Take the symmetrical form of the completed Zeta-function:

$\displaystyle \chi(s) \zeta(s) = \chi(1-s) \zeta(1-s)$


$\chi(s)=\pi^{-(\frac{s}{2})} \Gamma(\frac{s}{2})$.

For $s=\sigma + ti$, I conjecture that only for $\sigma=\frac12$:

$\Re(\chi(s)) = \Re(\zeta(s)) =0$ is always true, except when $s=\rho_n$ (assuming RH).

If $\sigma=\frac12$ and $t$ is real then $\Re(\chi(s))=0$ can be rewritten as:

$A(t) = \cot \left( \frac12 t\ln \left( \pi \right) \right) \dfrac{\Re \left( \Gamma \left(1/4+\frac{t i}{2} \right) \right)}{\Im \left(\Gamma \left( 1/4+\frac{t i}{2} \right) \right)} +1 = 0$

Similar to $s=\rho$ for non trivial zeros, let's call $s=\alpha$ when $A(t)=0$.

There seems to be a strong connection between the $\Re(\zeta(\alpha_m))$ and $\Re(\zeta(\rho_n))$. They exclusively come in an adjacent pair $(\alpha_m,\rho_n)$ or $(\rho_n,\alpha_m)$ that is connected via a 'sharp trough' of $\Re(\zeta(s))$ through the x-axis. The distance between the paired values appears to become smaller when $t$ grows.

If this paired pattern is true, then it would imply that when $\Re(\zeta(s))$ dives into a 'trough' through the x-axis and we find that $s =\alpha_m$, then one could predict with certainty that $s=\rho$ for the next time $\Re(\zeta(s)) = 0$ and that it must be located before $s=\alpha_{m+1}$. In that sense there would be information hidden in $\chi(s)$ that constrains the location of the $\rho$'s.

A bit complicated question, I know, but hope the picture below illustrates the thought.



  1. Has it been proven that $\Re(\chi(s)) = \Re(\zeta(s)) =0$ (and/or $\Im(\chi(s)) = \Im(\zeta(s)) =0$) only when $\sigma=\frac12$ and $s \ne \rho$ ?

  2. Is there anything known about $\alpha_m$ and $\rho_n$ always coming in connected pairs $(\alpha_m,\rho_n)$ or $(\rho_n,\alpha_m$) and/or that they converge when $t \rightarrow \infty$?


One additional afterthought:

Assuming RH and the adjacent paired values of $Re(\zeta(s))=0$ always having the shape:

$(\alpha,\rho)$ or $(\rho,\alpha)$,

then it would only require counting $\alpha$'s (from $t>1$) to establish the exact number of $\rho$'s $\pm1$ below a certain number $N$ (i.e. without even looking at $\zeta(s)$).


1 Answer 1


Your claim that "only for $\sigma=\frac12$: $\Re(\chi(s)) = \Re(\zeta(s)) =0$ is always true, except when $s=\rho_n$ (assuming RH)" appears false.

As seen on the X-Ray, $\Re(\chi(s)) = \Re(\zeta(s)) =0$ somewhere near $-4 + 8.5 i$. (The other common zero on the plot is indeed on 1/2).

Root finding in this case doesn't appear very easy.

Blue is $\Re(\zeta(s))=0$, red is $\Re(\chi(s))=0$ X-Ray

Another one:

  • $\begingroup$ Thanks Joro. This is a question I had almost forgotten about (or I did suppress it since it received two down votes...). You have convincingly shown that my conjecture is false, however I might restrict the claim to the critical strip only i.e.: $\Re(\chi(s)) = \Re(\zeta(s)) =0$ can only happen in the critical strip for $\sigma=\frac12$ and $s\ne\rho$ (the crossings in your graph between $0$ and $1$ appear to support this conjecture). $\endgroup$
    – Agno
    Commented Jun 6, 2013 at 21:31

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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