1

Or stated differently: for $s \in \mathbb{C}$ and with $\chi(s)= \pi^{-s}\,2^{1-s}\,\cos\left(\frac{\pi\,s} {2}\right)\,\Gamma(s)$, do all, except a finite few, of the complex (real ones exist as well) zeros of:

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

reside on the line $\Re(s)=1$ ?

The finite few lying off the line are:

  • $\pm = +$ the exceptional set: $(5.894... \pm 1.389...\,i)$ , $(2- 5.894... \pm 1.389...\,i)$
  • $\pm = -$ the exceptional set: $(3.006... \pm 2.438...\,i)$ , $(2- 3.006... \pm 2.438...\,i)$

Could a proof for this be within reach or is it just as hard as the RH?

Thanks!

Added a graph of the + version on request.

enter image description here

Agno
  • 4,179
  • You found some zeros off $Re(s) = 1$, but you also found some zeros on (or near) $Re(s) = 1$ ? And a possibly useful theorem (Titchmarsh p.292) – reuns Nov 21 '16 at 17:34
  • user1952009, thanks for the link and will review. I conjecture that all zeros are on the line $\Re(s)=1$ except for the finite few I listed. – Agno Nov 21 '16 at 18:02
  • Yes, but why ? Did you find many zeros on (or near) $Re(s) = 1$ ? Can you add a plot for illustrating this ? – reuns Nov 21 '16 at 18:05
  • Sure, have added the graph. I only find zeros on the line $\Re(s)=1$ wherever I probe (except for the finite small few). – Agno Nov 21 '16 at 19:37
  • and how do you know if they are on $Re(s) = 1$ or just near it ? – reuns Nov 21 '16 at 20:00
  • Great question, I can't. The only indication I have, is that for increased precisions of say 60, 120 , 300 or $n$ digits, I continue to get a $1.$ with $n$ zeros for the real part. I think to prove even one zero being exactly on the line requires a similar approach to what Riemann used by successfully detecting one zero rather than two in an appropriately tiny region in the complex plane. Not sure how to do that for this function though. – Agno Nov 21 '16 at 22:24
  • is it real on $Re(s) = 1$ ? (and how do you prove that $\xi(s) =s(s-1)\pi^{-s/2}\Gamma(s/2) \zeta(s)$ is real on $Re(s) = 1/2$ ?) – reuns Nov 21 '16 at 23:06
  • When I define for $\pm=-$, $$\Xi_-(s)=s,(s-1),(s-2),\big(\big(\zeta(s-1) -\zeta(s)\big) - \big(\zeta(1-s) - \zeta(2-s)\big)\big)$$ I do get: $$\Xi_-(s)=\Xi_-(2-s)$$ hence $$\Xi_-(1+t,i)=\Xi_-(1-t,i)$$ and this is real on $\Re(s)=1$. For $\pm=+$, I can do something similar by multiplying with $s, i$ instead of $s$ giving: $$\Xi_+(s)=-\Xi_+(2-s)$$ – Agno Nov 22 '16 at 12:56
  • What you wrote is unclear. If $F(\overline{s}) = \overline{F(s)}$ and $F(-s) = F(s)$ then $F(i t)$ is real. If $F(\overline{s}) = \overline{F(s)}$ and $F(-s) = -F(s)$ then $i F(i t)$ is real. – reuns Nov 22 '16 at 17:17

1 Answers1

2

I do not think this is so difficult as the Riemann hypothesis, I will only explain why this is so without giving complete proof.

First on the line $s=1+it$ the functions are $$(\zeta(it)-\zeta(1+it))\pm (\zeta(-it)-\zeta(1-it)).$$ In other words $2\Re(\zeta(it)-\zeta(1-it))$ and $2i\Im(\zeta(it)+\zeta(1-it))$.

