Explicit bounds on sums of the M\" obius functions

Abstract: Let $M(x)$ be the Mertens function $M(x) = \sum_{n\leq x} \mu(n)$. Most of us are used to thinking of the problem of estimating $M(x)$ as being essentially equivalent to the problem of estimating the number of primes up to $x$ (i.e., the Prime Number Theorem). If we want explicit bounds, however, the problem of bounding $M(x)$ becomes by far the harder of the two problems. The main reason is that, while the residue of $- \zeta'(s)/\zeta(s)$ at a zero of $\zeta(s)$ is just the order of the zero, we do not have a good way to control the residues of $1/\zeta(s)$. (There are bounds for $1/\zeta(s)$ inside the zero-free region, but their constants are very large.)

Up to now, the best bounds on $M(x)$ have been either (a) based on elementary methods that amount to delicate refinements of Chebyshev-Mertens, or (b) based on iterative processes that combine (a) with explicit versions of PNT. (The best results of these two kinds to date are due to Daval (unpublished) and Ramar\' e, respectively.) The same is true of sums of the form $m(x) = \sum_{n\leq x} \mu(n)/n$, $\check{m}(x) = \sum_{n\leq x} \mu(n) \log(x/n)/n$, etc., which appear often in analytic number theory.

We give considerably stronger bounds on $M(x)$, $m(x)$ and $\check{m}(x)$ by means of an analytic approach based on mean-value theorems.

Joint work with Andr\' es Chirre.

combinatoricsnumber theory

Audience: researchers in the topic

Combinatorial and additive number theory (CANY 2023)