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From parametric plots of $\zeta \left( \frac{1}{2} + it \right)$ it seems to be the case that:
(1) except for $\zeta(\frac{1}{2})$ the Riemann zeta function does not attain any negative real value on the critical line.
(2) the curve $(t, \zeta(1/2+it))$ is dense in the complex plane.

Are these statements known to be false, if not, is there any proof affirming them?

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(2) is an open problem and is conjectured to be true. This conjecture is usually attributed to Ramachandra. –  Micah Milinovich Aug 17 '11 at 20:19
For a recent paper supporting this conjecture see –  quid Aug 17 '11 at 20:27
The first conjecture seems to be disproved by the graph at… It doesn't say which of the curves shows the real part and which the imaginary, but either way there's a $t \in (10^{10}+53,10^{10}+54)$ where one of them crosses the $x$-axis and the other is negative. –  Noam D. Elkies Aug 17 '11 at 21:17
If the dotted line is the imaginary part, instead of the dashed line, then there is no point in this graph where the imaginary curve crosses 0 and meanwhile the dashed curve is negative. But I do get the point. –  Eren Mehmet Kiral Aug 17 '11 at 22:21
True — I somehow misread the graph. But see below for an actual counterexample. –  Noam D. Elkies Aug 17 '11 at 23:58

5 Answers 5

up vote 21 down vote accepted

A numerical counterexample to the first conjecture is $$ t = 282.4547208234621746108397940690599354\ldots $$ where both gp and Wolfram Alpha agree that $\zeta(\frac12 + it)$ has negative real part $\simeq -0.02763$ and negligible imaginary part, so the actual zero of ${\rm Im}(\zeta(\frac12+it))$ near $t=295.5839\ldots$ yields a negative value of $\zeta(\frac12+it)$.

This was found by approximating $\zeta'(\frac12+it)$ at each of the first "few" zeros of $\zeta$ tabulated by Odlyzko in and looking near the first zero (the 127th overall) at which $\zeta'$ has negative imaginary part. There are $22$ such zeros of the $649$ zeros whose imaginary part lies in $[0,1000]$; there's probably a counterexample near each of those, e.g. looking around the second such zero (#136) yields $$ t = 295.583906974228176092587915204356841\ldots $$ with $\zeta(\frac12+it) \simeq -0.0169004$.

EDIT 1) Henry Cohn (in a comment below) provides gp code that looks for solutions in an interval by dividing it into segments $(t_0, t_0 + 0.01)$, testing whether ${\rm Im}(\zeta(\frac12+it))$ changes sign between the endpoints, and if so whether the real part is negative at the crossing. Extending his computation to $0 \leq t \leq 1000$ finds the expected $22$ solutions; in particular $282.45472+$ seems to be the first.

2) Once one has calculated an answer one can ask Google for its previous appearances. Google recognizes $282.45472$ from J.Arias-de-Reyna's paper "X-Ray of Riemann zeta function" ( where it appears (to within $10^{-5}$) as the first counterexample to "Gram's law" — see the plot on page 26 (thick and thin curves show where $\zeta(s)$ is real and imaginary respectively).

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Incidentally, these seem to be the first two counterexamples. Here's some pari code: {for(i=1,300*100, if(imag(zeta(1/2+Ii/100))*imag(zeta(1/2+I*(i+1)/100))<0, x = solve(y=i/100,(i+1)/100,imag(zeta(1/2+Iy))); if(real(zeta(1/2+I*x))<-10^(-10),print(x))));} –  Henry Cohn Aug 18 '11 at 0:47
That would have looked a little nicer if markdown worked in the comments, but at least the code runs. –  Henry Cohn Aug 18 '11 at 0:49
@Henry: Thanks, that was what I guessed too. Extending the computation to imaginary part $10^3$, which takes a bit under 10 minutes here, finds 22 solutions, again corroborating the suggestion that they are near zeta zeros $\frac12 + i \gamma$ with negative ${\rm Im}(\zeta'(\frac12+i\gamma))$. [But eventually there are bound to be exceptions when the image of the critical line under $\zeta$ gets more complicated, or simply when some $\zeta'(\frac12+i\gamma)$ is very close to the real axis.] –  Noam D. Elkies Aug 18 '11 at 1:34
I do not get why a zero of $\zeta$ with $\Im(\zeta')$ negative at that point would imply that a counterexample to (1) is near, even heuristically. –  Eren Mehmet Kiral Aug 18 '11 at 8:19
@E.M.Kiral – to see why sign of ${\rm Im}(\zeta'(\frac12+i\gamma))$ might be relevant: look at a plot (such as the first plot in G.Helms' post) of the image under $\zeta$ of an initial segment of the critical line, and imagine what it would take for one of the arcs joining consecutive zeros to pass through the negative real axis. –  Noam D. Elkies Aug 18 '11 at 14:30

The reason (1) 'appears' to be true for small $t$ is related to Gram's Law for the zeros of $\zeta(s)$. Edwards' book Riemann's Zeta Function (Dover) has a good explanation starting on p.125. The short version is that the Euler Maclaurin formula for $\zeta(1/2+i t)$ starts with a $+1$, and,

"as long as it is not necessary to use too large a value of $N$, it will be unusual for the smaller terms which follow to combine to overwhelm this advantage on the plus side. As Gram puts it, equilibrium between plus and minus values of Re$\zeta$ will be achieved only very slowly as $t$ increases."

