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<channel>
	<title>zeta &amp;laquo; WordPress.com Tag Feed</title>
	<link>http://en.wordpress.com/tag/zeta/</link>
	<description>Feed of posts on WordPress.com tagged "zeta"</description>
	<pubDate>Sat, 05 Dec 2009 02:33:05 +0000</pubDate>

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<item>
<title><![CDATA[The ARAMAIC SQUARE an THE ZETA EffECT]]></title>
<link>http://codexsinaiticus.wordpress.com/2009/12/03/the-zeta-effect/</link>
<pubDate>Thu, 03 Dec 2009 18:14:27 +0000</pubDate>
<dc:creator>markthunderson</dc:creator>
<guid>http://codexsinaiticus.wordpress.com/2009/12/03/the-zeta-effect/</guid>
<description><![CDATA[In the Gospel of Jesus Christ, the Lord says many times: &#8220;He who has ears to hear, let Him hea]]></description>
<content:encoded><![CDATA[<div class='snap_preview'><p>In the Gospel of Jesus Christ, the Lord says many times: &#8220;He who has ears to hear, let Him hear!&#8221;</p>
<p>In the Church today the Pharisees say:</p>
<blockquote><p>&#8220;Though the New Testament is written in Greek, Jesus spoke Aramaic.&#8221;</p></blockquote>
<p>Hold Your Tongue, Preacher!  <em>You&#8217;re deceived</em>. Where else have we heard of something very similar to this?  Is it not that the Old Testament was written in Hebrew, even though the script is only ever written in Aramaic?  This is the exact same argument in antithesis.  How scandalous can one become, that a preacher would say to the Son of God: &#8220;Even though the WORD of GOD came in Greek, you still do not know the Language of God, nor do you know what it sounds like; because the Language of God and the Sound of God is only ever heard in Aramaic.&#8221;  This is the teaching of an unclean spirit; it is foul and it is demonic.</p>
<p>This insidious teaching regarding the Aramaic and Scripture is called &#8220;The Zeta Effect.&#8221;  It is extremely deceptive and destructive teaching that originates from the Scribes and is preached by the Pharisees.  At its core, this teaching maintains that the <em>sixth</em> (6th) letter of the Greek alphabet is &#8220;Z&#8221; but that this letter as a numeral is worth seven (7).  It is a very specific type of contradiction aimed at God and God&#8217;s Chosen.  The Zeta Effect teaches that the Path of God is crooked.  It claims a type of &#8220;rest,&#8221;  but it does not give rest; it only gives more contradiction, inconsistency, confusion, lies, and much more.  It is sinister and devilish. It is a game, and a game, moreover, that is played again and again by contemporary Christianity with the World.</p>
<p>To all such games I say &#8220;No.&#8221;  The sixth letter of the Language of God is  &#8221; C &#8221; (<em>Stau</em>), and as a numeral it is worth six (6).  STAU also doubles as SIGMA  &#8220;C&#8221; which looks exactly the same as the Stau.</p>
<p>Do not be conformed to this world!  Make straight the Way of the Lord!</p>
</div>]]></content:encoded>
</item>
<item>
<title><![CDATA[Finale!]]></title>
<link>http://sumidiot.wordpress.com/2009/11/27/finale/</link>
<pubDate>Sat, 28 Nov 2009 01:23:19 +0000</pubDate>
<dc:creator>sumidiot</dc:creator>
<guid>http://sumidiot.wordpress.com/2009/11/27/finale/</guid>
<description><![CDATA[I think I&#8217;m about ready to tie together Farey sequences and the Riemann zeta function (specifi]]></description>
