---
abstract: |
    In this Note, we start off with the primary representation of $e$ and from there present an elementary short proof for the Wallis formula for $\pi$.\
    \
author:
- Ali Sanayei
title: |
    From $e$ to $\pi$:\
    Derivation of the Wallis formula for $\pi$ from $e$ 
---

*Center for Optical Quantum Technologies, Institute for Laser Physics,*\
*University of Hamburg, Luruper Chaussee 149, D-22761 Hamburg, Germany*\
E-mail: `asanayei@physnet.uni-hamburg.de` \

A well-known pre-Newtonian formula for $\pi$, represented as an infinite product, was derived by John Wallis in 1655 using a method of successive interpolations [@Wallis; @Pi; @source; @book]:

$$\frac{\pi}{2}=\frac{2\,\,.\,\,2}{1\,\,.\,\,3}\,\,\frac{4\,\,.\,\,4}{3\,\,.\,\,5}\,\,\frac{6\,\,.\,\,6}{5\,\,.\,\,7}\,\,\frac{8\,\,.\,\,8}{7\,\,.\,\,9}\,\cdots\label{Wallis formula 1}$$

Different proofs have been offered from the diverse points of view in the recently past decades. For instance, proofs using the trigonometric integrals were proposed by Borwein, Borwein, and Stewart [@Pi; @AGM; @Stewart]. Sondow and Weisstein found a derivation by trigonometry [@Sondow; @and; @Weisstein; @Weisstein]. A proof from the point of view of the geometric means was offered by W[ä]{}stlund [@Wastlund]. Miller proposed a probabilistic proof [@Miller], and Kovalyova derived it from a combinatorial-probabilistic method [@Kovalyov]. More recently, Friedmann and Hagen simply applied the variational principle of quantum mechanics on the pure Coulomb potential (hydrogen atom) and showed a very interesting proof from the standpoint of physics [@Friedmann; @and; @Hagen].

Eventuated by the Euler’s identity $e^{i\pi}=-1$, that relates $e$ and $\pi$ together, one will be inspired to find a path from $e$ to $\pi$ and vice versa. For instance, given the Wallis formula (\[Wallis formula 1\]), Pippenger found an infinite product for $e$ [@Pippenger].

In this Note, we trace a path from $e$ to $\pi$ by offering an elementary proof for the Wallis formula (\[Wallis formula 1\]) only by starting off with the primary representation of $e$. To do that, we take the Bernoulli’s expression

$$\lim_{n\rightarrow\infty}\left(1+\frac{1}{n}\right)^{n}=e\label{e}$$

as the definition of $e$ (cf. Ref. [@Bernoulli]).

From Eq. (\[e\]) one can readily write

$$\lim_{n\rightarrow\infty}\,\frac{n}{n+\frac{1}{2}}\left(\frac{n}{n-\frac{1}{2}}\right)^{2n}=e\,,\label{e _2}$$

or equivalently

$$\lim_{n\rightarrow\infty}\,\frac{1}{n+\frac{1}{2}}\,\frac{n^{2n+1}}{\left(n-\frac{1}{2}\right)^{2n}}\,\frac{1}{e}=1\,.\label{e _3}$$

From Eq. (\[e \_3\]) one can then obtain

$$\lim_{n\rightarrow\infty}\,\frac{1}{n+\frac{1}{2}}\left[\frac{n^{n+\frac{1}{2}}\,e^{-n}}{\left(n-\frac{1}{2}\right)^{n}\,e^{-n+\frac{1}{2}}}\right]^{2}=1\,,\label{e _4}$$

which with the help of the Stirling’s formula (see Refs. [@Stirling] and [@Schopohl] for elementary proofs) and with the property of gamma functions, $z\,\Gamma\left(z\right)=\Gamma\left(z+1\right)$, leads to

$$\lim_{n\rightarrow\infty}\,\frac{\Gamma\left(n+1\right)}{\Gamma\left(n+\frac{1}{2}\right)}\,\frac{\Gamma\left(n+1\right)}{\Gamma\left(n+\frac{3}{2}\right)}=1\,.\label{Gamma 1}$$

Note that $\Gamma\left(n+\frac{1}{2}\right)$ can be written in terms of $\Gamma\left(2n\right)$ and $\Gamma\left(n\right)$ by the Legendre duplication formula, $\Gamma\left(2n\right)=\frac{1}{\Gamma\left(\frac{1}{2}\right)}\,2^{2n-1}\,\Gamma\left(n\right)\Gamma\left(n+\frac{1}{2}\right)$, see Ref. [@Weisstein; @_; @Legendre] for an elementary short proof. Since $\Gamma\left(\frac{1}{2}\right)=\sqrt{\pi}$ and $\Gamma\left(n\right)=\left(n-1\right)!$ for $n\in\mathbb{N}$, one can then write:

$$\frac{\Gamma\left(n+1\right)}{\Gamma\left(n+\frac{1}{2}\right)}=\frac{1}{\sqrt{\pi}}\,\frac{2^{n}\,n\,\Gamma\left(n\right)}{\frac{2n\,\Gamma\left(2n\right)}{2^{n}\,n\,\Gamma\left(n\right)}}=\frac{1}{\sqrt{\pi}}\,\prod_{j=1}^{n}\,\frac{2j}{2j-1}\,.\label{first part}$$

Applying the same argument and writing $\Gamma\left(n+\frac{3}{2}\right)$ in terms of $\Gamma\left(2n+2\right)$ and $\Gamma\left(n\right)$ by the Legendre duplication formula, one obtains:

$$\frac{\Gamma\left(n+1\right)}{\Gamma\left(n+\frac{3}{2}\right)}=\frac{2}{\sqrt{\pi}}\,\frac{2^{n}\,n\,\Gamma\left(n\right)}{\frac{\Gamma\left(2n+2\right)}{2^{n}\,n\,\Gamma\left(n\right)}}=\frac{2}{\sqrt{\pi}}\,\prod_{j=1}^{n}\,\frac{2j}{2j+1}\,.\label{second part}$$

Inserting (\[first part\]) and (\[second part\]) into (\[Gamma 1\]) yields

$$\lim_{n\rightarrow\infty}\,\prod_{j=1}^{n}\,\frac{\left(2j\right)\,\left(2j\right)}{\left(2j-1\right)\,\left(2j+1\right)}=\frac{\pi}{2}\,,\label{derivation}$$

which is the Wallis formula (\[Wallis formula 1\]).

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