Question

Consider a scaled sine function, $\sin(2\pi x/2^n)$, for some positive integer $n$. For this, I have the following linear combination.

$$ \sum_{x=1}^{2^{n-2}} c_x \sin(2\pi x/2^n).$$ (The upper limit to the sum is $2^{n-2}$.)

The question is whether there exist $c_x \in \{0, \pm 1, \pm 2\}$, not all $0$, that make the above expression $0$, for infinitely many $n$?

If it helps, the above came up in a computation concerning the discrete Fourier Transform.

Answer 1

No, such $c_x$ don't exist. Even if you replace $\lbrace 0,\pm 1,\pm 2\rbrace$ by $\mathbb Q$, this won't change. In fact, if they would exist, then, using the relation $\displaystyle \sin\frac{2\pi x}{2^n} = \frac{\zeta^x-\zeta^{-x}}{2i}$ (where $\zeta$ is a primitive $2^n$-th root of unity), the equation $\displaystyle \sum_{x=1}^{2^{n-2}}c_x\sin\frac{2\pi x}{2^n}=0$ would rewrite (after multiplication by $i\zeta^{2^{n-2}}$) as a polynomial equation (with rational coefficients!) for $\zeta$ of degree $\leq 2^{n-1}$. But the only (up to a scalar coefficient) polynomial equation of degree $\leq 2^{n-1}$ that $\zeta$ satisfies is $\zeta^{2^{n-1}}+1=0$ (since $X^{2^{n-1}}+1$ is the $2^n$-th cyclotomic polynomial, and the cyclotomic polynomials are known to be irreducible over $\mathbb Q$), and it is easy to see that this is not our equation (in fact, in our equation, $\zeta^{2^{n-1}}$ and $\zeta^0$ must occur with different signs, while in $\zeta^{2^{n-1}}+1=0$ they occur with the same sign).