule{0.3em}{0ex}}dx\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =\left(n-1\right){\int }_{0}^{\mathrm{\pi }∕2}{cos}^{n-2}\left(x\right)\left(1-{cos}^{2}\left(x\right)\right)\phantom{\rule{0.3em}{0ex}}\phantom{\rule{0.3em}{0ex}}dx\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =\left(n-1\right){\int }_{0}^{\mathrm{\pi }∕2}{cos}^{n-2}\left(x\right)\phantom{\rule{0.3em}{0ex}}\phantom{\rule{0.3em}{0ex}}dx-\left(n-1\right){\int }_{0}^{\mathrm{\pi }∕2}{cos}^{n}\left(x\right)\phantom{\rule{0.3em}{0ex}}\phantom{\rule{0.3em}{0ex}}dx.\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

Gathering terms, we get $n{J}_{n}=\left(n-1\right){J}_{n-2}$.

Now ${J}_{1}=1$ so recursively ${J}_{3}=\frac{2}{3}$, ${J}_{5}=\frac{2\cdot 4}{3\cdot 5}$ and inductively

 ${J}_{2n+1}=\frac{2\cdot 4\cdots \left(2n-2\right)\cdot \left(2n\right)}{1\cdot 3\cdots \left(2n-1\right)\cdot \left(2n+1\right)}.$ (4)

Likewise ${J}_{2}=\frac{\mathrm{\pi }}{2\cdot 2}$, and ${J}_{4}=\frac{3\cdot \mathrm{\pi }}{2\cdot 4\cdot 2}$, ${J}_{6}=\frac{3\cdot 5\cdot \mathrm{\pi }}{2\cdot 4\cdot 6\cdot 2}$ and inductively

${J}_{2n}=\frac{3\cdot 5\cdots \left(2n-3\right)\cdot \left(2n-1\right)\cdot \mathrm{\pi }}{2\cdot 4\cdots \left(2n-2\right)\cdot \left(2n\right)\cdot 2}.$

For $0\le x\le \mathrm{\pi }∕2$, $0\le cos\left(x\right)\le 1$, so ${cos}^{2n}\left(x\right)\ge {cos}^{2n+1}\left(x\right)\ge {cos}^{2n+2}\left(x\right)$, implying in turn that ${J}_{2n}\ge {J}_{2n+1}\ge {J}_{2n+2}$. Then

$1\ge \frac{{J}_{2n+1}}{{J}_{2n}}\ge \frac{{J}_{2n+2}}{{J}_{2n}}=\frac{2n+1}{2n+2}.$

Hence $\underset{n\to \infty }{lim}\frac{{J}_{2n+1}}{{J}_{2n}}=1$.

That is,

$\underset{n\to \infty }{lim}\left(\frac{2\cdot 2\cdot 4\cdot 4\cdot 6\cdot 6\dots \left(2n\right)\cdot \left(2n\right)}{1\cdot 3\cdot 3\cdot 5\cdot 5\dots \left(2n-1\right)\cdot \left(2n+1\right)}\frac{2}{\mathrm{\pi }}\right)=1$

or equivalently

$\underset{n\to \infty }{lim}\left(\frac{2\cdot 2\cdot 4\cdot 4\cdot 6\cdot 6\dots \left(2n\right)\cdot \left(2n\right)}{1\cdot 3\cdot 3\cdot 5\cdot 5\dots \left(2n-1\right)\cdot \left(2n+1\right)}\right)=\frac{\mathrm{\pi }}{2}.$

#### An elementary proof of Wallis formula

The following proof is adapted and expanded from . The proof uses only basic algebra, the Pythagorean Theorem, and the formula $\mathrm{\pi }{r}^{2}$ for the area of a circle of radius $r$. Another important property used implicitly is the completeness property of the reals.

Deﬁne a sequence of numbers by ${s}_{1}=1$ and for $n\ge 2$,

 ${s}_{n}=\frac{3}{2}\cdot \frac{5}{4}\cdots \frac{2n-1}{2n-2}.$ (5)

These are the reciprocals of the subsequence ${J}_{2n-1}$ deﬁned in equation (4).

The partial products of Wallis’ formula (1) with an odd number of terms in the numerator are

 ${o}_{n}=\frac{{2}^{2}\cdot {4}^{2}\cdots {\left(2n-2\right)}^{2}\cdot 2n}{1\cdot {3}^{2}\cdots {\left(2n-1\right)}^{2}}=\frac{2n}{{s}_{n}^{2}},$ (6)

while those with an even number of factors in the numerator are of the form

 ${e}_{n}=\frac{{2}^{2}\cdot {4}^{2}\cdots {\left(2n-2\right)}^{2}}{1\cdot {3}^{2}\cdots {\left(2n-3\right)}^{2}\cdot \left(2n-1\right)}=\frac{2n-1}{{s}_{n}^{2}}.$ (7)

