Wallis product: Definition from Answers.com

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

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

1 Proof
2 Relation to Stirling's approximation
3 Derivation of the Euler product for the sine
4 Finding Zeta(2)
5 Finding Zeta(4)
6 External links

Proof

Wallis derived his product as it is done in calculus books today, by comparing $\int_0^\pi \sin^nxdx$ for even and odd values of n, and noting that for large n, increasing n by 1 makes little change. Since infinitesimal calculus as we know it did not yet exist then, and mathematical analysis sufficient to discuss the convergence issue was inadequate, this was a harder piece of research than it sounds with hindsight, and more tentative. Wallis's product is, in retrospect, an easy corollary of the later Euler formula for the sine function:

$\frac{\sin(x)}{x} = \left(1 - \frac{x^2}{\pi^2}\right)\left(1 - \frac{x^2}{4\pi^2}\right)\left(1 - \frac{x^2}{9\pi^2}\right) \cdots = \prod_{n = 1}^\infty\left(1 - \frac{x^2}{n^2\pi^2}\right),$

in which we put x = π/2:

$\frac{2}{\pi} = \left(1 - \frac{1}{2^2}\right)\left(1 - \frac{1}{4^2}\right)\left(1 - \frac{1}{6^2}\right) \cdots = \prod_{n=1}^{\infty} \left(1 - \frac{1}{4n^2}\right),$

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

Q.E.D.

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:

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

p_k can be written as

$p_k ={1\over{2k+1}}\prod_{n=1}^{k} \frac{(2n)^4 }{((2n)(2n-1))^2}={1\over{2k+1}}\cdot {{2^{4k}\,(k!)^4}\over {((2k)!)^2}} \ .$

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 → ∞.

Derivation of the Euler product for the sine

If $α$ is a root of a polynomial then $(1 - x / α)$ is a factor. This leads to the conjecture that sin x might be of the form:

$\sin x = Ax(1 - x/\pi)(1 + x/\pi)(1 - x/2\pi)(1 + x/2\pi)\cdots = Ax(1-x^2/\pi^2)(1 - x^2/4\pi^2)(1 - x^2/9\pi^2)\cdots\ (1)$

If this formula is true then A must be 1 because if x is small, the left side is approximately x and the right side is approximately Ax.

Of course, this does not prove (1). Euler proved it in vol. 1 of his Introduction using De Moivre's formula

$\sin\ x = ((\cos x/n + i \sin x/n)^n - (\cos x/n -i \sin x/n)^n)/2i.$

He factored the difference of two n-th powers into linear factors using n-th roots of unity. He then used that for large n, cos x/n can be replaced by 1 and sin x/n by x/n. Later it was noted that Euler's use of complex numbers can be avoided as follows. For odd n, sin x is a polynomial of degree n in sin x/n and the roots of the polynomial are the numbers $\sin\ k\pi/n$ where k is any integer. Factoring the polynomial, making n go to infinity and replacing sin y by y for small y one gets (1).

Finding Zeta(2)

We can equate the above product for the sin x to the Taylor series for same:

$x\left(1 - \frac{x^2}{\pi^2}\right)\left(1 - \frac{x^2}{4\pi^2}\right)\left(1 - \frac{x^2}{9\pi^2}\right) \cdots = x - \frac{1}{3!}x^3 + \frac{1}{5!}x^5 - \cdots$ Equation 2

The next step is to equate the coefficients of x³ on both sides. The coefficient on the left hand side is the coefficient of x² in the infinite product (as the whole thing is multiplied by x). Terms in x² are produced when the '1' is chosen from all but one bracket and the x² term is chosen from the remainder. So the x³ term on the left hand side is:

$x\left(- \frac{x^2}{\pi^2} - \frac{x^2}{4\pi^2}-\frac{x^2}{9\pi^2} + \cdots\right)= -\frac{1}{\pi^2}\left(1+\frac{1}{4}+\frac{1}{9}+\cdots\right)x^3 =\left(-\frac{1}{\pi^2}\sum_{n=1}^\infty \frac{1}{n^2}\right)x^3.$

