Mertens conjecture

Image:Congettura Mertens.png $M(n)$ and the square roots $\pm \sqrt{n}$ for $n \le 10,000$ . After computing these values, Mertens conjectured that the absolute value of $M(n)$ is always bounded by $\sqrt{n}$ . This hypothesis, known as the Mertens conjecture, was disproved in 1985 by Andrew Odlyzko and Herman te Riele.]]

In mathematics, the Mertens conjecture is the statement that the Mertens function $M(n)$ is bounded by $\pm\sqrt{n}$ . Although now disproven, it had been shown to imply the Riemann hypothesis. It was conjectured by Thomas Joannes Stieltjes, in an 1885 letter to Charles Hermite (reprinted in {{harvs|txt|last=Stieltjes|authorlink=Thomas Joannes Stieltjes|year=1905}}), and again in print by {{harvs|txt|authorlink=Franz Mertens|last=Mertens|first=Franz|year=1897}}, and disproved by {{harvs|txt|last1=Odlyzko|first1=Andrew|authorlink1=Andrew Odlyzko|last2=te Riele|first2=Herman|authorlink2=Herman te Riele|year=1985}}.

It is a striking example of a mathematical conjecture proven false despite a large amount of computational evidence in its favor.

Definition

In number theory, the Mertens function is defined as

: $M(n) = \sum_{1 \le k \le n} \mu(k),$

where μ(k) is the Möbius function; the Mertens conjecture is that for all n > 1,

: $|M(n)| < \sqrt{n}.$

Disproof of the conjecture

Stieltjes claimed in 1885 to have proven a weaker result, namely that $m(n) := M(n)/\sqrt{n}$ was bounded, but did not publish a proof.{{cite book |editor1-last=Borwein |editor1-first=Peter |editor1-link=Peter Borwein |editor2-last=Choi |editor2-first=Stephen |editor3-last=Rooney |editor3-first=Brendan |editor4-last=Weirathmueller |editor4-first=Andrea |title=The Riemann hypothesis. A resource for the aficionado and virtuoso alike |series=CMS Books in Mathematics |location=New York, NY | publisher=Springer-Verlag |year=2007 |isbn=978-0-387-72125-5 |zbl=1132.11047 |page=69}} (In terms of $m(n)$ , the Mertens conjecture is that $-1 < m(n) < 1$ .)

In 1985, Andrew Odlyzko and Herman te Riele proved the Mertens conjecture false using the Lenstra–Lenstra–Lovász lattice basis reduction algorithm:{{Citation | last1=Odlyzko | first1=A. M. | author1-link=Andrew Odlyzko | last2=te Riele | first2=H. J. J. | author2-link=Herman te Riele | title=Disproof of the Mertens conjecture | url=http://www.dtc.umn.edu/~odlyzko/doc/arch/mertens.disproof.pdf | doi=10.1515/crll.1985.357.138 |mr=783538 | year=1985 | journal=Journal für die reine und angewandte Mathematik | volume=1985 | issue=357 | pages=138–160 | zbl=0544.10047 | s2cid=13016831 | issn=0075-4102 }}Sandor et al (2006) pp. 188–189.

: $\liminf m(n) < -1.009$ {{pad|4}} and {{pad|4}} $\limsup m(n) > 1.06.$

It was later shown that the first counterexample appears below $e^{3.21\times10^{64}} \approx 10^{1.39\times10^{64}}$ {{cite journal | last1 = Pintz | first1 = J. | year = 1987 | title = An effective disproof of the Mertens conjecture | url = http://www.numdam.org/article/AST_1987__147-148__325_0.pdf | journal = Astérisque | volume = 147–148 | pages = 325–333 | zbl=0623.10031 }}

but above 10¹⁶.{{cite arXiv |last=Hurst |first=Greg |date=2016 |title=Computations of the Mertens function and improved bounds on the Mertens conjecture |eprint=1610.08551 |class=math.NT}} The upper bound has since been lowered to $e^{1.59\times10^{40}}$ Kotnik and Te Riele (2006). or approximately $10^{6.91\times10^{39}},$ and then again to $e^{1.017\times10^{29}} \approx 10^{4.416\times10^{28}}$ .{{Cite arXiv |last1=Rozmarynowycz |first1=John |last2=Kim |first2=Seungki |year=2023 |title=A New Upper Bound On the Smallest Counterexample To The Mertens Conjecture |class=math.NT |eprint=2305.00345 }} In 2024, Seungki Kim and Phong Nguyen lowered the bound to $e^{1.96\times10^{19}} \approx 10^{8.512\times10^{18}}$ ,{{Cite web |last1=Seungki |first1=Kim |last2=Phong |first2=Nguyen |year=2024 |title=On counterexamples to the Mertens conjecture |url=https://antsmath.org/ANTSXVI/papers/KimNguyen.pdf}} but no explicit counterexample is known.