<content:encoded><![CDATA[<div class='snap_preview'><p>I think I&#8217;m about ready to tie together Farey sequences and the Riemann zeta function (specifically, the Riemann hypothesis). I know there were lots of gaps along the way, and lots more to explore from here, so that all gets to be the dénouement, if/when I get to it.</p>
<p><a href="http://sumidiot.wordpress.com/2009/11/25/mobius-inversion/">Yesterday</a> we ended with the identity</p>
<p style="text-align:center;"><img src='http://l.wordpress.com/latex.php?latex=%5Cdisplaystyle+%5Cfrac%7B1%7D%7B%5Czeta%28s%29%7D%3D%5Csum_%7Bn%7D%5Cfrac%7B%5Cmu%28n%29%7D%7Bn%5Es%7D.&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\displaystyle \frac{1}{\zeta(s)}=\sum_{n}\frac{\mu(n)}{n^s}.' title='\displaystyle \frac{1}{\zeta(s)}=\sum_{n}\frac{\mu(n)}{n^s}.' class='latex' /></p>
<p>Define <img src='http://l.wordpress.com/latex.php?latex=M%28x%29%3D%5Csum_%7Bn%5Cleq+x%7D%5Cmu%28n%29&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='M(x)=\sum_{n\leq x}\mu(n)' title='M(x)=\sum_{n\leq x}\mu(n)' class='latex' />, which you might call the cumulative Möbius function. If you are comfortable with your <a href="http://en.wikipedia.org/wiki/Measure_(mathematics)">measure</a> theory, you can decide that there is a measure, which I&#8217;ll denote <img src='http://l.wordpress.com/latex.php?latex=dM&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='dM' title='dM' class='latex' />, such that <img src='http://l.wordpress.com/latex.php?latex=%5Cint_a%5Eb+dM%3DM%28b%29-M%28a%29&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\int_a^b dM=M(b)-M(a)' title='\int_a^b dM=M(b)-M(a)' class='latex' />. And if you&#8217;re ok with that statement, you probably also don&#8217;t mind writing the identity for the reciprocal of zeta, above, as</p>
<p style="text-align:center;"><img src='http://l.wordpress.com/latex.php?latex=%5Cdisplaystyle+%5Cfrac%7B1%7D%7B%5Czeta%28s%29%7D%3D%5Cint_0%5E%7B%5Cinfty%7D+x%5E%7B-s%7D%5C+dM.&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\displaystyle \frac{1}{\zeta(s)}=\int_0^{\infty} x^{-s}\ dM.' title='\displaystyle \frac{1}{\zeta(s)}=\int_0^{\infty} x^{-s}\ dM.' class='latex' /></p>
<p>If you then integrate by parts you find that</p>
<p style="text-align:center;"><img src='http://l.wordpress.com/latex.php?latex=%5Cdisplaystyle+%5Cfrac%7B1%7D%7B%5Czeta%28s%29%7D%3Ds%5Cint_0%5E%7B%5Cinfty%7Dx%5E%7B-s-1%7DM%28x%29%5C+dx.&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\displaystyle \frac{1}{\zeta(s)}=s\int_0^{\infty}x^{-s-1}M(x)\ dx.' title='\displaystyle \frac{1}{\zeta(s)}=s\int_0^{\infty}x^{-s-1}M(x)\ dx.' class='latex' /></p>
<p>Now if <img src='http://l.wordpress.com/latex.php?latex=M%28x%29%5Cleq+x%5Ea&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='M(x)\leq x^a' title='M(x)\leq x^a' class='latex' /> (we could say something more like <img src='http://l.wordpress.com/latex.php?latex=M%28x%29%3DO%28x%5Ea%29&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='M(x)=O(x^a)' title='M(x)=O(x^a)' class='latex' />) for some positive real value <img src='http://l.wordpress.com/latex.php?latex=a&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='a' title='a' class='latex' />, then this most recent integral will converge for <img src='http://l.wordpress.com/latex.php?latex=%5Ctext%7BRe+%7Ds%26%2362%3Ba&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\text{Re }s&gt;a' title='\text{Re }s&gt;a' class='latex' />. Which means that <img src='http://l.wordpress.com/latex.php?latex=%5Czeta%28s%29&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\zeta(s)' title='\zeta(s)' class='latex' /> would