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        riddles >> putnam exam (pure math) >> Sine Product formula
        
(Message started by: Icarus on Sep 4^th, 2003, 6:12pm)

Title: Sine Product formula
Post by Icarus on Sep 4^th, 2003, 6:12pm

While perusing my copy of CRC standard Mathematical Tables (a holdover from my college days) for trigonometric formulas, I came across a formula that I had penciled in at some time. I no longer remember where this formula came from, but it is interesting, and it works at least for [smiley=n.gif] [le] 4:

[smiley=s.gif][smiley=i.gif][smiley=n.gif] [smiley=n.gif][theta] = 2[supn][supminus][sup1] [prod]_{[smiley=subk.gif]=0}^{[supn][supminus][sup1]} [smiley=s.gif][smiley=i.gif][smiley=n.gif] ([theta] + [smiley=k.gif][pi]/[smiley=n.gif])

Can anyone prove it?
Is there a corresponding formula for cosine?

Title: Re: Sine Product formula
Post by Barukh on Sep 5^th, 2003, 7:50am

Icarus, somehow it happens that I know the answers to some of your questions...

I've found this formula empirically a couple of years ago, and then found a derivation in an advanced textbook. Here's the sketch.

1. Start from deMoivre's formula: cos(n[phi]) + i*sin(n[phi]) = (cos[phi] + i*sin[phi])ⁿ. Use the binomial theorem to expand the right hand side, and get the following:

sin(n[phi]) = sin[phi] [sum]_{k <= n, k odd} ⁿC_k (-1)^j cos^n-k[phi] sin^k-1[phi]
where j = (k-1)/2.

2. Change the indexing in the above formula from k to j. Then, sin^k-1[phi] may be rewritten as (1-cos²[phi])^j, and it turns out that every term in the sum is a polynomial in cos[phi] of degree n-1. Therefore, sin(n[phi]) / sin[phi] is also a polynomial in cos[phi] of degree n-1 with the leading coefficient
[sum]_{k odd}ⁿC_k = 2^n-1, and we can write:

sin(n[phi]) / sin[phi] = 2^n-1 [prod]_k=1^n-1 (cos[phi] - cos[phi]_k)

3. [phi]_k are exactly the angles where sin(n[phi]_k) = 0, so [phi]_k = k[pi]/n. Plug this into the product, and group together terms k and n-k. Then, use the formula for a cosine of a double angle together with the identities: cos([pi]-x) = -cos(x), and cos(2x) - cos(2y) = 2sin(x+y)sin(y-x) to arrive at the desired result.

Hope this helps.

Title: Re: Sine Product formula
Post by Icarus on Sep 5^th, 2003, 3:32pm

Well done! I was not entirely truthful in saying I have no idea where it came from. I'm pretty sure it came out of complex analysis. Now that you've pointed out that sin(n[theta]) is zero at k[pi]/n, I believe I have a better idea, if I ever bother to chase it down.

I vaguely recall writing it down because it was a slick consequence of some very powerful (and very nice) CA results.

But, as I'm sure you know, CA is the part of mathematics where practically everything works right.

While I don't believe your derivation is as slick as the CA one, it is nice indeed, and certainly more elementary.

I'm not sure, but I believe that the CA procedure only worked for sine, though I don't remember why.

Title: Re: Sine Product formula
Post by SWF on Sep 23^rd, 2003, 10:15pm

Finding the most simple derivation for this one is more tricky than it looks. I found a pretty good approach which I believe is more simple than Barukh's. Turns out there is a comparable expression involving tangents than does not have the complication of the powers of 2:

[prod]_{[smiley=subk.gif]=0}^{[supn][supminus][sup1]} [smiley=t.gif][smiley=a.gif][smiley=n.gif] ([theta] + [smiley=k.gif][pi]/[smiley=n.gif])
equals (-1)^n/2 when n is even and (-1)^(n-1)/2tan(n[theta]) when n is odd.

