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Infinite Products and the Basel Problem

April 5th, 2009 |

Author: Merrin

We know the Maclaurin series of sine but we do not know the infinite product representation yet.

$\sin x=x-\cfrac{x^{3} }{3!} +\cfrac{x^{5} }{5!} =x\left(1-\cfrac{x^{2} }{6} +\cfrac{x^{4} }{120} -...\right)$

We want to find an infinite product representation of sine. We know that sine has zeroes at

$0,\pm \pi ,\pm 2\pi ,...$

so the basic idea is to write an infinite polynomial as a product of terms which have the same roots as sine.

$\sin x=Cx(\pi -x)(\pi +x)(2\pi -x)(2\pi +x)...$

We pull out the factor of x.

$\cfrac{\sin x}{x} =C(\pi -x)(\pi +x)(2\pi -x)(2\pi +x)...\quad ????$

We can adjust the constant prefactor C such that

$\mathop{\lim }\limits_{x\to 0} \cfrac{\sin x}{x} =1$
$\cfrac{\sin x}{x} =\left(1-\cfrac{x^{2} }{\pi ^{2} } \right)\left(1-\cfrac{x^{2} }{4\pi ^{2} } \right)\left(1-\cfrac{x^{2} }{9\pi ^{2} } \right)...$

And here we have the infinite product representation for sine.

$\sin x=x\prod _{n=1}^{\infty }\left(1-\cfrac{x^{2} }{n^{2} \pi ^{2} } \right)$

This is not actually a proof of the product series because there could still be an overall function multiplying our expression such as exp(x) which has a limit of one at x = 0. It turns out that this arbitrary multiple is just unity.

Example 1. Now we evaluate a curious sum.

$\sum _{n=1}^{\infty }\cfrac{1}{n^{2} \pi ^{2}}$

Solution 1. Now we try to add up the infinite product. We multiply all
the factors together and sum up the first two terms which have powers of
1 and x squared.

$(1-a^{2} )(1-b^{2} )(1-c^{2} )...=1-(a^{2} +b^{2} +c^{2} +...)+...$

If we do this for our expression of sine and collect the coefficient of x cubed we find

$-x^{3} \sum _{n=1}^{\infty }\cfrac{1}{n^{2} \pi ^{2} }$

But we also know that

$\sin x=x-\cfrac{x^{3} }{3!} +...$

Equating the coefficients for x cubed we have

$\sum _{n=1}^{\infty }\cfrac{1}{n^{2} } =\cfrac{\pi ^{2} }{6}$

This problem is known as the Basel problem and Euler first solved it using this method.

Example 2. What about the infinite product representation of cosine ?
Solution 2. By using the same method that we used for sine we find that.

$\cos x=\prod _{n=0}^{\infty }\left(1-\cfrac{4x^{2} }{\pi ^{2} (2n+1)^{2} } \right)$

A less sought after sum involving only the squares of positive odd integers can be found from the infinite product of cosine in the same way.

${\cfrac{x^{2} }{2} =\sum _{n=0}^{\infty }\cfrac{4x^{2} }{(2n+1)^{2} \pi ^{2} } } \\ {\cfrac{\pi ^{2} }{8} =\sum _{n=0}^{\infty }\cfrac{1}{(2n+1)^{2} } }$

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