Show that

$$\sum_{n\le x} \mu ^2(n)=\frac{x}{\zeta(2)}+o(\sqrt{x}) \; (x\to \infty)$$

I've proven so far that $\sum_{n\le x} \mu ^2(n)=\frac{x}{\zeta(2)}+O(\sqrt{x})$.

I want to reduce this error term $O(\sqrt{x})$ to $o(\sqrt{x})$, and I'm thinking of using the relation $M(x)=\sum_{n\le x}\mu(n)=o(x)$ which is equivalent to the Prime Number Theorem. How can I use this relation to reduce the error term?

I would greatly appreciate any help.

  • I am now writing down an elementary proof. I write this comment only because I expect to make errors on the fly, as I prove and write at the same time, so it might take me a few moments. – davidlowryduda Jun 13 '15 at 4:19
  • it is a basic general property of Dirichlet series that the abscissa of convergence is equal to the $b$ of the $\mathcal{O}(x^{b+\epsilon})$ term of $\sum_{n\le x} a_n$. I would advice to read : the "general theory of dirichlet series". then consider the Dirichlet series $\frac{\eta(s)}{\zeta(2s)}$. under the PNT its abscissa of convergence is $< 1/2$. done. – reuns Jul 18 '15 at 12:12
up vote 6 down vote accepted

First, let's see if we can understand why this "should" be true from an analytic perspective.

We know that $$ \sum_{n \geq 1} \frac{\mu(n)^2}{n^s} = \frac{\zeta(s)}{\zeta(2s)},$$ and a general way of extracting information from Dirichlet series is to perform a cutoff integral transform (or a type of Mellin transform). In this case, we get that $$ \sum_{n \leq X} \mu(n)^2 = \frac{1}{2\pi i} \int_{(2)} \frac{\zeta(s)}{\zeta(2s)} X^s \frac{ds}{s},$$ where the contour is the vertical line $\text{Re }s = 2$. By Cauchy's theorem, we shift the line of integration left and poles contribute terms or large order. The pole of $\zeta(s)$ at $s = 1$ has residue $$ \frac{X}{\zeta(2)},$$ so we expect this to be the leading order. Naively, since we know that there are no zeroes of $\zeta(2s)$ on the line $\text{Re } s = \frac{1}{2}$, we might expect to push our line to exactly there, leading to an error of $O(\sqrt X)$. But in fact, we know more. We know the zero-free region, which allows us to extend the line of integration ever so slightly inwards, leading to a $o(\sqrt X)$ result (or more specifically, something along the lines of $O(\sqrt X e^{-c (\log X)^\alpha})$ where $\alpha$ and $c$ come from the strength of our known zero-free region.

In this heuristic analysis, I have omitted bounding the top, bottom, and left boundaries of the rectangles of integration. But proceeding in a similar way as in the proof of the analytic prime number theorem, you could proceed here. So we expect the answer to look like $$ \frac{X}{\zeta(2)} + O(\sqrt X e^{-c (\log X)^\alpha})$$ using no more than the zero-free region that goes into the prime number theorem.

We will now prove this result, but in an entirely elementary way (except that I will refer to a result from the prime number theorem).


We do this in a series of steps.

Lemma

$$\sum_{d^2 \mid n} \mu(d) = \begin{cases} 1 & \text{if } n \text{ is squarefree} \\ 0 & \text{else} \end{cases}$$

Proof. This comes almost immediately upon noticing that this is a multiplicative function, and it's trivial to prove it for prime powers. $\spadesuit$

So to sum up the squarefree numbers up to $X$, we look at $$ \sum_{n \leq X} \sum_{d^2 \mid n} \mu(d) = \sum_{d^2e \leq X} \mu(d)= \sum_{d^2 \leq X} \mu(d) \left\lfloor \frac{X}{d^2} \right\rfloor.$$

This last expression is written in one of the links in Marty's answer, and they prove it with inclusion-exclusion. I happen to find this derivation more intuitive, but it's our launching point forwards.

We approximate the floored bit. Notice that $$ \left \lfloor \frac{X}{d^2} \right \rfloor = \frac{X}{d^2} + E(X,d)$$ with $\lvert E(x,d) \rvert \leq 1$ (we think of it as the Error of the approximation). So the number of squarefree numbers up to $X$ is $$ \sum_{d^2 \leq X} \mu(d) \frac{X}{d^2} + \sum_{d^2 \leq X}\mu(d) E(x,d).$$ We look at the two terms separately.

The first term

The first term can be approximated by the infinite series plus an error term. $$ \sum_{d^2 \leq X} \frac{\mu(d)}{d^2} = X\sum_{d \geq 1} \frac{\mu(d)}{d^2} - X\sum_{d > \sqrt X} \frac{\mu(d)}{d^2} = \frac{X}{\zeta(2)} - X\sum_{d > \sqrt X} \frac{\mu(d)}{d^2}.$$

We must now be a bit careful. If we perform the naive bound, by bounding $\mu(n) \leq 1$, then this last sum is of size $O(\sqrt X)$. That's too big!

