Last edited by hbghlyj at 2022-12-24 04:49:001. THE WALLIS PRODUCT FORMULA
In 1655, John Wallis wrote down the celebrated formula\begin{equation}\frac{2}{1} \cdot \frac{2}{3} \cdot \frac{4}{3} \cdot \frac{4}{5} \cdots=\frac{\pi}{2}\label{1}\end{equation}Most textbook proofs of \eqref{1} rely on evaluation of some definite integral like$$\int_{0}^{\pi / 2}(\sin x)^{n} d x$$by repeated partial integration. The topic is usually reserved for more advanced calculus courses. The purpose of this note is to show that \eqref{1} can be derived using only the mathematics taught in elementary school, that is, basic algebra, the Pythagorean theorem, and the formula $\pi \cdot r^{2}$ for the area of a circle of radius $r$.
Viggo Brun gives an account of Wallis' method in [1] (in Norwegian). Yaglom and Yaglom [2] give a beautiful proof of \eqref{1} which avoids integration but uses some quite sophisticated trigonometric identities.
A NUMBER SEQUENCE
We denote the Wallis product by\begin{equation}\label{2}W=\frac{2}{1} \cdot \frac{2}{3} \cdot \frac{4}{3} \cdot \frac{4}{5} \cdots\end{equation}The partial products involving an even number of factors form an increasing sequence, while those involving an odd number of factors form a decreasing sequence. We let $s_{0}=0, s_{1}=1$, and in general,$$s_{n}=\frac{3}{2} \cdot \frac{5}{4} \cdots \frac{2 n-1}{2 n-2}$$The partial products of \eqref{2} with an odd number of factors can be written as$$\frac{2 n}{s_{n}^{2}}=\frac{2^{2} \cdot 4^{2} \cdots(2 n)}{1 \cdot 3^{2} \cdots(2 n-1)^{2}}\gt\frac{2^{2} \cdot 4^{2} \cdots(2 n)}{1 \cdot 3^{2} \cdots(2 n-1)^{2}}\cdot\left(1-\frac1{(2n+1)^2}\right)\left(1-\frac1{(2n+3)^2}\right)⋯=W$$while the partial products with an even number of factors are of the form$$\frac{2 n-1}{s_{n}^{2}}=\frac{2^{2} \cdot 4^{2} \cdots(2 n-2)^{2}}{1 \cdot 3^{2} \cdots(2 n-3)^{2} \cdot(2 n-1)}\lt\frac{2^{2} \cdot 4^{2} \cdots(2 n-2)^{2}}{1 \cdot 3^{2} \cdots(2 n-3)^{2} \cdot(2 n-1)}\cdot\left(1-\frac1{(2n)^2}\right)^{-1}\left(1-\frac1{(2n+2)^2}\right)^{-1}⋯=W$$It follows that\begin{equation}\frac{2 n-1}{W}<s_{n}^{2}<\frac{2 n}{W}\end{equation}We denote the difference $s_{n+1}-s_{n}$ by $a_{n}$, and observe that$$a_{n}=s_{n+1}-s_{n}=s_{n}\left(\frac{2 n+1}{2 n}-1\right)=\frac{s_{n}}{2 n}=\frac{1}{2} \cdot \frac{3}{4} \cdots \frac{2 n-1}{2 n} .$$We first derive the identity\begin{equation}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} .\label4\end{equation}Proof. After the substitutions$$a_{i+1}=\frac{2 i+1}{2(i+1)} a_{i}$$and$$a_{j+1}=\frac{2 j+1}{2(j+1)} a_{j}$$the right hand side of \eqref{4} becomes$$a_{i} a_{j}\left(\frac{2 j+1}{2(j+1)} \cdot \frac{j+1}{i+j+1}+\frac{2 i+1}{2(i+1)} \cdot \frac{i+1}{i+j+1}\right)=a_{i} a_{j} .$$$\square$If we start from $a_{0}^{2}$ and repeatedly apply \eqref{4}, we obtain the identities\begin{equation}\begin{aligned}
1=a_{0}^{2}=a_{0} a_{1}+a_{1} a_{0}=a_{0} a_{2}+a_{1}^{2}+a_{2} a_{0} &=\ldots \\
\cdots &=a_{0} a_{n}+a_{1} a_{n-1}+\cdots+a_{n} a_{0}
\end{aligned}\end{equation}
Proof. By applying \eqref{4} to every term, the sum $a_{0} a_{n-1}+\cdots+a_{n-1} a_{0}$ becomes$$\left(a_{0} a_{n}+\frac{1}{n} a_{1} a_{n-1}\right)+\left(\frac{n-1}{n} a_{1} a_{n-1}+\frac{2}{n} a_{2} a_{n-2}\right)+\cdots+\left(\frac{1}{n} a_{n-1} a_{1}+a_{n} a_{0}\right) .$$After collecting terms, this simplifies to $a_{0} a_{n}+\cdots+a_{n} a_{0}$.
