What happens to $\zeta(s)$ when all its $\Im(\rho_n)$ are "scaled" linearly? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-26T08:35:20Z http://mathoverflow.net/feeds/question/115128 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/115128/what-happens-to-zetas-when-all-its-im-rho-n-are-scaled-linearly What happens to $\zeta(s)$ when all its $\Im(\rho_n)$ are "scaled" linearly? Agno 2012-12-02T00:32:03Z 2012-12-16T13:11:33Z <p>I found that the following infinite product with $\mu = a +n b i$ and a,b real, $s \in \mathbb{C}$:</p> <p>$$\displaystyle \prod_{n=1}^\infty \left(1- \frac{s}{\mu} \right) \left(1- \frac{s}{1-\mu} \right)$$</p> <p>can be expressed in a closed form (with poles at $a,b = 0$ or $a=1$ and when $a=s$):</p> <p>$$\dfrac{\left( {a}^{2}-a \right)} {\left( {a} ^{2}-a+s-{s}^{2} \right)} \dfrac{\Gamma \left( {\frac {-ia}{b}} \right) \Gamma \left( {\frac {-i \left( a-1 \right) }{b}} \right)}{\Gamma \left( {\frac {-i \left( a-s \right) }{b}} \right) \Gamma \left( {\frac {-i \left( a+s-1 \right) }{b}} \right)}$$</p> <p>When $a=\frac12$ this could be further reduced to (poles at $s=\frac12$ and $b=0$):</p> <p>$$\dfrac{1}{(2s-1)} \dfrac{\sinh \left( {\frac { \left( 2s-1 \right) \pi }{2b}} \right)} { \sinh \left({\frac {\pi }{2b}} \right)}$$</p> <p>Encouraged by this result, my wish was to use it to find new hints about the Hadamard product:</p> <p>$$\displaystyle \prod_\rho \left(1- \frac{s}{\rho} \right) \left(1- \frac{s}{1-\rho} \right) = \dfrac{2(s-1)\Gamma(1+\frac{s}{2}){\zeta(s)}}{ \pi^{\frac{s}{2}}}$$</p> <p>but it is quite obviously an impossible task to transform the linear element $nb$ into to very random imaginary parts of the $\rho$s. However, it still triggered a follow up question:</p> <p>With $\rho = \sigma + ti$ and $t,x$ real, the following product:</p> <p>$$Had(s,x):=\displaystyle \prod_\rho \left(1- \frac{s}{\sigma + xti} \right) \left(1- \frac{s}{1-(\sigma + xti)} \right)$$</p> <p>allows for "scaling" of the imaginary parts of the $\rho$s.</p> <p>Since $Had(s,1)$ has a closed form and is entire, does this imply that the (linearly) scaled $\Im(\rho_n)$ must also induce entire functions and have closed forms (possibly related to $\zeta(s)$ and assuming RH is true)?</p> <p>Edit: Extra question:</p> <p>To take it a step further: similar for the infinite products with $n$ above, could the known closed form:</p> <p>$$\dfrac{2(s-1)\Gamma(1+\frac{s}{2}){\zeta(s)}}{ \pi^{\frac{s}{2}}}$$</p> <p>just be the 'reduced' version for $\sigma=\frac12$ and be extended with $\sigma$ and $x$ to express $Had(s,x,\sigma)$? </p> <p>P.S.:</p> <p>I wrote a program to calculate $Had(s,x)$ by using the first 2 mln $\rho$s from Andrew Odlyzko's table, however when calibrating the results with the known $Had(2,1) =\dfrac{\pi}{3}$, I found that the accuracy is limited to 5 decimals max. (i.e. too few to link it to known constants). Are there any larger $\rho$-files available on the web? </p> http://mathoverflow.net/questions/115128/what-happens-to-zetas-when-all-its-im-rho-n-are-scaled-linearly/115161#115161 Answer by joro for What happens to $\zeta(s)$ when all its $\Im(\rho_n)$ are "scaled" linearly? joro 2012-12-02T09:09:37Z 2012-12-02T09:09:37Z <p>About database of zeros. mpmath can compute the zeta zeros to arbitrary precision and sage has optional package containing a lot of zeros (though IIRC with not much precision).