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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?

P.S.:

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?

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