have no zeroes in the half-plane <img src='http://l.wordpress.com/latex.php?latex=%5Ctext%7BRe+%7Ds%26%2362%3Ba&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\text{Re }s&gt;a' title='\text{Re }s&gt;a' class='latex' />. Since there are zeroes on the critical line, where <img src='http://l.wordpress.com/latex.php?latex=%5Ctext%7BRe+%7Ds%3D1%2F2&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\text{Re }s=1/2' title='\text{Re }s=1/2' class='latex' />, we know that no value of <img src='http://l.wordpress.com/latex.php?latex=a&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='a' title='a' class='latex' /> less than 1/2 will work. But if you can show that <img src='http://l.wordpress.com/latex.php?latex=M%28x%29%5Cleq+x%5E%7B%281%2F2%29%2B%5Cepsilon%7D&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='M(x)\leq x^{(1/2)+\epsilon}' title='M(x)\leq x^{(1/2)+\epsilon}' class='latex' /> for every positive <img src='http://l.wordpress.com/latex.php?latex=%5Cepsilon&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\epsilon' title='\epsilon' class='latex' />, then you&#8217;ll have proven the Riemann hypothesis (and earned yourself a <a href="http://www.claymath.org/millennium/">cool million</a>).</p>
<p>Ok, so now we just have to relate the Farey sequences to this <img src='http://l.wordpress.com/latex.php?latex=M%28x%29&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='M(x)' title='M(x)' class='latex' /> somehow.</p>
<p>It&#8217;s <a href="http://sumidiot.wordpress.com/2009/11/03/farey-sequences/">been a while</a>, so perhaps we should recall some basics about the Farey sequences. The <img src='http://l.wordpress.com/latex.php?latex=n&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='n' title='n' class='latex' />-th Farey sequence, denoted <img src='http://l.wordpress.com/latex.php?latex=F_n&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='F_n' title='F_n' class='latex' /> consists of the reduced fractions <img src='http://l.wordpress.com/latex.php?latex=p%2Fq&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='p/q' title='p/q' class='latex' /> in <img src='http://l.wordpress.com/latex.php?latex=%5B0%2C1%5D&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='[0,1]' title='[0,1]' class='latex' /> with <img src='http://l.wordpress.com/latex.php?latex=q%5Cleq+n&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='q\leq n' title='q\leq n' class='latex' />, ordered as usual. For today, let&#8217;s actually remove 0 from the sequences. Thinking about how many terms are in <img src='http://l.wordpress.com/latex.php?latex=F_n&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='F_n' title='F_n' class='latex' />, we quickly determine that there are <img src='http://l.wordpress.com/latex.php?latex=%5Csum_%7Bi%5Cleq+n%7D%5Cphi%28i%29&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\sum_{i\leq n}\phi(i)' title='\sum_{i\leq n}\phi(i)' class='latex' />, where <img src='http://l.wordpress.com/latex.php?latex=%5Cphi%28i%29&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\phi(i)' title='\phi(i)' class='latex' /> is the number of positive integers less than <img src='http://l.wordpress.com/latex.php?latex=i&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='i' title='i' class='latex' /> that are relatively prime to <img src='http://l.wordpress.com/latex.php?latex=i&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='i' title='i' class='latex' /> (with <img