User draks - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-25T18:15:58Z http://mathoverflow.net/feeds/user/11856 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/119210/an-expression-for-log-zetans-derived-from-the-limit-of-the-truncated-prime An Expression for $\log\zeta(ns)$ derived from the Limit of the truncated Prime $\zeta$ Function draks 2013-01-17T20:47:55Z 2013-01-24T20:57:09Z <p>I think, <a href="http://math.stackexchange.com/q/115230/19341" rel="nofollow">here</a>, I found $$P_x(s)=\sum_{p &lt; x} \frac{1}{p^s} =\sum_{n=1}^{\infty}\frac{ \mu (n)}{n} \sum_{z\in{1,\rho}}(-1)^{1-\delta_{1z}} \left[ {\rm li}(t^{\frac zn-s}) \right]^{x}_2 \tag{7}$$</p> <p>where $\rho$ are <strong>all</strong> the zeros (trivial and non-trivial) of $\zeta$ function. See the linked question for more detail, corrections are welcome. Further we know, that $$P(s)=\sum_{n> 0}\frac {\mu(n)}n{\log\zeta(ns)} .$$</p> <p>So my question is </p> <blockquote> <p>If $\lim_{x\to \infty} P_x(s)=P(s)$ then $$\log\zeta(ns)=\lim_{x\to \infty} \sum_{z\in{1,\rho}}(-1)^{1-\delta_{1z}} \left[ {\rm li}(t^{\frac1n( z-ns)}) \right]^{x}_2 ?$$</p> </blockquote> <p>What I got so far is:</p> <ul> <li><p>Could $\displaystyle \log \zeta(s) = s \int_0^\infty \frac{\pi(x)}{x(x^s-1)}\,dx$ be useful somehow?</p></li> <li><p><a href="http://math.stackexchange.com/a/285413/19341" rel="nofollow">Thanks to robjohn</a> it possible to see that both coincide at least some special values:<br> If $ns=1$ or $ns=\rho$, one addend in the sum diverges like $\lim_{x\to\infty} \log\left(\frac{\log(x)}{\log(2)}\right)=\infty$. So we get $$\begin{eqnarray} ns=1: &amp; \log(\zeta(1)) &amp;=&amp; +\infty\ ns=\rho: &amp; \log(\zeta(\rho)) &amp;=&amp; -\infty \end{eqnarray}$$</p></li> <li><p>I looked at the series expansion at $s=0$ for ${\rm li}(x^{\frac1n(z-ns)})=$ <a href="http://tinyurl.com/b57ga7j" rel="nofollow">${\rm Ei}((\frac zn-s)\ln(x))$</a> and <a href="http://tinyurl.com/a3tvnml" rel="nofollow">$\log\zeta(ns)$</a>. Assuming I'm not wrong, you'll get the following when you compare the linear terms $$\color{grey}{ins}\log(2\pi) \overset{?}{=} \color{grey}{ins} \lim_{x\to \infty}\sum_{z\in{1,\rho}}(-1)^{\delta_{1z}} \left[ \frac{ t^{\frac{z}n}}{z}\right]_2^x ,$$ which looks a little irritating, since the RHS has to be independent of $x$ an $n$. Does this show that it's wrong at all?</p></li> </ul> http://mathoverflow.net/questions/109269/partly-obscured-rubiks-cube/109274#109274 Answer by draks for partly obscured Rubik's cube draks 2012-10-10T05:39:26Z 2012-10-10T05:39:26Z <p>You might be interested in this:</p> <p><a href="http://math.stackexchange.com/q/127577/19341" rel="nofollow">How to tell if a Rubik's cube is solvable</a></p> <p>and Henning's answer.</p> http://mathoverflow.net/questions/119210/an-expression-for-log-zetans-derived-from-the-limit-of-the-truncated-prime Comment by draks draks 2013-01-17T22:01:52Z 2013-01-17T22:01:52Z Yes but I wanted a compact, notation. http://mathoverflow.net/questions/119210/an-expression-for-log-zetans-derived-from-the-limit-of-the-truncated-prime Comment by draks draks 2013-01-17T20:50:24Z 2013-01-17T20:50:24Z Cross-posted at <a href="http://math.stackexchange.com/q/280984/19341" rel="nofollow">math.stackexchange.com/q/280984/19341</a>