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SUMMARY:Harald Helfgott (Universit\\" at G\\" ottingen\, Germany and CNRS\
 , France)
DTSTART:20230525T160000Z
DTEND:20230525T162500Z
DTSTAMP:20260423T041654Z
UID:CANT2023/42
DESCRIPTION:Title: <a href="https://researchseminars.org/talk/CANT2023/42/
 ">Explicit bounds on sums of the M\\" obius functions</a>\nby Harald Helfg
 ott (Universit\\" at G\\" ottingen\, Germany and CNRS\, France) as part of
  Combinatorial and additive number theory (CANT 2023)\n\nLecture held in R
 oom 4102 in CUNY Grad Center.\n\nAbstract\nLet $M(x)$ be the Mertens funct
 ion $M(x) = \\sum_{n\\leq x} \\mu(n)$. Most of us are used to thinking of 
 the problem of estimating $M(x)$ as being essentially equivalent to the pr
 oblem of estimating the number of primes up to $x$ (i.e.\, the Prime Numbe
 r 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 t
 he residues of $1/\\zeta(s)$.\n(There are bounds for $1/\\zeta(s)$ inside 
 the zero-free region\, but their constants are very large.)\n\nUp to now\,
  the best bounds on $M(x)$ have been either (a) based on elementary method
 s that amount to delicate refinements of Chebyshev-Mertens\, or (b) based 
 on iterative processes that combine (a) with explicit versions of PNT. (Th
 e best results of these two kinds to date are due to Daval (unpublished) a
 nd 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.\n\nWe
  give considerably stronger bounds on $M(x)$\, $m(x)$ and $\\check{m}(x)$ 
 by means of an analytic approach based on mean-value theorems. \n\nJoint w
 ork with Andr\\' es Chirre.\n
LOCATION:https://researchseminars.org/talk/CANT2023/42/
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