src='http://l.wordpress.com/latex.php?latex=%5Cphi%281%29%3D1&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\phi(1)=1' title='\phi(1)=1' class='latex' />). Let me call this sum <img src='http://l.wordpress.com/latex.php?latex=%5CPhi%28n%29&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\Phi(n)' title='\Phi(n)' class='latex' />, so that <img src='http://l.wordpress.com/latex.php?latex=F_n&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='F_n' title='F_n' class='latex' /> has <img src='http://l.wordpress.com/latex.php?latex=%5CPhi%28n%29&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\Phi(n)' title='\Phi(n)' class='latex' /> terms.</p>
<p>To keep my notation under control, fix an <img src='http://l.wordpress.com/latex.php?latex=n&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='n' title='n' class='latex' />. Then for <img src='http://l.wordpress.com/latex.php?latex=1%5Cleq+v%5Cleq+%5CPhi%28n%29&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='1\leq v\leq \Phi(n)' title='1\leq v\leq \Phi(n)' class='latex' />, let <img src='http://l.wordpress.com/latex.php?latex=r_v&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='r_v' title='r_v' class='latex' /> be the <img src='http://l.wordpress.com/latex.php?latex=v&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='v' title='v' class='latex' />-th term in the Farey sequence <img src='http://l.wordpress.com/latex.php?latex=F_n&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='F_n' title='F_n' class='latex' />. The terms in <img src='http://l.wordpress.com/latex.php?latex=F_n&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='F_n' title='F_n' class='latex' /> are not equally spaced in <img src='http://l.wordpress.com/latex.php?latex=%5B0%2C1%5D&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='[0,1]' title='[0,1]' class='latex' />, so let us also consider the sequence of <img src='http://l.wordpress.com/latex.php?latex=%5CPhi%28n%29&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\Phi(n)' title='\Phi(n)' class='latex' /> equidistant points in <img src='http://l.wordpress.com/latex.php?latex=%5B0%2C1%5D&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='[0,1]' title='[0,1]' class='latex' />. These will be the values <img src='http://l.wordpress.com/latex.php?latex=s_v%3Dv%2F%5CPhi%28n%29&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='s_v=v/\Phi(n)' title='s_v=v/\Phi(n)' class='latex' />. We&#8217;re interested in how far the Farey sequence is from this equidistant sequence, so let&#8217;s set <img src='http://l.wordpress.com/latex.php?latex=%5Cdelta_v%3Dr_v-s_v&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\delta_v=r_v-s_v' title='\delta_v=r_v-s_v' class='latex' />.</p>
<p>If <img src='http://l.wordpress.com/latex.php?latex=f%3A%5B0%2C1%5D%5Cto+%5Cmathbb%7BR%7D&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='f:[0,1]\to \mathbb{R}' title='f:[0,1]\to \mathbb{R}' class='latex' />, then you can show (essentially by Möbius inversion) that</p>
<p style="text-align:center;"><img src='http://l.wordpress.com/latex.php?latex=%5Cdisplaystyle+%5Csum_%7Bv%3D1%7D%5E%7B%5CPhi%28n%29%7Df%28r_v%29%3D%5Csum_%7Bk%3D1%7D%5E%7B%5Cinfty%7D%5Csum_%7Bj%3D1%7D%5E%7Bk%7D+f%28j%2Fk%29M%28n%2Fk%29.&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\displaystyle \sum_{v=1}^{\Phi(n)}f(r_v)=\sum_{k=1}^{\infty}\sum_{j=1}^{k} f(j/k)M(n/k).' title='\displaystyle \sum_{v=1}^{\Phi(n)}f(r_v)=\sum_{k=1}^{\infty}\sum_{j=1}^{k} f(j/k)M(n/k).' class='latex' /></p>