A GEOMETRIC CONSTRUCTION
We divide the positive quarter of the xy-plane into rectangles by drawing the straight lines $x=s_{n}$ and $y=s_{n}$ 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 $a_{i} a_{j}$. Therefore the identity (5) states that the total area of the rectangles $R_{i, j}$ for which $i+j=n$ is 1 . We let $P_{n}$ be the polygonal region consisting of all rectangles $R_{i, j}$ for which $i+j<n$. Hence the area of $P_{n}$ is $n$ (see Figure 1 ).
The outer corners of $P_{n}$ are the points $\left(s_{i}, s_{j}\right)$ for which $i+j=n+1$. By the Pythagorean theorem, the distance of such a point to the origin is$$\sqrt{s_{i}^{2}+s_{j}^{2}}$$By $(3)$, this is bounded from above by$$\sqrt{\frac{2(i+j)}{W}}=\sqrt{\frac{2(n+1)}{W}}$$Similarly, the inner corners of $P_{n}$ are the points $\left(s_{i}+s_{j}\right)$ for which $i+j=n$. The distance of such a point to the origin is bounded from below by$$\sqrt{\frac{2(i+j-1)}{W}}=\sqrt{\frac{2(n-1)}{W}}$$
\begin{tikzpicture}[scale=3]
\foreach \i in {0,1,3/2,15/8,35/16}\draw[dashed](\i,0)node[below]{\i}--(\i,2.7) (0,\i)node[left]{\i}--(2.7,\i);\draw[dashed](2.5,0)node[below,yshift=-1]{$\cdots$}--(2.5,2.7) (0,2.5)node[left,xshift=-3]{$\vdots$}--(2.7,2.5);\draw[thick](0,0)--(0,35/16)--(1,35/16)--(1,15/8)--(3/2,15/8)--(3/2,3/2)--(15/8,3/2)--(15/8,1)--(35/16,1)--(35/16,0)--(0,0);\draw(1/2,32.5/16)node{$R_{0,3}$};\draw(1/2,13.5/8)node{$R_{0,2}$};\draw(1/2,5/4)node{$R_{0,1}$};\draw(1/2,1/2)node{$R_{0,0}$};\draw(5/4,1/2)node{$R_{1,0}$};\draw(13.5/8,1/2)node{$R_{2,0}$};\draw(32.5/16,1/2)node{$R_{3,0}$};\draw(13.5/8,5/4)node{$R_{2,1}$};\draw(5/4,5/4)node{$R_{1,1}$};\draw(5/4,13.5/8)node{$R_{1,2}$};
\end{tikzpicture}
Figure 1. The region $P_{4}$ of area 4.
Therefore $P_{n}$ contains a quarter circle of radius $\sqrt{2(n-1) / W}$, and is contained in a quarter circle of radius $\sqrt{2(n+1) / W}$. Since the area of a quarter circle of radius $r$ is equal to $\pi r^{2} / 4$, we obtain the following bounds for the area of $P_{n}$ :$$\frac{\pi(n-1)}{2 W}<n<\frac{\pi(n+1)}{2 W} \text {. }$$Since this holds for every $n$, we conclude that$$W=\frac{\pi}{2}$$
REFERENCES
[1] Brun, Viggo., Wallis's og Brounckers formler for $\pi$ (in Norwegian), Norsk matematisk tidskrift, 33 (1951) 73-81.
[2] Yaglom, A. M. and Yaglom, I. M., An elementary derivation of the formulas of Wallis, Leibnitz and Euler for the number $\pi$ (in Russian), Uspechi matematiceskich nauk. (N. S.) 8 , no. 5 (57) (1953), 181-187. |