</p> <p>You are asking about products over zeros with scaled imaginary parts, but I suppose such sums are much easier, including finding closed form assuming RH (computing such sums without RH will be interesting to me).</p> <p><a href="http://ipht.cea.fr/Docspht/articles/t03/078/public/publi.pdf" rel="nofollow">More Zeta Functions for the Riemann Zeros</a> explains how to compute:</p> <p>$$Z_1(\sigma,v) = \sum_{k=1}^\infty (\tau_k^2+v)^{-\sigma}$$</p> <p>$$Z_2(s,x) = \sum_{k=1}^\infty (x-\rho)^{-s}$$</p> <p>where {$\rho$} = { $\frac12 \pm i \tau_k$ } k=1,2,... = {the Riemann zeros}</p> <p>Assume RH (this means $\tau_k \in \mathbb{R}$) and suppose you want to compute $\sum_\rho \frac{1}{1/2+ 2 i \tau_k}$. </p> <p>Grouping $\rho, 1-\rho$ gives $\sum_\rho \frac{1/4}{\tau_k^2+1/16}$ which is directly computable by $Z_1(1,1/16)$.</p> <p>Modulo errors the last sum is:</p> <p>$$1/12\,{\frac {-16\,\zeta \left( 3/4 \right) +3\,\Psi \left( 3/8 \right) \zeta \left( 3/4 \right) -3\,\ln \left( \pi \right) \zeta \left( 3/4 \right) +6\,\zeta' \left(3/4 \right) }{\zeta \left( 3/ 4 \right) }}$$</p> <p>The same approach works for other scalings.</p> http://mathoverflow.net/questions/115128/what-happens-to-zetas-when-all-its-im-rho-n-are-scaled-linearly/115171#115171 Answer by juan for What happens to $\zeta(s)$ when all its $\Im(\rho_n)$ are "scaled" linearly? juan 2012-12-02T11:05:03Z 2012-12-02T12:37:37Z <p>I assume Riemann hypothesis on all this answer.<br> I want a closed form for </p> <p>$$f(s,x):= \prod_{n=1}^\infty\Bigl(1-\frac{s}{\frac12+i x\gamma_n}\Bigr) \Bigl(1-\frac{s}{\frac12-i x\gamma_n}\Bigr).$$ Of course $\gamma_n$ runs here over the ordinates of the zeros of $\zeta(s)$ but only those with $\gamma>0$.</p> <p>We know that $$\Xi(t)=\Xi(0)\prod_\gamma\Bigl(1-\frac{t^2}{\gamma^2}\Bigr).\qquad (1)$$ Therefore $$\Xi(t/x)=\Xi(0)\prod_\gamma\Bigl(1-\frac{t^2}{x^2\gamma^2}\Bigr).$$ Substitute here $s=\frac12+it$ then $it=s-\frac12$ $$\Xi(t/x)=\Xi(0)\prod_\gamma\Bigl(1+\frac{(s-\frac12)^2}{x^2\gamma^2}\Bigr)=$$ $$= \Xi(0)\prod_\gamma\Bigl(\frac{(s-\frac12-ix\gamma)(s-1/2+ix\gamma)}{x^2\gamma^2}\Bigr).$$ Now we call $\rho=\frac12+ix\gamma$ and we get $$\Xi(t/x)=\Xi(0) \prod_\gamma\Bigl(\frac{\rho(1-\rho)}{x^2\gamma_n^2}\Bigr)\cdot \prod_\gamma\Bigl(1-\frac{s}{\rho}\Bigr)\Bigl(1-\frac{s}{1-\rho}\Bigr).$$ By (1), this is equal to $$\Xi(t/x)=\Xi(0) \prod_\gamma\Bigl(\frac{\frac14+x^2\gamma^2}{x^2\gamma^2}\Bigr)\cdot \prod_\gamma\Bigl(1-\frac{s}{\rho}\Bigr)\Bigl(1-\frac{s}{1-\rho}\Bigr)=$$ $$= \Xi(i/2x)\prod_\gamma\Bigl(1-\frac{s}{\rho}\Bigr)\Bigl(1-\frac{s}{1-\rho}\Bigr)$$</p> <p>Therefore we have $$\prod_\gamma\Bigl(1-\frac{s}{\rho}\Bigr)\Bigl(1-\frac{s}{1-\rho}\Bigr)= \frac{\Xi(t/x)}{\Xi(i/2x)}$$</p> <p>By definition we have</p> <p>`$$\Xi(t)=\frac{s(s-1)}{2}\pi^{-s/2}\Gamma(s/2)\zeta(s), \qquad \text{if} \quad s=\frac12+it.