non-trivial zeros of partial zeta functions - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-22T17:01:12Z http://mathoverflow.net/feeds/question/97174 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/97174/non-trivial-zeros-of-partial-zeta-functions non-trivial zeros of partial zeta functions Hugo Chapdelaine 2012-05-17T00:46:26Z 2012-05-19T16:08:27Z <p>Let <code>$N,a\in\mathbf{Z}_{\geq 1}$</code>. Define a partial $\zeta$-function as $$\zeta(s;N,a):=\sum_{\substack{n\geq 1\newline n\equiv a\pmod{N}}} \frac{1}{n^s}$$ where $Re(s)>1$. Let $\omega$ be either the trivial character or the sign character i.e. $x\mapsto sign(x)$. Define a partial $\Psi$-function as $$\Psi(s,\omega;N,a):=\sum_{\substack{0\neq n\in\mathbf{Z}\newline n\equiv a\pmod{N}}}\frac{\omega(n)}{|n|^s},$$ for $Re(s)>1$. Then it is well known that $\zeta(s;N,a)$ and $\Psi(s,\omega;N,a)$ admit a meromorphic continuation to $\mathbf{C}$ with at most of pole of order $1$ at $s=1$. If $\omega$ is the sign character then $$\Psi(s,\omega;N,a)+\Psi(s,1;N,a)=2\zeta(s;N,a)$$ and $$\zeta(s;N,a)-\zeta(s;N,-a)=\Psi(s,\omega;N,a).$$ Note that when $N>1$, the functions $\Psi$ and $\zeta$ do not have an Euler product. So here are 2 natural questions:</p> <p>Q1: For a fixed $N$ do we know if there exists a constant $C_N>1$ such that if $Re(s)>C_N$ then $\Psi$ and $\zeta$ do not vanish (if the answer is yes then how to prove it)?</p> <p>Q2: What do we know in general about the nontrivial zeros of $\Psi$ and $\zeta$?</p> http://mathoverflow.net/questions/97174/non-trivial-zeros-of-partial-zeta-functions/97179#97179 Answer by Micah Milinovich for non-trivial zeros of partial zeta functions Micah Milinovich 2012-05-17T02:04:50Z 2012-05-17T02:04:50Z <p>The answer to question 1 is classical: Any Dirichlet series which has a finite abscissa of absolute convergence has a zero-free half-plane.</p> <p>Suppose the Dirichlet series $$A(s)=\sum_{n=1}^\infty \frac{a_n}{n^s}$$ has an abscissa of absolute convergence $\sigma_a$. If $a_m$ is the first non-zero coefficient, then for $\Re(s)>\sigma_a$ $$m^s A(s) = a_m + a_{m+1} \Big(\frac{m}{m+1}\Big)^s + a_{m+2} \Big(\frac{m}{m+2}\Big)^s +\cdots \to a_m \ \text{ as } \ \Re(s) \to +\infty.$$ Hence, there exists an absolute constant $C$ such that for $\Re(s)>C$ we have $$\left|a_{m+1} \Big(\frac{m}{m+1}\Big)^s + a_{m+2} \Big(\frac{m}{m+2}\Big)^s +\cdots \right| \leq \frac{|a_m|}{2}.$$ Consequently, $m^s A(s)$ has no zeros in the half-plane $\Re(s)>C$.</p> http://mathoverflow.net/questions/97174/non-trivial-zeros-of-partial-zeta-functions/97220#97220 Answer by joro for non-trivial zeros of partial zeta functions joro 2012-05-17T12:54:00Z 2012-05-17T14:04:09Z <p><strong>EDIT</strong> Experimentally the zeros of $\zeta(s;5,1)$ and $\zeta(s;6,1)$ are those of zeta (the previous revision incorrectly included wrong zeros caused by insufficient precision).</p> <p>The zeros of $\zeta(s;6,2)$ (if computed correctly) don't appear related to those of zeta.