# Improved upper bound for second moment of reduced residues modulo $q$?

The question asked here is a follow up to this question, which was answered by user GH from MO. [EDIT: The question is edited for clarification after receiving a comment on the original posting.]

As there, let $$q$$ be a natural number, let $$P = \phi(q)/q$$ be the "probability" that a randomly chosen integer is relatively prime to q. Then, following Montgomery and Vaughan in On the distribution of reduced residues, the second moment of the number of reduced residues modulo $$q$$ in an interval of length $$h$$ about its mean, $$hP$$, can be stated as $$M_2(q;h) = qP^2 \sum_{\substack{{r \mid q }\\{r > 1}}} \mu(r)^2 \left( \prod_{\substack{ {p \mid q }\\{p \nmid r} }} \frac{p(p-2)}{(p-1)^2} \right) r^2 \phi(r)^{-2} \left\{ \frac{h}{r}\right\}\left( 1 - \left\{ \frac{h}{r}\right\}\right).$$ Applying the inequality $$\{\alpha\}(1 - \{\alpha\}) \leq \alpha$$ then gives the upper bound $$\tag{1} M_2(q;h)\leq qhP,$$ while a lower bound is given by $$M_2(q; h) \geq qhP - qhPQ + O(qhP^2)$$ where $$Q=\prod_{\substack{{p \mid q}\\{p>h}}} (1-1/p)$$.

Since $$\{\alpha\}\leq 1$$, we also have that $$\{\alpha\}(1 - \{\alpha\}) \leq 0.25$$. This allows for a second upper bound, $$\tag{2} M_2(q;h) \leq 0.25 \, qP^2 \sum_{\substack{{r \mid q }\\{r > 1}}} \mu(r)^2 \left( \prod_{\substack{ {p \mid q }\\{p \nmid r} }} \frac{p(p-2)}{(p-1)^2} \right) r^2 \phi(r)^{-2}.$$

As pointed out by user Ofir Gorodetsky in the comment below, the upper bound in (1) is already optimal for those choices of $$q$$ such that the upper and lower bounds are of same order. For example, taking $$q=\prod_{p\leq h^C} p$$ for large enough $$C$$, we have that $$M_2(q;h)\asymp qhP$$.

Here I'd like to consider the rather opposite situation, when $$C$$ is small. Specifically, take $$q=\prod_{p\leq h^C} p$$, where $$C<1$$. Then we have that $$Q=1$$ and the lower bound takes the form $$M_2(q; h) \geq O(qhP^2),$$ which leaves some more wiggle room between the upper and lower bounds, and I'd like to know whether the upper bound in (2) this case provides some improvement on (1).

# Question

Given $$q=\prod_{p\leq h^C} p$$, where $$C<1$$. Does the upper bound for $$M_2(q;h)$$ stated in (2) provide a sharper bound compared to (1) in this case? And is it possible to express the upper bound in (2) a bit more simplified or asymptotically?

• At least for some families of $q$, the upper bound is already optimal in the sense that there is a matching lower bound of the same order magnitude; this follows from the general lower bound given at the end of your linked question. To be concrete, taking $q = \prod_{p \le h^C} p$ for sufficiently large $C$, we have $M_2(q;h) \asymp qhP$. May 14, 2020 at 9:27
• Thanks for pointing to an important clarification! I've edited the question to focus on the case when $q=\prod_{p\leq h^C} p$ and $C<1$, as in this case the lower bound should be $M_2(q;h)\geq O(qhP^2)$ (is it a better or more correct way to write this inequality?). May 14, 2020 at 11:28
• Unfortunately, if you just use the fact that $\{ \alpha\}(1-\{\alpha\})$ is bounded, you do not obtain a superior upper bound. Indeed, assuming $q$ is squarefree (as we may), both $\mu(r)^2$ and $\prod_{p\mid q, \, p \nmid r} p(p-2)/(p-1)^2$ are $\asymp 1$, and your idea leads to an upper bound of order $qP^2 \sum_{r \mid q, \, r> 1} r^2/\phi(r)^2$, which is much larger qualitatively than $qPh$ for $q=\prod_{p \le x} p$ with $x \ge \log^2 h$... May 14, 2020 at 12:27
• A heuristic for using $\{ \alpha \} (1-\{\alpha\}) \le \alpha$ is that $r^2/\phi(r)^2$ being large is correlated with $r$ being large which is correlated with $\{ h/r \}(1-\{h/r \})$ being of order $h/r$. May 14, 2020 at 12:29
• Thanks for the effort! I'd be happy to accept that as an answer if you'd want to write it up as such. May 14, 2020 at 13:07