Sum Simplification

Question

For $q,k\in\mathbb{N}$ and $1\leq q\leq N$, is the following simplification $$ f(q,k)=\sum_{j=1}^Ne^{-(2k\pi\text{i})jq/N}=N\sum_{i=1}^k\delta_{iq,N} $$ correct? Here, $\delta_{i,j}$ is the Kronecker delta. Any better way to represent $f$?

Answer 1

Suppose $ \frac{kq}{N}\in\mathbb{Q}\setminus\mathbb{N} $, then : \begin{aligned}\sum_{j=1}^{N}{\operatorname{e}^{-\operatorname{i}\frac{2k\pi jq}{N}}}=\operatorname{e}^{-\operatorname{i}\frac{2k\pi q}{N}}\frac{1-\operatorname{e}^{-\operatorname{i}2k\pi jq}}{1-\operatorname{e}^{-\operatorname{i}\frac{2k\pi q}{N}}}=0\end{aligned}

Otherwise : $$ \sum_{j=1}^{N}{\operatorname{e}^{-\operatorname{i}\frac{2k\pi jq}{N}}}=N $$

Thus : $$ \sum_{j=1}^{N}{\operatorname{e}^{-\operatorname{i}\frac{2k\pi jq}{N}}}=N\operatorname{\mathbf{1}_{\mathbb{N}}}\left(\frac{kq}{N}\right) $$

$ \operatorname{\mathbf{1}_{A}}:X\rightarrow\mathbb{R} $ denotes the usual indicator function : $ \operatorname{\mathbf{1}_{A}}:x\mapsto\left\lbrace\begin{aligned}1,\ \ \ \ \ &x\in A\\ 0,\ \ \ \ \ &x\notin A\end{aligned}\right. $.