Wallis product

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

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

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

[edit] Proof

Starting with the Euler-Wallis formula for sine:

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

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.

[edit] 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 → ∞.

[edit] Derivation of the Euler-Wallis product for the sine

The thinking behind this infinite product is that there might be an equation for the sin x such as the following:

$\sin x = Ax(x - \pi)(x + \pi)(x - 2\pi)(x + 2\pi)\cdots$ (Equation 1)

This will be useful if we can find a value for A. We proceed as follows:

$\lim_{x \to 0}\frac{\sin x}{x} = \lim_{x \to 0}A(x - \pi)(x + \pi)(x - 2\pi)(x + 2\pi)\cdots$

$1 = \lim_{x \to 0}A(x - \pi)(x + \pi)(x - 2\pi)(x + 2\pi)\cdots$

$A = \lim_{x \to 0}\frac{1}{(x - \pi)(x + \pi)(x - 2\pi)(x + 2\pi)\cdots}$

$A = \frac{1}{(-\pi)(\pi)(-2\pi)(2\pi)\cdots}$

Thus, the terms in the denominator of this expression for A act as divisors for the corresponding terms in the product to the right of A in Equation 1. For example,

$\frac{(x - \pi)}{-\pi} = 1 - \frac{x}{\pi}$

Similarly,

$\frac{(x + \pi)}{\pi} = 1 + \frac{x}{\pi}$

Carrying this logic forward through all terms, we can write:

$\sin x = x(1 - \frac{x}{\pi})(1 + \frac{x}{\pi})(1 - \frac{x}{2\pi})(1 + \frac{x}{2\pi})\cdots$

The principle of the difference between two squares can now be employed with consecutive terms of the expression. For example,

$(1 -\frac{x}{\pi})(1 + \frac{x}{\pi}) = 1 - \frac{x^2}{\pi^2}$

And,

$(1 -\frac{x}{2\pi})(1 + \frac{x}{2\pi}) = 1 - \frac{x^2}{4\pi^2}$

Generalizing,

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

and

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

[edit] Finding Zeta(2)

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

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

The next step is very hard to visualize. To simplify, pretend that the above equation read like this:

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

Distributing the x,

$x - \frac{x^3}{\pi^2} = x - \frac{1}{6}x^3 + \frac{1}{5\!}x^5 - \cdots$

From this, we would conclude that the coefficients of x cubed are equal:

$-\frac{1}{\pi^2} = -\frac{1}{6}$

Of course, if the left-hand side of Equation 2 were fully expanded we would acquire an infinite number of terms of 3rd degree plus a lot more "garbage" terms. We need to see what pattern they fall into. The process is made somewhat easy by the proliferation of 1's in the expression.

The $1 - \frac{x^2}{\pi^2}$ will be preserved when it gets multiplied by the 1 in the next term. It also has to be multiplied by the $\frac{-x^2}{4\pi^2}$ . This will yield two more terms: $\frac{-x^2}{4\pi^2}$ and a term of 4th degree which doesn't interest us. The result:

$1 - \frac{x^2}{\pi^2} - \frac{x^2}{4\pi^2} + garbage$

If we multiply this by $1 - \frac{x^2}{9\pi^2}$ , the above expression will replicate itself, $\frac{-x^2}{9\pi^2}$ will be replicated and we will gather other terms of various other degrees. Summarizing our results so far,

$1 - \frac{x^2}{\pi^2} - \frac{x^2}{4\pi^2}-\frac{x^2}{9\pi^2} + garbage$

The trend will continue for ever. For instance, the next term would be $\frac{-x^2}{16\pi^2}$ , which is taken from the next factor to the right in Equation 2. In general we can write:

$1 - \frac{x^2}{\pi^2} - \frac{x^2}{4\pi^2}-\frac{x^2}{9\pi^2} + \cdots + garbage$

When we distribute the x, as Equation 2 requires, and disregard the 1 and the "garbage" terms, both of which are of too high or too low a degree to generate an x cubed, and equate this to the cubic term in the Taylor series we derive:

$\frac{-x^3}{\pi^2} - \frac{x^3}{4\pi^2}-\frac{x^3}{9\pi^2} + \cdots = \frac{-x^3}{6}$

Dividing both sides by x cubed:

$\frac{-1}{\pi^2} - \frac{1}{4\pi^2}-\frac{1}{9\pi^2} + \cdots = \frac{-1}{6}$

Now we multiply through by $- π 2$ :

$1 + \frac{1}{4}+\frac{1}{9} + \cdots = \frac{\pi^2}{6}$

Or,

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

This expression is commonly known as zeta(2). See Riemann zeta function.

[edit] Finding Zeta(4)

Zeta(4) is defined as

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 $(1 - \frac{x^2}{\pi^2}) (1 - \frac{x^2}{4\pi^2})$ . 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:

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

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 $\frac{x^5}{5!}$ :

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

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

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

$\cdots$

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

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

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

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

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

$\cdots$

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

More succintly,

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

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

$\sum_{n=0}^{\infty}\frac{1}{n^4}$ .

We would like to derive it this way:

$\sum_{m=0}^{\infty}\frac{1}{m^2}\sum_{n=0}^{\infty}\frac{1}{n^2}$ ,

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

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

and

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

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

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: