# Is the product of two equidistributed power series equidistributed?

Let $$f(T) = \sum a_n T^n \in \mathbf{F}_p [[ T ]]$$ be a power series. We'll say that the coefficients of $$f(T)$$ are equidistributed modulo $$p$$ if for every residue class $$a$$ modulo $$p$$, we have $$\lim_{X \to \infty} \dfrac{1}{X} \cdot \# \{n
Suppose that $$f(T), g(T) \in \mathbf{F}_p [[ T ]]$$ are power series whose coefficients are equidistributed modulo $$p$$ in the above sense. Is it true that the coefficients of the product $$f(T) \cdot g(T)$$ are also equidistributed modulo $$p$$?

• Taking $g(T)=f(-T)$, the product will be biased towards $0$. Aug 21, 2023 at 23:18

No. Let $$f(T)$$ be chosen uniformly at random from all power series with constant term nonzero and let $$g(T) = 1/ f(T)$$. Then $$g(T)$$ is also chosen uniformly at random from all power series with constant term nonzero. (Here uniformly means each possible sequence of first $$n$$ coefficients has equal probability, and the proof is just that the first $$n$$ coefficients of $$f$$ determine the first $$n$$ coefficients of $$g$$ and vice versa.)
So $$f(T)$$ and $$g(T)$$ are each equidistributed with probability $$1$$ but $$f(T)g(T)=1$$ is not equidistributed.
The same argument shows that any condition on $$f$$ and $$g$$ that is "generic" in the sense of holding for a random power series is not sufficient.
• I see - are there conditions on $f(T)$ and $g(T)$ that would ensure that $f(T) \cdot g(T)$ is equidistributed? Perhaps some kind of independence condition that excludes the possibility of $g$ being determined by $f$ as in your counterexample? Aug 21, 2023 at 23:22