<p>The idea is that the function <img src='http://l.wordpress.com/latex.php?latex=D%28m%29&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='D(m)' title='D(m)' class='latex' /> that is 1 for <img src='http://l.wordpress.com/latex.php?latex=m%26%2362%3B1&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='m&gt;1' title='m&gt;1' class='latex' /> and 0 for <img src='http://l.wordpress.com/latex.php?latex=m%26%2360%3B1&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='m&lt;1' title='m&lt;1' class='latex' /> is also</p>
<p style="text-align:center;"><img src='http://l.wordpress.com/latex.php?latex=D%28m%29%3D%5Csum_n+M%28m%2Fn%29%2C&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='D(m)=\sum_n M(m/n),' title='D(m)=\sum_n M(m/n),' class='latex' /></p>
<p>because you can write (fairly cleverly, I feel)</p>
<p style="text-align:center;"><img src='http://l.wordpress.com/latex.php?latex=M%28m%29%3D%5Csum_n+%5Cmu%28n%29D%28m%2Fn%29.&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='M(m)=\sum_n \mu(n)D(m/n).' title='M(m)=\sum_n \mu(n)D(m/n).' class='latex' /></p>
<p>By the way, I&#8217;m being a bit loose with my values of these stepwise things when they actually make their steps. Apparently the convention is to define the value at the change to be the midpoint of the values on either side. I feel like it&#8217;s a technicality I&#8217;m not hugely interested in right now.</p>
<p>Now we apply this identity to <img src='http://l.wordpress.com/latex.php?latex=f%28u%29%3De%5E%7B2%5Cpi+iu%7D&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='f(u)=e^{2\pi iu}' title='f(u)=e^{2\pi iu}' class='latex' />, obtaining</p>
<p style="text-align:center;"><img src='http://l.wordpress.com/latex.php?latex=%5Cdisplaystyle+%5Csum_v+e%5E%7B2%5Cpi+ir_v%7D%3D%5Csum_k+M%28n%2Fk%29%5Csum_j+f%28j%2Fk%29.&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\displaystyle \sum_v e^{2\pi ir_v}=\sum_k M(n/k)\sum_j f(j/k).' title='\displaystyle \sum_v e^{2\pi ir_v}=\sum_k M(n/k)\sum_j f(j/k).' class='latex' /></p>
<p>That inner sum on the right-hand side is the sum of <img src='http://l.wordpress.com/latex.php?latex=k&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='k' title='k' class='latex' /> unit vectors equally distributed around the unit circle, and so is 0 except when <img src='http://l.wordpress.com/latex.php?latex=k%3D1&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='k=1' title='k=1' class='latex' />. So we obtain</p>
<p style="text-align:center;"><img src='http://l.wordpress.com/latex.php?latex=M%28n%29%3D%5Csum_v+e%5E%7B2%5Cpi+ir_v%7D.&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='M(n)=\sum_v e^{2\pi ir_v}.' title='M(n)=\sum_v e^{2\pi ir_v}.' class='latex' /></p>
<p>Finally, replace <img src='http://l.wordpress.com/latex.php?latex=r_v&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='r_v' title='r_v' class='latex' /> with <img src='http://l.wordpress.com/latex.php?latex=s_v%2B%5Cdelta_v&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='s_v+\delta_v' title='s_v+\delta_v' class='latex' />. After shuffling some symbols around, you can write</p>
<p style="text-align:center;"><img src='http://l.wordpress.com/latex.php?latex=M%28n%29%3D%5Csum_v+e%5E%7B2%5Cpi+s_v%7D%28e%5E%7B2%5Cpi+i%5Cdelta_v%7D-1%29%2B%5Csum_v+e%5E%7B2%5Cpi+is_v%7D.&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='M(n)=\sum_v e^{2\pi s_v}(e^{2\pi i\delta_v}-1)+\sum_v e^{2\pi is_v}.' title='M(n)=\sum_v e^{2\pi s_v}(e^{2\pi i\delta_v}-1)+\sum_v e^{2\pi is_v}.' class='latex' /></p>
<p>Taking absolute values on each side, using the triangle inequality, and the identity we already used about sums of equally spaced unit vectors, we can write</p>
<p style="text-align:center;"><img src='http://l.wordpress.com/latex.php?latex=%5Cbegin%7Barray%7D%7Bl%7D%26%23124%3BM%28n%29%26%23124%3B%5Cleq+%5Cdisplaystyle+%5Csum_v+%5Cleft%26%23124%3Be%5E%7B2%5Cpi+i%5Cdelta_v%7D-1%5Cright%26%23124%3B+%3D%5Csum_v+%5Cleft%26%23124%3B+e%5E%7B%5Cpi+i%5Cdelta_v%7D-e%5E%7B-%5Cpi+i%5Cdelta_v%7D%5Cright%26%23124%3B%3D2%5Csum_v+%5Cleft%26%23124%3B%5Csin+%5Cpi%5Cdelta_v%5Cright%26%23124%3B%5C%5C+%5Cdisplaystyle+%5Cqquad+%5Cleq+2%5Cpi+%5Csum_v+%5Cleft%26%23124%3B%5Cdelta_v%5Cright%26%23124%3B.%5Cend%7Barray%7D&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\begin{array}{l}&#124;M(n)&#124;\leq \displaystyle \sum_v \left&#124;e^{2\pi i\delta_v}-1\right&#124; =\sum_v \left&#124; e^{\pi i\delta_v}-e^{-\pi i\delta_v}\right&#124;=2\sum_v \left&#124;\sin \pi\delta_v\right&#124;\\ \displaystyle \qquad \leq 2\pi \sum_v \left&#124;\delta_v\right&#124;.\end{array}' title='\begin{array}{l}&#124;M(n)&#124;\leq \displaystyle \sum_v \left&#124;e^{2\pi i\delta_v}-1\right&#124; =\sum_v \left&#124; e^{\pi i\delta_v}-e^{-\pi i\delta_v}\right&#124;=2\sum_v \left&#124;\sin \pi\delta_v\right&#124;\\ \displaystyle \qquad \leq 2\pi \sum_v \left&#124;\delta_v\right&#124;.\end{array}' class='latex' /></p>
<p>Using our lemma from earlier, about how to obtain the Riemann hypothesis from <img src='http://l.wordpress.com/latex.php?latex=M&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='M' title='M' class='latex' />, we have, at last:</p>
<p style="text-align:left;padding-left:30px;">If <img src='http://l.wordpress.com/latex.php?latex=%5Csum_v+%26%23124%3B%5Cdelta_v%26%23124%3B&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\sum_v &#124;\delta_v&#124;' title='\sum_v &#124;\delta_v&#124;' class='latex' /> is <img src='http://l.wordpress.com/latex.php?latex=O%28n%5E%7B%281%2F2%29%2B%5Cepsilon%7D%29&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='O(n^{(1/2)+\epsilon})' title='O(n^{(1/2)+\epsilon})' class='latex' /> for every <img src='http://l.wordpress.com/latex.php?latex=%5Cepsilon%26%2362%3B0&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\epsilon&gt;0' title='\epsilon&gt;0' class='latex' />, then the Riemann hypothesis is true.</p>
<p>Hurray! The result that encouraged me to wander down this path!</p>
<p>So, that was fun. It turns out that both of today&#8217;s results that imply the Riemann hypothesis are actually <em>equivalent</em> to the Riemann hypothesis, but perhaps that&#8217;s a topic for another day&#8230;</p>
<p>[<strong>Update</strong> 20091203: Noticed I had some absolute value bars in the wrong places in the line before the final theorem, and corrected them (hopefully)]</p>
</div>]]></content:encoded>
</item>
<item>
<title><![CDATA[Möbius Inversion]]></title>
<link>http://sumidiot.wordpress.com/2009/11/25/mobius-inversion/</link>
<pubDate>Thu, 26 Nov 2009 01:35:40 +0000</pubDate>
<dc:creator>sumidiot</dc:creator>
<guid>http://sumidiot.wordpress.com/2009/11/25/mobius-inversion/</guid>
<description><![CDATA[Yesterday I mentioned Möbius inversion, and I decided that it was worth looking at a little bit more]]></description>
<content:encoded><![CDATA[<div class='snap_preview'><p>Yesterday I mentioned Möbius inversion, and I decided that it was worth looking at a little bit more. First of all, it&#8217;s pretty fun. Secondly, it&#8217;ll give us a useful identity concerning <img src='http://l.wordpress.com/latex.php?latex=%5Czeta%28s%29&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\zeta(s)' title='\zeta(s)' class='latex' />.</p>
