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MC III HW4 help

 
 
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Re: MC III HW4 help
WangDr. Kevin - 2022年07月16日 Saturday 14:06
 

2.35(a): I will give a full solution here.  2.35(b) is similar.

The question is: Find the product: \[ \prod_{k=1}^{n-1}\sin\frac{k\pi}{2n}. \] Solution: Let $\displaystyle\epsilon=\text{cis}\frac{2\pi}{2n}=\text{cis}\frac{\pi}{n}$ be the imaginary root of unity of degree $2n$. Then, \[ 1, \epsilon, \epsilon^2, \ldots, \epsilon^{n-1}, -1, \epsilon^{n+1}, \ldots,\epsilon^{2n-1} \] are all the $2n$ roots of unity of degree $2n$, and \[ z^{2n}-1 = (z-1)(z-\epsilon)(z-\epsilon^2)\cdots(z-\epsilon^{n-1})(z+1) \cdots(z-\epsilon^{2n-1}). \] For the root $\epsilon^k$ ($1\leq k\leq n-1$), match it with its conjugate, which is $\epsilon^{2n-k}$, and we get \[ \begin{array}{rcl} (z-\epsilon^k)(z-\epsilon^{2n-k}) &=&\displaystyle \left({z-\cos\frac{k\pi}{n}-i\sin\frac{k\pi}{n}}\right) \left({z-\cos\frac{k\pi}{n}+i\sin\frac{k\pi}{n}}\right)\\ \\ &=& \displaystyle z^2 - 2z\cos\frac{k\pi}{n} + 1. \end{array} \] Hence in the following factorization, \[ \begin{array}{cl} & z^{2n}-1 \\ =& (z-1)(z-\epsilon)(z-\epsilon^2)\cdots(z-\epsilon^{n-1})(z+1) \cdots(z-\epsilon^{2n-1}) \\ \\ =&\displaystyle (z-1)(z+1)\left({(z-\epsilon)(z-\epsilon^{2n-1})}\right) \left({(z-\epsilon^2)(z-\epsilon^{2n-2})}\right)\cdots \left({(z-\epsilon^{n-1})(z-\epsilon^{n+1})}\right) \\ \\ =& \displaystyle(z^2-1)\left({z^2 - 2z\cos\frac{\pi}{n} + 1}\right) \left({z^2 - 2z\cos\frac{2\pi}{n} + 1}\right) \cdots\left({z^2 - 2z\cos\frac{(n-1)\pi}{n} + 1}\right) \\ \\ =& \displaystyle (z^2-1)\prod_{k=1}^{n-1}\left({z^2 - 2z\cos\frac{k\pi}{n} + 1}\right). \end{array} \] Therefore, \[ \frac{z^{2n}-1}{z^2-1} = \prod_{k=1}^{n-1}\left({z^2 - 2z\cos\frac{k\pi}{n} + 1}\right), \] so \[ z^{2n-2} + z^{2n-4} + \cdots + z^4 + z^2 + 1 = \prod_{k=1}^{n-1}\left({z^2 - 2z\cos\frac{k\pi}{n} + 1}\right). \] Let $z=1$ in the above identity, \[ n = \prod_{k=1}^{n-1}\left({2-2\cos\frac{k\pi}{n}}\right) = \prod_{k=1}^{n-1} 4\sin^2\frac{k\pi}{2n} = 2^{2n-2}\prod_{k=1}^{n-1}\sin^2\frac{k\pi}{2n}. \] Therefore \[ \prod_{k=1}^{n-1}\sin\frac{k\pi}{2n} = \frac{\sqrt{n}}{2^{n-1}}. \]
 
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