# Hausdorff dimension and Mertens function

Hello,

when one plots the Mertens function, it really looks like a fractal. So does anyone know the (approximate) value of the Hausdorff dimension of the set $\{(x,y),y=M(x),x\in\mathbb{R}^+\}$? Moreover, would there be a link between this Hausdorff dimension and the asymptotic behavior of the Mertens function? Thank you in advance.

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"would there be a link between this Hausdorff dimension and the asymptotic behavior of the Mertens function" - why do you think there might be? – Yemon Choi May 18 '11 at 22:52
Because in some sense the Hausdorff dimension of the set $\{(x,y),y=f(x),x\in\mathbb{R}^+\}$ measures the irregularity of the function $f$. And as $M(x)$ counts the number of squarefree integers with an even number of prime factors minus the number of squarefree integers with an odd number of prime factors, the maximal deviation between these two numbers might reflect this irregularity. Maybe one could expect something like $M(x)=O(x^{D-1+\varepsilon})$, with $D$ the Hausdorff dimension I'm talking about. – Sylvain JULIEN May 18 '11 at 23:06
Sylvain, your enthusiasm and curiosity are admirable. However, based on your various questions, I would like to gently recommend that you spend some time with a pile of number theory books. If you want to think seriously about things like the Selberg class or zeros of L-functions, you need to acquire a pretty strong mastery of analysis and modular forms. Henryk Iwaniec's books are an excellent place to start. – David Hansen May 19 '11 at 6:54

## 1 Answer

Well, as I understand this "Mertens function": it is constant on each interval $[n,n+1)$ where $n$ is a positive integer. Such a graph has Hausdorff dimension $1$. So perhaps you want some other kind of dimension...

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Spoilsport........ – Igor Rivin May 19 '11 at 1:59
"...many a beautiful theory was killed by an ugly fact.” (Attrib. T H Huxley) – Gerry Myerson May 19 '11 at 5:49