<p>Before I get too far, I should add another reference to my growing list. With Edwards&#8217; book, and Hardy and Wright, include Apostol&#8217;s &#8220;Introduction to Analytic Number Theory.&#8221;</p>
<p>Anyway, yesterday I defined the Möbius function</p>
<p style="text-align:center;"><img src='http://l.wordpress.com/latex.php?latex=%5Cdisplaystyle+%5Cmu%28n%29%3D%5Cbegin%7Bcases%7D1+%26%2338%3B+n%3D1+%5C%5C+0+%26%2338%3B+n%5Ctext%7B+not+squarefree%7D+%5C%5C+%28-1%29%5Ek+%26%2338%3B+n%3Dp_1%5Ccdots+p_k%5Ctext%7B+distinct+primes%7D%5Cend%7Bcases%7D&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\displaystyle \mu(n)=\begin{cases}1 &amp; n=1 \\ 0 &amp; n\text{ not squarefree} \\ (-1)^k &amp; n=p_1\cdots p_k\text{ distinct primes}\end{cases}' title='\displaystyle \mu(n)=\begin{cases}1 &amp; n=1 \\ 0 &amp; n\text{ not squarefree} \\ (-1)^k &amp; n=p_1\cdots p_k\text{ distinct primes}\end{cases}' class='latex' /></p>
<p>and stated Möbius inversion as follows:</p>
<p style="padding-left:30px;"><strong>Thm</strong>: If <img src='http://l.wordpress.com/latex.php?latex=f%28n%29%3D%5Csum%5Climits_%7Bd%26%23124%3Bn%7Dg%28d%29&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='f(n)=\sum\limits_{d&#124;n}g(d)' title='f(n)=\sum\limits_{d&#124;n}g(d)' class='latex' /> then <img src='http://l.wordpress.com/latex.php?latex=g%28n%29%3D%5Csum%5Climits_%7Bd%26%23124%3Bn%7D%5Cmu%28d%29f%28n%29&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='g(n)=\sum\limits_{d&#124;n}\mu(d)f(n)' title='g(n)=\sum\limits_{d&#124;n}\mu(d)f(n)' class='latex' />.</p>
<p>Let&#8217;s begin today with a proof of this statement. <strong>Pf</strong>: Suppose <img src='http://l.wordpress.com/latex.php?latex=f%28n%29&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='f(n)' title='f(n)' class='latex' /> is as stated, and consider <img src='http://l.wordpress.com/latex.php?latex=%5Csum%5Climits_%7Bd%26%23124%3Bn%7D%5Cmu%28d%29f%28n%29&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\sum\limits_{d&#124;n}\mu(d)f(n)' title='\sum\limits_{d&#124;n}\mu(d)f(n)' class='latex' />. By assumption on <img src='http://l.wordpress.com/latex.php?latex=f&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='f' title='f' class='latex' />, this is <img src='http://l.wordpress.com/latex.php?latex=%5Csum%5Climits_%7Bd%26%23124%3Bn%7D%5Cmu%28d%29%5Csum%5Climits_%7Bd%27%26%23124%3B%28n%2Fd%29%7D+g%28d%27%29&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\sum\limits_{d&#124;n}\mu(d)\sum\limits_{d&#039;&#124;(n/d)} g(d&#039;)' title='\sum\limits_{d&#124;n}\mu(d)\sum\limits_{d&#039;&#124;(n/d)} g(d&#039;)' class='latex' /> which I&#8217;ll write as <img src='http://l.wordpress.com/latex.php?latex=%5Csum%5Climits_%7Bdd%27%26%23124%3Bn%7D%5Cmu%28d%29g%28d%27%29&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\sum\limits_{dd&#039;&#124;n}\mu(d)g(d&#039;)' title='\sum\limits_{dd&#039;&#124;n}\mu(d)g(d&#039;)' class='latex' />. Now in this sum, what is the coefficient of <img src='http://l.wordpress.com/latex.php?latex=g%28m%29&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='g(m)' title='g(m)' class='latex' />? Well, it is <img