The conjecture Are all zeros of $\zeta(0+s) \pm \zeta(0-s)$ except a finite few on the line $\Re(s)=0$? was shown to be unconditionally true.

The proof can even be extended towards the domain $\sigma_0 \le 0$ with $\zeta(\sigma_0 + s) \pm \zeta(\sigma_0 -s)$.

Building on this, I started to experiment with the sum/difference of finite Euler products and now like to conjecture that with:

$$E(s,X) := \prod_{p \le X} \left( \dfrac{1}{1-p^{-s}} \right)$$

and $\sigma_0 \le 0$ and $X \ge 3$, all zeros (i.e. no exceptions) of:

$$E(\sigma_0 + s,X) \pm E(\sigma_0 -s,X)$$

are on the line $\Re(s)=0$.

Obviously couldn't test all values of $\sigma_0$, but is this provable?


I also experimented with the zeros of:

$$E(s,X) \pm E(1-s,X)$$

and found that in the critical strip and with $X \ge 2$, by far most of the zeros lie on the line $\Re(s)=\frac12$. For lower values of $X$, I did also find zeros off the critical line (within and outside the strip), but with an increasing $X$ these appear to 'crawl' towards the lines $\Re(s)=0,\frac12$ or $1$. However, I struggle to figure out what the exact final destiny of these zeros is when $X \rightarrow \infty$.

There are also real zeros for this formula, however only for any other prime. So, $E(s,X) + E(1-s,X)$ only has a real root for $s$ at $X=2,5,11,17,23,\dots$, whereas $E(s,X) - E(1-s,X)$ only vanishes for $s=\frac12$ and values of $s$ at $X=3,7,13,19,29,\dots$. Guess this is just something trivial that stems from the odd/even number of factors in the finite Euler product?


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.