Wallis product

In mathematics, Wallis' product for π, written down in 1655 by John Wallis, states that

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

Comparison of the convergence of the Wallis product (purple asterisks) and several historical infinite series for π. S_n is the approximation after taking n terms. Each subsequent subplot magnifies the shaded area horizontally by 10 times. (click for detail)

Derivation

Wallis derived this infinite product as it is done in calculus books today, by comparing $\scriptstyle \int_0^\pi \sin^nx dx$ for even and odd values of n, and noting that for large n, increasing n by 1 results in a change that becomes ever smaller as n increases. Since infinitesimal calculus as we know it did not yet exist then, and the mathematical analysis of the time was inadequate to discuss the convergence issues, this was a hard piece of research, and tentative as well.

Wallis' product is, in retrospect, an easy corollary of the later Euler formula for the sine function. In 2015 researchers from the University of Rochester, in a surprise discovery, found the same formula in quantum mechanical calculations of the energy levels of a hydrogen atom.^{[1][2][3][4][5][6]}

Proof using Euler's infinite product for the sine function^[7]

$\frac{\sin x}{x} = \prod_{n=1}^\infty\left(1 - \frac{x^2}{n^2\pi^2}\right)$

Let x = ^π⁄₂:

$\begin{align} \Rightarrow\frac{2}{\pi} &= \prod_{n=1}^{\infty} \left(1 - \frac{1}{4n^2}\right) \\ \Rightarrow\frac{\pi}{2} &= \prod_{n=1}^{\infty} \left(\frac{4n^2}{4n^2 - 1}\right) \\ &= \prod_{n=1}^{\infty} \left(\frac{2n}{2n-1}\cdot\frac{2n}{2n+1}\right) = \frac{2}{1} \cdot \frac{2}{3} \cdot \frac{4}{3} \cdot \frac{4}{5} \cdot \frac{6}{5} \cdot \frac{6}{7} \cdots \end{align}$

Proof using integration^[8]

Let:

$I(n) = \int_0^\pi \sin^nxdx$

(a form of Wallis' integrals). Integrate by parts:

$\begin{align} u &= \sin^{n-1}x \\ \Rightarrow du &= (n-1) \sin^{n-2}x \cos x dx \\ dv &= \sin x dx \\ \Rightarrow v &= -\cos x \end{align}$

$\begin{align} \Rightarrow I(n) &= \int_0^\pi \sin^nxdx=\int_0^\pi u dv = uv |_{x=0}^{x=\pi}-\int_0^\pi v du \\ {} &= -\sin^{n-1}x\cos x |_{x=0}^{x=\pi} - \int_0^\pi - \cos x(n-1) \sin^{n-2}x \cos x dx \\ {} &= 0 - (n-1) \int_0^\pi -\cos^2x \sin^{n-2}x dx, n > 1 \\ {} &= (n - 1) \int_0^\pi (1-\sin^2 x) \sin^{n-2}x dx \\ {} &= (n - 1) \int_0^\pi \sin^{n-2}x dx - (n - 1) \int_0^\pi \sin^{n}x dx \\ {} &= (n - 1) I(n-2)-(n-1) I(n) \\ {} &= \frac{n-1}{n} I(n-2) \\ \Rightarrow \frac{I(n)}{I(n-2)} &= \frac{n-1}{n} \\ \Rightarrow \frac{I(2n-1)}{I(2n+1)} &=\frac{2n+1}{2n} \end{align}$

This result will be used below:

$\begin{align} I(0) &= \int_0^\pi dx = x|_0^\pi = \pi \\ I(1) &= \int_0^\pi \sin xdx = -\cos x|_0^\pi = (-\cos \pi)-(-\cos 0) = -(-1)-(-1) = 2 \\ I(2n) &= \int_0^\pi \sin^{2n}xdx = \frac{2n-1}{2n}I(2n-2) = \frac{2n-1}{2n} \cdot \frac{2n-3}{2n-2}I(2n-4) \end{align}$

Repeating the process,

$=\frac{2n-1}{2n} \cdot \frac{2n-3}{2n-2} \cdot \frac{2n-5}{2n-4} \cdot \cdots \cdot \frac{5}{6} \cdot \frac{3}{4} \cdot \frac{1}{2} I(0)=\pi \prod_{k=1}^n \frac{2k-1}{2k}$

$I(2n+1)=\int_0^\pi \sin^{2n+1}xdx=\frac{2n}{2n+1}I(2n-1)=\frac{2n}{2n+1} \cdot \frac{2n-2}{2n-1}I(2n-3)$

Repeating the process,

$=\frac{2n}{2n+1} \cdot \frac{2n-2}{2n-1} \cdot \frac{2n-4}{2n-3} \cdot \cdots \cdot \frac{6}{7} \cdot \frac{4}{5} \cdot \frac{2}{3} I(1)=2 \prod_{k=1}^n \frac{2k}{2k+1}$

$\sin^{2n+1}x \le \sin^{2n}x \le \sin^{2n-1}x, 0 \le x \le \pi$

$\Rightarrow I(2n+1) \le I(2n) \le I(2n-1)$

$\Rightarrow 1 \le \frac{I(2n)}{I(2n+1)} \le \frac{I(2n-1)}{I(2n+1)}=\frac{2n+1}{2n}$, from above results.

