Extrema of $\xi(1/2+it)$ for the Landau Riemann xi function

Question

"Relations and positivity results for derivatives of the Riemann xi function" by Coffey characterizes the Riemann xi function and its relation to the Riemann zeta function, and there has been much ado about their zeros. $\Omega(t) = \xi(1/2+it)$ is a real-valued, entire, even funcion for $t$ real. Its zeros are at the imaginary part of the nontrivial zeros of the Riemann zeta.

I'm looking for references on the extrema of $\Omega(t)$, those of truncations $\Omega_n(t)$ to lower orders of its Taylor series, and the zeros and extrema of the Appell polynomials with the exponential generating functions $\Omega_n(t)e^{xt}$ and $\Omega(t)e^{xt}$.

Stat mech models of zeros of the Riemann zeta have been of perennial interest--GUE models, for example. Just last month Rodgers and Tao in "The de Bruijn-Newma constant is non-negative" (see also the presentation by Terry Tao "Vaporizing and freezing the Riemann zeros") explored such in a modified rep of zeta, and in "Dynamic behavior of the roots of the Taylor polynomials of the Riemann xi function with growing degree" Jenkins and Mclaughlin look at another kind of perturbation of the zeros. I'm interested in whether the local extrema provide additional info since they (for omega) have a height as well as an abscissa and, of course, are correlated with the locations of the zeros of zeta/xi/omega and associated Jensen/Appell polynomials.

Answer 1

• Proving that $\xi(s)$ is fast decaying on $\Re(s)=1/2$ and giving an upper bound is easy, from that $\Gamma(s)$ is Schwartz (and $O(e^{-(\pi/2-\epsilon) t})$) on vertical lines and that $\zeta(s)$ is polynomially bounded. Then from the explicit fast decreasing bound you can find the maximum and minimum of $\xi(1/2+it)$ numerically.

• Under the RH the local maximum/minimum are located between the non-trivial zeros and we can find there are plenty of conjectures regarding how the two are related.

• The truncation of its Taylor series at $s_0$ is a polynomial it doesn't make much sense to ask for extrema. Since $\xi$ is entire, the zeros of the truncation polynomials converge to those of $\xi$.

  Because $\xi(1/2+it)$ is real, its Taylor series at $t=0$ has real coefficients. Thus for a simple critical line zero, because a complex root of the polynomial will come in pair, then for $n$ large enough the approximating zero of the truncation will be on the critical line.