From: artfldodgr@my-deja.com
Subject: Re: Infinite Product representing Sine
Date: Tue, 05 Sep 2000 20:44:19 GMT
Newsgroups: sci.math
Summary: Weierstrass's factorization theorem (of complex-analytic functions)

In article <8opi4g$vab$1@nnrp1.deja.com>,
  Jose Capco <jcapco@my-deja.com> wrote:
> Dear NG,
>
> Can someone prove to me how the infinite product representing the sine
> function is obtained and if possible who first obtained this expression:
>
> sinx = x Pi_{n=1}^{inf}(1-x^2/(pi^2*n^2))

This "factorization" is due to Leonhard Euler, and dates to ca. 1735.
It played a key role in Euler's solution of the "Basel problem", which
was to compute the sum of the reciprocals of the squares of the positive
integers.  The episode is discussed in the recent *EULER, The Master of
Us All* by Wm. Dunham.

It appears that Euler viewed the infinite product expansion of sin(x) as
the factorization of (the infinite degree polynomial!) sin(x) into
monomials, as indicated by its roots at x=0, and x=(+/-)k*pi for
k=1,2,....

Nowadays one may view the expansion as an instance of
Weierstrass's factorization theorem which (in the present situation) says
that

(*)  sin(z) = z*exp(g(z))*Pi_{n=1}^{infty} [1-(z^2)/(pi*n)^2]

where g is an entire function.  Because sin is of genus 1, g is a
polynomial of degree at most 1, say g(z) = a+bz.  Because sin(z)/z
converges to 1 as z goes to 0, we must have a = 0.  To evaluate b,
logarithmically differentiate in (*) and then multiplty through by z to
obtain

cot(z) = (1/z) + b - sum_{n=1}^{infty} [2z/(pi^2n^2-z^2)].

Now then send z to 0; the sum of the series on the right goes to 0 as
does cot(z)-(1/z) .  Thus b=0. QED

--
A.


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