A book I'm reading claims the following:
Let $a > 0$, and $\{x_m\}_{m=1}^{2n} \subset [0,a)$.
Consider the function $$g(x) = \prod_{m=1}^{2n}\sin\left(\frac{\pi (x-x_m)}{a}\right)$$
Then $g(x)$ has a finite Fourier expansion of the form;
$$\sum_{r = -n}^{r = n}a_r\exp\left(\frac{2\pi rix}{a}\right)$$
I tried the calculate the Fourier coefficients for $|m| > |n|$, and from what I found they're not necessarily zero. Can you help me understand this?
Note that the $m$ in the claimed formula is not the $m$ in the givens! The stated formula reads $$g(x)=a_{-n}e^{-2n\pi i x/a}+a_{-n+1}e^{-2(n-1)\pi i x/a}+\ldots+a_{n-1}e^{2(n-1)\pi i x/a}+a_ne^{2n\pi i x/a}\ .\tag{1}$$ It follows that lower and upper bounds of the "degrees" are as expected. What is surprising however, is that all odd terms are missing in this expansion. This can be explained as follows: The formula $$2\sin(x-\alpha)\sin(x-\beta)=\cos(\beta-\alpha)-\cos(2x-\alpha-\beta)$$ shows that each product of two sines $\sin{\pi(x-x_m)\over a}$ can be written in the form $$p+q\cos{2\pi(x- x')\over a}\ ,$$ and the product of $n$ such terms can be written in the form $(1)$.