Negative values of Riemann zeta function on the critical line. - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-20T14:26:16Z http://mathoverflow.net/feeds/question/73098 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/73098/negative-values-of-riemann-zeta-function-on-the-critical-line Negative values of Riemann zeta function on the critical line. Eren Mehmet Kiral 2011-08-17T20:12:11Z 2011-09-13T12:43:45Z <p>From parametric plots of $\zeta \left( \frac{1}{2} + it \right)$ it seems to be the case that:<br> (1) except for $\zeta(\frac{1}{2})$ the Riemann zeta function does not attain any negative real value on the critical line.<br> (2) the curve $(t, \zeta(1/2+it))$ is dense in the complex plane.</p> <p>Are these statements known to be false, if not, is there any proof affirming them?</p> http://mathoverflow.net/questions/73098/negative-values-of-riemann-zeta-function-on-the-critical-line/73100#73100 Answer by Stopple for Negative values of Riemann zeta function on the critical line. Stopple 2011-08-17T21:31:14Z 2011-08-17T21:31:14Z <p>The reason (1) 'appears' to be true for small $t$ is related to Gram's Law for the zeros of $\zeta(s)$. Edwards' book <em>Riemann's Zeta Function</em> (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, </p> <blockquote> <p>"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."</p> </blockquote> http://mathoverflow.net/questions/73098/negative-values-of-riemann-zeta-function-on-the-critical-line/73109#73109 Answer by Noam D. Elkies for Negative values of Riemann zeta function on the critical line. Noam D. Elkies 2011-08-17T23:57:06Z 2011-08-18T04:13:51Z <p>A numerical counterexample to the first conjecture is $$t = 282.4547208234621746108397940690599354\ldots$$ where both <strong>gp</strong> 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 $\zeta(\frac12+it)$.</p> <p>This was found by approximating $\zeta'(\frac12+it)$ at each of the first "few" zeros of $\zeta$ tabulated by Odlyzko in <a href="http://www.dtc.umn.edu/~odlyzko/zeta_tables/zeros1" rel="nofollow">http://www.dtc.umn.edu/~odlyzko/zeta_tables/zeros1</a> 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 $\zeta(\frac12+it) \simeq -0.0169004$.</p> <p><strong>EDIT</strong> 1) Henry Cohn (in a comment below) provides <strong>gp</strong> 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.</p> <p>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" (http://arxiv.org/abs/math/0309433) where it appears (to within $10^{-5}$) as the first counterexample to "Gram's law" &mdash; see the plot on page 26 (thick and thin curves show where $\zeta(s)$ is real and imaginary respectively).</p> http://mathoverflow.net/questions/73098/negative-values-of-riemann-zeta-function-on-the-critical-line/73113#73113 Answer by Gottfried Helms for Negative values of Riemann zeta function on the critical line. Gottfried Helms 2011-08-18T01:14:13Z 2011-08-18T07:28:31Z <p>@ Q1: After the counterexample of Noam Elkies I used Pari/GP to draw that parametric plot to get more visual impression;<br> [update] The visual impression in the <em>1:1000</em> zoomed picture had <strong>artifacts</strong>; I deleted the picture and provide a more precise one and corrected in my original answer [/update]</p> <p>Plot 1 shows the known curve in the complex plane, when <em>t</em> increases from <em>0</em> to <em>100</em>: </p> <pre> \\ Pari/GP: ri_zeta(t)=local(tmp);tmp=zeta(1/2+I*t);return([real(tmp),imag(tmp)]) ploth(x=0,100,ri_zeta(x),1) </pre> <p><img src="http://go.helms-net.de/math/images/mo_110814_z1.png" alt="plot 1"> </p> <p>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: </p> <pre> \\ Pari/GP: ri_zeta(t)=local(tmp);tmp=10*zeta(1/2+I*t);return([tanh(real(tmp)),tanh(imag(tmp))]) ploth(x=0,100,ri_zeta(x),1) </pre> <p><img src="http://go.helms-net.de/math/images/mo_110814_z2.png" alt="plot 2"> </p> <p>and then a strong scaling factor of <em>1:1000</em> 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 <em>no</em> crossings of the negative real axis can be seen [/update] </p> <pre> \\ Pari/GP: ri_zeta(t)=local(tmp);tmp=1000*zeta(1/2+I*t);return([tanh(real(tmp)),tanh(imag(tmp))]) </pre> <p><img src="http://go.helms-net.de/math/images/mo_110814_z4.png" alt="plot 4"> </p> <p>I used internal precision of 200 dec digits, [update] so I think the computation of the <strong>single points</strong> 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.</p> http://mathoverflow.net/questions/73098/negative-values-of-riemann-zeta-function-on-the-critical-line/73141#73141 Answer by quid for Negative values of Riemann zeta function on the critical line. quid 2011-08-18T12:35:07Z 2011-09-13T12:43:45Z <p>Update, some recent information on (1): </p> <p><a href="http://arxiv.org/pdf/1109.2224" rel="nofollow">Kalpokas, Korolev, Steuding</a> 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). </p> <hr> <p>Since (1) already received several answers, I expand and upgrade the comments on (2):</p> <p>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 $1/2 &lt; \sigma &lt; 1$.]</p> <p>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. </p> <p>There is very recent work on this problem due to Delbaen, Kowalski, and Nikeghbali. See in particular <a href="http://arxiv.org/pdf/0912.3237" rel="nofollow">this preprint</a> by the latter two and <a href="http://arxiv.org/pdf/1107.5657" rel="nofollow">this</a> 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). </p> http://mathoverflow.net/questions/73098/negative-values-of-riemann-zeta-function-on-the-critical-line/73143#73143 Answer by Juan Arias de Reyna for Negative values of Riemann zeta function on the critical line. Juan Arias de Reyna 2011-08-18T12:38:41Z 2011-08-18T13:49:32Z <p>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). </p> <p>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).</p> <p>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.<br> 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, ...</p> <p>g_126 = 282.45472082346217461077</p> <p>In fact it is proved there are infinite bad Gram points. </p> <p>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</p> <p>zeta(0.5+i t) = -119.6304321077241661374</p> <p>This is easily confirmed in mpmath (or Mathematica) ( grampoint(2601005843707) = 669980906189.53552206792 ).</p>