6
$\begingroup$

Numerical evidence suggests that all complex zeros (real ones exist as well) of:

$$\frac{\zeta'}{\zeta}(s) \pm \frac{\zeta'}{\zeta}(1-s)$$

reside on the critical line with $\Re(s)=\frac12$.

I made some progress by taking:

(1) $\zeta(s):=\chi(1-s)\,\zeta(1-s), \, \, \chi(s)= \Gamma \left( s \right) \cos \left(\frac12\pi s \right) 2 \left( 2\pi \right) ^{-s}$

(2) $\zeta(s):= -\dfrac{\zeta'(1-s)+\chi(s)\,\zeta'(s)}{\chi'(s)}$ (derived from Apostol's paper found here).

and then the formulae can be rewritten into:

\begin{align*} \frac{\zeta'}{\zeta}(s) + \frac{\zeta'}{\zeta}(1-s) &= -\dfrac{\chi'}{\chi}(s) \\ \\ \frac{\zeta'}{\zeta}(s) - \frac{\zeta'}{\zeta}(1-s) &= -\dfrac{\chi'}{\chi}(s) \cdot \dfrac{\chi(s)\,\zeta'(s)-\zeta'(1-s)}{\chi(s)\,\zeta'(s)+\zeta'(1-s)} \\ \end{align*}

The (+) version only has a single pair of zeros at $\frac12 \pm 6.2898359888369027796...$ and shows a monotonically increasing absolute value from that point onwards (note that the expected poles at the non-trivial zeros $\rho$ are all annihilated by $\zeta'(1-s)+\chi(s)\,\zeta'(s)$) that also induces the $\rho$s).

The (-) version does have a pole at each $\rho$, however these appear to be always separated by a single new zero that apparently always resides on the critical line. Could the latter be proved?

Thanks.

$\endgroup$
  • $\begingroup$ Remotely related: mathoverflow.net/questions/134029/… $\endgroup$ – joro May 10 '15 at 18:47
  • $\begingroup$ Thanks Joro. This one mathoverflow.net/questions/93323/are-all-zeros-of-ζks±ζk1−s-on-the-critical-line-k-k-th-derivative is also related, although in that question there are zeros outside the strip (contrary to the question above). $\endgroup$ – Agno May 10 '15 at 19:50
  • $\begingroup$ What do you about the expected poles of "+"? Zero off the line will induce pole in the LHS and in the RHS. Or am I missing something? $\endgroup$ – joro May 11 '15 at 6:50
  • $\begingroup$ I am not sure the RHS have poles at zeros. If this is true, this will give alternative computation of zeros -- look for poles of RHS, which is much simpler than zeta. $\endgroup$ – joro May 11 '15 at 11:52
  • $\begingroup$ Formula not involving zeta having zeros/poles at zeta zeros might be quite interesting. Agno, do you get poles of the "+" RHS at zeta zeros? $\endgroup$ – joro May 11 '15 at 12:21
3
$\begingroup$

See the following papers which treat this and similar questions.

  • MR1986257 (2004c:11152) Reviewed Saidak, Filip(3-CALG-MS); Zvengrowski, Peter(3-CALG-MS) On the modulus of the Riemann zeta function in the critical strip. (English summary) Math. Slovaca 53 (2003), no. 2, 145–172.

  • MR3277049 Reviewed Matiyasevich, Yu.(RS-AOS2); Saidak, F.(1-NCG); Zvengrowski, P.(3-CALG-MS) Horizontal monotonicity of the modulus of the zeta function, L-functions, and related functions. Acta Arith. 166 (2014), no. 2, 189–200.

$\endgroup$
  • $\begingroup$ Many thanks for these references, Peter. Very helpful. I have reviewed both documents now and immediately recognised the threshold at $|t|=6.2898..i$ (they put this at $2\,\pi+1$) below which the monotonicity breaks down (but of course that lower region can be inspected on the location of zeros by computer). The proof of the horizontal monotonicity for $\sigma < 0$ is nice, however it now also becomes clear to me that an extended proof towards the region $0 \le \sigma \le \frac12$ actually implies the RH. Could therefore a proof of my conjecture be constructed assuming the RH a priori? $\endgroup$ – Agno May 10 '15 at 20:04

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.