Average of Sums of Exponentials

Question

Suppose you have a function of the form $f(x)=\sum_{\mathcal{A}}e(nx)$, where $\mathcal{A}$ is some subset of the naturals and $e(x)=e^{2\pi ix}$. Now let $b$ and $q$ be coprime integers. I think that in some situations, $$\sum_{\substack{0 \lt a \lt q \\ (a,q)=1}}\Big|f\Big(\frac{b a}{q}\Big)\Big|= \sum_{\substack{0 \lt a \lt q \\ (a,q)=1}}\Big|f\Big(\frac{a}{q}\Big)\Big|.$$ But I don't know what conditions to add so that this is true. What I tried was using the fact that $f$ is periodic with period $1$, but I don't know how to make things work in general. For instance, when $b=15$ and $q=8$, we have by reducing $ba$ modulo $8$, $$ \sum_{\substack{0 \lt a \lt 8\\ (a,8)=1}}|f\Big(\frac{15 a}{8}\Big)|=\Big|f\Big(\frac{7}{8}\Big)\Big|+\Big|f\Big(\frac{5}{8}\Big)\Big|+\Big|f\Big(\frac{6}{8}\Big)\Big|+\Big|f\Big(\frac{1}{8}\Big)\Big|, $$ but $$ \sum_{\substack{0 \lt a \lt 8\\ (a,8)=1}}|f\Big(\frac{a}{8}\Big)|=\Big|f\Big(\frac{1}{8}\Big)\Big|+\Big|f\Big(\frac{3}{8}\Big)\Big|+\Big|f\Big(\frac{5}{8}\Big)\Big|+\Big|f\Big(\frac{7}{8}\Big)\Big| $$ In this case they differ by one term. If it is not possible to say this quantities are the same, is there any way to compare them? Maybe a bound in terms of the other quantity.

Answer 1

The integers between 1 and q relatively prime with q form a group G with respect to multiplication modulo q. The element b mod q lies in this group. The map f: G -> G defined by f(a) = ab is a bijection, the inverse map being the multiplication by the inverse of b.

Therefore your two summations are equivalent to each other modulo q and will give the same result for any periodic function.

In the example you gave the sums are actually equal, the term 6/8 is wrong.