From: wramey@home.removethis.com (Wade Ramey)
Subject: Re: Infinite Product representing Sine
Date: Sat, 02 Sep 2000 04:11:56 GMT
Newsgroups: sci.math
Summary: [missing]

In article <8opi4g$vab$1@nnrp1.deja.com>, Jose Capco <jcapco@my-deja.com> wrote:

> sinx = x Pi_{n=1}^{inf}(1-x^2/(pi^2*n^2))

That's a beautiful result. One way to get it is through complex analysis.
So in fact,

sin z = z Pi_{n=1}^{inf}(1-z^2/(pi^2*n^2))

for all complex z.

Now how do you obtain this rigorously? First, the right hand side (z times
the infinite product) defines an entire function whose zeros are precisely
those of sin z. Call the right hand side f(z). The function  (sin z)/f(z)
is then entire. If you could show this function is bounded, then
Liouville's Theorem would imply it is constant. And because (sin z)/f(z) =
1 when z = 0, we would have (sin z)/f(z) = 1 for all z, which is what you
want. Now how to get this function bounded ...... hmmmm, I'm stuck at the
moment.

Anyway, that's sort of the idea. I'm away from my math books, but there's
a whole theory of factoring entire functions of "slow" growth (like sin z)
which should polish this off. The name Weierstrass (him again) comes up in
this theory.

And of course there could be a more elementary way to get your formula.

Wade
==============================================================================

From: ullrich@math.okstate.edu (David C. Ullrich)
Subject: Re: Infinite Product representing Sine
Date: Sat, 02 Sep 2000 14:08:07 GMT
Newsgroups: sci.math

On Sat, 02 Sep 2000 04:11:56 GMT, wramey@home.removethis.com (Wade
Ramey) wrote:

>In article <8opi4g$vab$1@nnrp1.deja.com>, Jose Capco <jcapco@my-deja.com> wrote:
>
>> sinx = x Pi_{n=1}^{inf}(1-x^2/(pi^2*n^2))
>
>That's a beautiful result. One way to get it is through complex analysis.
>So in fact,
>
>sin z = z Pi_{n=1}^{inf}(1-z^2/(pi^2*n^2))
>
>for all complex z.
>
>Now how do you obtain this rigorously? First, the right hand side (z times
>the infinite product) defines an entire function whose zeros are precisely
>those of sin z. Call the right hand side f(z). The function  (sin z)/f(z)
>is then entire. If you could show this function is bounded, then
>Liouville's Theorem would imply it is constant. And because (sin z)/f(z) =
>1 when z = 0, we would have (sin z)/f(z) = 1 for all z, which is what you
>want. Now how to get this function bounded ...... hmmmm, I'm stuck at the
>moment.

	It's not too hard to show directly that f(z+2Pi) = f(z); now 
you just have to show sin(z)/f(z) is bounded in a vertical strip of
width 2Pi. And you can do this "directly": For example log(|f|)
is a sum, and you can estimate that sum by a certain integral
in the obvious way and get what you need, I think. Um, could
be that showing f(z)/sin(z) is bounded works better: Taking 
Pi=2Pi=1 because it's Saturday, log(|f(z)|) is less than the sum 
of log(1+|z|^2/n^2). Replace that with the integral from 1 to
infinity of log(1+|z|^2/t^2), (estimating the error and showing
it's small enough not to matter), then make a change of
variables in the integral to figure out how it behaves for
large z...
	I think the last time I actually did this I got
f(z)/sin(z) to have polynomial growth instead of
being bounded. Hence it's a polynomial of period
2Pi, hence constant.

>Anyway, that's sort of the idea. I'm away from my math books, but there's
>a whole theory of factoring entire functions of "slow" growth (like sin z)
>which should polish this off. The name Weierstrass (him again) comes up in
>this theory.
>
>And of course there could be a more elementary way to get your formula.
>
>Wade