Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Recently, I ran across a product that seems interesting.

Does anyone know how to get to the closed form:

$$\prod_{k=1}^{n}\cos\left(\frac{k\pi}{n}\right)=-\frac{\sin(\frac{n\pi}{2})}{2^{n-1}}$$

I tried using the identity $\cos(x)=\frac{\sin(2x)}{2\sin(x)}$ in order to make it "telescope" in some fashion, but to no avail. But, then again, I may very well have overlooked something.

This gives the correct solution if $n$ is odd, but of course evaluates to $0$ if $n$ is even.

So, I tried taking that into account, but must have approached it wrong.

How can this be shown? Thanks everyone.

share|improve this question
1  
It's finite, isn't it? –  draks ... Jul 20 '12 at 12:20
    
You should have $sin(n\pi/2)$. –  Mhenni Benghorbal Jul 20 '12 at 12:35
1  
In the argument of the $\sin$, instead of $k$, it should be $n$. –  Mhenni Benghorbal Jul 20 '12 at 12:38
    
Yes, I know. It is a careless typo. Sorry. –  Cody Jul 20 '12 at 13:52
2  
This is almost identical to the second part of this question –  robjohn Jul 20 '12 at 18:03

4 Answers 4

The roots of the polynomial $X^{2n}-1$ are $\omega_j:=\exp\left(\mathrm i\frac{2j\pi}{2n}\right)$, $0\leq j\leq 2n-1$. We can write \begin{align} X^{2n}-1&=(X^2-1)\prod_{j=1}^{n-1}\left(X-\exp\left(\mathrm i\frac{2j\pi}{2n}\right)\right)\left(X-\exp\left(-\mathrm i\frac{2j\pi}{2n}\right)\right)\\ &=(X^2-1)\prod_{j=1}^{n-1}\left(X^2-2\cos\left(\frac{j\pi}n\right)X+1\right). \end{align} Evaluating this at $X=i$, we get $$(-1)^n-1=(-2)(-2\mathrm i)^{n-1}\prod_{j=1}^{n-1}\cos\left(\frac{j\pi}n\right),$$ hence \begin{align} \prod_{j=1}^n\cos\left(\frac{j\pi}n\right)&=-\prod_{j=1}^{n-1}\cos\left(\frac{j\pi}n\right)\\ &=\frac{(-1)^n-1}{2(-2\mathrm i)^{n-1}}\\ &=\frac{(-1)^n-1}2\cdot \frac{\mathrm i^{n-1}}{2^{n-1}}. \end{align} The RHS is $0$ if $n$ is even, and $-\dfrac{(-1)^m}{2^{2m}}=-\dfrac{\sin(n\pi/2)}{2^{n-1}}$ if $n$ is odd with $n=2m+1$.

share|improve this answer
    
You lost a factor $-(-i)^n$ in the last step; your result is non-negative whereas the result in the question oscillates. –  joriki Jul 20 '12 at 12:35
1  
I think the problem is that the roots $-1/1$ don't come in complementary pairs. So it's really $X^{2n}-1 = (X^2-1)\prod_{i=1}^{n-1} ...$ –  Thomas Andrews Jul 20 '12 at 12:58
    
Also, you need a $i$ in the definition of $\omega_j$: $$\omega_j = exp(\frac{2\pi j}{2n} i)$$ –  Thomas Andrews Jul 20 '12 at 13:00
    
@ThomasAndrews There are indeed indeed couple of mistake. I will correct them. Thanks! –  Davide Giraudo Jul 20 '12 at 13:48
3  
Corrected some misprints, missing occurrences of $\mathrm i$, and wrong indexations, and added a one-sentence at the end. Hope the result suits you. –  Did Jul 20 '12 at 14:24

The idea of using $\cos x=\frac{1}{2}\frac{\sin 2x}{\sin x}$ is a nice one, and works quickly. We should not use this when $x=\frac{n\pi}{n}$,because of the $\frac{0}{0}$ issue. Also, as you observe, the product is $0$ if $n$ is even, so we needn't bother. So let $n$ be odd.

Look at the product from $k=1$ to $k=n-1$. As $k$ ranges over these values, the numbers $2k$ range, modulo $n$, over all numbers from $1$ to $n-1$. So the $\cos(2k\pi/n)$ (apart from sign) range in some order over the $\cos(\pi j/n)$. Thus, apart from sign, there is cancellation and the product has absolute value $\frac{1}{2^{n-1}}$.