Equating the coefficients of x³ on both sides of equation 2 now gives

$-\frac{1}{\pi^2}\sum_{n=1}^\infty \frac{1}{n^2}=-\frac{1}{6}.$

Finally, multiplying through by $- π 2$ gives the value of $ζ(2)$ (see Riemann zeta function):

$\zeta(2) = \sum_{n=1}^\infty \frac{1}{n^2} = \frac{\pi^2}{6}.$

Finding Zeta(4)

Zeta(4) is defined as

$\sum_{n=1}^{\infty}\frac{1}{n^4}$ .

To derive a value in terms of $π$ for this sum, we start out with Equation 2, but focus on products of neighboring factors, such as $\left(1 - \frac{x^2}{\pi^2}\right) \left(1 - \frac{x^2}{4\pi^2}\right)$ . These products supply us with terms that are fourth degree in x, and when multiplied by x, will give us fifth degree terms that can be summed up and equated to $\frac{x^5}{5!}$ . We restate Equation 2:

$x\left(1 - \frac{x^2}{\pi^2}\right)\left(1 - \frac{x^2}{4\pi^2}\right)\left(1 - \frac{x^2}{9\pi^2}\right) \cdots = x - \frac{1}{6}x^3 + \frac{1}{120}x^5 - \cdots$ Equation 2

For the expression to the right of the x, the following observations need to be made:

Every $\frac{x^2}{n^2\pi^2}$ gets multiplied by every other such term to form an infinite sum of $\frac{x^4}{m^2n^2\pi^2}$ terms.

Identical terms are not multiplied by each other, e.g. $\frac{-x^2}{4\pi^2}$ is not multiplied by itself.

These are the only ways that fourth degree terms that will appear.

Using the logic of the previous section, the sum of all these products, when multiplied by the x must equal $\frac{x^5}{5!}$ :

$x \Bigl[ \left(\frac{-x^2}{\pi^2}\right)\left(\frac{-x^2}{4\pi^2}\right) + \left(\frac{-x^2}{\pi^2}\right)\left(\frac{-x^2}{9\pi^2}\right) + \left(\frac{-x^2}{\pi^2}\right)\left(\frac{-x^2}{16\pi^2}\right) + \cdots +$

$\left(\frac{-x^2}{4\pi^2}\right)\left(\frac{-x^2}{9\pi^2}\right)+ \left(\frac{-x^2}{4\pi^2}\right)\left(\frac{-x^2}{16\pi^2}\right)+ \left(\frac{-x^2}{4\pi^2}\right)\left(\frac{-x^2}{25\pi^2}\right)+\cdots +$

$\left(\frac{-x^2}{9\pi^2}\right)\left(\frac{-x^2}{16\pi^2}\right)+ \left(\frac{-x^2}{9\pi^2}\right)\left(\frac{-x^2}{25\pi^2}\right)+ \left(\frac{-x^2}{9\pi^2}\right)\left(\frac{-x^2}{36\pi^2}\right)+\cdots +$

$\cdots$

$\cdots \Bigr] = \frac{x^5}{5!}$

Next, we divide both sides by $x 5$ , and multiply the same by $π 4$ :

$\left(\frac{1}{1}\right)\left(\frac{1}{4}\right)+ \left(\frac{1}{1}\right)\left(\frac{1}{9}\right)+ \left(\frac{1}{1}\right)\left(\frac{1}{16}\right)+ \cdots +$

$\left(\frac{1}{4}\right)\left(\frac{1}{9}\right)+ \left(\frac{1}{4}\right)\left(\frac{1}{16}\right)+ \left(\frac{1}{4}\right)\left(\frac{1}{25}\right)+\cdots +$

$\left(\frac{1}{9}\right)\left(\frac{1}{16}\right)+ \left(\frac{1}{9}\right)\left(\frac{1}{25}\right)+ \left(\frac{1}{9}\right)\left(\frac{1}{36}\right)+\cdots +$

$\cdots$

$\cdots = \frac{\pi^4}{5!}$ ....................................................................Equation 3

More succinctly,

$\sum_{m<n}^{\infty}\frac{1}{m^2n^2} = \frac{\pi^4}{5!} = \frac{\pi^4}{120}$ .

