Closed Form of Generalized Trigonometric Products

a)\text{a)} Find a closed form for the following Trigonometric Products

r=1n1sin(krπn) \bullet \prod_{r=1}^{n-1} \sin \left(k \cdot \dfrac{r\pi}{n} \right)

r=1n1cos(krπn) \bullet \prod_{r=1}^{n-1} \cos \left(k \cdot \dfrac{r\pi}{n}\right)

Here k>1k > 1 is a positive integer.

b) \text{b)} Generalize for all complex kk


This is a part of the set Formidable Series and Integrals.

Note by Ishan Singh
3 years, 5 months ago

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@Otto Bretscher @Mark Hennings What are your thoughts on this?

Ishan Singh - 3 years, 5 months ago

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I'm at work now; just a quick reply. Those products are not hard to do with complex numbers. The sine product, obviously, is 0 when gcd(k,n)1\gcd(k,n)\neq 1, and it is n2n1\frac{n}{2^{n-1}} in absolute value when gcd(n,k)=1\gcd(n,k)=1... the roots of unity get permuted in that case. For the cosine it gets just a bit messier but it is also very doable.

You can play with my solution here to get those results.

Otto Bretscher - 3 years, 5 months ago

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Okay. Sorry to disturb and thanks for the input. Btw, I had tried using roots of unity for cosine but was getting a contradiction (that was a year ago, I'll have to re-check my work now).

Ishan Singh - 3 years, 5 months ago

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@Ishan Singh No problem at all. I will write the solutions up in the evening unless somebody else does the work ;)

Otto Bretscher - 3 years, 5 months ago

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@Ishan Singh Try it yourself! If you use the complex expressions for sin and cosine, the computations are quite straightforward, and fun.

Otto Bretscher - 3 years, 5 months ago

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@Otto Bretscher I'll try it again, maybe I did some calculation mistake when I last tried it a year ago.

Ishan Singh - 3 years, 5 months ago

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@Ishan Singh We know that roots of unity work for the cosine when k=1k=1... look at my solution here. It works the exact same way when gcd(n,k)=1\gcd(n,k)=1. Now tackle the case gcd(n,k)1\gcd(n,k)\neq 1.

Otto Bretscher - 3 years, 5 months ago

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@Otto Bretscher Hmm. I found my flaw. It's pretty straight forward with roots of unity. I was earlier committing a calculation mistake and getting a contradiction when I tried using roots of unity. I also tried using Chebyshev Polynomials on the lines of my solution here but it got somewhat messy.

Ishan Singh - 3 years, 5 months ago

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@Ishan Singh I love Chebyshev polynomials, but here roots of unity seem like the way to go.

Are we done with this, or do you want to pursue it further?

Otto Bretscher - 3 years, 5 months ago

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@Otto Bretscher POST SOLUTION PLEASE!

Pi Han Goh - 3 years, 5 months ago

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@Pi Han Goh Finding the signs is a straightforward but tedious book keeping issue; let's just go for the absolute values for now since I have limited time.

In the sign product, the rr should go from 1 to n1n-1; the way it's written now the product is always 0 ;)

I will essentially be copying my solutions from here and here

If gcd(k,n)1\gcd(k,n)\neq 1 then the product is 0 as krkr will be a multiple of nn for some rr.

If gcd(k,n)=1\gcd(k,n)=1 then

r=1n12sin(krπ/n)=r=1n1eikrπ/neikrπ/n=r=1n11e2ikrπ/n=n\prod_{r=1}^{n-1}2|\sin(kr\pi/n)|=\prod_{r=1}^{n-1}|e^{ikr\pi/n}-e^{-ikr\pi/n}|=\prod_{r=1}^{n-1}|1-e^{2ikr\pi/n}|=n

so that r=1n1sin(krπ/n)=n2n1\prod_{r=1}^{n-1}|\sin(kr\pi/n)|=\frac{n}{2^{n-1}}

Likewise, if gcd(k,n)=1\gcd(k,n)=1 and nn is odd, then

r=1n2cos(krπ/n)=r=1neikrπ/n+eikrπ/n=r=1n1e2ikrπ/n=2\prod_{r=1}^{n}2|\cos(kr\pi/n)|=\prod_{r=1}^{n}|e^{ikr\pi/n}+e^{-ikr\pi/n}|=\prod_{r=1}^{n}|-1-e^{2ikr\pi/n}|=2

so that r=1ncos(krπ/n)=12n1\prod_{r=1}^{n}|\cos(kr\pi/n)|=\frac{1}{2^{n-1}}

If k=pqk=pq where p=gcd(k,n)p=\gcd(k,n), let n=mpn=mp. Then

r=1ncos(krπ/n)=r=1pmcos(qrπ/m)=(12m1)p=12ngcd(n,k)\prod_{r=1}^{n}|\cos(kr\pi/n)|=\prod_{r=1}^{pm}|\cos(qr\pi/m)|=\left(\frac{1}{2^{m-1}}\right)^p=\frac{1}{2^{n-\gcd(n,k)}}

Otto Bretscher - 3 years, 5 months ago

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@Otto Bretscher (+1) Nice solution! My solution is essentially the same as yours, except instead of using cyclotomic polynomials, if gcd(k,n)=1\gcd (k,n) =1 , note that r=1n1sin(rkπn)=(1)n1r=1n1sin(rπn)=(1)n1n2n1\displaystyle \prod_{r=1}^{n-1} \sin \left(\dfrac{rk \pi}{n} \right) = (-1)^{n-1} \prod_{r=1}^{n-1} \sin \left(\dfrac{r \pi}{n} \right) = (-1)^{n-1} \dfrac{n}{2^{n-1}} since the set rk(modn)rk \pmod {n} is exhaustive for 1rn11 \leq r \leq n-1. Similarly for cos\cos

Ishan Singh - 3 years, 5 months ago

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@Ishan Singh Yes, of course, I wasn't using the cyclotomic polynomials in this problem either. I was using them for the other problem because there we were restricting ourselves to the rr with gcd(r,n)=1\gcd(r,n)=1.

Otto Bretscher - 3 years, 5 months ago

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@Otto Bretscher @Otto Bretscher Sorry to disturb again, but when I inspected my earlier work, I found that I was trying to generalize the result for all complex kk. I tried to use transformation of roots on the nthn^{th} roots of unity, but to no avail. Can this be done?

Ishan Singh - 3 years, 5 months ago

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@Ishan Singh It's a nice sunny spring day here, and I don't feel like thinking about messy stuff like this right now. Maybe on a rainy day... ;)

Otto Bretscher - 3 years, 5 months ago

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@Otto Bretscher Yeah sure :) Thanks for the help.

Ishan Singh - 3 years, 5 months ago

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@Pi Han Goh I have solved it for both cases. I'll post solution by night.

Ishan Singh - 3 years, 5 months ago

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