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Is it known whether the apparent bias in the value distribution of $\mathrm{Re}\zeta(1/2+it)$ dissolves as $t \to \infty$? –  David Hansen Aug 18 '11 at 0:11
Paul, the zeta function is not real-valued on the critical line. –  David Hansen Aug 18 '11 at 0:12
Here is the graph of real and imaginary parts along the critical line, from wikipedia: with $t$ (they call it $x$) between -30 to 30 –  Will Jagy Aug 18 '11 at 1:20
I agree that Gram's Law is the thing that is relevant. –  Junkie Aug 18 '11 at 4:47
Sorry, my earlier comment that zeta itself is real-valued on the criticial line is of course false, but the normalized version from the functional equation is. The gamma factor is non-vanishing, etc. –  paul garrett Aug 18 '11 at 12:41

The zeta function is real on the critical line only at the zeros and at Gram points, this is because zeta(1/2+it)=exp(-ivartheta(t)) Z(t).

At the Gram point g_k we have by definition vartheta(g_k)=pi k. so that zeta(1/2+ig_k) =(-1)^k Z(g_k).

Now a Gram point g_k is said a good Gram point if (-1)^k Z(g_k) >0. In other case it is said a bad Gram point.
Since it appear improbable a zero just at a Gram point. You are asking if there exists bad Gram points, there are plenty. The first few bad Gram points are g_126, g_134, g_195, g_211, ...

g_126 = 282.45472082346217461077

In fact it is proved there are infinite bad Gram points.

Also we may easily obtain large negative values. For example using data of T. Kotnik "Computational estimation of the order of zeta(1/2+it) Math of Comp. (2003) we easily locate the point t = grampoint(2601005843707) were we have

zeta(0.5+i t) = -119.6304321077241661374

This is easily confirmed in mpmath (or Mathematica) ( grampoint(2601005843707) = 669980906189.53552206792 ).

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Update, some recent information on (1):

Kalpokas, Korolev, Steuding recently released a preprint showing that $\zeta(1/2 + it)$ takes aribtrarily large positive and negative (real) values; and also show analog statements for the other lines through the origin, that is positive and negative (real) values of arbitary says of $e^{-i \phi} \zeta(1/2 + it)$ for any $\phi$. The paper contains also more quantitative results along these lines (cf. in particular Corollary 3 and the preceeding discussion).

Since (1) already received several answers, I expand and upgrade the comments on (2):

Yes, indeed it is conjectured, but unproved, that $\zeta(1/2 + i t)$ for $t \in \mathbb{R}$ is dense in the complex plane. [Side note: It is well-known that this is so for the lines $\sigma +it$ with $1/2 < \sigma < 1$.]

It seems that this conjecture was first formulated by Ramachandra (Durham, 1979), however only appeared in print in the second edition of Titchmarsh's book (note's by Heath-Brown), see the articles below for details.

There is very recent work on this problem due to Delbaen, Kowalski, and Nikeghbali. See in particular this preprint by the latter two and this by all three. Among others: in the former, they show how this result would follow "from a suitable version of the Keating--Snaith moment conjectures"; in the latter, they propose a refinement of the density conjecture, a quantitative version (see Conj. 1, in Sec. 3.9).

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@ Q1: After the counterexample of Noam Elkies I used Pari/GP to draw that parametric plot to get more visual impression;
[update] The visual impression in the 1:1000 zoomed picture had artifacts; I deleted the picture and provide a more precise one and corrected in my original answer [/update]

Plot 1 shows the known curve in the complex plane, when t increases from 0 to 100:

 \\ Pari/GP:

plot 1

From the drawing one cannot discern, whether there is some crossing of the negative real axis. Here is a rescaling; the values of the zeta-function are scaled by the tanh-function:

 \\ Pari/GP:

plot 2

and then a strong scaling factor of 1:1000 applied. [update] To remove artifacts, there is an option "recursive" in the plot-routine to scatter the coordinates more regularly; the strong zoom separated the dots of the plot too much so that artifacts are likely to occur. With an improvement of the sampling no crossings of the negative real axis can be seen [/update]

 \\ Pari/GP:

plot 4

I used internal precision of 200 dec digits, [update] so I think the computation of the single points do not introduce artefacts, but the connection by lines may do due to the strong scaling required. This type of plotting seems to require much resources; I'll see whether it can verify the crossing in the near of t=282 visually; I'll update then this answer again.

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This is suspicious because it would mean that Henry's computation missed some solutions with $t<100$. That is possible but unlikely: there would have to be at least two solutions in an interval $(t,t+0.01)$. Even more suspicious are angles in what purports to be the graph of a smooth function ;-) It must be that only the vertices, not the line segments between them, come from actual data points accurate to about $10^{-200}$. I'd expect that each of the crossings of the real axis, whether positive or negative in the above picture, is exactly at zero in reality. –  Noam D. Elkies Aug 18 '11 at 2:39
[that is, the crossings might not be artifacts, but their position relative to the origin is.] –  Noam D. Elkies Aug 18 '11 at 3:26
@Noam: you'right, that were artifacts. When I plotted without joining lines then I could see, that the single dots were far apart of each other so the joining lines crossed the axis at artificial points. An improvement of the plot routine showed then no more such crossings. Unfortunately the applied method is still too rough to show the actual crossing to which you refer in the even improved picture for the range t=200 to 300. So that all has only been "a good idea"... –  Gottfried Helms Aug 18 '11 at 7:26

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