0
$\begingroup$

Let $M(N) := \sum_{n=1}^N \mu(n)$

It is known that bounding $M(N)$ by $N^{1/2+\epsilon}$ implies R.H.

A bound of $M(N)$ by $K\sqrt N$ for say $ K\ge2 $ is also sufficient. $K = 1$ is excluded as Mertens's conjecture is false.

Suppose you show that for each $\epsilon > 0$ there are infinitely many values N such that $M(N) > N^{1-\epsilon}$.

Question 1: Does this invalidate R.H.?

Question 2: How does a result like $ M(N)>N^{0,99} $ for any $N>N_o$ for some $N_o$ stand with R.H.?

In the second question does it kill some well accepted heuristics about R.H.?

Most texts about number theory or R.H. give sufficient conditions and discuss chains of necessity leading to R.H., But I found little about falsity of R.H. (the "heretic side" as called somewhere in Tao's blog).

Notational remark: Mathematicians should use $\dot \mu(N)$ instead of $M(N)$, saving symbols and brain but not eyes (because LaTeX point is too small).

Improve this question
$\endgroup$
3
  • 2
    $\begingroup$ Yes to Q1, no to Q2, of course, because then the statement that $M(x) \ll_{\varepsilon} x^{1/2 + \varepsilon}$ would be false, and this is equivalent to RH. Basically, if $\limsup_{x \to \infty} |M(x)|x^{-\Theta} > 0$ for some $\Theta > 1/2$, then RH is false. $\endgroup$
    – Peter Humphries
    Commented Jun 28, 2016 at 17:53
  • $\begingroup$ OK I agree that my question are a bit simplistic, I thought a mere specific epsilon was enough. Considering how far we are ( N/logN) I feel justified in my mistake. Sorry for the disturbance though. $\endgroup$
    – Jérôme JEAN-CHARLES
    Commented Jun 29, 2016 at 21:14
  • $\begingroup$ If $M(n) = O(n^{a})$ then there is $N,C$ such that for every $n > N$ : $|M(n)| < C n^{a}$. Of course $M(n) > n^{1-\epsilon}$ for infinitely many $n$ contradicts this statement $\endgroup$
    – reuns
    Commented Feb 25, 2017 at 22:45

2 Answers 2

Reset to default
5
$\begingroup$

In fact many texts give equivalent statements to RH in terms of $M(N)$ and related arithmetic functions. In particular, RH is equivalent to the following: for any $\epsilon>0$, there is $N_0$ depending on $\epsilon$, such that for any $N>N_0$ we have $|M(N)|<N^{1/2+\epsilon}$.

Informally, this means that if RH fails, then $|M(N)|$ is occasionally much larger than $\sqrt{N}$, and vice versa. It is a subtle question how large $|M(N)|$ can really get if RH holds. For the state-of-the-art in this direction and for what more can be expected, see Soundararajan's paper.

Improve this answer
$\endgroup$
1
  • $\begingroup$ Perfect: thanks for the paper and the removal of a misconception of mine. $\endgroup$
    – Jérôme JEAN-CHARLES
    Commented Jun 29, 2016 at 21:13
4
$\begingroup$

If you have $M(N) > N^{0.99}$ for infinitely many $N$ then this means there is a an exceptional zero with real part $\geq 0.99$, that therefore invalidates RH.

As to the order of growth of $M(x)$ a well substantiated conjectured due to Steve Gonek and Nathan Ng, is that $$ \limsup_{x \rightarrow \infty} \frac{|M(x)|}{\sqrt{x} (\log\log\log x)^{5/4}} = B $$ for some finite constant $B > 0$, see http://www.cs.uleth.ca/~nathanng/RESEARCH/mobius2b.pdf on p. 9. In particular this implies that there is no finite constant $K > 0$ such that $|M(x)| < K \sqrt{x}$ for all $x$.

Finally your notational suggestion is bad, since it will lead to endless confusion between $\mu$ (the function) and $\mu$ (the summatory function). What happens if you subtract the last value of $\mu$ (the function) from the summatory function $\mu(N)$ of Moebius up to $N$? $\mu(N) - \mu(N) = \mu(N - 1)$? I would argue this strains both my brain and eyes!

Improve this answer
$\endgroup$
6
  • 1
    $\begingroup$ I think you missed the dot over the mu in the suggested notation for the summatory function. $\endgroup$
    – Gerry Myerson
    Commented Jun 28, 2016 at 23:02
  • 3
    $\begingroup$ which shows that the idea of having a dot did not quite make things clearer :-) $\endgroup$
    – Suvrit
    Commented Jun 28, 2016 at 23:48
  • $\begingroup$ @aosjckajsd You must have a problem with your browser : I have the point on my screen. As I said the point should be must bigger, I still think this is nice and not ambiguous but habits die hard...Now a dot is mot discrete and as generic as capitalizing a letter but save the M ... $\endgroup$
    – Jérôme JEAN-CHARLES
    Commented Jun 29, 2016 at 20:46
  • 1
    $\begingroup$ One problem with a dot is that in some contexts it stands for derivative, but you want to use it for a summation, which is pretty nearly the opposite concept. $\endgroup$
    – Gerry Myerson
    Commented Jun 30, 2016 at 2:57
  • 1
    $\begingroup$ I think Newton used it long before Einstein. $\endgroup$
    – Gerry Myerson
    Commented Jun 30, 2016 at 13:15

Not the answer you're looking for? Browse other questions tagged .