5

Note that $i$ here refers to indexing variable, not $\sqrt{-1}$.

$$\prod_{i=1}^{n-1} \sin\left(\frac{i \pi}{n}\right) = 2^{1-n} n$$ This formula was used here to give an 'elementary' proof of product of diagonals = N. Mathworld is the only place I can find the formula and that website cites a personal communication with 'T. Drane'

I have tried to prove it on my own but this is the only progress I've made.

Letting $n=2k+1$, I can show that

$$\prod_{i=1}^{2k} \sin\left(\frac{i \pi}{2k+1}\right) = \prod_{i=1}^{k} \cos^2\left(\frac{(2i-1) \pi}{4k+2}\right)$$

I used the sine product formula $\quad\sin(a)\sin(b)=\dfrac{\cos(a-b)-\cos(a+b)}{2}\quad$ to pair of the first/last terms, etc. Then used the cosine double-angle formula $\quad \cos(2a)+1 = 2\cos^2(a)\quad$ on the resulting product.

I don't know where to go on from on here and I was just investigating the special case of odd $n$. Any ideas on what to do next? Simpler approach that would also apply to $n=2k$?

PS: It would be preferred if the proof avoided complex functions in order to complete the answer to the diagonals question reference earlier.

Community
  • 1
genepeer
  • 1,700
  • Both answers use complex functions. – genepeer Jul 02 '14 at 12:29
  • 1
    @amWhy how has this been answered?! I specifically asked for no complex functions. – genepeer Jul 02 '14 at 17:04
  • 2
    Have you looked at Chebyshev polynomials? They might help you. The idea I'm suggesting (for odd $n$) is to consider the polynomial $R_{n-1}(u)$ of degree $n-1$ defined by the equation $$\sin nx=\sin x R_{n-1}(\sin u).$$ For example, when $n=3$ we have the formula $$\sin 3x=3\sin x-4\sin^3x,$$ so $R_2(u)=3-4u^2$. Anyway, the idea is that the zeros of $R_{n-1}(u)$ are clearly $u_k=\sin(k\pi/n)$ for $k=1,2,\ldots,n-1$. Therefore, if $$R_{n-1}(u)=a_0+\cdots+a_{n-1}u^2,$$ we have $$u_1u_2\cdots u_{n-1}=(-1)^{n}a_0/a_{n-1}.$$ – Jyrki Lahtonen Jul 02 '14 at 18:12
  • (cont'd) Known facts about Chebyshev polynomials probably imply that $a_0=n$ and $a_{n-1}=-2^{n-1}$ giving your claim. I'm afraid I'm too rusty on Chebyshev stuff to dig all that out. The formulas about Chebyshev polynomials can be proven without resorting to complex exponential. But I don't see right away how to handle the even case... – Jyrki Lahtonen Jul 02 '14 at 18:17
  • I had never heard of them but I'm reading about them. It seems promising. Thanks for the suggestion. – genepeer Jul 02 '14 at 18:51
  • Typos above: $$\sin x=\sin x,R_{n-1}(\sin x)$$ and $$R_{n-1}(u)=a_0+\cdots+a_{n-1}u^{n-1}.$$ – Jyrki Lahtonen Jul 02 '14 at 18:55
  • 1
    Also, $u_1 u_2 \cdots u_{n-1} = (-1)^{n-1} a_0/a_{n-1}$. But I understood what you meant :) – genepeer Jul 02 '14 at 19:01
  • But $u_k = \sin(k\pi/n)$ for $k=1,2$ does not account for all roots of $R_2(u)$. I think $u_k$ for $k=-1,1$ does however. Assuming that the polynomials are always odd functions (they seem to be, haven't looked into it), then I can show that $$\prod_{k=1}^{n-1} \sin(k\pi/n) = a_0/|a_{n-1}|$$. – genepeer Jul 02 '14 at 20:33
  • My progress on the case for even $n$. Let $F(n) = \prod_{k=1}^{n-1} \sin(k\pi/n)$. I think I can show the following recurrences using induction: $$F(4m+2) = \frac{1}{2^{2m}} F(2m+1).$$ And $$F(4m) = \frac{1}{2^{2m-1}} F(2m)$$ with $F(4) = \frac{1}{2}.$ It's very late right now, so I'll attempt the induction proofs tomorrow. – genepeer Jul 03 '14 at 00:00
  • What I don't understand is why people would want to avoid complex numbers. I won't be surprised if all proofs can be rephrased in much simpler terms using complex numbers. – user21820 Jul 03 '14 at 10:10
  • I'm avoiding complex numbers 'cause that's been done before. In fact, from all that I've found, that's the only thing that's been done before. So is everyone using complex numbers just because they simplify it or because it is necessary? I don't know why but I think that's a question worth my time. – genepeer Jul 03 '14 at 10:30

1 Answers1

3

Thanks to Lahtonen's suggestion to use Chebyshev's polynomials, I was able to answer the question. I used this paper as a reference on the properties of the polynomials.

