24
$\begingroup$

The number theory identity $\phi(1) + \phi(2) + \dots + \phi(n) \approx \frac{3n^2}{\pi^2}$ can be interpreted as counting relatively prime pairs of numbers $0 \leq \{ x,y \} \leq n$ .

      

Has anyone studied the distribution of error term? $$\displaystyle \frac{1}{n} \left[\sum_{k=1}^n \phi(k) - \frac{3n^2}{\pi^2}\right]$$ It looks like white noise:

      

The histogram has a distinctive shape, maybe hard to prove. I suspect it's the Gaussian Unitary Ensemble (a Hermite polynomial times a Gaussian).

      

Similar questions:

Question concerning the arithmetic average of the Euler phi function:

averages of Euler-phi function and similar

Improve this question
$\endgroup$
3
  • 2
    $\begingroup$ Sorry to contribute more noise to this post, but somehow I feel that this question admits an easy (fourier) answer, but unfortunately, I lack the knowledge to turn my feeling into a concrete answer. $\endgroup$
    – Suvrit
    Commented May 3, 2012 at 19:37
  • $\begingroup$ @Suvrit Since $\sum_{k=1}^\infty \phi(k) k^{-s} = \frac{\zeta(s-1}{\zeta(s)}$ then $\sum_{k \le n} \phi(k)$ has an explicit formula (which converges in the sense of distributions) in term of the zeros of $\zeta(s)$ $\endgroup$
    – reuns
    Commented Oct 25, 2018 at 1:28
  • $\begingroup$ @reuns, this is surprisingly wrong: the main contribution of the error term for this formula does not come from the zeroes of $\zeta(s)$. This is explained quite well in doi.org/10.1016/j.jnt.2010.06.010 $\endgroup$
    – Peter Humphries
    Commented Oct 25, 2018 at 11:59

1 Answer 1

Reset to default
17
$\begingroup$

This is an interesting question. I don't think anyone has worked out what the distribution of the error term $$\frac{E(x)}{x} = \frac{1}{x}\left(\sum_{n \leq x}{\phi(n)} - \frac{3x^2}{\pi^2}\right)$$ actually looks like in any useful sense; from what I can make out, it seems to essentially be the same as the distribution of $$\sum_{n = 1}^{\infty}{\frac{\mu(n)}{n} \left\{\frac{x}{n}\right\}}$$ where $\mu(n)$ is the Möbius function and $\{x\}$ is the fractional part of $x$, but this doesn't really seem to tell you anything particularly useful.

Nevertheless, it certainly is known that $E(x)/x$ has a distribution function; this is proved on p.13 of "On the Existence of Limiting Distributions of Some Number-Theoretic Error Terms" by Yuk-Kam Lau. This also follows quite easily from the fact that $$\frac{E(x)}{x} = H(x) + O\left((\log x)^{-4}\right)$$ where $$H(x) = \sum_{n \leq x}{\frac{\phi(n)}{n}} - \frac{6 x}{\pi^2}$$ combined with the main result of the paper "The Existence of a Distribution Function for an Error Term Related to the Euler Function" by Erdős and Shapiro. What is better known is the average behaviour of $E(x)$, in the form of the asymptotics $$\sum_{n \leq x}{E(x)} \sim \frac{3 x^2}{2 \pi^2}$$ and $$\int^{x}_{0}{E(t)^2 dt} \sim \frac{x^3}{6 \pi^2}.$$

Probably the best reference for what is currently known about this error term is the paper "Oscillations of the remainder term related to the Euler totient function" by Kaczorowski and Wiertelak (but unfortunately this paper isn't available for free online).

EDIT: Very recently, Lemke Oliver and Soundararajan have reproven the existence of the limiting distribution of $E(x)/x$, as well as quantitative bounds for the tails of this distribution; see Theorem 1.3 of their paper.

Improve this answer
$\endgroup$
1
  • $\begingroup$ I like your first expression with the Mobius function. Since my plot has three peaks, they may be related to the three values of $\mu(x) = -1,0,1$. $\endgroup$
    – john mangual
    Commented May 5, 2012 at 12:00

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .