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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

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2 Answers

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

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Juan, thanks. Barry –  Barry Brent Jun 6 '12 at 20:51
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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 $1$.

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This isn't exactly what I was asking. The linked paper already mentions the precision issues in the discussion of $\Re(s) \approx 222.48$. In view of the graphics included in the PDF (linked above) the geometry of the graph of zeta, not the numerical estimates, suggested the proposal. So I asked, not whether it's true, but whether it's obviously wrong. –  Barry Brent Jun 6 '12 at 15:29
Haven't read the paper. Just pointed out I can't find any zero of zeta(s)=1 near your line Re(s)=54 and my results depend on the precision as explained in the answer. –  joro Jun 6 '12 at 15:36
Not sure what is the "illusion" you are asking about. –  joro Jun 6 '12 at 15:53
By illusion I meant 'optical illusion'. I was asking, were my observations ruled out by theory. Circles, or curves very near circles, appear in the PDF graphics. I see above that Juan has answered me saying that the phenomenon actually is ruled out. You're right though in the sense that precision issues at 'microscopic' scales in my plots may have caused a visual illusion. Barry –  Barry Brent Jun 6 '12 at 21:03
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