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Jan 23, 2019 at 12:26 comment added Basj Oh you mean this proof on math.stackexchange (that I linked in my question), I forgot it was yours @reuns ;)
Jan 23, 2019 at 4:07 comment added kodlu MSE (Mathematics Stack Exchange) is what @reuns is referring to.
Jan 22, 2019 at 16:44 comment added reuns ?? My answer on MSE is elementary. I don't think elementary arguments can count the number of zeros crossings, that's why i mentioned the explicit formula
Jan 22, 2019 at 16:41 comment added Basj Good to know @reuns. Do you think this could help to find an elementary argument? (better than my wrong answer here!)
Jan 22, 2019 at 16:34 comment added reuns $-1=\sum_{d \le x} M(x/d) (-1)^{d+1} = \sum_{n \in (\sqrt{x},x]} M(n)(-1)^{x+n} +\sum_{n \le \sqrt{x}} M(n) (\lfloor \frac{x}{n}\rfloor-\lfloor \frac{x}{n+1} \rfloor \bmod 2)$ for $x \ge 2$
Jan 22, 2019 at 16:11 comment added Basj You're right @reuns: $\sum_{n \le \sqrt{x}} |M(n)| (\lfloor \frac{x}{n}\rfloor-\lfloor \frac{x}{n+1} \rfloor) \gg x$ even if we took $|M(n)| \ll n^{1/2+\epsilon}$ (that would require RH...). Something else: what did you mean with $\sum_{d \le x} M(x/d)(-1)^d$ and $\lfloor \frac{x}{n}\rfloor\bmod 2$?
Jan 22, 2019 at 15:30 history edited Basj CC BY-SA 4.0
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Jan 22, 2019 at 15:28 comment added Emil Jeřábek I don't see how you could fix that. For constant $A$, $\sum_{x/A<d\le x}M(x/d)=c_Ax+O(1)$ where $c_A$ is a (usually nonzero) constant depending on $A$, and all your argument shows is that, assuming $M$ does not change sign infinitely often, $|c_A|\ge1/A$ for all sufficiently large $A$. This is likely even harder to disprove than the original assumption.
Jan 22, 2019 at 15:25 comment added reuns The key of the (elementary) argument in my linked answer is that $\zeta(s)$ is meromorphic with no zeros on $(0,\infty)$ (not only that the coefficients of $\zeta(s)$ are close to $1$,), and that $M(x)$ stays non-negative for $x > N$ would imply $1/\zeta(s)$ has a singularity at $\sigma$ its abscissa of convergence, so your argument should use it
Jan 22, 2019 at 15:21 comment added Basj @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?
Jan 22, 2019 at 15:19 comment added reuns $\sum_{d \le x} M(x/d) = \sum_{n \in (\sqrt{x},x]} M(n)+\sum_{n \le \sqrt{x}} M(n) (\lfloor \frac{x}{n}\rfloor-\lfloor \frac{x}{n+1} \rfloor)$, 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)
Jan 22, 2019 at 13:47 comment added Basj 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 $x/d < \sqrt x$.
Jan 22, 2019 at 13:34 comment added Emil Jeřábek 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 $\sum_{x/2<d\le x}M(x/d)=x/2$.
Jan 22, 2019 at 10:42 history edited Basj CC BY-SA 4.0
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Jan 22, 2019 at 10:34 history answered Basj CC BY-SA 4.0