Solutions of $\zeta(s) = 1$, $\zeta(\zeta(s)) = 1$ near a line and a circle, respectively? - MathOverflow

Solutions of $\zeta(s) = 1$, $\zeta(\zeta(s)) = 1$ near a line and a circle, respectively?

2012-06-05T21:21:32Z

Are solutions of $\zeta(s) = 1$ very near a line $\Re(s) = 54$ and solutions of $\zeta(\zeta(s)) = 1$ either on or very near a circle with center $\approx .00936$ tangent to $\Re(s) = 1$, known to exist? I seem to be observing these things in computer experiments. The solutions to the first equation (zeta images of the solutions to the second one) look like they lie on circles of radius $\approx 4 \times 10^{-10}$ centered on points of the form $54 + i k \pi/\log(2), k$ odd.

Is this ridiculous? What might produce such an illusion?

Edit: To illustrate, here is a link to a PDF with graphics of the phenomenon: http://barrybrent.9f.com/zeta=1.pdf. (~ a meg.)

There are other such apparent circles for higher zeta iterates, which I'll show in a later draft, if the observations aren't knocked down.

Barry Brent

Answer by joro for Solutions of $\zeta(s) = 1$, $\zeta(\zeta(s)) = 1$ near a line and a circle, respectively?

2012-06-06T08:00:30Z

About $\zeta(s)=1$. Your "illusion" about zeros close to $\Re(s)=54$ well might be caused by working with insufficient precision. Probably zeta is so close to $1$ your precision believes it is exactly $1$.

Does the "illusion" disappear if you work with more precision?

My results with sage/mpmath:

With precision 16 decimal digits find a lot of zeros close to $\Re(s)=77$.

With precision 40 digits the zeros move to $\Re(s)=82$.

With precision 100 can't find large zeros fast, only with $\Re(s)<2$.

The sum for zeta for $\Re(s)>1$ explains why you get close to $1$ for large $\Re(s)$.

For $\zeta(\zeta(s))=1$ the solution again appear to be related to insufficient precision. The solutions I found are with large $\Re(\zeta(s))$, again tending to $1$.

Answer by juan for Solutions of $\zeta(s) = 1$, $\zeta(\zeta(s)) = 1$ near a line and a circle, respectively?

2012-06-06T19:21:04Z

Any solution to $\zeta(s) = 1$ must have real part $\sigma \le \sigma(1)$ where $\sigma(1)$ is equal to the unique solution $\sigma>1$ of the equation $$\zeta(\sigma)=\frac{2^\sigma+1}{2^\sigma-1}$$ Numerically $\sigma(1)=1.9401016837\dots$

$\sigma(1)$ is the best possible constant here.

See the paper arXiv:1107.5134 where other problems of this type are considered.

About the solutions of $\zeta(\zeta(s))=1$. There are a countable set of solutions to $\zeta(s) = 1$. $s_1$, $s_2$, $s_3$, $\dots$ These solutions are situated mainly very near the critical line. Near the pole $\zeta (s)\sim (s-1)^{-1}+ \gamma $ if you solve $(s-1)^{-1}+\gamma = 1/2+it$ you find a circle with center at $1+(1/2-\gamma)^{-1} /2=-5.47537$ and radius 6.47537.
Therefore one expects to find solutions of $\zeta(s) = s_k$ for points simultaneously near this circle and near the point 1. With Mathematica you find for example one of this in this way

a = s /. FindRoot[Zeta[s] - 1, {s, N[ZetaZero[10000]]}, WorkingPrecision -> 50]; FindRoot[Zeta[s] - a, {s, 1 + (a - EulerGamma)^(-1)}, WorkingPrecision -> 50]

That gives you the solution

s= 1.0000000002505104088470167417362938109319852591130 - 0.00010123550290056930653742177989540980110048808885363 I

That satisfies $\zeta(\zeta(s))=1$