We have by the functional equation $$\zeta(it)-\zeta(1-it)=(\chi(it)-1)\zeta(1-it).$$ For $t$ real and $t\to+\infty$ we have $$\chi(it)-1=\Bigl(\frac{t}{2\pi}\Bigr)^{1/2}e^{i(-t\log\frac{t}{2\pi}+t+\frac{\pi}{4})}(1+O(t^{-1/2})).$$ The argument of $\zeta(1-it)$ is $O(\log t)$ [$O(\log\log\log t)$ under RH] and is zero at points $t_k\to+\infty$ (If taken $-\pi/2$ at $z=1$).

Therefore function $2\Re(\zeta(it)-\zeta(1-it))$ has approximately $\frac{T}{\pi}\log\frac{T}{2\pi}-\frac{T}{\pi}$ zeros in the interval $0<t<T$.

That these are essentially all zeros of this functions need a little work, but I think it is possible to prove it by counting the number of all zeros and comparing.

The other function is treated analogously.

juan
  • 6,976
  • I don't understand what you wrote. the function we are considering is real (resp. purely imaginary) on $Re(s) = 1$, and since its sign oscillates at $t \to \infty$, it has infinitely many zeros on $Re(s) = 1$. You suggest its number of zeros up to height $T$ is approximatively $\frac{T}{\pi} \log\frac{T}{2\pi}-\frac{T}{\pi}$, but you have no argument for saying it doesn't have infinitely many zeros off $Re(s) = 1$ – reuns Nov 22 '16 at 23:02
  • For proving $F(s)$ has a finite number of zeros off $Re(s) = 1$ you'll need a function $\Phi(s)$ such that $\text{arg} (\frac{F(s)}{\Phi(s)})$ is bounded on $Re(s) = a<1$ and $Re(s)= b > 1$ – reuns Nov 22 '16 at 23:21
  • @user1952009 But it is easy to obtain asymptotic expansion for vertical lines far of the critical strip. I said I will not give the details. – juan Nov 23 '16 at 07:18
  • I meant $\text{arg}(\frac{F(s)}{\Phi(s)})$ on $Re(s) = 1 \pm \epsilon$. With your method, I think you will be able to show that all but finitely many zeros are in the strip say $Re(s) \in [-2,4]$ – reuns Nov 23 '16 at 07:38
  • @user1952009 I have localized (T/pi) log(T/2pi)-T/pi in the line sigma=1. – juan Nov 23 '16 at 15:08
  • @user1952009 I have localized (T/pi) log(T/2pi)-T/pi in the line sigma=1. That these are all need a work. That the missing zeros are at most O( log T) is easy. I think it can be proved that there is at most a finite number off sigma=1. – juan Nov 23 '16 at 15:21
  • You have shown $F(s)$ has $\frac{T}{\pi}\log\frac{T}{2\pi}-\frac{T}{\pi}+ \mathcal{O}(\log T)$ zeros ($\mathcal{O}(\log \log T)$ under the RH) on $Re(s) = 1$, and on $Re(s) \le -2$ or $Re(s) \ge 4$ : $\text{arg}(F(s)) = \text{arg}(\chi(s)+\chi(2-s))+ \mathcal{O}(1) = \frac{T}{\pi}\log\frac{T}{2\pi}-\frac{T}{\pi}+ \mathcal{O}(1)$, so there is $ \frac{T}{\pi}\log\frac{T}{2\pi}-\frac{T}{\pi}+ \mathcal{O}(1)$ zeros in $Re(s) \in (-2,4)$ and at most $\mathcal{O}(\log T)$ zeros in $Re(s) \in (-2,4), Re(s) \ne 1$. But why you think it is possible to obtain a better bound, I have no idea. – reuns Nov 24 '16 at 04:56
  • Muchas gracias, Juan. Your proof would be very interesting especially since the spacing between the zeros seems quite irregular (contrary to for instance the proof of this question http://mathoverflow.net/questions/89324/are-all-zeros-of-gammas-pm-gamma1-s-on-a-line-with-real-part-frac12). Wondered whether for your proof you did apply similar techniques as used in this http://web.yonsei.ac.kr/haseo/p21-reprint.pdf or this https://arxiv.org/pdf/0712.1266v1.pdf paper? – Agno Nov 24 '16 at 10:14