So instead, we integrate by parts (Riemann-Stieltjes integration) or equivalently we perform summation by parts to see that $$ \sum_{d > \sqrt X} \frac{\mu(d)}{d^2} = O\left( \int_{\sqrt X}^\infty \frac{M(t)}{t^3} dt \right)$$ where $$ M(t) = \sum_{n \leq t} \mu(n).$$ By the prime number theorem (and as you mention in your question), we know that $M(t) = o(t)$. (In fact, the analytic prime number theorem in one of the easy forms is that $M(X) = O(X e^{-c (\log X)^{1/9}})$, which we might use here). This means that this last term is bounded by $$ o(\sqrt X)$$ if we just use that $M(t) = o(t)$, or $$ O(\sqrt X e^{-c (\log X)^{1/9}})$$ if we use more. This completes the first term $\spadesuit$

Second term

This is easier now. We again use integration by parts. Notice that $$ \begin{align} \sum_{d \leq \sqrt X} \mu(d) E(X,d) &= \sum_{d \leq \sqrt X} (M(d) - M(d - 1))E(X,d) \\ &= M(\lfloor \sqrt X \rfloor) E(X, \lfloor \sqrt X \rfloor) + \sum_{d \leq \sqrt X - 1} M(d) (E(X, d) - E(X, d+1)). \end{align}$$

By using that $E(X,d) \leq 1$ and $M(\sqrt X) = o(\sqrt X)$ (or the stronger version), we match the results from the first part. $\spadesuit$

Putting these two results together, we have proven that the number of squarefree integers up to $X$ is $$\frac{X}{\zeta(2)} + o(\sqrt X)$$ using only that $M(X) = o(\sqrt X)$, and alternately $$ \frac{X}{\zeta(2)} + O(\sqrt X e^{-c (\log X)^{1/9}})$$ using a bit of the zero-free region. This completes the proof. $\diamondsuit$

  • 1
    Very nice. If you just wrote this up, I am awestruck; if you had this previously written, I am merely extremely impressed. Upvoted. – marty cohen Jun 13 '15 at 5:08
  • @mixedmath Two questions. First, in the proof of the first term, how do we get that the last term is bounded by $o(\sqrt{X})$ from $M(t)=o(t)$. I think you omitted some details, but I don't follow this part. Second, in the proof of the second term, you use $M(X)=o(\sqrt{X})$. Where did this come from? I thought we only had $M(x)=o(x)$. – takecare Jun 13 '15 at 7:14
  • 1
    @takecare to your first question, you can just stick that in the integral and integrate it. If $f < g$ then $\int_a^b f dx < \int_a^b g dx$. For the second, it was a typo. We're only summing up to $\sqrt X$, so I used that $M(\sqrt X) = o(\sqrt X)$. I've corrected that now. – davidlowryduda Jun 13 '15 at 8:49
  • @mixedmath Sorry but I don't follow some parts of the proof. How can we integrate $o(t)/t^3$? Also in the second term, how do we show $\sum_{d\le \sqrt{X}-1} M(d)(E(X,d)-E(X,d+1))=o(\sqrt{X})$? I don't see how these results follow easily. – takecare Jun 13 '15 at 14:30
  • 1
    @takecare to be $o(t)$ means to be strictly smaller than $t$. So $o(t)/t^3 = o(1/t^2)$. Then $\int o(1/t^2)dt = o( \int 1/t^2 dt) = o(1/t)$. We're evaluating from $\sqrt X$ to $\infty$. At $\infty$, it's zero. So we are left with $o(1/ \sqrt X)$. But remember that we're multiplying by $X$ on the outside, so we're left with $o(\sqrt X)$. – davidlowryduda Jun 13 '15 at 20:50

I did a search for "number of squarefree integers". Almost all results give the error as $O(\sqrt{x})$, so it seems that doing better is hard.

In this paper (http://arxiv.org/pdf/1107.4890v1.pdf) they quote this result:

Under the assumption of the Riemann hypothesis the error term can be further reduced [3]: $S(n) = \frac{6}{π^2}n + O(n^{17/54+ε}) $.

This is the paper referenced: 3. Jia, C.H.: The distribution of square-free numbers. Science in China Series A: Mathematics 36(2), 154–169 (1993)

If I find anything more, I'll edit this answer.


Here's a paper with a proof that the error is $O(n^{2/5+\epsilon})$:

"On the order of the error function of the square-free numbers" at http://www.new.dli.ernet.in/rawdataupload/upload/insa/INSA_1/20005abb_196.pdf .

  • $S(n) = \frac{n}{\zeta(2)} + \mathcal{O}(n^{\sigma_0/2+\epsilon})$ where $\sigma_0$ is the greater real part of the zeta's zeros. so that if you prove something like $S(n) = \frac{n}{\zeta(2)} + \mathcal{O}(n^{a+\epsilon})$ then $\sigma_0 \le 2a$ and you proved a zero-free strip for zeta which NOBODY DID UNTIL TODAY !!!!! $\displaystyle\frac{\zeta(s)}{s \zeta(2s)} = \int_1^\infty S(x) x^{-s-1}dx$, $\displaystyle\frac{\eta(s)}{s \zeta(2s)} = \int_1^\infty (S(x)-2S(x/2)) x^{-s-1}dx$, so that if $S(x)-2S(x/2) = \mathcal{O}(x^a)$ you immediatly get a zero free strip for $\zeta(2s)$ ... – reuns Jul 18 '15 at 12:03
  • under the Riemann hypothesis of course $S(n) = \frac{n}{\zeta(2)} + \mathcal{O}(n^{1/4+\epsilon})$ no more no less – reuns Jul 18 '15 at 12:07

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