Interpret ${e}_{1}=1$ as an empty product. Since $\frac{{\left(2n\right)}^{2}}{\left(2n-1\right)\left(2n+1\right)}>1$, clearly ${e}_{n}<{e}_{n+1}$. Since $\frac{\left(2n\right)\left(2n+2\right)}{{\left(2n+1\right)}^{2}}<1$, clearly ${o}_{n}>{o}_{n+1}$. Also, ${e}_{n}<{o}_{n}$. Therefore

${e}_{1}<{e}_{2}<{e}_{3}<\cdots {e}_{n}<{o}_{n}<\cdots <{o}_{3}<{o}_{2}<{o}_{1}$

for any $n$. Furthermore, for $1\le i\le n$,

$\frac{2i}{{s}_{i}^{2}}={o}_{i}\ge {o}_{n}$

and

$\frac{2i-1}{{s}_{i}^{2}}={e}_{i}\le {e}_{n}$

from which it follows that

 $\frac{2i-1}{{e}_{n}}\le {s}_{i}^{2}\le \frac{2i}{{o}_{n}}.$ (8)

For convenience, deﬁne ${s}_{0}=0$ so that inequality (8) holds also for $i=0$. Denote the successive diﬀerence ${a}_{n}={s}_{n+1}-{s}_{n}$ so ${a}_{0}={s}_{1}-{s}_{0}=1$ and for $n\ge 1$

${a}_{n}={s}_{n+1}-{s}_{n}={s}_{n}\left(\frac{2n+1}{2n}-1\right)=\frac{{s}_{n}}{2n}=\frac{1}{2}\cdot \frac{3}{4}\cdots \frac{2n-1}{2n}.$

Lemma 2. For the sequence ${a}_{i}$ we have the identity

 ${a}_{i}{a}_{j}=\frac{j+1}{i+j+1}{a}_{i}{a}_{j+1}+\frac{i+1}{i+j+1}{a}_{i+1}{a}_{j}$ (9)

for any $i$ and $j$.

Proof. Make the substitutions

${a}_{i+1}=\frac{2i+1}{2\left(i+1\right)}{a}_{i}$

and

${a}_{j+1}=\frac{2j+1}{2\left(j+1\right)}{a}_{j}$

the right side of the proposed identity (9) becomes

${a}_{i}{a}_{j}\left(\frac{2j+1}{2\left(j+1\right)}\cdot \frac{j+1}{i+j+1}+\frac{2i+1}{2\left(i+1\right)}\cdot \frac{i+1}{i+j+1}\right)={a}_{i}{a}_{j}.$

Lemma 3.

$\begin{array}{llll}\hfill 1={a}_{0}^{2}& ={a}_{0}{a}_{1}+{a}_{1}{a}_{0}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & ={a}_{0}{a}_{2}+{a}_{1}^{2}+{a}_{2}{a}_{0}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & \dots \phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & ={a}_{0}{a}_{n}+{a}_{1}{a}_{n-1}+\cdots +{a}_{n}{a}_{0}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

Proof. Start from ${a}_{0}^{2}=1$ and repeatedly apply the identity (9). At stage $n$ applying the identity (9) to every term the sum

${a}_{0}{a}_{n}+{a}_{1}{a}_{n-1}+\cdots +{a}_{n}{a}_{0}$

becomes

$\left({a}_{0}{a}_{n}+\frac{1}{n}{a}_{1}{a}_{n-1}\right)+\left(\frac{n-1}{n}{a}_{1}{a}_{n-1}+\frac{2}{n}{a}_{2}{a}_{n-2}\right)+\cdots +\left(\frac{1}{n}{a}_{n-1}{a}_{1}+{a}_{n}{a}_{0}\right).$

Collecting terms, this simpliﬁes to ${a}_{0}{a}_{n}+{a}_{1}{a}_{n-1}+\cdots +{a}_{n}{a}_{0}$. □

Now divide the positive quadrant of the $xy$-plane into rectangles by drawing the vertical lines $y={s}_{n}$ and the horizontal lines $y={s}_{n}$ for all $n$. Let ${R}_{i,j}$ be the rectangle with lower left corner $\left({s}_{i},{s}_{j}\right)$ and upper right corner $\left({s}_{i+1},{s}_{j+1}\right)$. The area of ${R}_{i,j}$ is ${a}_{i}{a}_{j}$. Thus the identity $1={a}_{0}{a}_{n}+{a}_{1}{a}_{{n}_{1}}+\cdots +{a}_{n}{a}_{0}$ states that the total area of the rectangles ${R}_{i,j}$ for which $i+j=n$ is $1$. Let ${P}_{n}$ be the polygonal region consisting of all rectangles ${R}_{i,j}$ for which $i+j. Hence the area of ${P}_{n}$ is $n$.

