7
$\begingroup$

In this question Agno asked about the zeros of $\zeta\left(\frac{s}{a}\right) \pm \zeta\left(\frac{1-s}{a}\right)$.

I fixed $a=2$ and the minus sign and defined:

$$ f(s)=\Re \left( \zeta\left(\frac{s}{2}\right) - \zeta\left(\frac{1-s}{2}\right)\right) $$

The X-Ray of $f(s)$ looks like this:

image1

The poles at $-1$ and $2$ distort it a bit.

This looked very close to ellipse or a degree $2$ algebraic curve to me. Tried to fit it to the form $Ax^2+Bxy+Cy^2+Dx+Ey+F=0$.

Numerical computations suggested:

A,B,C,D,E,F=1,0,1.1800298285336532, -0.9962733514416349,0,-1.997900275186291

Plotting the algebraic curve dotted blue we get:

enter image description here

This appears very good approximation to me.

Q1 Is the approximation good enough?

Q2 Is it expected expression involving zeta to be well approximated by algebraic curve?

Q3 would someone try to find high precision approximation and check if $f(s)$ vanishes?

$\endgroup$
  • $\begingroup$ The zeta function is given by a power series to the left of the critical strip. Doesn't this power series converges pretty quickly, which is why Riemann was able to calculate the first few zeroes by hand? So the real part is a two-variable power series that converges fairly quickly. You may be looking at a good polynomial approximation to the power series - the leading term, plus a small correction for some of the higher terms. Note that the approximation fails near the pole, which is causing the little nicks. $\endgroup$ – Will Sawin Feb 18 '15 at 14:55
  • $\begingroup$ @WillSawin Thanks, why not expand this as answer? I am not sure the approximation fails near the poles -- sage fails to plot the correct zeta extremely near the poles because of them. $\endgroup$ – joro Feb 18 '15 at 15:16
  • $\begingroup$ I might if I figure out how to show that the approximation is very good, and that this explains your observation. It has to fail near the poles because every number is a poor approximation of $\infty$. $\endgroup$ – Will Sawin Feb 18 '15 at 15:33
  • $\begingroup$ @WillSawin I continue to believe that the red curve of zeta is distorted in both my plots, almost sure because of the poles. $\endgroup$ – joro Feb 18 '15 at 17:02

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.