An expression containing cosines is

2[supn][supminus][sup1] [prod]_{[smiley=subk.gif]=0}^{[supn][supminus][sup1]} [smiley=c.gif][smiley=o.gif][smiley=s.gif] ([theta] + [smiley=k.gif][pi]/[smiley=n.gif])
equals (-1)^n/2*sin(n[theta]) when n is even and (-1)^(n-1)/2cos(n[theta]) when n is odd.

Title: Re: Sine Product formula
Post by Cster on May 18^th, 2004, 3:01pm

Here's a formula that popped up when looking at Legendre's relation on the Gamma function

n = 2^n-1[smiley=prod.gif]₁^n-1 sin(k*pi/n)

I have not been able to prove it. Anybody know this one?

Title: Re: Sine Product formula
Post by towr on May 18^th, 2004, 3:08pm

You could use the formula in the post above yours for cosine.
But I doubt your equation is true, which would make it difficult to prove..

Title: Re: Sine Product formula
Post by Icarus on May 18^th, 2004, 10:16pm

Take the formula I originally posted, and which Barukh has proved, divide both sides by [smiley=s.gif][smiley=i.gif][smiley=n.gif] [theta].

Then let [theta] [to] 0. The result is Cster's formula.

Title: Re: Sine Product formula
Post by towr on May 18^th, 2004, 11:41pm

ah.. I see I missed that one starts the product at k=0, and the other at k=1..

Title: Re: Sine Product formula
Post by CSter on May 19^th, 2004, 9:09am

Thanks, Icarus!
I could not see how to wrangle my product out of yours.
That's pretty cool.

Title: Re: Sine Product formula
Post by grimbal on May 23^rd, 2004, 3:52pm

on 05/18/04 at 15:01:19, Cster wrote:

Here's a formula that popped up when looking at Legendre's relation on the Gamma function

n = 2^n-1[smiley=prod.gif]₁^n-1 sin(k*pi/n)

I have not been able to prove it. Anybody know this one?

I don't remember the details, but I could prove that formula using the eigenvalues of the matrix describing the movement of a string (in finite elements). The matrix is tridiagonal, a_ij = -2 if i=j, 1 if i=j±1. The eigenvalues are 1/2*sin(k*pi/n). They are related to the harmonics of a swinging string. If you compute the characteristic polynomial, there is a n as next-to-last coefficient, which also is the product of the eigenvalues. There was some tweaking needed depending if n is odd or even, but it worked.

Sorry about the vagueness, but it is a long time ago.

Title: Re: Sine Product formula
Post by SWF on May 31^st, 2004, 6:01pm

Below, is the "simple" solution I mentioned back in September. Complex conjugate of z will be denoted by <z>, and the [prod] symbol means product for k=0 to n-1.

The 2n roots of z²ⁿ-1=0, will be denoted by [omega]^k, k=0,1,2,...,2n-1 and equal exp(i[pi]k/n).

A couple of results that follow from [omega]²ⁿ=1 and [omega]ⁿ=-1:
(1) [prod] [omega]^k+1 = [omega]^n(n+1)/2 = (-1)^(n+1)/2 = iⁿ⁺¹
(2) <[omega]^k> = [omega]^2n-k

Since the roots of zⁿ - <a>ⁿ = 0, are <a>*[omega]^2k, k=0,1,2,...,n-1,
zⁿ - <a>ⁿ = [prod] (z - <a>[omega]^2k)
Using (1), this equals -i/iⁿ * [prod] [omega]^n-k*(z - <a>[omega]^2k)
From (2) it becomes -i/iⁿ * [prod] (z*[omega]^n-k - <a*[omega]^n-k>).

Setting z to a=exp(i*[theta]), and strategically introducing 2's gives:
(aⁿ - <aⁿ>)/(2*i) = -2^n-1 [prod] (a*[omega]^n-k - <a*[omega]^n-k>)/(2*i)

The negative sign goes away when [omega]ⁿ (= -1) is factored from the k=0 term, and the result follows from
sin([phi]) = (exp(i*[phi])-<exp(i*[phi])>)/(2*i)