Wallis product

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

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

Proof

Starting with the Euler-Wallis formula for sine:

:$frac\{sin(x)\}\{x\}\; =\; left(1\; -\; frac\{x^2\}\{pi^2\}\; ight)left(1\; -\; frac\{x^2\}\{4pi^2\}\; ight)left(1\; -\; frac\{x^2\}\{9pi^2\}\; ight)\; cdots\; =\; prod\_\{n\; =\; 1\}^inftyleft(1\; -\; frac\{x^2\}\{n^2pi^2\}\; ight),$

we put "x" = π/2:

:$frac\{2\}\{pi\}\; =\; left(1\; -\; frac\{1\}\{2^2\}\; ight)left(1\; -\; frac\{1\}\{4^2\}\; ight)left(1\; -\; frac\{1\}\{6^2\}\; ight)\; cdots\; =\; prod\_\{n=1\}^\{infty\}\; left(1\; -\; frac\{1\}\{4n^2\}\; ight),$

:

Q.E.D.

Relation to Stirling's approximation

Stirling's approximation for "n"! asserts that:$n!\; =\; sqrt\; \{2pi\; n\}\; \{left(frac\{n\}\{e\}\; ight)\}^n\; left(\; 1\; +\; Oleft(frac\{1\}\{n\}\; ight)\; ight)$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\; =\{1over\{2k+1prod\_\{n=1\}^\{k\}\; frac\{(2n)^4\; \}\{((2n)(2n-1))^2\}=\{1over\{2k+1cdot$2^{4k},(k!)^4}over {((2k)!)^2 .

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

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:

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

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

:$lim\_\{x\; o\; 0\}frac\{sin\; x\}\{x\}\; =\; lim\_\{x\; o\; 0\}A(x\; -\; pi)(x\; +\; pi)(x\; -\; 2pi)(x\; +\; 2pi)cdots$

:$1\; =\; lim\_\{x\; o\; 0\}A(x\; -\; pi)(x\; +\; pi)(x\; -\; 2pi)(x\; +\; 2pi)cdots$

:$A\; =\; lim\_\{x\; o\; 0\}frac\{1\}\{(x\; -\; pi)(x\; +\; pi)(x\; -\; 2pi)(x\; +\; 2pi)cdots\}$

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

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:

:$sin\; x\; =\; xleft(1\; -\; frac\{x\}\{pi\}\; ight)\; left(1\; +\; frac\{x\}\{pi\}\; ight)\; left(1\; -\; frac\{x\}\{2pi\}\; ight)\; left(1\; +\; frac\{x\}\{2pi\}\; ight)cdots$

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

:$left(1\; -\; frac\{x\}\{pi\}\; ight)\; left(1\; +\; frac\{x\}\{pi\}\; ight)\; =\; 1\; -\; frac\{x^2\}\{pi^2\}$

And,

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

Generalizing,

:$sin(x)\; =\; xleft(1\; -\; frac\{x^2\}\{pi^2\}\; ight)left(1\; -\; frac\{x^2\}\{4pi^2\}\; ight)left(1\; -\; frac\{x^2\}\{9pi^2\}\; ight)\; cdots\; =x\; prod\_\{n\; =\; 1\}^inftyleft(1\; -\; frac\{x^2\}\{n^2pi^2\}\; ight),$

and

:$frac\{sin(x)\}\{x\}\; =\; left(1\; -\; frac\{x^2\}\{pi^2\}\; ight)left(1\; -\; frac\{x^2\}\{4pi^2\}\; ight)left(1\; -\; frac\{x^2\}\{9pi^2\}\; ight)\; cdots\; =\; prod\_\{n\; =\; 1\}^inftyleft(1\; -\; frac\{x^2\}\{n^2pi^2\}\; ight),$

Finding Zeta(2)

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

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

The next step is to equate the coefficients of "x"³ on both sides. The coefficient on the left hand side is the coefficient of "x"² in the infinite product (as the whole thing is multiplied by "x"). Terms in "x"² are produced when the '1' is chosen from all but one bracket and the "x"² term is chosen from the remainder. So the "x"³ term on the left hand side is:

:$xleft(-\; frac\{x^2\}\{pi^2\}\; -\; frac\{x^2\}\{4pi^2\}-frac\{x^2\}\{9pi^2\}\; +\; cdots\; ight)=\; -frac\{1\}\{pi^2\}left(1+frac\{1\}\{4\}+frac\{1\}\{9\}+cdots\; ight)x^3\; =left(-frac\{1\}\{pi^2\}sum\_\{n=1\}^infty\; frac\{1\}\{n^2\}\; ight)x^3.$

Equating the coefficients of "x"³ 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 $-pi^2$ gives the value of $zeta(2)$ (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 $pi$ for this sum, we start out with Equation 2, but focus on products of neighboring factors, such as $(1\; -\; frac\{x^2\}\{pi^2\})\; (1\; -\; frac\{x^2\}\{4pi^2\})$. 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:

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

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

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

* Identical terms are not multiplied by each other, e.g. $frac\{-x^2\}\{4pi^2\}$ 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 $frac\{x^5\}\{5!\}$:

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

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

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

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

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

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

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

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

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

More succinctly,

:

The

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 2"ab" would have to be removed from the expansion of the latter. Think of 2"ab" 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 &mdash; twice the succint opposite side of Equation 3:

:$sum\_\{n=0\}^\{infty\}frac\{1\}\{n^4\}\; =\; sum\_\{m=0\}^\{infty\}frac\{1\}\{m^2\}sum\_\{n=0\}^\{infty\}frac\{1\}\{n^2\}\; -\; 2\; left(frac\{pi^4\}\{120\}\; ight)$

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

:$sum\_\{n=0\}^\{infty\}frac\{1\}\{n^4\}\; =\; left(frac\{pi^2\}\{6\}\; ight)\; left(frac\{pi^2\}\{6\}\; ight)\; -\; 2left(frac\{pi^4\}\{120\}\; ight)\; =\; frac\{pi^4\}\{90\}.$

External links

* [http://planetmath.org/encyclopedia/WeierstrassProductTheorem.html PlanetMath page on complex analysis, including a proof of the infinite product]