The law of the iterated logarithm states that if {{mvar|μ}} is replaced by a random sequence of +1s and −1s then the order of growth of the partial sum of the first {{mvar|n}} terms is (with probability 1) about {{nowrap|{{math|{{sqrt| n log log n}}}},}} which suggests that the order of growth of {{math|m(n)}} might be somewhere around {{math|{{sqrt|log log n}}}}. The actual order of growth may be somewhat smaller; in the early 1990s Steve Gonek conjectured that the order of growth of {{math|m(n)}} was $(\log\log\log n)^{5/4},$ which was affirmed by Ng (2004), based on a heuristic argument, that assumed the Riemann hypothesis and certain conjectures about the averaged behavior of zeros of the Riemann zeta function.{{cite web |url=http://www.cs.uleth.ca/~nathanng/RESEARCH/mobius2b.pdf |title=The distribution of the summatory function of the Möbius function |first=Nathan |last=Ng |year=2004}}

In 1979, Cohen and DressCohen, H. and Dress, F. 1979. “Calcul numérique de Mx)” 11–13. [Cohen et Dress 1979], Rapport, de I'ATP A12311 ≪ Informatique 1975 ≫ found the largest known value of $m(n) \approx 0.570591$ for {{math|M(7766842813) {{=}} 50286,}} and in 2011, Kuznetsov found the largest known negative value (largest in the sense of absolute value) $m(n) \approx -0.585768$ for {{math|M(11609864264058592345) {{=}} −1995900927.}}{{cite arXiv |last=Kuznetsov |first=Eugene |date=2011 |title=Computing the Mertens function on a GPU |eprint=1108.0135 |class=math.NT}} In 2016, Hurst computed {{math|M(n)}} for every {{math|n ≤ 10¹⁶}} but did not find larger values of {{math|m(n)}}.

In 2006, Kotnik and te Riele improved the upper bound and showed that there are infinitely many values of {{mvar|n}} for which {{nowrap|{{math|m(n) > 1.2184}},}} but without giving any specific value for such an {{mvar|n}}.Kotnik & te Riele (2006). In 2016, Hurst made further improvements by showing

: $\liminf m(n) < -1.837625$ {{pad|4}} and {{pad|4}} $\limsup m(n) > 1.826054.$

Connection to the Riemann hypothesis

The connection to the Riemann hypothesis is based on the Dirichlet series

for the reciprocal of the Riemann zeta function,

: $\frac{1}{\zeta(s)} = \sum_{n=1}^\infty \frac{\mu(n)}{n^s},$

valid in the region $\mathcal{Re}(s) > 1$ . We can rewrite this as a

Stieltjes integral

: $\frac{1}{\zeta(s)} = \int_0^\infty x^{-s} dM(x)$

and after integrating by parts, obtain the reciprocal of the zeta function

as a Mellin transform

: $\frac{1}{s \zeta(s)} = \left\{ \mathcal{M} M \right\}(-s)
= \int_0^\infty x^{-s} M(x)\, \frac{dx}{x}.$

Using the Mellin inversion theorem we now can express {{mvar|M}} in terms of {{frac|1|{{mvar|ζ}}}} as

: $M(x) = \frac{1}{2 \pi i} \int_{\sigma-i\infty}^{\sigma+i\infty} \frac{x^s}{s \zeta(s)}\,ds$

which is valid for {{math|1 < σ < 2}}, and valid for {{math|{{frac|1|2}} < σ < 2}} on the Riemann hypothesis.

From this, the Mellin transform integral must be convergent, and hence

{{math|M(x)}} must be {{math|O(x^e)}} for every exponent e greater than {{sfrac|1|2}}. From this it follows that

: $M(x) = O\Big(x^{\tfrac{1}{2} + \epsilon}\Big)$

for all positive {{mvar|ε}} is equivalent to the Riemann hypothesis, which therefore would have followed from the stronger Mertens hypothesis, and follows from the hypothesis of Stieltjes that

: $M(x) = O\Big(x^\tfrac{1}{2}\Big).$

References

External links

{{cite web |title=A Prime Surprise (Mertens Conjecture) |work=Numberphile |date=January 23, 2020 |via=YouTube |url=https://www.youtube.com/watch?v=uvMGZb0Suyc |archive-url=https://ghostarchive.org/varchive/youtube/20211221/uvMGZb0Suyc |archive-date=2021-12-21 |url-status=live}}{{cbignore}}

Category:Analytic number theory

Category:Disproved conjectures