There is no sign issue. If $n\equiv 1\pmod {4}$, the product from $1$ to $n-1$ has an even number of negative terms. The term $\cos(\pi n/n)$ then gives us an extra $-1$, and the product is negative. For the same reason, if $n\equiv 3\pmod{4}$ then the product is positive. The $-\sin(n\pi/2)$ term captures these sign facts, and also produces the right answer of $0$ when $n$ is even.

share|improve this answer
    
Thanks Andre. I was on the right path afterall. I noticed the 0/0 issue you mention but failed to see the forest for the trees. Thanks. I managed to see the $2^{n-1}$ from this method, but did not see how the $\sin(\pi n/2)$ was revealed. –  Cody Jul 20 '12 at 13:55
    
@Cody: Corrected a few typos. One would think such a short argument would be typo-free! As you can see, there is basically term by term cancellation if one looks at things right. –  André Nicolas Jul 20 '12 at 14:01
    
@AndréNicolas: now that I read your answer, I see that mine is just a paraphrase into $\prod$s. (+1) –  robjohn Jul 20 '12 at 16:55

The monic Chebyshev polynomial of the second kind,

$$\hat{U}_n(x)=\frac{\sin((n+1)\arccos\,x)}{2^n \sqrt{1-x^2}}$$

can be easily seen to have the roots $x_k=\cos\dfrac{\pi k}{n+1}$ for $k=1,\dots,n$. By Vieta, the constant term of $\hat{U}_n(x)$ is equal to

$$\hat{U}_n(0)=\prod_{k=1}^n \cos\dfrac{\pi k}{n+1}$$

and thus

$$\prod_{k=1}^n \cos\dfrac{\pi k}{n}=-\hat{U}_{n-1}(0)=-\frac{\sin\frac{n\pi}{2}}{2^{n-1}}$$

share|improve this answer
    
Thanks JM. I certainly would have not thought of this. I have heard of Chebyshev polynomials, but that's about it. I will have to look into them. –  Cody Jul 20 '12 at 14:15
    
Big guns work fast :-) (+1) –  robjohn Jul 20 '12 at 16:52
    
We both have seen this before. –  robjohn Jul 20 '12 at 20:34
    
@rob, oh dear, precisely the same argument, except I hadn't mentioned Vieta in the other answer... :) –  J. M. Jul 20 '12 at 22:03

If $n$ is even, then the term with $k=n/2$ makes the product on the left $0$ and $\sin\left(\frac{n}{2}\pi\right)=0$. So assume that $n$ is odd. $$ \begin{align} \prod_{k=1}^n\cos\left(\frac{k\pi}{n}\right) &=-\prod_{k=1}^{n-1}\cos\left(\frac{k\pi}{n}\right)\tag{1}\\ &=-\prod_{k=1}^{n-1}\frac{\sin\left(\frac{2k\pi}{n}\right)}{2\sin\left(\frac{k\pi}{n}\right)}\tag{2}\\ &=\frac{-1}{2^{n-1}}\frac{\prod\limits_{k=\frac{n+1}{2}}^{n-1}\sin\left(\frac{2k\pi}{n}\right)}{\prod\limits_{k=1}^{\frac{n-1}{2}}\sin\left(\frac{(2k-1)\pi}{n}\right)}\tag{3}\\ &=\frac{(-1)^{\frac{n+1}{2}}}{2^{n-1}}\frac{\prod\limits_{k=1}^{\frac{n-1}{2}}\sin\left(\frac{(2k-1)\pi}{n}\right)}{\prod\limits_{k=1}^{\frac{n-1}{2}}\sin\left(\frac{(2k-1)\pi}{n}\right)}\tag{4}\\ &=-\frac{\sin\left(n\frac\pi2\right)}{2^{n-1}}\tag{5} \end{align} $$ $(1)$: $\cos(\pi)=-1$

$(2)$: $\sin(2x)=2\sin(x)\cos(x)$

$(3)$: cancel $\sin\left(\frac{j\pi}{n}\right)$ in the numerator and denominator for even $j$ from $2$ to $n-1$

$(4)$: in the numerator, change variable $k\mapsto k+\frac{n-1}{2}$ and use $\sin(x+\pi)=-\sin(x)$

$(5)$: for odd $n$, $\sin\left(n\frac\pi2\right)=(-1)^{\frac{n-1}{2}}$

share|improve this answer
    
Thanks RobJohn. That is how I started it, but was too dumb to finish :(. Thanks a bunch. –  Cody Jul 20 '12 at 17:19
    
@Cody: It is not easy to see all the cancellations. I had to picture where the points on the circle were and note where the terms from the numerator and denominator overlapped to get step $(3)$. Good question! (+1) –  robjohn Jul 20 '12 at 17:55
    
Thanks RobJohn. I noticed in the link you provided this has been asked before. I always look before I post to see if the same question as been asked. But, I did not find this one. Often, it is difficult to locate something in particular. I thought this was a cool problem. Thanks for all the input and different methods. –  Cody Jul 21 '12 at 10:31

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.