MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

A comment in OEIS about Mertens function claims:

The graph seems to show a negative bias for the Mertens function which is eerily similar to the Chebyshev bias (described in A156749 and A156709). The purported bias seems to be empirically approximated (by looking at the graph) $- (6 / \pi^2) (\sqrt{n} / 4)$ where $6 / \pi^2 = 1 / \zeta(2)$ is the asymptotic density of squarefree numbers (the squareful numbers having Moebius $\mu$ of 0). This would be a growth pattern akin to the Chebyshev bias. - Daniel Forgues, Jan 23 2011

Let $M(n)$ be the Mertens function.

Q1 Is there a bias (negative or positive) in Mertens function?

Q2 Is there a provable bias in some generalizations of Mertens function, e.g. in number fields or in arithmetic progression $M(an+d)$?

Plot of $M(x)$ with coloured bias:

Mertens function with coloured bias

share|cite|improve this question
It is certainly not the case that $\sum_1^{\infty}\mu(n)=-2$. There may be some summation method giving that value, but that's a different matter. – Gerry Myerson May 28 '12 at 7:18
Thank you Gerry deleted this sum. – joro May 28 '12 at 7:51
I've posted a lnk to an approach to the mertens-sum via Eulersummation. See the link to an image here . This was in the thread where there was also a very simple answer... – Gottfried Helms May 28 '12 at 9:16
Thanks Gottfried Helms. I linked to the answer in your question but deleted it after Gerry's comment. – joro May 28 '12 at 9:25
In a recent paper by R. P. Brent and J. van de Lune (arXiv:1112.4911) you may find the proof of $$\sum_{n=1}^\infty \mu(n)\frac{x^n}{1+x^n} = x-2x^2$$ Therefore $$\lim_{x\to1^+} \sum_{n=1}^\infty \mu(n)\frac{x^n}{1+x^n} = -1.$$ This appear to indicate a certain bias to the negative of Merten's function. – juan May 28 '12 at 17:41

Assuming some well-known conjectures, Nathan Ng established the existence of a limiting distribution for $e^{-y/2}M(e^{y})$. This paper can also be found on the arXiv:

The paper on summatory function of the Louiville function, mentioned in kolik's answer, uses many of the same techniques.

share|cite|improve this answer
Interesting, thanks. This paper doesn't seem to answer any of the questions? – joro May 28 '12 at 12:30
The point is that if you want to see if the mertens function is negatively biased it suffices to plot the graph of the distribution function – kolik May 28 '12 at 20:36
Assuming rh and Li there is in that paper an explicit formula for the distribution function in terms of the probability distribution of some random variables . Now, it might be computationally difficult to get a hold of it , but that's a different story. Note that the paper I posted below is more directly focused on biases – kolik May 28 '12 at 20:39

Somehow I missed this question when it was originally asked. I'm not entirely sure what you mean by a negative bias; the bits you've highlighted in the graph of $M(x)$ aren't when $M(x)$ is negative, but rather where is is, in some sense, "decreasing on average", and I don't know how to formalise that notion. If you're actually interested in the set of $x$ for which $M(x)$ is negative, then it seems clear from the graph that this happens roughly half the time. Here's how to make that notion formal.

In Nathan Ng's paper linked in Micah Milinovich's answer, Ng (very conditionally!) proves the existence of a limiting logarithmic distribution of $M(x)/\sqrt{x}$, so that there exists a measure $\nu$ satisfying $$\lim_{X \to \infty} \frac{1}{\log X} \int_{1}^{X}{f\left(\frac{M(x)}{\sqrt{x}}\right) \ \frac{dx}{x}} = \int_{\mathbb{R}}{f(x) \ d\nu(x)}$$ for every continuous bounded $f : \mathbb{R} \to \mathbb{R}$; equivalently, for every Borel $B \subset \mathbb{R}$ whose boundary has $\nu$-measure zero, $$\lim_{X \to \infty} \frac{1}{\log X} \int\limits_{\{x \in [1,X] : M(x)/\sqrt{x} \in B\}}{ \ \frac{dx}{x}} = \nu(B).$$ Furthermore, he calculates the Fourier transform of $\nu$ explicitly and shows, among other things, that $\widehat{\nu}$ is even about the origin. This implies that $$\lim_{X \to \infty} \frac{1}{\log X} \int\limits_{\{x \in [1,X] : M(x) < 0\}}{ \ \frac{dx}{x}} = \nu((-\infty,0)) = \nu((0,\infty)) = \lim_{X \to \infty} \frac{1}{\log X} \int\limits_{\{x \in [1,X] : M(x) > 0\}}{ \ \frac{dx}{x}} = \frac{1}{2}.$$ To show this, we need to know that the set $\{0\}$ has $\nu$-measure zero, but this follows as Ng's explicit formula for $\widehat{\nu}$ is in $L^1(\mathbb{R})$, and hence that $\nu$ is absolutely continuous with respect to the Lebesgue measure on $\mathbb{R}$. (I don't think Ng actually includes this argument, but it's in my paper that kolik linked in his answer.)

So this shows (conditionally) that the median on $\nu$ is $0$, and hence that $M(x)$ is unbiased, in the sense that the logarithmic density of the set of points where $M(x)$ is negative is the same as that of the set of points where $M(x)$ is positive.

Interestingly, if you consider instead the weighted sum $$M_{1/2}(x) = \sum_{n \leq x}{\frac{\mu(n)}{\sqrt{n}}},$$ then there is a negative bias; this follows from the same methods in my article that kolik linked to, together with the explicit expression $$M_{1/2}(x) = \frac{1}{\zeta(1/2)} + \sum_{\rho}{\frac{x^{\rho - 1/2}}{(\rho - 1/2) \zeta'(\rho)}} + R(x),$$ where the sum is over the nontrivial zeroes of the Riemann zeta function, and $R(x)$ is some small error term. Other than this, I don't know much about other biases, though it should be the same in number fields, and I recently wrote a paper about this kind of thing in function fields, based on previous work of Byungchul Cha (in which he does not explicitly state the analogous result of there being no bias, though it is clear from the results).

Edit: I forgot to mention Brent and van de Lune's recent paper, where they look at a form of the Lambert series generated by $\mu(n)$ and show that it is negative as $x$ tends to $1$ from below (as juan mentioned in a comment above). But this really isn't telling you anything other than that the Riemann zeta function is negative at $s = 0$, which is a much weaker statement than, say, a pole whose residue is negative (as is the case with the summatory function of the Liouville function).

share|cite|improve this answer

I believe that this paper

By one of the members of MO, will have some relevant answers for the closely related case of the Liouville function.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.