5
$\begingroup$

It is known that with $M(x) = \sum_{n\le x}\mu(n)$, there are infinitely many $x$ s.t. $|M(x)|\ge x^{\frac{1}{2} - \varepsilon}$ (see Chapter 15 of Montgomery-Vaughan, for example). Is there any way to make this result effective in the following sense: show that for large $X$, there exists some $x\in [X, f(X)]$ s.t. $|M(x)|\ge x^{\frac{1}{2} - \varepsilon}$ (or $x^{\sigma - \varepsilon}$ if $\zeta(\sigma + it) = 0$ for some $\sigma > 1/2$), where $f(X)$ is some explicit function in $X$? (Potentially naive) heuristics suggest one can take $f(X) = X + X^{2\sigma}$.

$\endgroup$

1 Answer 1

3
$\begingroup$

It was proved by Kaczorowski and Pintz (Acta Math. Hungar. 48 (1986), 173-185, doi: 10.1007/BF01949062) that there exists $x\in[X,X^{1+o(1)}]$ such that $M(x)\geq x^{\sigma-\varepsilon}$, and there also exists $x\in[X,X^{1+o(1)}]$ such that $M(x)\leq -x^{\sigma-\varepsilon}$. See Corollary 1 in their paper.

$\endgroup$
2
  • $\begingroup$ Does the result or method quickly lead to any uniformity in the height of the zero (so if there is a zero at $\sigma + it$, can one make the dependence of the result on $|t|$ explicit?). $\endgroup$ Sep 1, 2021 at 21:15
  • $\begingroup$ @MayankPandey: I don't know, I have not studied the details. $\endgroup$
    – GH from MO
    Sep 2, 2021 at 1:00

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.