The outer corners of ${P}_{n}$ are the points $\left({s}_{i},{s}_{j}\right)$ for which $i+j=n+1$ and $1\le i,j,\le n$. By the Pythagorean theorem, the distance of such a point to the origin is $\sqrt{{s}_{i}^{2}+{s}_{j}^{2}}$. By (8)

$\sqrt{\frac{2\left(i+j\right)}{{o}_{n}}}=\sqrt{\frac{2\left(n+1\right)}{{o}_{n}}}.$

bounds this distance from above. Similarly the inner corners of ${P}_{n}$ are the points $\left({s}_{i},{s}_{j}\right)$ for which $i+j=n$ and $0\le i,j\le n$. The distance of such a point to the origin is bounded from below by

$\sqrt{\frac{2\left(i+j-1\right)}{{e}_{n}}}=\sqrt{\frac{2\left(n-1\right)}{{e}_{n}}}.$

Therefore, ${P}_{n}$ contains a quarter circle of radius $\sqrt{2\left(n-1\right)∕{e}_{n}}$ and is contained in a quarter circle of radius $\sqrt{2\left(n+1\right)∕{o}_{n}}$. See Figure 2 for a diagram of the polygonal region ${P}_{n}$ and the corresponding inner and outer quarter circles for $n=4$. Figure 2: A diagram of the regions ${R}_{i,j}$ and the inner and outer quarter circles for the case $n=4$.

The area of a quarter circle of radius $r$ is $\mathrm{\pi }{r}^{2}$ and the area of ${P}_{n}$ is $n$. This leads to the bounds

 $\frac{\left(n-1\right)\mathrm{\pi }}{2{e}_{n}} (10)

from which it follows that

$\frac{\left(n-1\right)\mathrm{\pi }}{2n}<{e}_{n}<{o}_{n}<\frac{\left(n+1\right)\mathrm{\pi }}{2n}.$

Then as $n\to \infty$, ${e}_{n}$ and ${o}_{n}$ both approach $\mathrm{\pi }∕2$.

#### Wallis’ Formula and the Central Binomial Coeﬃcient

This subsection gives a detailed proof that Wallis’ Formula gives an explicit inequality bound on the central binomial term that in turn implies the asymptotic formula for the central binomial coeﬃcient. The derivation in . motivates this proof.

Start from the expansion (2) for the Central Binomial Coeﬃcient:

${w}_{n}=\prod _{j=1}^{n}\frac{{\left(2j\right)}^{2}}{\left(2j-1\right)\left(2j+1\right)}=\frac{{2}^{4n}}{{\left(\genfrac{}{}{0.0pt}{}{2n}{n}\right)}^{2}\left(2n+1\right)}.$

Rearrange it to

${\left(\genfrac{}{}{0.0pt}{}{2n}{n}\right)}^{2}=\frac{1{6}^{n}}{2n+1}\prod _{j=1}^{n}\frac{\left(2j-1\right)\left(2j+1\right)}{{\left(2j\right)}^{2}}$

and use the deﬁnitions (6) and (7) of the partial products of the Wallis formula to obtain

${\left(\genfrac{}{}{0.0pt}{}{2n}{n}\right)}^{2}=\frac{1{6}^{n}}{\left(2n+1\right)}\frac{2n+2}{\left(2n+1\right){o}_{n+1}}$

and

${\left(\genfrac{}{}{0.0pt}{}{2n}{n}\right)}^{2}=\frac{1{6}^{n}}{\left(2n+1\right)}\frac{1}{{e}_{n+1}}.$

Rearrange the inequality (10) to obtain

Then

$\frac{1{6}^{n}}{\left(2n+1\right)}\frac{{\left(2n+2\right)}^{2}}{\left(2n+1\right)\left(n+2\right)\mathrm{\pi }}<{\left(\genfrac{}{}{0.0pt}{}{2n}{n}\right)}^{2}<\frac{1{6}^{n}}{\left(2n+1\right)}\frac{\left(2n+2\right)}{n\mathrm{\pi }}$

and taking square roots and slightly rearranging again

$\frac{{4}^{n}}{\sqrt{\left(2n+1\right)\left(\mathrm{\pi }∕2\right)}}\sqrt{\frac{2{\left(n+1\right)}^{2}}{\left(2n+1\right)\left(n+2\right)}}<\left(\genfrac{}{}{0.0pt}{}{2n}{n}\right)<\frac{{4}^{n}}{\sqrt{\left(2n+1\right)\left(\mathrm{\pi }∕2\right)}}\sqrt{\frac{\left(n+1\right)}{n}}.$

To simplify, take a series expansion of the square roots of the rational expressions and truncate, leaving

$\frac{{4}^{n}}{\sqrt{\left(2n+1\right)\left(\mathrm{\pi }∕2\right)}}\left(1-\frac{1}{2n}\right)<\left(\genfrac{}{}{0.0pt}{}{2n}{n}\right)<\frac{{4}^{n}}{\sqrt{\left(2n+1\right)\left(\mathrm{\pi }∕2\right)}}\left(1+\frac{1}{2n}\right).$

#### A proof using the product expansion of the sine function

Theorem 4. The expansion of $sin\left(z\right)$ as an inﬁnite product is

$sin\left(z\right)=z\prod _{m=1}^{\infty }\left(1-\frac{{z}^{2}}{{m}^{2}{\mathrm{\pi }}^{2}}\right).$

Proof. See [7, page 312] for the proof. The article in the Mathworld.com article on http://mathworld.wolfram.com/Sine.html. also gives references to Edwards 2001, pages 18 and 47; and Borwein et al. 2004, page 5. □

Although the common proof uses complex analysis, as in the texts cited above, a proof using only elementary analysis is possible. The following proof is adapted from .

