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It is well known that the Mertens function

$$M(x)=\sum _{n\leq x}\mu(n)$$

has infinitely many zeros, and this seems to be a short proof.

Are there known results about how often the Mertens function is 0? (i.e. how many times on average between $1$ and $x$)

Also, is it possible to prove that it vanishes infinitely often with elementary techniques only, and no complex analysis / Zeta function? (In the same way the PNT has been proved elementary by Selberg/Erdös around 1950).

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  • $\begingroup$ Regarding your second question: we can start by asking whether there is an elementary proof of classical oscillation theorems, for example that $\psi(x)-x$ changes sign infinitely often. $\endgroup$
    – Greg Martin
    Commented Jan 9, 2019 at 18:09

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Quoting an answer to the question https://mathoverflow.net/questions/273845/oscillation-of-the-summatory-möbius-function

Let $c=14.1347251…$. Then there are at least $(c/\pi-o(1))\log y$ sign changes in $M(x)$ in the interval $[1,y]$. This was proved by Kaczorowski and Pintz (Acta Math. Hungar. 48 (1986), 173-185, doi 10.1007/BF01949062).

This may well be the state of the art, but any comments on further results would of course be welcome. If no more is known then this question is a duplicate of the linked question.

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    $\begingroup$ I guess c is the imaginary part of the first non trivial zero of zeta ? $\endgroup$
    – Sylvain JULIEN
    Commented Jan 9, 2019 at 17:49
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    $\begingroup$ To the best of my knowledge, nothing stronger than $\Omega(\log y)$ is known for the number of sign changes of any of the usual number-theoretic functions and their error terms. Even showing that the number of sign changes grows faster than any constant multiple of $\log y$ seems quite difficult. Daniel Fiorilli pointed out in his thesis that one of the reasons we can't do better (despite the truth probably being around $\sqrt y$) is that these proofs use a many-times-averaged version of these functions, which actually do have only $O(\log y)$ sign changes. $\endgroup$
    – Greg Martin
    Commented Jan 9, 2019 at 18:12
  • $\begingroup$ @SylvainJULIEN, yes. Basj, I will see if I can track down a PDF. $\endgroup$
    – kodlu
    Commented Jan 9, 2019 at 21:24
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    $\begingroup$ @Basj, how can I get the PDF to you? I don't have a webpage I can upload to. $\endgroup$
    – kodlu
    Commented Jan 10, 2019 at 11:28
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    $\begingroup$ Teaching you to fish: :) Typing "Fiorilli" into The Mathematics Genealogy Project uncovers the title of his thesis, Irrégularités dans la distribution des nombres premiers et des suites plus générales dans les progressions arithmétiques; the first Google hit gets you the PDF. It definitely uses complex methods, specifically the explicit formula. $\endgroup$
    – Greg Martin
    Commented Jan 15, 2019 at 17:13
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Edit: Wrong, kept for the comments below.

I'm trying an answer with an elementary argument:

Let's assume there exists $A > 0$ such that $M(x) \neq 0$ for $x \geq A$. We can assume that $M(x) \geq 1$ for $x \geq A$, as the proof would be the same in the negative case. Let:

$$K = \sum_{k \leq A} |M(k)|.$$

Then the well-known identity

$$1 = \sum_{d \leq x} M(x/d)$$

gives $$1 = \left|\sum_{d \leq x/A} M(x/d) + \sum_{x/A < d\leq x} M(x/d)\right|\geq \sum_{d \leq x/A} M(x/d)-K \geq \lfloor{x/A}\rfloor -K \xrightarrow[x\to\infty]\, +\infty$$

which is a contradiction.

As a conclusion, the Mertens function $M(x)$ has infinitely many zeros.

I wonder if a similar (but more clever) elementary argument can lead to an estimation of how often the Mertens function vanishes between $1$ and $y$.

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  • $\begingroup$ The estimate is wrong, because each of the values of $M(k)$ is included many times. For example, take $A=2$. Then $K=1$, but $\sum_{x/2<d\le x}M(x/d)=x/2$. $\endgroup$
    – Emil Jeřábek
    Commented Jan 22, 2019 at 13:34
  • $\begingroup$ My bad, that's right... Do you think there's a way to fix this @EmilJeřábek? Because repetitions of $M(x/d)$ in $\sum_{d\leq x} M(x/d)$ only seem to appear when $x/d < \sqrt x$. $\endgroup$
    – Basj
    Commented Jan 22, 2019 at 13:47
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    $\begingroup$ $\sum_{d \le x} M(x/d) = \sum_{n \in (\sqrt{x},x]} M(n)+\sum_{n \le \sqrt{x}} M(n) (\lfloor \frac{x}{n}\rfloor-\lfloor \frac{x}{n+1} \rfloor)$, then see what you get with $M(n)=n^{\sigma}$, that the second sum is not small is one of the main difficulty for the RH. Looking instead at $\sum_{d \le x} M(x/d)(-1)^d$ means $\lfloor \frac{x}{n}\rfloor\bmod 2$. Anyway to estimate the number of sign changes you should start with the density of zeros and an effective explicit formula such as $\psi(x)-x-\log 2\pi = \sum_{|Im(\rho)| < T} \frac{x^\rho}{\rho}+ \mathcal{O}(x^{1/2} \log^2x/T)$ (under the RH) $\endgroup$
    – reuns
    Commented Jan 22, 2019 at 15:19
  • $\begingroup$ @reuns Thank you. I have already looked at complex methods / involving zeros of zeta function, etc., but here I was curious if there could be a fully elementary method. So do you mean fixing my (wrong) argument in this answer is hopeless? $\endgroup$
    – Basj
    Commented Jan 22, 2019 at 15:21
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    $\begingroup$ $-1=\sum_{d \le x} M(x/d) (-1)^{d+1} = \sum_{n \in (\sqrt{x},x]} M(n)(-1)^{x+n} +\sum_{n \le \sqrt{x}} M(n) (\lfloor \frac{x}{n}\rfloor-\lfloor \frac{x}{n+1} \rfloor \bmod 2)$ for $x \ge 2$ $\endgroup$
    – reuns
    Commented Jan 22, 2019 at 16:34

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