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Take the symmetrical form of the completed Zeta-function:

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

with

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

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.

http://imageshack.us/photo/my-images/855/riemannzero.jpg/

Questions:

  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 $t \rightarrow \infty$?

Thanks.


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 $\zeta(s)$).

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1 Answer 1

up vote 2 down vote accepted

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:

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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). –  Agno Jun 6 '13 at 21:31

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