Proof. Start with the deﬁnition of the Chebyshev polynomials ${T}_{n}\left(x\right)$ from the trigonometric identity $cos\left(nx\right)={T}_{n}\left(cos\left(x\right)\right)$. Then

$cos\left(2kx\right)={T}_{k}\left(cos\left(2x\right)\right)={T}_{k}\left(1-2{sin}^{2}x\right).$

Together with the sine product identity

$sin\left(\left(2k+1\right)x\right)-sin\left(\left(2k-1\right)x\right)=2sin\left(x\right)\cdot cos\left(2kx\right)$

this inductively shows that there is a polynomial ${F}_{m}$ of degree $m$ such that

$sin\left(\left(2m+1\right)x\right)=sin\left(x\right)\cdot {F}_{m}\left({sin}^{2}\left(x\right)\right).$

(In fact, using the deﬁnition ${U}_{n}\left(cos\left(x\right)\right)=\frac{sin\left(\left(n+1\right)x\right)}{sin\left(x\right)}$ for the Chebyshev polynomials of the second kind, it is easy to show that ${F}_{m}\left(x\right)={U}_{2n}\left(\sqrt{1-x}\right)$.) Substituting ${x}_{k}=\frac{k\mathrm{\pi }}{2m+1}$ and noting that $sin\left(\left(2m+1\right){x}_{k}\right)=0$ shows that ${F}_{m}$ has zeros at ${sin}^{2}\left(\left(\frac{k\mathrm{\pi }}{2m+1}\right)$ for $k=1,2,\dots ,m$. These zeros are distinct, so ${F}_{m}$ has no other zeros, then

${F}_{m}\left(y\right)={F}_{m}\left(0\right)\prod _{k=1}^{m}\left(1-\frac{y}{{sin}^{2}\left(\frac{k\mathrm{\pi }}{2m+1}\right)}\right)$

and

${F}_{m}\left(0\right)=\underset{x\to 0}{lim}\frac{sin\left(\left(2m+1\right)x\right)}{sin\left(x\right)}=2m+1.$

Therefore

$sin\left(\left(2m+1\right)x\right)=\left(2m+1\right)sin\left(x\right)\prod _{k=1}^{m}\left(1-\frac{{sin}^{2}\left(x\right)}{{sin}^{2}\left(\frac{k\mathrm{\pi }}{2m+1}\right)}\right)$

and changing variables

 $sin\left(x\right)=\left(2m+1\right)sin\left(\frac{x}{2m+1}\right)\prod _{k=1}^{m}\left(1-\frac{{sin}^{2}\left(\frac{x}{2m+1}\right)}{{sin}^{2}\left(\frac{k\mathrm{\pi }}{2m+1}\right)}\right).$ (11)

The goal now is to estimate the product terms. For all real $t$, ${\mathrm{e}}^{t}\ge 1+t$ and therefore ${\mathrm{e}}^{-t}\le \frac{1}{1+t}$. For $u<1$, the choice $t=u∕\left(1-u\right)$ gives ${\mathrm{e}}^{-u∕\left(1-u\right)}\le 1-u$. Then for every collection of numbers ${u}_{k}\in \left[0,1\right)$, we have

 $1-\sum _{k}\frac{{u}_{k}}{1-{u}_{k}}\le {\mathrm{e}}^{-\sum _{k}\frac{{u}_{k}}{1-{u}_{k}}}\le \prod _{k}\left(1-{u}_{k}\right)\le {\mathrm{e}}^{-\sum _{k}{u}_{k}}\le 1.$ (12)

If also ${\sum }_{k}{u}_{k}<1$, then we also know that

 ${\mathrm{e}}^{-\sum _{k}{u}_{k}}\le \frac{1}{1+\sum _{k}{u}_{k}}$ (13)

and subtracting the ﬁrst and third terms of (12) from $1$ and using (13)

 $\frac{\sum _{k}{u}_{k}}{1+\sum _{k}{u}_{k}}\le 1-\prod _{k}\left(1-{u}_{k}\right)\le \sum _{k}\frac{{u}_{k}}{1-{u}_{k}}.$ (14)

Let $m$ and $N$ be positive integers with $m>N$. Take $x$ such that $|x|<\frac{1}{4}N\mathrm{\pi }$ and $\frac{x}{\mathrm{\pi }}\notin ℤ$. Then deﬁne ${u}_{k}$ by

${u}_{k}={\left(\frac{sin\left(\frac{x}{2m+1}\right)}{sin\left(\frac{k\mathrm{\pi }}{2m+1}\right)}\right)}^{2},\phantom{\rule{2em}{0ex}}k=1,2,\dots ,m.$

Use (11) by dividing the leading factor and the ﬁrst $N$ factors onto the left side to obtain

$\frac{sin\left(x\right)}{\left(2m+1\right)sin\left(\frac{x}{2m+1}\right)\prod _{k=1}^{N}\left(1-{u}_{k}\right)}=\prod _{k=N+1}^{m}\left(1-{u}_{k}\right).$

