Re: Simplify infinite series with Sines???

On Nov 19, 3:01 pm, r...@xxxxxxxxxxxxxx (Rob Johnson) wrote:
In article <cbc3ac47-c875-4794-84e5-42de3f192...@xxxxxxxxxxxxxxxxxxxxxxxxxxx>,



TechnoBuddhist <goo...@xxxxxxxxxx> wrote:

I can't say that makes sense to me. I can understand why you might
want to know if there is a simpler form for the original expression,
but if you're asking for something else then I don't understand what
the question is.

For this example, the resultant waveform is periodic but it's
obviously not sinusoidal.

Thanks for the help guys, I really appreciate the time you've spent on
this, given my mathematical ineptitude.

Why would you need to understand why I want check for volume or quiet
every whole second(integer on x-axis) and not be bothered about the
volume at every fraction of a second(real value)? I could easily use a
sawtooth wave and it would tell me the same thing. I could actually
use a square wave as well and I'd get the same result too. In fact it
it makes it easier to simplify; then using a saw/square wave for k=2
to n then I'm happy with that too.

I really thought it was a simple question; is it possible to simplify
the expression;

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

Imagine if you have n=20; you have 20 terms in the expression. if
n=500; you have an expression with 500 terms, when you start to get to
n=1000, n=2000, you get my point. The expression becomes unwieldy.
Therefore my question was can it be simplified.

Can sin(x pi/2) * sin(x pi/3) be simplified?
Can sin(x pi/2) * sin(x pi/3) * sin(x pi/4) be simplified?

Gerry;
I appreciate that if any single term is zero then the whole product
will return zero(even my maths grasps that concept), but when you are
dealing with 100's or 1000's of terms this is unwieldy as you have to
evaluate each term in turn to test for zero.

I apologise for any frustrations I may be causing.

Using the trig identities sin(a) sin(b) = 1/2 (cos(a-b) - cos(a+b))
and cos(a) = 1 - 2 sin^2(a), we get the identity

2 a+b 2 a-b
sin(a) sin(b) = sin ( --- ) - sin ( --- )
2 2

Thus, sin(x pi/2) sin(x pi/3) = sin^2(x 5pi/12) - sin^2(x pi/12).

Additionally using cos(a) sin(b) = 1/2 (sin(a+b) - sin(a-b)), we get

sin(a) sin(b) sin(c) = 1/2 (cos(a-b) - cos(a+b)) sin(c)

= 1/4 (sin(a-b+c) - sin(a-b-c) - sin(a+b+c) + sin(a+b-c))

= 1/4 (sin(s-2a) + sin(s-2b) + sin(s-2c) - sin(s))

where s = a + b + c. The last form is symmetric in a, b, and c.

Thus, we get (using s = x 13pi/12),

sin(x pi/2) sin(x pi/3) sin(x pi/4)

= 1/4 (sin(x pi/12) + sin(x 5pi/12) + sin(x 7pi/12) - sin(x 13pi/12))

I don't know if you would consider these simplified, but identities
like these can probably be found for any n.

Let m = LCM(2, ..., n) and for each j in {1, ..., n}
write m_j = m/j, which is a positive integer.
Let z = exp(i x pi / m) where i = sqrt(-1). Then

sin(x pi / 2) sin(x pi / 3) ... sin(x pi / (n-1)) sin(x pi / n)

is equal to

(2i)^{1-n} (z^(m_2) - z^(-m_2)) (z^(m_3) - z^(-m_3)) ... (z^(m_n) -
z^(-m_n))

which is clearly a Laurent polynomial P in z. When
one substitutes z by z^{-1}, P is multiplied
by (-1)^(n-1). It follows immediately from
this that it is equal to a sum. This means that
the coefficient a(i) of z^i in P is
(-1)^(n-1) a(-i). This means that either
P can be written as a sum of terms of
the form

a (z^k + z^{-1})

with integer a, or as a sum of terms of
the form

a (z^k - z^{-1})

with, again, integer a. Remembering that
z = exp(pi x i / m_j), one sees at once
that, depending on the parity of n, we
can write P as a sum of cosines or of sines
of angles which look like

pi x r / m

with r the integers which can be written as

r = e_2 m_2 + e_3 m_3 + ... + e_n m_n

with e_2, ..., e_n either +1 or -1.
Moreover, the coefficients of the sines/cosines
are more or less (and upto the factor (2i)^{1-n})
count the number of such expressions for r.

-- m



.

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