$$</p> <p>I shall continue in other answer because the TeX do not runs well</p> http://mathoverflow.net/questions/115128/what-happens-to-zetas-when-all-its-im-rho-n-are-scaled-linearly/115172#115172 Answer by juan for What happens to $\zeta(s)$ when all its $\Im(\rho_n)$ are "scaled" linearly? juan 2012-12-02T11:06:39Z 2012-12-02T12:38:14Z <p>... continue the above answer </p> <p>Therefore $$\prod_\gamma\Bigl(1-\frac{s}{\rho}\Bigr)\Bigl(1-\frac{s}{1-\rho}\Bigr)= \frac{\frac14+\frac{t^2}{x^2}}{\frac14-\frac{1}{4x^2}}\frac{\pi^{-\frac{it}{2x}}} {\pi^{\frac{1}{4x}}}\frac{\Gamma(\frac14+\frac{it}{2x})} {\Gamma(\frac14-\frac{1}{4x})}\frac{\zeta(\frac12+i\frac{t}{x})} {\zeta(\frac12-\frac{1}{2x})},$$ or equivalently $$\prod_\gamma\Bigl(1-\frac{s}{\rho}\Bigr)\Bigl(1-\frac{s}{1-\rho}\Bigr)=$$ $$\frac{x^2-4(s-\frac12)^2}{x^2-1}\pi^{-s/2x} \frac{\Gamma(\frac{1}{4}-\frac{1}{4x}+\frac{s}{2x})} {\Gamma(\frac14-\frac{1}{4x})}\frac{\zeta(\frac12+\frac{s}{x}-\frac{1}{2x})} {\zeta(\frac12-\frac{1}{2x})}.$$ If my computation are correct. </p> http://mathoverflow.net/questions/115128/what-happens-to-zetas-when-all-its-im-rho-n-are-scaled-linearly/116523#116523 Answer by Agno for What happens to $\zeta(s)$ when all its $\Im(\rho_n)$ are "scaled" linearly? Agno 2012-12-16T13:11:33Z 2012-12-16T13:11:33Z <p>Have not given up yet on whether or not there exists a closed form for:</p> <p>$$Had(s, \sigma, x):=\displaystyle \prod_\rho \left(1- \frac{s}{\sigma + xti} \right) \left(1- \frac{s}{1-(\sigma + xti)} \right)$$</p> <p>that, as Juan proved above, reduces to (assuming RH):</p> <p>$$Had(s,\frac12,x):=\frac{x^2-4(s-\frac12)^2}{x^2-1}\pi^{-s/2x} \frac{\Gamma(\frac{1}{4}-\frac{1}{4x}+\frac{s}{2x})} {\Gamma(\frac14-\frac{1}{4x})}\frac{\zeta(\frac12+\frac{s}{x}-\frac{1}{2x})} {\zeta(\frac12-\frac{1}{2x})}$$</p> <p>and for $x=1$, further reduces to the Hadamard product: </p> <p>$$Had(s, \frac12, 1):=\dfrac{2(s-1)\Gamma(1+\frac{s}{2}){\zeta(s)}}{ \pi^{\frac{s}{2}}}$$</p> <p>Assuming RH, a closed form for $Had(s, \sigma, x)$ requires:</p> <ul> <li>$Had(0, \sigma, x)=1$ and $Had(1, \sigma, x)=1$.</li> <li>$Had(s, \sigma, x)= Had(s, 1-\sigma, x)$</li> <li>$Had(\frac12, \sigma, x)$ is the function's minimum.</li> <li>$Had(s, \sigma, x)$ to reduce to the closed forms for $Had(s,\frac12,x)$ and $Had(s, \frac12, 1)$</li> <li>the Zeta function's non-trivial zeros to be the 'source' for all (horizontally shifted) zeros.</li> <li>the function to be entire (all poles annihilated by zeros).</li> </ul> <p>The following function does meet all the criteria, except for the second:</p> <p>$$\displaystyle {\frac {{x}^{2}-4 \left( \sigma-s \right) ^{2}}{{x}^{2}-4 \left( 2s\sigma- s-\sigma \right) ^{2}}}{\pi }^{{\frac {s \left( \sigma-1 \right) }{x}}} \dfrac{\Gamma \left( \dfrac{\frac12-{\frac {\sigma}{x}}+{\frac {s}{x}}}{2}\right)}{\Gamma \left( \dfrac{\frac12-{\frac {\sigma}{x}}+{\frac {s(2\sigma-1)}{x}}}{2}\right)} \dfrac{\zeta \left( \frac12-{\frac {\sigma}{x}}+{\frac {s}{x}} \right)}{\zeta \left( \frac12-{\frac {\sigma}{x}}+{\frac {s(2\sigma-1)}{x}} \right)}$$</p> <p>My 'brute force' infinite product calculations (based on the first 2 mln zeros) show that the shapes of the curves are close (but not equal), however the results for $Had(s, \sigma, x)$ and $Had(s, 1-\sigma, x)$ differ (slightly, yet consistently) from each other.</p> <p>Could there be any way to improve this?</p>