Then use the ﬁrst, third and ﬁfth terms of (12) to see that

 $1-\sum _{k=N+1}^{m}\frac{{u}_{k}}{1-{u}_{k}}\le \frac{sin\left(x\right)}{\left(2m+1\right)sin\left(\frac{x}{2m+1}\right)\prod _{k=1}^{N}\left(1-{u}_{k}\right)}\le 1.$ (15)

For $0\le t\le \frac{\mathrm{\pi }}{2}$ we have $sin\left(t\right)\ge \frac{2}{\mathrm{\pi }}t$, so we see that

${u}_{k}\le {\left(\frac{\left(2m+1\right)sin\left(\frac{x}{2m+1}\right)}{2k}\right)}^{2}\le {\left(\frac{x}{2k}\right)}^{2};$

thus

$\frac{{u}_{k}}{1-{u}_{k}}\le \frac{{x}^{2}}{{\left(2k\right)}^{2}-{x}^{2}}\phantom{\rule{2em}{0ex}}\text{for}k>N.$

Hence

$\sum _{k=N+1}^{m}\frac{{u}_{k}}{1-{u}_{k}}\le \frac{{x}^{2}}{2N-|x|}.$

Thus it follows from (15) that

$1-\frac{{x}^{2}}{2N-|x|}\le \frac{sin\left(x\right)}{\left(2m+1\right)sin\left(\frac{x}{2m+1}\right)\prod _{k=1}^{N}\left(1-{u}_{k}\right)}\le 1.$

Let $m\to \infty$ so that

$1-\frac{{x}^{2}}{2N-|x|}\le \frac{sin\left(x\right)}{x\prod _{k=1}^{N}\left(1-{u}_{k}\right)}\le 1.$

Now let $N\to \infty$ and we obtain for $\frac{x}{\mathrm{\pi }}\notin ℤ$

$sinx=x\prod _{k=1}^{\infty }\left(1-\frac{{x}^{2}}{{k}^{2}{\mathrm{\pi }}^{2}}\right).$

For $\frac{x}{\mathrm{\pi }}\in ℤ$ this equality is also true. □

The following somewhat probabilistic proof of Euler’s inﬁnite product formula is adapted from .

${\int }_{0}^{n}{x}^{t-1}{\left(1-\frac{x}{n}\right)}^{n}\phantom{\rule{0.3em}{0ex}}dx$

and integrate by parts once:

$\begin{array}{llllllll}\hfill u& ={\left(1-\frac{x}{n}\right)}^{n}\phantom{\rule{2em}{0ex}}& \hfill v& =\frac{1}{t}{x}^{t}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill \phantom{\rule{0.3em}{0ex}}du& =-{\left(1-\frac{x}{n}\right)}^{n-1}\phantom{\rule{0.3em}{0ex}}dx\phantom{\rule{2em}{0ex}}& \hfill \phantom{\rule{0.3em}{0ex}}dv& ={x}^{t-1}\phantom{\rule{0.3em}{0ex}}dx.\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & \phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}& \hfill \end{array}$

Then

$\begin{array}{llll}\hfill {\int }_{0}^{n}{x}^{t-1}{\left(1-\frac{x}{n}\right)}^{n}\phantom{\rule{0.3em}{0ex}}dx& ={{\left(1-\frac{x}{n}\right)}^{n}|}_{0}^{n}+{\int }_{0}^{n}\frac{1}{t}{x}^{t}{\left(1-\frac{x}{n}\right)}^{n-1}\phantom{\rule{0.3em}{0ex}}dx\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =\frac{1}{t}{\int }_{0}^{n}{x}^{t}{\left(1-\frac{x}{n}\right)}^{n-1}\phantom{\rule{0.3em}{0ex}}dx\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

Integrate by parts again:

$\begin{array}{llllllll}\hfill u& ={\left(1-\frac{x}{n}\right)}^{n-1}\phantom{\rule{2em}{0ex}}& \hfill v& =\frac{1}{t+1}{x}^{t+1}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill \phantom{\rule{0.3em}{0ex}}du& =-\frac{n-1}{n}{\left(1-\frac{x}{n}\right)}^{n-2}\phantom{\rule{0.3em}{0ex}}dx\phantom{\rule{2em}{0ex}}& \hfill \phantom{\rule{0.3em}{0ex}}dv& ={x}^{t}\phantom{\rule{0.3em}{0ex}}dx\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & \phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}& \hfill \end{array}$

so

${\int }_{0}^{n}{x}^{t-1}{\left(1-\frac{x}{n}\right)}^{n}\phantom{\rule{0.3em}{0ex}}dx=\frac{\left(n-1\right)}{t\left(t+1\right)n}{\int }_{0}^{n}{x}^{t+1}{\left(1-\frac{x}{n}\right)}^{n-2}\phantom{\rule{0.3em}{0ex}}dx.$

After integration by parts $n$ times:

