0
$\begingroup$

It is known that the Riemann conjecture is equivalent to $$M(x) = O(x^{\frac12+\epsilon}),$$ where M(x) is the Mertens function.

Does there exist an analogue to this equivalence for the generalized Riemann conjecture ? (say a generalized Mertens function defined somehow with characters, and an asymptotic behavior)

$\endgroup$
  • 1
    $\begingroup$ Sure $L(s,\chi)$ has no zeros for $\Re(s) > 1/2$ iff $\sum_{n \le x} \mu(n) \chi(n) = O(x^{1/2+\epsilon})$. For other L-functions it works quite the same way (but this is special to them, see $\eta(s-1)$ whose series only converges for $\Re(s) > 1$) $\endgroup$ – reuns Nov 18 '18 at 15:25
  • $\begingroup$ do you have a source for this ? $\endgroup$ – MikeTeX Nov 18 '18 at 15:30
  • $\begingroup$ The proof is the same as for $\zeta(s)$ and is quite similar to the PNT. Search about the PNT in arithmetic progressions. $\endgroup$ – reuns Nov 18 '18 at 15:31
  • 1
    $\begingroup$ math.stackexchange.com/questions/1619762/… $\endgroup$ – Peter Humphries Nov 22 '18 at 8:32
  • $\begingroup$ Yes, this completely answers my question. Thanks. $\endgroup$ – MikeTeX Nov 25 '18 at 8:39

Your Answer

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

Browse other questions tagged or ask your own question.