The $m < n$ operator indicates that every mth term has been multiplied by every nth term that occurs to the right of it in Equation 3. For example, we find 1 multiplied by 1/4 then by 1/9, then by 1/16, etc.

This is a very compact expression, but what we want is

$\sum_{n=1}^{\infty}\frac{1}{n^4}.$

We would like to derive it this way:

$\sum_{m=1}^{\infty}\frac{1}{m^2}\sum_{n=1}^{\infty}\frac{1}{n^2}$ ,

so that we can exploit what we obtained in the last section.

The problem with this is that it's a little like trying to find $a 2 + b 2$ , by calculating $(a + b) 2$ . An error factor of 2ab would have to be removed from the expansion of the latter. Think of 2ab as a heterogeneous product, because a and b are different elements.

When we specify the summation from n = 0 to infinity of $\frac{1}{n^4}$ we are just asking for homogeneous products such as

$\frac{1}{2^2} \cdot \frac{1}{2^2} = \frac{1}{2^4}$

and

$\frac{1}{3^2} \cdot \frac{1}{3^2} = \frac{1}{3^4}$ .

Imagine the way that m and n interact when both track from 0 to infinity. Along the way they generate products like $\frac{1}{2^2} \cdot \frac{1}{2^2}$ as well as $\frac{1}{2^2} \cdot \frac{1}{3^2}$ .

What, in aggregate, are the heterogeneous products that need to be removed? Twice all of Equation 3's left side, which contains all the possible coefficients of fifth degree heterogeneous products. Why twice? Because commutes exist in the product of the two summations. For every 1/4 times 1/9 there exists a 1/9 times a 1/4. The m < n index excluded commutes, as did the mass multiplication that it notates. This means that twice $\frac{\pi^4}{120}$ has to be removed — twice the succint opposite side of Equation 3:

$\sum_{n=1}^{\infty}\frac{1}{n^4} = \sum_{m=1}^{\infty}\frac{1}{m^2}\sum_{n=1}^{\infty}\frac{1}{n^2} - 2 \left(\frac{\pi^4}{120}\right)$

The two summations on the right are each equal to $\frac{\pi^2}{6}$ , as was shown in the previous section. We write:

$\sum_{n=1}^{\infty}\frac{1}{n^4} = \left(\frac{\pi^2}{6}\right) \left(\frac{\pi^2}{6}\right) - 2\left(\frac{\pi^4}{120}\right) = \frac{\pi^4}{90}.$

External links

PlanetMath page on complex analysis, including a proof of the infinite product

This entry is from Wikipedia, the leading user-contributed encyclopedia. It may not have been reviewed by professional editors (see full disclaimer)

Donate to Wikimedia

	List of topics related to π
	Pietro Mengoli
	List of formulae involving π

	Does Johnson wax make a product called Big Wally and you would like to know if it is still sold? Read answer...
	Is there such thing as the wally wally tree? Read answer...
	Where is wally in 'Wally in Hollywood'- Wally the musical? Read answer...

	Why is joanne a wally?
	Who are laren and wally?
	Were the f is wally?

		Sci-Tech Dictionary. McGraw-Hill Dictionary of Scientific and Technical Terms. Copyright © 2003, 1994, 1989, 1984, 1978, 1976, 1974 by McGraw-Hill Companies, Inc. All rights reserved. Read more
		Wikipedia. This article is licensed under the Creative Commons Attribution/Share-Alike License. It uses material from the Wikipedia article "Wallis product". Read more

Wallis product

Contents