$\begin{array}{llll}\hfill {\int }_{0}^{n}{x}^{t-1}{\left(1-\frac{x}{n}\right)}^{n}\phantom{\rule{0.3em}{0ex}}dx& =\frac{\left(n-1\right)!}{t\left(t+1\right)\dots \left(t+n-1\right){n}^{n-1}}{\int }_{0}^{n}{x}^{t+n-1}\phantom{\rule{0.3em}{0ex}}dx\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =\frac{\left(n-1\right)!}{t\left(t+1\right)\dots \left(t+n-1\right){n}^{n-1}}{\frac{{x}^{t+n}}{t+n}|}_{0}^{n}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =\frac{n!\phantom{\rule{0.3em}{0ex}}{n}^{t}}{t\left(t+1\right)\dots \left(t+n\right)}.\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

Then by dominated convergence it follows that

$\Gamma \left(t\right)={\int }_{0}^{\infty }{x}^{t-1}{\mathrm{e}}^{-x}\phantom{\rule{0.3em}{0ex}}dx=\underset{n\to \infty }{lim}\frac{n!\phantom{\rule{0.3em}{0ex}}{n}^{t}}{t\left(t+1\right)\dots \left(t+n\right)}.$

Note that limit deﬁnition of the Gamma function is also http://dlmf.nist.gov/5.8 E1.. It follows that for $-1 that

$\Gamma \left(1+t\right)\Gamma \left(1-t\right)=\prod _{k=1}^{\infty }\frac{1}{1-{t}^{2}∕{k}^{2}}.$

For $X$ and $Y$ independent exponential random variables, both with expectation $1$ we have that

$𝔼\left[{X}^{t}\right]={\int }_{0}^{\infty }{x}^{t}{\mathrm{e}}^{-x}\phantom{\rule{0.3em}{0ex}}dx=\Gamma \left(1+t\right)$

and likewise $𝔼\left[{Y}^{-t}\right]=\Gamma \left(1-t\right)$. Then using the independence and symmetry

$\Gamma \left(1+t\right)\Gamma \left(1-t\right)=𝔼\left[{X}^{t}\right]𝔼\left[{Y}^{-t}\right]=𝔼\left[{\left(X∕Y\right)}^{t}\right]=𝔼\left[{\left(X∕Y\right)}^{|t|}\right]$

The distribution function for the random variable $X∕Y$ is

$ℙ\left[X∕Y\le u\right]=ℙ\left[Y\ge X∕u\right]=𝔼\left[{\mathrm{e}}^{-X∕u}\right]=\frac{u}{u+1}$

for $u>0$. Then the probability density is

$\frac{\phantom{\rule{0.3em}{0ex}}d}{\phantom{\rule{0.3em}{0ex}}du}ℙ\left[X∕Y\le u\right]=\frac{1}{{\left(1+u\right)}^{2}},\phantom{\rule{2em}{0ex}}u>0.$

This gives by integration by parts, for $-1, that

$𝔼\left[{\left(X∕Y\right)}^{|t|}\right]={\int }_{0}^{\infty }\frac{{u}^{|t|}}{{\left(u+1\right)}^{2}}\phantom{\rule{0.3em}{0ex}}du={\int }_{0}^{\infty }\frac{|t|{u}^{|t|-1}}{u+1}\phantom{\rule{0.3em}{0ex}}du$

The last integral is $\mathrm{\pi }t∕sin\left(\mathrm{\pi }t\right)$. For example, this is formula 613, page 445 in the 18th Edition of the CRC Standard Mathematical Tables. It can also be done with contour integration in the complex plane

${\int }_{0}^{+\infty }\frac{|t|{u}^{|t|-1}}{u+1}\phantom{\rule{0.3em}{0ex}}du-2\mathrm{\pi }\mathrm{i}{\left(-1\right)}^{|t|-1}+{\int }_{+\infty }^{0}\frac{|t|{u}^{2\mathrm{\pi }\mathrm{i}\left(|t|-1\right)}}{u+1}\phantom{\rule{0.3em}{0ex}}du=0$

so

${\int }_{0}^{\infty }\frac{|t|{u}^{|t|-1}}{u+1}\phantom{\rule{0.3em}{0ex}}du\left(1-{\mathrm{e}}^{2\mathrm{\pi }\mathrm{i}|t|}\right)=-2\mathrm{\pi }\mathrm{i}{\mathrm{e}}^{\mathrm{\pi }\mathrm{i}|t|}.$

Thus

${\int }_{0}^{\infty }\frac{|t|{u}^{|t|-1}}{u+1}\phantom{\rule{0.3em}{0ex}}du=\frac{|t|2\mathrm{\pi }\mathrm{i}{\mathrm{e}}^{\mathrm{\pi }\mathrm{i}|t|}}{{\mathrm{e}}^{2\mathrm{\pi }\mathrm{i}|t|}-1}=\frac{\mathrm{\pi }|t|}{sin\left(\mathrm{\pi }|t|\right)}=\frac{\mathrm{\pi }t}{sin\left(\mathrm{\pi }t\right)}.$