</p> <p><img src="http://s14.postimage.org/icoicob4h/zeta62.png"></p> <p>sage code:</p> <pre><code>import mpmath mpmath.mp.pretty=True mpmath.mp.dps=100 N=5 a=1 an=[0]*N an[a-1]=1 def L(x): return mpmath.dirichlet(x,an) def search1(): cac={} rr=list(range(-20,0)) P=[] for k in xrange(1,40): rr += [0.1 + mpmath.j*k] for x in rr: try: r=mpmath.findroot(L,[x],maxsteps=1000) except: continue print r ks=str(r) if ks in cac: continue cac[ks]=1 P += [(RR(r.real),RR(r.imag))] return P P=search1() pt=points(P) pt.show() </code></pre> http://mathoverflow.net/questions/97174/non-trivial-zeros-of-partial-zeta-functions/97388#97388 Answer by Johan Andersson for non-trivial zeros of partial zeta functions Johan Andersson 2012-05-19T11:35:05Z 2012-05-19T11:35:05Z <p>Micah's answer answers your Q1. My reply gives some information on your Q2. Let us assume that $1 \leq a\leq N$ for simplicity (this condition can be removed). Your zeta-functions $\zeta$ and $\Psi$ can be expressed by the Hurwitz zeta-function <a href="http://en.wikipedia.org/wiki/Hurwitz_zeta_function" rel="nofollow">http://en.wikipedia.org/wiki/Hurwitz_zeta_function</a>. For your first function $$\zeta(s;N,a)=N^{-s} \zeta \left(s,\frac a N \right).$$ Davenport and Heilbronn proved that for any $c>0$, $a/N \neq 1/2,1$ there exists $\gg T$ zeros of this function in the strip $1&lt;\Re(s)&lt;1+c, | \Im (s)| &lt; T$. By Voronin universality for the Hurwitz zeta-function the same can be said in any strip $1/2&lt;\sigma_0&lt; \Re (s)&lt;\sigma_1&lt;1$ (this can be found for example in Jörn Steuding's SLN or Garunkstis-Laurincikas book on the Lerch zeta-function).</p> <p>Your second function can also be expressed in terms of the Hurwitz zeta-function $$\Psi(s,\omega;N,a)=N^{-s} \left(\zeta\left(s,\frac a N\right)+\omega(-1)\zeta \left( s,1-\frac a N \right) \right).$$ In this case joint universality of the Hurwitz zeta-function (see same references as above) can be used to show that except when $\Psi$ has an Euler-product (which only occurs for some small special cases when $a / N$ equals 1/6, 1/4, 1/3, 1/2, 2/3, 3/4, 5/6, or 1.) there are $\gg T$ zeros in any strip $1/2&lt;\sigma_0&lt; \Re (s) &lt; \sigma_1 &lt; 1$ and $|\Im(s)| &lt; T$. Davenport-Heilbronn's result for $\Re(s)>1$ can be obtained in this case as well.</p> <p>For the special cases where your function has an Euler-product it will essentially be a Dirichlet L-function or the Riemann zeta-function, and its zeros will be the same.</p> http://mathoverflow.net/questions/97174/non-trivial-zeros-of-partial-zeta-functions/97406#97406 Answer by Hugo Chapdelaine for non-trivial zeros of partial zeta functions Hugo Chapdelaine 2012-05-19T16:08:27Z 2012-05-19T16:08:27Z <p>This is only a <strong>comment</strong> regarding the functional equation of the $\Psi$-function. So define $$\Psi^*(s,\omega;N,a):=\sum_{0\neq n\in\mathbf{Z}}\frac{\omega(n)}{|n|^s}e^{-2\pi i\frac{a}{N}n}$$ Then one can show using Riemann's classical idea (see for example <a href="http://www.mat.ulaval.ca/fileadmin/Pages_personnelles_des_profs/hchapd/Papers/Zeta_function.pdf" rel="nofollow">1</a>) that $$(-1)^p N^s\cdot\Gamma_{\infty}(s)\Psi(s,\omega;N,a)=\Gamma_{\infty}(1-s)\Psi^*(1-s,\omega;N,a),$$ where $p=0$ (resp. p=$1$) if $\omega=1$ (resp. $\omega=sign$) and $\Gamma_{\infty}(s):=\pi^{-s/2}\Gamma(\frac{s+p}{2})$.</p> <p>So it seems that many of the non-trivial zeros of the $\Psi$-functions are located along the symmetry axis of the functional equation. Is there some heuristic argument that could explain this phenomenon? </p>