Re: Simplify infinite series with Sines???

In article
<e97bdea8-41af-4acc-b2c9-4c1c55251293@xxxxxxxxxxxxxxxxxxxxxxxxxxx>,
galathaea <galathaea@xxxxxxxxx> wrote:

On Nov 17, 12:01 pm, 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!

although it's still a bit unclear
if you are looking at finite or infinite products
(since you say infinite
but all your examples have been finite)
there is a simple plan of attack for these types of problems

just plug in the terms pairwise
into the the product rule for trigonometrics

sin(x) sin(y) = 1/2(cos(x + y) - cos(x - y))

and subsequently

cos(x) sin(y) = 1/2(sin(x + y) + sin(x - y))

folding these into the products for the finite case
should give you a final sum
that may be closer to what you want as a simplification

Depending, of course, on your definition of "simplification."
It seems to me that this will turn a product of, say, 10 terms
into a sum of 512 terms.

Is there any reason to think that there is an expression simpler
than the original product?

--
Gerry Myerson (gerry@xxxxxxxxxxxxxxx) (i -> u for email)
.

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