3
$\begingroup$

Let $\mu(n)$ be the Mobius function, how to estimate $$\sum_{1\le i<j\le x}\mu(i)\mu(j) $$ as $x$ goes to $\infty$? Are there some references on this?

$\endgroup$
12
$\begingroup$

If $S$ is your sum then

$$ \left\lvert \sum_{1\leq n\leq x}\mu(n)\right\rvert^2 = 2S+ \sum_{1\leq n\leq x}\mu(n)^2.$$

The second sum on the right is $(\frac{6}{\pi^2}+o(1))x$, and hence estimating $S$ is equivalent to estimating $\lvert \sum_{n\leq x}\mu(n)\rvert$, a classical problem of analytic number theory.

In particular, assuming the Riemann Hypothesis, the left-hand side is $O(x^{1+o(1)}$), and hence (assuming RH)

$$ S \ll x^{1+o(1)}.$$

Unconditionally we can show that $S=o(x^2)$, but cannot show $S\ll x^{2-\epsilon}$ for any $\epsilon>0$.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ "The best we can conclude unconditionally is $S = o(x^2)$". What do you mean by this? I'm no analytic number theorist, but I'd be surprised if $|\sum_{n \le x} \mu(n)| \ll \frac{x}{\log\log x}$ wasn't known unconditionally. $\endgroup$ – mathworker21 Jul 18 at 2:43
  • 1
    $\begingroup$ Yes, sorry, I was being imprecise - I meant that in terms of improving the exponent nothing better is known. The standard proof of the prime number theorem can be adapted to show that $S\ll x^2 \exp(-O(\sqrt{\log x}))$, for example, and I believe better quantitative results are known. $\endgroup$ – Thomas Bloom Jul 18 at 9:43

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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