How to transform the product to sum?

Question

I just wonder that how to prove that $$ \prod_{m=1}^{n}\Big(x-2\cos\frac{m\pi}{n+1}\Big)=\sum_{k=0}^{[n/2]}(-1)^{k}\binom{n-k}{k}x^{n-2k}. $$ Similarly, how to transform the product $$ \prod_{m=1}^{n}\Big(x+2\cos\frac{2m\pi}{n}\Big)\overset{?}{=}\sum_{k}b_k x^k? $$

Answer 1

Let, $\displaystyle P(x) = \sum\limits_{k=0}^{[n/2]} (-1)^k\binom{n-k}{k}x^{n-2k}$

We show, $\displaystyle P(\omega + \omega^{-1}) = 0$ where, $\displaystyle\omega = \exp{\frac{2\pi i}{2n+2}}$ is a $(2n+2)^{th}$ roots of unity.

$\begin{align} P(\omega+ \omega^{-1}) &= \sum\limits_{k=0}^{[n/2]} (-1)^k\binom{n-k}{k}(\omega+ \omega^{-1})^{n-2k} \\& = \sum\limits_{k=0}^{[n/2]}\sum\limits_{l=0}^{n-2k} (-1)^k\binom{n-k}{k}\binom{n-2k}{l}\omega^{n-2k-2l} \end{align}$

If we collect the coefficients of $\omega^{n-2r}$, where, $r = k+l$ together,

$\displaystyle \begin{align} \sum\limits_{k=0}^{r}(-1)^k\binom{n-k}{k}\binom{n-2k}{r-k} = \sum\limits_{k=0}^{r}(-1)^k\binom{r}{k}\binom{n-k}{r} = 1\end{align} \tag{1}$

Where, the identity $(1)$ can be derived from the general form:

$\displaystyle \sum\limits_{k=0}^{r}(-1)^k\binom{r}{k}\binom{n-k}{s} = \begin{cases} 1 & \textrm{ if } r = s \\ 0 &\textrm{ otherwise } \end{cases}$

which can be proved by induction or otherwise.

Thus, $\displaystyle P(\omega+\omega^{-1}) = \sum\limits_{r=0}^{n} \omega^{n-2r} = 0$

Hence, $\displaystyle P(x) = \prod\limits_{r=1}^{n} \left(x - \omega^{r} - \omega^{-r}\right) = \prod\limits_{r=1}^{n} \left(x - 2\cos \frac{\pi r}{n+1}\right)$

As @Roger209 points out in the comments, $\displaystyle P(x) = U_n\left(\frac{x}{2}\right)$, where $U_n$ in the Chebyshev Polynomial of $2^{nd}$ Kind. So, $\displaystyle P(2\cos \tau) = \frac{\sin (n+1)\tau}{\sin \tau}$, which has roots at $\displaystyle \tau = \dfrac{m\pi}{n+1}$, for $m = 1(n)n$.

Second Part:

We take the second polynomial: $\displaystyle Q(x) = \prod\limits_{m=1}^{n}\left(x + 2\cos \frac{2m\pi}{n}\right)$

Keeping in mind that the Chebyshev Polynomial of $1^{st}$ Kind $T_n$ satisfies, $$T_n(\cos \theta) = \cos n\theta$$

Thus, $\displaystyle T_n\left(\cos \frac{2m\pi}{n}\right) = \cos 2 m\pi = 1$, for $m = 1,2,\cdots,n$.

Thus, $\displaystyle T_n(x) - 1 = - 1 + \frac{n}{2}\sum\limits_{r=0}^{[n/2]}\frac{(-1)^r}{n-r}\binom{n-r}{r}(2x)^{n-2r} = 2^{n-1}\prod\limits_{m=1}^{n}\left(x - \cos \frac{2m\pi}{n}\right)$

I.e., $\displaystyle Q(x) = (-1)^n2\left(T_n\left(-\frac{x}{2}\right) - 1\right)$.