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Icarus
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Sine Product formula
« on: Sep 4th, 2003, 6:12pm » |
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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?
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Barukh
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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 nCk (-1)j cosn-k[phi] sink-1[phi]
where j = (k-1)/2.
2. Change the indexing in the above formula from k to j. Then, sink-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 oddnCk = 2n-1, and we can write:
sin(n[phi]) / sin[phi] = 2n-1 [prod]k=1n-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.
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Icarus
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Re: Sine Product formula
« Reply #2 on: Sep 5th, 2003, 3:32pm » |
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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.
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SWF
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Re: Sine Product formula
« Reply #3 on: Sep 23rd, 2003, 10:15pm » |
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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.
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Cster
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Here's a formula that popped up when looking at Legendre's relation on the Gamma function
n = 2n-1[smiley=prod.gif]1n-1 sin(k*pi/n)
I have not been able to prove it. Anybody know this one?
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towr
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Re: Sine Product formula
« Reply #5 on: May 18th, 2004, 3:08pm » |
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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..
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Icarus
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Re: Sine Product formula
« Reply #6 on: May 18th, 2004, 10:16pm » |
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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.
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towr
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Re: Sine Product formula
« Reply #7 on: May 18th, 2004, 11:41pm » |
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ah.. I see I missed that one starts the product at k=0, and the other at k=1..
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CSter
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Re: Sine Product formula
« Reply #8 on: May 19th, 2004, 9:09am » |
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Thanks, Icarus!
I could not see how to wrangle my product out of yours.
That's pretty cool.
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Grimbal
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Re: Sine Product formula
« Reply #9 on: May 23rd, 2004, 3:52pm » |
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on May 18th, 2004, 3:01pm, Cster wrote:Here's a formula that popped up when looking at Legendre's relation on the Gamma function
n = 2n-1[smiley=prod.gif]1n-1 sin(k*pi/n)
I have not been able to prove it. Anybody know this one? |
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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, aij = -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.
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SWF
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Re: Sine Product formula
« Reply #10 on: May 31st, 2004, 6:01pm » |
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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) = -2n-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)
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