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 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. Sent via Deja.com http://www.deja.com/ Before you buy.