Here is a plot of partial sums of Liouville Lambda and Moebius Mu: enter image description here

Notice the differences (in green) are tantalizingly close to $-n^{\frac{1}{2}}$.

Does this have any correlation to Merten's function?

Edit $$M(x)=O(x^{ \frac{1}{2}+\epsilon})$$ for some $\epsilon<\frac{1}{2}$

Edit tested to 50 million with $\epsilon=\frac{1}{32},$ no exceptions.

  • $\begingroup$ Mertens' function has no limit. Could you clarify what your question exactly is? $\endgroup$
    – Wojowu
    Aug 15, 2015 at 22:11
  • 1
    $\begingroup$ Did you plot any further? These functions are notoriously erratic, so this small sample may mean nothing at all... $\endgroup$ Aug 16, 2015 at 13:11
  • $\begingroup$ @MarkusSchepke, retested to 3M, found exceptions. Retested using $\epsilon=\frac{1}{32}$, no exceptions. $\endgroup$ Aug 16, 2015 at 15:43
  • $\begingroup$ @MarkusSchepke, It seems that the more $\lambda$ and $\mu$ become erratic, the less erratic are the differences. $\endgroup$ Aug 16, 2015 at 17:19

1 Answer 1


Let $\mu(n)$ denote the Mobius function, let $\lambda(n)$ denote the Liouville function, and let $M(x) = \sum_{n \leq x} \mu(n)$ and $L(x) = \sum_{n \leq x} \lambda(n)$ denote their summatory functions. The Riemann hypothesis implies $M(x) = O_{\varepsilon}(x^{1/2 + \varepsilon})$ and $L(x) = O_{\varepsilon}(x^{1/2 + \varepsilon})$.

In fact, if one assumes RH and the simplicity of the zeroes of $\zeta(s)$, then one can show that \[\frac{M(x)}{\sqrt{x}} = \sum_{\rho} \frac{1}{\zeta'(\rho)} \frac{x^{i\gamma}}{\rho} \] for $x$ a positive real that is not an integer. Here the sum is over the nontrivial zeroes $\rho = 1/2 + i\gamma$ of $\zeta(s)$.

Similarly, one can show under the same hypotheses that \[\frac{L(x)}{\sqrt{x}} = \frac{1}{\zeta(1/2)} + \sum_{\rho} \frac{\zeta(2\rho)}{\zeta'(\rho)} \frac{x^{i\gamma}}{\rho} \]

This explains why the functions look similar: as $x^{i\gamma} = \cos(\gamma \log x) + i \sin(\gamma \log x)$, these functions basically look like sums of trigonometric waves of decreasing amplitude and frequency.

So if one plots the difference, one finds that \[\frac{L(x) - M(x)}{\sqrt{x}} = \frac{1}{\zeta(1/2)} + \sum_{\rho} \frac{(\zeta(2\rho) - 1)}{\zeta'(\rho)} \frac{x^{i\gamma}}{\rho}. \] What you are seeing in your plots is the presence of the "main term" $1/\zeta(1/2)$. However, the contribution of the trigonometric waves cannot be ignored: although usually the "secondary terms" are not very large, very occasionally they can be big (and if one assumes a conjecture called the linear independence hypothesis, one can show that the secondary terms can get arbitrarily large).

For more on this phenomenon, see e.g. my paper here: http://arxiv.org/abs/1108.1524


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