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My conjecture is that all zeros in the strip $-1 \le \Re(s) \le 1$ of $\zeta(s) \pm \zeta(-s)$ are on the line $\Re(s)=0$.

I did find three complex zeros for $\pm =+$ (i.e. 12 in total) and two complex zeros for $\pm = -$ (i.e. 8 in total) that lie outside this strip, however their imaginary part seems capped at $|\Im(s)| < 6$ and also rapidly converges to the real zeros that in their turn converge to $s=2k, k \in \mathbb{Z}, |k| > 9$. The root-finding graph below for $\pm = -$ illustrates the point:

enter image description here

I could not find a function on the web that directly links $\zeta(s)$ to $\zeta(-s)$, but via this formula (derived here),

$$\displaystyle \frac{\xi(s)}{\xi(-s)} = \prod_{n=1}^\infty \left(1- \frac{1}{(\rho_n+s)} \right) \left(1- \frac{1}{(1-\rho_n +s)} \right)$$

with $\xi(s) = \frac12 s(s-1) \pi^{-\frac{s}{2}} \Gamma(\frac{s}{2}) \zeta(s)$ being the Riemann Xi-function and $\rho_n$ the n-th nontrivial zero of $\zeta(s)$, I derived that $\zeta(s) \pm \zeta(-s)=0$ is equivalent to:

$$\displaystyle -\frac{(s+1) \, 2 \pi^{s+1}}{s(s-1) \sin \left(\frac{\pi \, s}{2} \right) \Gamma \left(\frac{s}{2}\right)^2} \prod_{n=1}^\infty \left(1- \frac{1}{(\rho_n+s)} \right) \left(1- \frac{1}{(1-\rho_n +s)} \right) = \pm 1 $$

This function suggests that the non-trivial zeros could play a role in the conjecture (and maybe the RH even implies it), however grateful for any thoughts on how I might proceed in proving (or falsifying) this.

Many thanks.

ADDITIONAL QUESTION:

The Haseo Ki paper in Joro's answer below, proves that all but finitely many complex zeros of $\zeta(0+s) \pm \zeta(0-s)$ are on the line $\Re(s)=0$.

I now wondered if this implies that the same must be true for:

$$\zeta^{k}(0+s) \pm \zeta^{k}(0-s)$$

with $k=k^{th}$ derivative of $\zeta(s)$.

I tested it for $k=1..12$ and up to $\Im(s)=99$ and in the strip $-1 \le \Re(s) \le 1$. I again only find zeros lying on $\Re(s)=0$ (note there are always a finite few zeros outside this strip for lower $\Im(s)$). Is this a consequence of Haseo Ki's proof?

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1 Answer 1

up vote 9 down vote accepted

Finiteness of counterexamples follows from the paper On the zeros of sums of the Riemann zeta function, Haseo Ki

From p.1:

For fixed real number $\sigma_0$ define

$$ H(\sigma_0,s) = \zeta(\sigma_0 + s) + \zeta(\sigma_0 -s) \text { or } \zeta(\sigma_0 + s) - \zeta(\sigma_0 -s) $$

Your question is about $\sigma_0 = 0$.

From p. 3, Theorem 1:

2) If $\sigma_0 \le 0$, then all but finitely many complex zeros of $H(\sigma_0,s)$ are on $\Re(s)=0$.

This is unconditional.

A generalization:

1) If $\sigma_0 \le \frac12$, then all but finitely many complex zeros of $H(\sigma_0,s)$ are on $\Re(s)=0$ provided that $\zeta(s)$ has only finitely many complex zeros in $\Re(s) < \sigma_0$.

Related Theorem 3, p.4.

(1) Let $\sigma_0 < \frac12$ . Then, all zeros of $H(\sigma_0,s)$ in $|\Im(s)| \ge 100$ are on $\Re s = 0$ provided that $\zeta(s)$ in $\Re(s) < \sigma_0$ and $\Im(s) \ne 0$ has no zeros.

(2) The Riemann hypothesis holds if and only if for any $\sigma_0 <\frac12$ , all zeros of $H(\sigma_0,s)$ in $|\Im(s)| \ge 100$ are on $\Re(s) = 0$.

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Many thanks, Joro. Impressive answer! –  Agno Jan 26 at 22:04

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