Hence, for $-1

$\Gamma \left(1+t\right)\Gamma \left(1-t\right)=𝔼\left[{\left(X∕Y\right)}^{t}\right]=\frac{\mathrm{\pi }t}{sin\left(\mathrm{\pi }t\right)}.$

A standard integration-by-parts gives the fundamental Gamma function recursion $\Gamma \left(t\right)=\Gamma \left(1+t\right)∕t$. Using this to deﬁne $\Gamma \left(t\right)$ for $t<0$, it follows that for all real $t$,

$sin\left(\mathrm{\pi }t\right)=\frac{\mathrm{\pi }}{\Gamma \left(t\right)\Gamma \left(1-t\right)}.$

Combining all results above,

$sin\left(\mathrm{\pi }t\right)=\mathrm{\pi }t\prod _{k=1}^{\infty }\left(1-\frac{{t}^{2}}{{k}^{2}}\right).$

Corollary 1 (Wallis’ Formula).

$\underset{n\to \infty }{lim}\frac{2\cdot 2\cdot 4\cdot 4\cdot 6\cdot 6\dots \left(2n\right)\cdot \left(2n\right)}{1\cdot 3\cdot 3\cdot 5\cdot 5\dots \left(2n-1\right)\cdot \left(2n-1\right)\cdot \left(2n+1\right)}=\frac{\mathrm{\pi }}{2}.$

Proof. Substitute $z=\mathrm{\pi }∕2$ in the product expansion of $sin\left(z\right)$. □

Remark. In  Ciaurri uses Tannery’s Theorem for Inﬁnite Products, a trig identity for cotangent and tangent, and Wallis formula to provide an elementary proof of product expansion of the sine function. In this sense, the product expansion of the sine function is equivalent to Wallis’ formula.

#### A Geometric Interpretation of Wallis’ Formula

This section gives a geometric interpretation of Wallis’ formula, constructing a subset of the unit square with area $\frac{8}{9}\cdot \frac{24}{25}\cdot \frac{48}{49}\cdots =\frac{2}{3}\cdot \frac{4}{3}\cdot \frac{4}{5}\cdot \frac{6}{5}\cdot \frac{6}{7}\cdot \frac{8}{7}\cdots =\frac{1}{2}\cdot \frac{\pi }{2}$. The construction is reminiscent of the construction of the Sierpinski Triangle.

Take a unit square ${W}_{0}$ and remove a square of area $\frac{1}{3}\cdot \frac{1}{3}$ from the center. Label the result as ${W}_{1}$ with area $\frac{8}{9}$. Next remove a square of area $\frac{1}{5}\cdot \frac{1}{5}$ from the center of each of the $8$ small sub-squares of area $\frac{1}{9}$ constituting ${W}_{1}$ and call the result ${W}_{2}$ with area $\frac{8}{9}\cdot \frac{24}{25}$. Continuing by removing squares of area $\frac{1}{7}\cdot \frac{1}{7}$ from each sub-square of area $\frac{1}{5}\cdot \frac{1}{5}$ surrounding the hole removed at stage $2$, construct region ${W}_{3}$ with area $\frac{8}{9}\cdot \frac{24}{25}\cdot \frac{48}{49}$. Continue to create ${W}_{4}$, and so on. Call the limiting region ${W}_{\infty }$ which has area $\frac{8}{9}\cdot \frac{24}{25}\cdot \frac{48}{49}\cdots =\frac{1}{2}\cdot \frac{\pi }{2}$.

The side length of the removed squares at stage $n$ is $\frac{{2}^{n}n!}{\left(2n+1\right)!}$. After $3$ stages, $201$ squares have been removed, and the stage $3$ holes have side length $\frac{1}{3}\cdot \frac{1}{5}\cdot \frac{1}{7}=\frac{1}{105}$. It will be impractical to illustrate more than $3$ stages of this iterative process. Figure 3: Three stages in the construction of the Wallis sieve.

Here area means the Lebesgue measure of the limiting region ${W}_{\infty }$. However ${W}_{\infty }$ does not contain any product set $A×B$ with $A,B\subset ℝ$, each with positive Lebesgue measure. To see this, consider a maximal product subset of ${W}_{1}$, for example $\left[0,1\right]×\left(\left[0,\frac{1}{3}\right]\cup \left[\frac{2}{3},1\right]\right)$ (or equivalently $\left(\left[0,\frac{1}{3}\right]\cup \left[\frac{2}{3},1\right]\right)×\left[0,1\right]$ although from here on, consider only the product sets with ﬁrst factor $\left[0,1\right]$). This product subset has measure $\frac{2}{3}$. A similar maximal product subset of ${W}_{2}$ has measure $\frac{2}{3}\cdot \frac{4}{5}$. Continuing, a maximal product subset of ${W}_{\infty }$ has measure $\frac{2}{3}\cdot \frac{4}{5}\cdot \frac{6}{7}\cdots =\prod _{n=1}^{\infty }\left(1-\frac{1}{2n+1}\right)=0$.

