4
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ Both answers use complex functions. $\endgroup$ – genepeer Jul 2 '14 at 12:29
  • 1
    $\begingroup$ @amWhy how has this been answered?! I specifically asked for no complex functions. $\endgroup$ – genepeer Jul 2 '14 at 17:04
  • 2
    $\begingroup$ 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 $$u_1u_2\cdots u_{n-1}=(-1)^{n}a_0/a_{n-1}.$$ $\endgroup$ – Jyrki Lahtonen Jul 2 '14 at 18:12
  • $\begingroup$ (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... $\endgroup$ – Jyrki Lahtonen Jul 2 '14 at 18:17
  • 1
    $\begingroup$ Also, $u_1 u_2 \cdots u_{n-1} = (-1)^{n-1} a_0/a_{n-1}$. But I understood what you meant :) $\endgroup$ – genepeer Jul 2 '14 at 19:01
2
$\begingroup$

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 $n<4m$,

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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