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Mertens function to n=10000
Mertens function to n=10,000,000

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 values −1, 0, and +1, it is 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 + \varepsilon}). 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 Big O notation.

Contents

[edit] Representations

[edit] As an integral

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 the second 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) + \cdots

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

 \oint_C F(s)e^{st} \, ds \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 \frac{x^s}{s \zeta (s)} \, ds = \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 \left(\frac{x}{t+1}\right) + x\int_0^x du \frac{y(u)} {u^{2}} = x^{-1}H(\log x)

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

Titchmarsh (1960) provided a Trace formula involving a sum over mobius function and zeros of Riemann Zeta in the form

 \sum_{n=1}^{\infty}\mu(n)n^{-1/2}g(logn) = \sum_t \frac{h(t)}{\zeta'(1/2+it)}+2\sum_{n=1}^\infty \frac{ (-1)^{n} (2\pi )^{2n}}{(2n)! \zeta(2n+1)}\int_{-\infty}^{\infty}dxg(x)exp(-x(2n+1/2))

here 't' sums over the imaginary parts of nontrivial zeros, and (g, h) are related by a Fourier transform so  \pi g(x)= \int_{0}^{\infty}du h(u) cos(ux)

[edit] As a sum over Farey sequences

Another formula for the Mertens function is

M(n)= \sum_{a\in \mathcal{F}_n} e^{2\pi i a}   where    \mathcal{F}_n   is the Farey sequence of order n.

This formula is used in the proof of the Franel–Landau theorem.[1]

[edit] As a determinant

M(n) is the determinant of the n × n Redheffer matrix, a (0,1) matrix in which aij is 1 if either j is 1 or i divides j.

[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 × 105
von Sterneck 1901 5 × 105
von Sterneck 1912 5 × 106
Neubauer 1963 108
Cohen and Dress 1979 7.8 × 109
Dress 1993 1012
Lioen and van der Lune 1994 1013
Kotnik and van der Lune 2003 1014

[edit] Notes

  1. ^ Edwards, Ch. 12.2

[edit] References




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