Thus, ${W}_{\infty }$ is an example of a subset of the plane ${ℝ}^{2}$ with positive Lebesgue measure, but not admitting any product subset with positive Lebesgue measure. Note also that ${W}_{\infty }$ is a “square-like” region with area equal to a circle with radius $\frac{1}{2}$. In that sense, this is a solution to the ancient problem of “squaring the circle”.

#### Sources

This section is adapted from a sketch of the proof of Wallis’ Formula in Kazarinoﬀ  and the short note by Wästlund . The elementary proofs of the sine product are from  and . The geometric interpretation of Wallis’ Formula is adapted from .

_______________________________________________________________________________________________ ### Problems to Work for Understanding

1. Show that
$\frac{{2}^{4n}\cdot {\left(n!\right)}^{4}}{{\left(\left(2n\right)!\right)}^{2}\left(2n+1\right)}=\frac{2\cdot 2\cdot 4\cdot 4\cdot 6\cdot 6\dots \left(2n\right)\cdot \left(2n\right)}{1\cdot 3\cdot 3\cdot 5\cdot 5\dots \left(2n-1\right)\cdot \left(2n+1\right)}.$

2. Numerically ﬁnd the rate at which the sequence
$\frac{2\cdot 2\cdot 4\cdot 4\cdot 6\cdot 6\dots \left(2n\right)\cdot \left(2n\right)}{1\cdot 3\cdot 3\cdot 5\cdot 5\dots \left(2n-1\right)\cdot \left(2n+1\right)}$

approaches $\mathrm{\pi }∕2$ by calculating values and using regression.

3. Use induction to prove
${J}_{2n+1}=\frac{\left(2n\right)\cdot \left(2n-2\right)\dots 4\cdot 2}{\left(2n+1\right)\cdot \left(2n-1\right)\dots 3\cdot 1}$

and

${J}_{2n}=\frac{\left(2n-1\right)\cdot \left(2n-3\right)\dots 3\cdot \mathrm{\pi }}{\left(2n\right)\cdot \left(2n-2\right)\dots 4\cdot 2\cdot 2}.$

4. The simple proof of the sine product formula uses the polynomial ${F}_{m}$ of degree $m$ such that
$sin\left(\left(2m+1\right)x\right)=sin\left(x\right)\cdot {F}_{m}\left({sin}^{2}\left(x\right)\right).$

Explicitly ﬁnd ${F}_{0}\left(x\right),{F}_{1}\left(x\right),{F}_{2}\left(x\right),{F}_{3}\left(x\right),{F}_{4}\left(x\right)$ and plot them on $\left[-1,1\right]$.

5. Give a detailed proof that
${\int }_{0}^{n}{x}^{t-1}{\left(1-\frac{x}{n}\right)}^{n}\phantom{\rule{0.3em}{0ex}}dx=\frac{n!\phantom{\rule{0.3em}{0ex}}{n}^{t}}{t\left(t+1\right)\dots \left(t+n\right)}.$

6. Provide a careful and detailed proof that if $X$ and $Y$ are independent exponential random variables, both with expectation $1$ we have that
$𝔼\left[{\left(X∕Y\right)}^{t}\right]=𝔼\left[{\left(X∕Y\right)}^{|t|}\right].$

7. Provide a detailed proof using contour integration in the complex plane to show that
${\int }_{0}^{\infty }\frac{|t|{u}^{|t|-1}}{u+1}\phantom{\rule{0.3em}{0ex}}du=\frac{|t|2\mathrm{\pi }\mathrm{i}{\mathrm{e}}^{\mathrm{\pi }\mathrm{i}\phantom{\rule{1em}{0ex}}|t|}}{{\mathrm{e}}^{2\mathrm{\pi }\mathrm{i}|t|}-1}=\frac{\mathrm{\pi }t}{sin\left(\mathrm{\pi }t\right)}.$

__________________________________________________________________________ ### References

   Michael D. Hirschhorn. Wallis’s product and the central binomial coeﬃcient. American Mathematical Monthly, 122(7):689, August-September 2015.

   Lars Holst. A proof of Euler’s inﬁnite product for the sine. American Mathematical Monthly, 119:518–521, June-July 2012.

   Nicholaus D. Kazarinoﬀ. Analytic Inequalities. Holt, Rinehart and Winston, 1961.

   R. A. Kortram. Simple proofs for ${\sum }_{k=1}^{\infty }\frac{1}{{k}^{2}}=\frac{{\pi }^{2}}{6}$ and $sinx=x{\prod }_{k=1}^{\infty }\left(1-\frac{{x}^{2}}{{k}^{2}{\pi }^{2}}\right)$. Mathematics Magazine, 69(2):123–125, April 1996.

   Óscar Ciaurri. Euler’s product expansion for the sine: An elementary proof. American Mathematical Monthly, 122(7):693–695, August-September 2015.

   Hansklaus Rummler. Squaring the circle with holes. American Mathematical Monthly, 100(9):858–860, November 1993.

   S. Saks and A. Zygmund. Analytic Functions. Elsevier Publishing, 1971.

   Johan Wästlund. An elementary proof of the wallis product formula for pi. American Mathematical Monthly, 114(10):914–917, December 2007.

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