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Mertens function

From Wikipedia, the free encyclopedia

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

In number theory, the Mertens function is defined for all positive integers n as

where μ(k) is the Möbius function. The function is named in honour of Franz Mertens. This definition can be extended to positive real numbers as follows:

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

The first 143 M(n) is: (sequence A002321 in the OEIS)

M(n) +0 +1 +2 +3 +4 +5 +6 +7 +8 +9 +10 +11
0+ 1 0 −1 −1 −2 −1 −2 −2 −2 −1 −2
12+ −2 −3 −2 −1 −1 −2 −2 −3 −3 −2 −1 −2
24+ −2 −2 −1 −1 −1 −2 −3 −4 −4 −3 −2 −1
36+ −1 −2 −1 0 0 −1 −2 −3 −3 −3 −2 −3
48+ −3 −3 −3 −2 −2 −3 −3 −2 −2 −1 0 −1
60+ −1 −2 −1 −1 −1 0 −1 −2 −2 −1 −2 −3
72+ −3 −4 −3 −3 −3 −2 −3 −4 −4 −4 −3 −4
84+ −4 −3 −2 −1 −1 −2 −2 −1 −1 0 1 2
96+ 2 1 1 1 1 0 −1 −2 −2 −3 −2 −3
108+ −3 −4 −5 −4 −4 −5 −6 −5 −5 −5 −4 −3
120+ −3 −3 −2 −1 −1 −1 −1 −2 −2 −1 −2 −3
132+ −3 −2 −1 −1 −1 −2 −3 −4 −4 −3 −2 −1

The Mertens function slowly grows in positive and negative directions both on average and in peak value, oscillating in an apparently chaotic manner passing through zero when n has the values

2, 39, 40, 58, 65, 93, 101, 145, 149, 150, 159, 160, 163, 164, 166, 214, 231, 232, 235, 236, 238, 254, 329, 331, 332, 333, 353, 355, 356, 358, 362, 363, 364, 366, 393, 401, 403, 404, 405, 407, 408, 413, 414, 419, 420, 422, 423, 424, 425, 427, 428, ... (sequence A028442 in the OEIS).

Because the Möbius function only takes the values −1, 0, and +1, the Mertens function moves slowly and there is no x such that |M(x)| > x. The Mertens conjecture went further, stating that there would be no x where the absolute value of the Mertens function exceeds the square root of x. The Mertens conjecture was proven false in 1985 by Andrew Odlyzko and Herman te Riele. However, the Riemann hypothesis is equivalent to a weaker conjecture on the growth of M(x), namely M(x) = O(x1/2 + ε). Since high values for M(x) grow at least as fast as the square root of x, this puts a rather tight bound on its rate of growth. Here, O refers to Big O notation.

The true rate of growth of M(x) is not known. An unpublished conjecture of Steve Gonek states that

Probabilistic evidence towards this conjecture is given by Nathan Ng.[1] In particular, Ng gives a conditional proof that the function has a limiting distribution on . That is, for all bounded Lipschitz continuous functions on the reals we have that

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Transcription

Contents

Representations

As an integral

Using the Euler product one finds that

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

where c > 1.

Conversely, one has the Mellin transform

which holds for .

A curious relation given by Mertens himself involving the second Chebyshev function is

Assuming that the Riemann zeta function has no multiple non-trivial zeros, one has the "exact formula" by the residue theorem:

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

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

There is also a trace formula involving a sum over the Möbius function and zeros of the Riemann zeta function in the form

where the first sum on the right-hand side is taken over the non-trivial zeros of the Riemann zeta function, and (g,h) are related by the Fourier transform, such that

As a sum over Farey sequences

Another formula for the Mertens function is

  where     is the Farey sequence of order n.

This formula is used in the proof of the Franel–Landau theorem.[2]

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.

As a sum of the number of points under n-dimensional hyperboloids[citation needed]

This formulation expanding the Mertens function suggests asymptotic bounds obtained by considering the Piltz divisor problem which generalizes the Dirichlet divisor problem of computing asymptotic estimates for the summatory function of the divisor function.

Calculation

Neither of the methods mentioned previously leads to practical algorithms to calculate the Mertens function. Using sieve methods similar to those used in prime counting, the Mertens function has been computed for all integers up to an increasing range of x.[3][4]

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 de Lune 1994 1013
Kotnik and van de Lune 2003 1014
Hurst 2016 1016

The Mertens function for all integer values up to x may be computed in O(x log log x) time. Combinatorial based algorithms can compute isolated values of M(x) in O(x2/3(log log x)1/3) time, and faster non-combinatorial methods are also known.[5]

See OEISA084237 for values of M(x) at powers of 10.

Known upper bounds

Ng notes that the Riemann hypothesis (RH) is equivalent to

for some positive constant . Other upper bounds have been obtained by Maier, Montgomery, and Soundarajan assuming the RH including

Other explicit upper bounds are given by Kotnik as

See also

Notes

  1. ^ Ng
  2. ^ Edwards, Ch. 12.2
  3. ^ Kotnik, Tadej; van de Lune, Jan (November 2003). "Further systematic computations on the summatory function of the Möbius function". MAS-R0313.
  4. ^ Hurst, Greg (2016). "Computations of the Mertens Function and Improved Bounds on the Mertens Conjecture". arXiv:1610.08551 [math.NT].
  5. ^ Rivat, Joöl; Deléglise, Marc (1996). "Computing the summation of the Möbius function". Experimental Mathematics. 5 (4): 291–295. ISSN 1944-950X.

References

This page was last edited on 4 January 2019, at 20:49
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