# Discrete Fourier Transforms, showing the transform of a product of a series

If $$a=(a_{1}, \ldots,a_{n})$$. Define $$F_{a}( \lambda)= n^{-1/2} \sum\limits_{t=1}^{n} a_{t}e^{-it \lambda}$$.

Let $$\lbrace x_{1}, \ldots,x_{n} \rbrace$$ and $$\lbrace y_{1},\ldots,y_{n}\rbrace$$ be real numbers. Let $$z_{t}=x_{t}y_{t}$$. I want to show the following equality, $$F_{z}(2 \pi j /n)= n^{-1/2} \sum\limits_{ k \in D_{n}} F_{X}( 2 \pi j/n)F_{Y}(2 \pi (j-k)/n)$$, where $$D_{n} = \lbrace j \in \mathbb{Z} : 2 \pi j/ n \in (- \pi, \pi] \rbrace$$.

• See how the terms $e^{-i(t_1-t_2) \lambda}$ appear in $F_x(\lambda)F_y(u-\lambda)$ then for $0 \ne t_1-t_2$ dividing $n!$ : $\sum_{k= 1}^{n!} e^{-i(t_1-t_2) 2\pi k/n!} =0$. You can replace $n!$ by $lcm(1,\ldots,n)$ but not by $n$. – reuns Dec 9 '18 at 1:17
• I tried to go from either ways to the other, but miserabely failed. Would you be able to help more ? – mich95 Dec 9 '18 at 1:20
• Where are you stuck in what I said ? – reuns Dec 9 '18 at 1:21
• I do not get any $n!$... – mich95 Dec 9 '18 at 2:19