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 Feb 20 revised Is the absolute value of the intersection of two functions related to the nontrivial zeros always equal to $\sqrt{2}$? Slightly rephrased the second question Feb 6 asked Is the absolute value of the intersection of two functions related to the nontrivial zeros always equal to $\sqrt{2}$? Sep 11 revised Integral representations of $\zeta(s)$ using the floor/frac functions. How could this one be derived? added the simple integral for zeta(s) in the domain Re(s) >1. Sep 11 asked Integral representations of $\zeta(s)$ using the floor/frac functions. How could this one be derived? Aug 29 revised Is there anything known about the zeros of $\displaystyle \sum_{n=1}^{\infty} \left(\frac{1}{\rho_n^s} +\frac{1}{\overline{\rho_n}^s}\right)$? Added comment to the graph (n=99) Aug 29 asked Is there anything known about the zeros of $\displaystyle \sum_{n=1}^{\infty} \left(\frac{1}{\rho_n^s} +\frac{1}{\overline{\rho_n}^s}\right)$? Jun 30 comment Does $\zeta(s)^2 \pm \zeta(1-s)^2$ have roots at the $\rho$s? Thanks mixedmath. It must indeed be the root finding algorithm that struggles here. I found the same issue in Mathematica that also only at a precision of $<40$ digits could find the $\rho$. I did indeed find the many other zeros as well, but given some earlier questions I raised on MO (e.g. this one mathoverflow.net/questions/89518/…), I believe these can all be proven to be on the critical line (which appears to be slightly more difficult to prove for the $\rho$s... :-) ). Jun 30 accepted Does $\zeta(s)^2 \pm \zeta(1-s)^2$ have roots at the $\rho$s? Jun 29 comment Does $\zeta(s)^2 \pm \zeta(1-s)^2$ have roots at the $\rho$s? Yes, all other root finding works fine up to very high accuracies and also for the exponent $1$ I find all the $\rho$s. However, when I plot the zeros for exponent $2$ at $1/2+s*i$ it can be clearly seen that the expected roots "at the $\rho$s" have a "parabolic dip", whilst the other zeros (that I all find at all accuracies) have a sharp downward "spike". You might be right that it is the shape of the curve at the zeros that confuses Newton's method. Jun 29 comment Does $\zeta(s)^2 \pm \zeta(1-s)^2$ have roots at the $\rho$s? Steven, I am using the RootFinding[Analytic] function in Maple and I believe this is based on Newton's method. Jun 29 asked Does $\zeta(s)^2 \pm \zeta(1-s)^2$ have roots at the $\rho$s? Jun 16 comment Does the Euler product for the Dirichlet $\beta$-function converge for all $\Re(s)>\frac12$? Chappers, you are probably right, although this is one of the very few Euler products that I know of that converges for $s=1$ ($\beta(1)$ is a well-known formula). Hence it could be an indication that further convergence exists. Numercal evidence does indeed suggest (very slow) convergence continues in the strip, however certainly not $\le \frac12$. Jun 16 revised Does the Euler product for the Dirichlet $\beta$-function converge for all $\Re(s)>\frac12$? Added elegant formula just for info. Jun 16 asked Does the Euler product for the Dirichlet $\beta$-function converge for all $\Re(s)>\frac12$? Jun 9 revised Are all complex zeros of ${\frac {\zeta \left( s+1 \right) }{\zeta \left( s-1 \right) }}\pm\, 2\,\pi\frac{2-s}{s\,(s+1)}$ on the critical line? removed additional question since it doesn't work. Jun 9 revised Are all complex zeros of ${\frac {\zeta \left( s+1 \right) }{\zeta \left( s-1 \right) }}\pm\, 2\,\pi\frac{2-s}{s\,(s+1)}$ on the critical line? added bit more text under addition and included a follow up question. Jun 7 revised Are all complex zeros of ${\frac {\zeta \left( s+1 \right) }{\zeta \left( s-1 \right) }}\pm\, 2\,\pi\frac{2-s}{s\,(s+1)}$ on the critical line? Added a further simplification for the equation inducing zeros on the line $\Re(s)=\frac12$ Jun 6 asked Are all complex zeros of ${\frac {\zeta \left( s+1 \right) }{\zeta \left( s-1 \right) }}\pm\, 2\,\pi\frac{2-s}{s\,(s+1)}$ on the critical line? May 18 comment Can the Riemann Zeta derivative be expressed in terms of Riemann Zeta? Anixx, it actually is defined for all indices in $\mathbb{C}$ of the polygamma. See link here: maplesoft.com/support/help/Maple/view.aspx?path=Psi I tested it in a Maple and it works fine. Was surprised to learn a function exists that links $\zeta(s)$ to its derivative. Through the functional equation you can then also link $\zeta'(s)$ and $\zeta'(1-s)$. However $\Psi^{s}(1)$ is defined in terms of $\zeta(s)$, so we won't gain much. May 14 revised Can the Riemann Zeta derivative be expressed in terms of Riemann Zeta? Added formula linking zeta'(s) and zeta(s) (but there is a snag).