Mertens function

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Mertens function to n=10 thousand

Mertens function to n=10 million

In number theory, the Mertens function is

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

where μ(k) is the Möbius function. The function is named in honour of Franz Mertens.

Less formally, M(n) is the count of square-free integers up to n that have an even number of prime factors, minus the count of those that have an odd number. M(n) = 0 for the n values

2, 39, 40, 58, 65, 93, 101, 145, 149, 150, 159, 160, 163, 164, 166, 214, 231, 232, 235, 236, 238, 254, ... (sequence A028442 in OEIS)

Because the Möbius function has only the return values -1, 0 and +1, it's obvious that the Mertens function moves slowly and that there is no k such that |M(k)| > k. The Mertens conjecture went even further, stating that there would be no k where the absolute value of the Mertens function exceeds the square root of k. The Mertens conjecture was disproven in 1985. However, the Riemann hypothesis is equivalent to a weaker conjecture on the growth of M(k), namely $M(k) = o(k^{\frac12 + \epsilon})$ . Since high values for M grow at least as fast as the square root of k, this puts a rather tight bound on its rate of growth. Here, $o$ refers to little-o notation.

[edit] Integral representations

Using the Euler product one finds that

$\frac{1}{\zeta(s) }= \prod_{p} (1-p^{-s})= \sum_{n=1}^{\infty}\mu (n)n^{-s}$

where $ζ(s)$ is the Riemann zeta function and the product is taken over primes. Then, using this Dirichlet series with Perron's formula, one obtains:

$\frac{1}{2\pi i}\oint_{C}ds \frac{x^{s}}{s\zeta(s) }=M(x)$

where "C" is a closed curve encircling all of the roots of $ζ(s).$

Conversely, one has the Mellin transform

$\frac{1}{\zeta(s)} = s\int_1^\infty \frac{M(x)}{x^{s+1}}\,dx$

which holds for $R e (s) > 1$ .

A curious relation given by Mertens himself involving Chebyshev function is:

$\Psi (x) = -M\left(\frac{x}{2}\right)\log(2)-M\left(\frac{x}{3}\right)\log(3)-M\left(\frac{x}{4}\right)\log(4)+......$

A good evaluation, at least asymptotically, would be to obtain, by the method of steepest descent, an inequality:

$\oint_{C}dsF(s)e^{st} \sim M(e^{t})$

assuming that there are not multiple non-trivial roots of $ζ(ρ)$ you have the "exact formula" by residue theorem:

$\frac{1}{2 \pi i} \oint _ {C}ds \frac{x^s}{s \zeta (s)} = \sum _ {\rho} \frac{x^{\rho}}{\rho \zeta '(\rho)}-2+\sum_{n=1}^{\infty} \frac{ (-1)^{n-1} (2\pi )^{2n}}{(2n)! n \zeta(2n+1)x^{2n}}$

Weyl conjectured that Mertens function satisfied the approximate functional-differential equation

$(1/2)y(x)-\sum_{r=1}^{N} \frac{B_{2r}}{(2r)!}D_{t}^{2r-1}y(\frac{x}{t+1})+x\int_{0}^{x}du \frac{y(u)}{u^{2}}=x^{-1}H(logx)$

where H(x) is the Heaviside step function, B are Bernoulli numbers and all derivatives with respect to t are evaluated at t = 0.

[edit] Calculation

The Mertens function has been computed for an increasing range of n.

Person	Year	Limit
Mertens	1897	10⁴
von Sterneck	1897	1.5 x 10⁵
von Sterneck	1901	5 x 10⁵
von Sterneck	1912	5 x 10⁶
Neubauer	1963	10⁸
Cohen and Dress	1979	7.8 x 10⁹
Dress	1993	10¹²
Lioen and van der Lune	1994	10¹³
Kotnik and van der Lune	2003	10¹⁴

[edit] References

F. Mertens, "Über eine zahlentheoretische Funktion", Akademie Wissenschaftlicher Wien Mathematik-Naturlich Kleine Sitzungsber, IIa 106, (1897) 761-830.
A. M. Odlyzko and H.J.J. te Riele, "Disproof of the Mertens Conjecture", Journal für die reine und angewandte Mathematik 357, (1985) pp. 138-160.
Eric W. Weisstein, Mertens function at MathWorld.
Values of the Mertens function for the first 10,000 n are given by A002321.

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