Mertens function

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Mertens function to n=10 thousand
Mertens function to n=10 thousand
Mertens function to n=10 million
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 Re(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