By the squeeze theorem,

$\Rightarrow \lim_{n\rightarrow\infty} \frac{I(2n)}{I(2n+1)}=1$

$\lim_{n\rightarrow\infty} \frac{I(2n)}{I(2n+1)}=\frac{\pi}{2} \lim_{n\rightarrow\infty} \prod_{k=1}^n \left(\frac{2k-1}{2k} \cdot \frac{2k+1}{2k}\right)=1$

$\Rightarrow \frac{\pi}{2}=\prod_{k=1}^\infty \left(\frac{2k}{2k-1} \cdot \frac{2k}{2k+1}\right)=\frac{2}{1} \cdot \frac{2}{3} \cdot \frac{4}{3} \cdot \frac{4}{5} \cdot \frac{6}{5} \cdot \frac{6}{7} \cdot \cdots$

Relation to Stirling's approximation

Stirling's approximation for n! asserts that

$n! = \sqrt {2\pi n} {\left(\frac{n}{e}\right)}^n \left[1 + O\left(\frac{1}{n}\right) \right]$

as n → ∞. Consider now the finite approximations to the Wallis product, obtained by taking the first k terms in the product:

$p_k = \prod_{n=1}^{k} \frac{2n}{2n - 1}\frac{2n}{2n + 1}$

p_k can be written as

$\begin{align} p_k &= {1 \over {2k + 1}} \prod_{n=1}^{k} \frac{(2n)^4}{[(2n)(2n - 1)]^2} \\ &= {1 \over {2k + 1}} \cdot {{2^{4k}\,(k!)^4} \over {[(2k)!]^2}} \end{align}$

Substituting Stirling's approximation in this expression (both for k! and (2k)!) one can deduce (after a short calculation) that p_k converges to ^π⁄₂ as k → ∞.

ζ'(0)^[7]

The Riemann zeta function and the Dirichlet eta function can be defined:

$\begin{align} \zeta(s) &= \sum_{n=1}^\infty \frac{1}{n^s}, \Re(s)>1 \\ \eta(s) &= (1-2^{1-s})\zeta(s) \\ &= \sum_{n=1}^\infty \frac{(-1)^{n-1}}{n^s}, \Re(s)>0 \end{align}$

Applying an Euler transform to the latter series, the following is obtained:

$\begin{align} \eta(s) &= \frac{1}{2}+\frac{1}{2} \sum_{n=1}^\infty (-1)^{n-1}\left[\frac{1}{n^s}-\frac{1}{(n+1)^s}\right], \Re(s)>-1 \\ \Rightarrow \eta'(s) &= (1-2^{1-s})\zeta'(s)+2^{1-s} (\ln 2) \zeta(s) \\ &= -\frac{1}{2} \sum_{n=1}^\infty (-1)^{n-1}\left[\frac{\ln n}{n^s}-\frac{\ln (n+1)}{(n+1)^s}\right], \Re(s)>-1 \end{align}$

$\begin{align} \Rightarrow \eta'(0) &= -\zeta'(0) - \ln 2 = -\frac{1}{2} \sum_{n=1}^\infty (-1)^{n-1}\left[\ln n-\ln (n+1)\right] \\ &= -\frac{1}{2} \sum_{n=1}^\infty (-1)^{n-1}\ln \frac{n}{n+1} \\ &= -\frac{1}{2} \left(\ln \frac{1}{2} - \ln \frac{2}{3} + \ln \frac{3}{4} - \ln \frac{4}{5} + \ln \frac{5}{6} - \cdots\right) \\ &= \frac{1}{2} \left(\ln \frac{2}{1} + \ln \frac{2}{3} + \ln \frac{4}{3} + \ln \frac{4}{5} + \ln \frac{6}{5} + \cdots\right) \\ &= \frac{1}{2} \ln\left(\frac{2}{1}\cdot\frac{2}{3}\cdot\frac{4}{3}\cdot\frac{4}{5}\cdot\cdots\right) = \frac{1}{2} \ln\frac{\pi}{2} \\ \Rightarrow \zeta'(0) &= -\frac{1}{2} \ln\left(2 \pi\right) \end{align}$

See also

• John Wallis - English mathematician who is given partial credit for the development of infinitesimal calculus and pi.
• Viète's formula, a different infinite product formula for π.
• Leibniz formula for π, an infinite sum that can be converted into an infinite Euler product for π.
• Wallis sieve
• C. R. Hagen - American professor of particle physics at the University of Rochester. Co-discoverer of Higgs Mechanism and Higgs Boson in 1964. In 2015 discovered the Wallis formula for π existed in physics while calculating energy levels of the Hydrogen atom.

Notes

1. T. Friedmann ; C.R. Hagen. "Quantum mechanical derivation of the Wallis formula for pi." Journal of Mathematical Physics Vol. 56 (2015)
2. "Quantum mechanical derivation of the Wallis formula for pi." Journal of Mathematical Physics Vol. 56 (arxiv)
3. "Discovery of classic pi formula a ‘cunning piece of magic'." University of Rochester (November 10, 2015
4. "New derivation of pi links quantum physics and pure math." American Institute of Physics (November 10, 2015)
5. "New derivation of pi links quantum physics and pure math." Phys.org (November 10, 2015)
6. "Revealing the hidden connection between pi and Bohr's hydrogen model." Physics World (November 17, 2015)
7. "Wallis Formula".
8. "Integrating Powers and Product of Sines and Cosines: Challenging Problems".

External links

• Hazewinkel, Michiel, ed. (2001), "Wallis formula", Encyclopedia of Mathematics, Springer, ISBN 978-1-55608-010-4