Possible locations for non trivial zeroes lying off the critical line - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-22T23:48:07Z http://mathoverflow.net/feeds/question/85351 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/85351/possible-locations-for-non-trivial-zeroes-lying-off-the-critical-line Possible locations for non trivial zeroes lying off the critical line Agno 2012-01-10T18:39:25Z 2012-01-10T21:47:08Z <p>It has been proven that:</p> <p>1) if $s$ is a non trivial zero $\rho$ of $\zeta(s)$ then so is $1−s$.</p> <p>2) $\zeta(s) = 2^s \pi^{s-1} \sin(\frac{\pi s}{2}) \Gamma(1-s) \zeta(1-s)$</p> <p>3) $0 &lt; \Re(\rho) &lt;1$</p> <p>From this it follows that when $s \to \rho$:</p> <p>$\displaystyle \lim_{s \to \rho} |\dfrac{\zeta(s)}{\zeta(1-s)}| = |2^s \pi^{s-1} \sin(\frac{\pi s}{2}) \Gamma(1-s)|=1$</p> <p>It is easy to see that the outcome will be $1$ for all $y$ in $s=\frac12 + y i$.</p> <p>But if a $\rho$ would lie off this critical line, it also must reside in 'spots' where $\displaystyle \lim_{s \to \rho} |\dfrac{\zeta(s)}{\zeta(1-s)}|=1$.</p> <p>On which points off the critical line could this occur? I found a surprisingly small domain (no proof).</p> <p>The blue line shows the only values where:</p> <p>$\displaystyle |2^s \pi^{s-1} \sin(\frac{\pi s}{2}) \Gamma(1-s)|=1$, $s=x + y i$, $0 \le x \le 1$.</p> <p>Note that $y \to 2\pi$ for both $x=0$ and $x=1$. The $y$ rises only a little in the middle.</p> <p>This doesn't say anything about whether or not off-line $\rho$'s are actually hiding on this curve. There still is an infinite number to check. However, I wondered if anything more is known about this curve? </p> <p>[IMG]http://img822.imageshack.us/img822/3065/riemanntest.jpg[/IMG]</p> <p><a href="http://img822.imageshack.us/img822/3065/riemanntest.jpg" rel="nofollow">Please click here for the picture</a></p> http://mathoverflow.net/questions/85351/possible-locations-for-non-trivial-zeroes-lying-off-the-critical-line/85363#85363 Answer by Stopple for Possible locations for non trivial zeroes lying off the critical line Stopple 2012-01-10T20:23:27Z 2012-01-10T20:23:27Z <p>I believe you're mistaken that $$\lim_{s\to\rho}\left|\frac{\zeta(s)}{\zeta(1-s)}\right|=1.$$ Write $\zeta(1-s)=\zeta(s)f(s)$ with $f(s)$ as implied by your equation (2). The series expansion for $\zeta(s)$ at $s=\rho$ is $$\zeta(s)=\zeta^\prime(\rho)(s-\rho)+O(s-\rho)^2.$$ The series expansion for $\zeta(1-s)$ at $s=\rho$ is $$\zeta(1-s)=\zeta(s)f(s)=\zeta^\prime(\rho)f(\rho)(s-\rho)+O(s-\rho)^2.$$ By standard manipulation of series, $$\frac{\zeta(s)}{\zeta(1-s)}=\frac{1}{f(\rho)}+O(s-\rho),$$ so the limit should equal $1/f(\rho)$.</p> http://mathoverflow.net/questions/85351/possible-locations-for-non-trivial-zeroes-lying-off-the-critical-line/85366#85366 Answer by David Hansen for Possible locations for non trivial zeroes lying off the critical line David Hansen 2012-01-10T20:48:37Z 2012-01-10T20:48:37Z <p>There is some mild confusion here. Yes, for $s$ with real part $1/2$ the function $f(s)=2^{s}\pi^{s-1} \sin(\pi s/2)\Gamma(1-s)$ has magnitude $1$, which is an easy consequence of $|\Gamma(1/2+it)|=\sqrt{\frac{\pi}{\cosh {\pi t}}}$, but $|f(s)| \neq 1$ in general since a meromorphic function whose magnitude is constant on any open set is necessarily a constant. The limit $\lim_{s \to \rho} \frac{\zeta(s)}{\zeta(1-s)}$ is $f(\rho)$, but this doesn't put any constraint on $\rho$...</p> http://mathoverflow.net/questions/85351/possible-locations-for-non-trivial-zeroes-lying-off-the-critical-line/85370#85370 Answer by Micah Milinovich for Possible locations for non trivial zeroes lying off the critical line Micah Milinovich 2012-01-10T21:39:30Z 2012-01-10T21:47:08Z <p>Let $\chi(s)=2^s \pi^{s-1}\sin(\pi s/2) \Gamma(1-s)$ so that $\zeta(s)=\chi(s)\zeta(1-s)$. You are asking about the curve $|\chi(s)|=1$.</p> <p>As you have observed, $|\chi(1/2+it)|=1$ for real $t$. There is a partial converse to this statement, namely that there is a positive absolute constant $C_0$ such that if $|\chi(\sigma+it)| = 1$ with $0 \le \sigma \le 1$ and $|t| \ge C_0$, then $\sigma=1/2$. </p> <p>A simple proof can be found in Lemma 6.1 of S. M. Gonek "Finite Euler products and the Riemann hypothesis" <em>Trans. Amer. Math. Soc.</em> 364 (2012), 2157-2191. This paper is also on the arXiv. Gonek states that $C_0&lt;6.3$ so it seems that phenomena in your pictures stops shortly after the ranges you plotted.</p>