Re: Simplify infinite series with Sines???

On Mon, 17 Nov 2008 12:47:26 -0800 (PST), TechnoBuddhist
<google@xxxxxxxxxx> wrote:

On Nov 17, 8:39 pm, Cary <c...@xxxxxxxxxxxxxx> wrote:
On Mon, 17 Nov 2008 12:01:47 -0800 (PST), TechnoBuddhist

<goo...@xxxxxxxxxx> wrote:
Yeah I found the following site;
http://www.hostsrv.com/webmab/app1/MSP/quickmath/02/pageGenerate?site...

and put in my formula and it gave me the following;

  sin((x*pi)/2) * sin((x*pi)/3) = (1/6)sin^2(x pi)

I double checked(AGAIN) and found that it didn't give the same result
(possible typo on my behalf) and it certainly doesn't give me the same
graph!! I removed the posting, but obviously not before some of you
guys read it!

I'm still stuck!

OK, so is the following the expression you're asking about?

   Product[ sin(x pi/k) ] for k = 2 to n  (x a real variable)

For n = 3, this is
   sin(x pi/2) * sin(x pi/3)

For n = 5, this is
   sin(x pi/2) * sin(x pi/3) * sin(x pi/4) * sin(x pi/5)

absolutely, you got it, but I'm not bothered about Reals, +ve Integers
only interest me.

So x takes on only integer values? Then for some values of x the
product obviously equals 0. How about giving us an some "typical"
value for x and n that you're actually working with?

.

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