I'll prove $\frac{\sin((n+1)\theta)}{\sin(\theta)} = U_n(\cos(\theta))$ using induction (the paper used complex numbers to prove it).

$$ \frac{\sin(\theta)}{\sin(\theta)} = 1 = U_0(\cos(\theta)), \qquad \frac{\sin(2\theta)}{\sin(\theta)} = 2\cos(\theta) = U_1(\cos(\theta)), \\ {~}\\ {~}\\ \begin{aligned} \frac{\sin((n+1)\theta)}{\sin(\theta)} &= 2\cos(\theta)\frac{\sin(n\theta)}{\sin(\theta)} - \frac{\sin((n-1)\theta}{\sin(\theta)} \\ &= 2\cos(\theta)U_{n-1}(\cos(\theta))-U_{n-2}(\cos(\theta)) \\ &= U_n(\cos(\theta)) \end{aligned}$$

Using Identities 7/9 in the linked paper, it followed that $$\sin((2m+1)\theta) = (-1)^mT_{2m+1}(\sin(\theta))$$

Since $T_{2m+1}$ is an odd polynomial, I can factor it as such: $$\sin((2m+1)\theta) = \sin(\theta)R_{2m}(\sin(\theta))$$

where $R_{2m}$ is an even polynomial of degree $2m$ with the following properties:

  • unique roots are $u_k = \sin\left(\frac{k\pi}{2m+1}\right)$ for $k =\pm 1, \ldots, \pm m$.
  • constant term is $2m+1$
  • leading coefficient is $(-1)^m 2^{2m}$

Therefore, $$\prod_{k=1}^m u_k u_{-k} = (-1)^m \frac{2m+1}{2^{2m}}$$

By symmetry properties of the sine function, $$\prod_{k=1}^{2m} \sin\left(\frac{k\pi}{2m+1}\right) = \frac{2m+1}{2^{2m}} \qquad \square$$

This proves the identity for when $n$ is an odd integer. I'll introduce an abbreviation for simplicity's sake: let $F(n) = \prod_{k=1}^{n-1} \sin(k\pi/n)$. We have shown that $F(2m+1) = 2^{-2m}(2m+1)$ Also note that $$F(2m+1) = \prod_{k=1}^{m} \sin^2\left(\frac{k\pi}{2m+1}\right) = \prod_{k=1}^{m} \cos^2\left(\frac{(2m+1-2k)\pi}{4m+2}\right)$$ This will be useful soon. Now, let $n=2m$. $$F(2m) = \prod_{k=1}^{m-1} \sin^2\left(\frac{k\pi}{2m}\right) = \prod_{k=1}^{m-1} \cos^2\left(\frac{(m-k)\pi}{2m}\right) = \prod_{k=1}^{m-1} \cos^2\left(\frac{k\pi}{2m}\right)$$

I'll need to split this in two cases again: $n=4m$ and $n=4m+2$.

$$\begin{aligned} F(4m+2) &= \prod_{k=1}^{2m} \cos^2\left(\frac{k\pi}{4m+2}\right) \\ &= \prod_{k=1}^m \left(\cos\left(\frac{k\pi}{4m+2}\right)\cos^2\left(\frac{(2m+1-k)\pi}{4m+2}\right) \right)^2 \\ &= \prod_{k=1}^m \frac{1}{4} \cos^2 \left(\frac{(2m+1-2k)\pi}{4m+2}\right) \\ &= \frac{1}{2^{2m}}F(2m+1) \\ &= \frac{1}{2^{2m}}\cdot\frac{2m+1}{2^{2m}} &\square \end{aligned}$$

and finally, assume that $F(n)=\frac{n}{2^{n-1}}$ for all $n<4m$,

$$\begin{aligned} F(4m) &= \prod_{k=1}^{2m-1} \cos^2\left(\frac{k\pi}{4m}\right) \\ &= \frac{1}{2}\prod_{k=1}^{m-1} \left(\cos\left(\frac{k\pi}{4m}\right)\cos\left(\frac{(2m-k)\pi}{4m}\right) \right)^2 \\ &= \frac{1}{2}\prod_{k=1}^{m-1} \frac{1}{4}\cos^2\left(\frac{(m-k)\pi}{2m}\right) \\ &= \frac{1}{2^{2m-1}} F(2m) \\ &= \frac{1}{2^{2m-1}}\cdot \frac{2m}{2^{2m-1}} & \blacksquare \end{aligned}$$

genepeer
  • 1,700
  • It is becoming more and more apparent, that there is no elementary derivation of http://math.stackexchange.com/questions/835278/simple-prove-that-product-of-the-diagonals-of-a-polygon-n/852839#852839 – Bobby Ocean Jul 03 '14 at 20:22