Upper bound of $\sum_{j=0}^{n-1}\cos((2\pi j)/n + 2\pi\theta)^2$

Question

I am interested in finding a closed form/upper bound on

$$ \sum_{j=0}^{n-1}\cos^2\left(\frac{2\pi j}n + 2\pi\theta\right) $$ Where $\theta\in[0,1]$. As part of my problem I am taking the supremum over the choice of $\theta$, so if this simplifies things it is fine. Clearly there is a trivial bound of $\leq n$. I am mostly interested if there is some asymptotic improvement to this bound, e.g. if it can be shown to be $o(n)$, though improved constants are of course not bad.

Answer 1

Note \begin{eqnarray} &&\sum_{j=0}^{n-1}\cos^2((2\pi j)/n + 2\pi\theta)\\ &=&\frac12\sum_{j=0}^{n-1}\bigg[1+\cos\bigg(\frac{4\pi j}n + 2\pi\theta\bigg)\bigg]\\ &=&\frac n2+\frac12\sum_{j=0}^{n-1}\bigg[\cos\bigg(\frac{4\pi j}n\bigg)\cos(2\pi\theta)-\sin\bigg(\frac{4\pi j}n\bigg)\sin(2\pi\theta)\bigg]\\ &=&\frac n2+\frac12\cos(2\pi\theta)\sum_{j=0}^{n-1}\cos\bigg(\frac{4\pi j}n\bigg)-\frac12\sin(2\pi\theta)\sum_{j=0}^{n-1}\sin\bigg(\frac{4\pi j}n\bigg). \end{eqnarray} Using $$ \sum_{j=0}^n\cos(jx)=\frac{\sin \left(\frac{1}{2} (n+1) x\right) \cos \left(\frac{n x}{2}\right)}{\sin \left(\frac{x}{2}\right)}$$ and $$ \sum_{j=0}^n\sin(x)=\frac{\sin \left(\frac{nx}{2}\right) \sin \left(\frac{1}{2} (n+1)x\right)}{\sin \left(\frac{x}{2}\right) }$$ one has $$ \sum_{j=0}^{n-1}\cos\bigg(\frac{4\pi j}n\bigg)=0, \quad\sum_{j=0}^n\sin\bigg(\frac{4\pi j}n\bigg)=0. $$ So $$ \sum_{j=0}^{n-1}\cos^2((2\pi j)/n + 2\pi\theta)=\frac n2 $$ and hence $$ \sup_{\theta\in[0,1]}\sum_{j=0}^n\cos^2((2\pi j)/n + 2\pi\theta)=\frac {n}2. $$

Answer 2

$$\begin{align} L&:=\frac{n}{2}+\frac{1}{2}\sum_{j=0}^{n-1}\cos\left(\frac{4\pi}{n}j+4\pi\theta \right)\\ &=\frac{n}{2} +\frac{1}{2}\mathcal{Re}\left(\sum_{j=0}^{n-1}\exp\left(\mathbf{i}\frac{4\pi}{n}j+\mathbf{i}4\pi\theta \right) \right)\\ &=\frac{n}{2} +\frac{1}{2}\mathcal{Re}\left(\exp(\mathbf{i}4\pi\theta)\cdot\sum_{j=0}^{n-1}\exp\left(\mathbf{i}\frac{4\pi}{n}j \right) \right)\\ &=\frac{n}{2} +\frac{1}{2}\mathcal{Re}\left(\exp(\mathbf{i}4\pi\theta)\cdot\frac{\exp\left(\mathbf{i}4\pi \right) - 1}{\exp\left(\mathbf{i}\frac{4\pi}{n} \right) - 1}\right)\\ &=\frac{n}{2} +\frac{1}{2}\mathcal{Re}\left(\exp(\mathbf{i}4\pi\theta)\cdot\frac{1 - 1}{\exp\left(\mathbf{i}\frac{4\pi}{n} \right) - 1}\right)\\ &=\frac{n}{2} \end{align}$$