src='http://l.wordpress.com/latex.php?latex=%5Csum%5Climits_%7Bdm%26%23124%3Bn%7D%5Cmu%28d%29%3D%5Csum%5Climits_%7Bd%26%23124%3B%28n%2Fm%29%7D%5Cmu%28d%29&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\sum\limits_{dm&#124;n}\mu(d)=\sum\limits_{d&#124;(n/m)}\mu(d)' title='\sum\limits_{dm&#124;n}\mu(d)=\sum\limits_{d&#124;(n/m)}\mu(d)' class='latex' />. We saw yesterday that this sum is 0 unless <img src='http://l.wordpress.com/latex.php?latex=n%2Fm%3D1&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='n/m=1' title='n/m=1' class='latex' />, i.e., unless <img src='http://l.wordpress.com/latex.php?latex=m%3Dn&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='m=n' title='m=n' class='latex' />. So <img src='http://l.wordpress.com/latex.php?latex=%5Csum%5Climits_%7Bd%26%23124%3Bn%7D%5Cmu%28d%29f%28n%29%3Dg%28n%29&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\sum\limits_{d&#124;n}\mu(d)f(n)=g(n)' title='\sum\limits_{d&#124;n}\mu(d)f(n)=g(n)' class='latex' />, as claimed.</p>
<p>That&#8217;s pretty fun, I think.</p>
<p>There&#8217;s another way to go about these things, and it is by way of what are called Dirichlet series (according to Wikipedia, <a href="http://en.wikipedia.org/wiki/Dirichlet">Dirichlet</a> had many names). Suppose <img src='http://l.wordpress.com/latex.php?latex=f&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='f' title='f' class='latex' /> is a function defined on the positive integers (such a function seems to sometimes get called an arithmetic function). I&#8217;ll say that the Dirichlet series for <img src='http://l.wordpress.com/latex.php?latex=f&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='f' title='f' class='latex' /> is the series <img src='http://l.wordpress.com/latex.php?latex=F%28s%29%3D%5Csum_n+f%28n%29n%5E%7B-s%7D&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='F(s)=\sum_n f(n)n^{-s}' title='F(s)=\sum_n f(n)n^{-s}' class='latex' />. I&#8217;ve written it this way, so that I have a function defined whenever the series converges, but I&#8217;ll pretty much ignore convergence aspects in what follows.</p>
<p>Suppose <img src='http://l.wordpress.com/latex.php?latex=F%28s%29&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='F(s)' title='F(s)' class='latex' /> and <img src='http://l.wordpress.com/latex.php?latex=G%28s%29&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='G(s)' title='G(s)' class='latex' /> are the Dirichlet series associated to some <img src='http://l.wordpress.com/latex.php?latex=f%2Cg&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='f,g' title='f,g' class='latex' />, respectively. What does the product of these two series look like? Well, since <img src='http://l.wordpress.com/latex.php?latex=f%28n%29n%5E%7B-s%7D%5Ccdot+g%28m%29m%5E%7B-s%7D%3Df%28n%29g%28m%29%28nm%29%5E%7B-s%7D&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='f(n)n^{-s}\cdot g(m)m^{-s}=f(n)g(m)(nm)^{-s}' title='f(n)n^{-s}\cdot g(m)m^{-s}=f(n)g(m)(nm)^{-s}' class='latex' />, you can determine that the product of these two Dirichlet series is another Dirichlet series, this one associated to the function <img src='http://l.wordpress.com/latex.php?latex=c%28n%29%3D%5Csum%5Climits_%7Bd%26%23124%3Bn%7Df%28d%29g%28n%2Fd%29&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='c(n)=\sum\limits_{d&#124;n}f(d)g(n/d)' title='c(n)=\sum\limits_{d&#124;n}f(d)g(n/d)' class='latex' />. Indeed, just ask yourself, &#8220;what is the coefficient of <img src='http://l.wordpress.com/latex.php?latex=n%5E%7B-s%7D&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='n^{-s}' title='n^{-s}' class='latex' /> in the product of the series <img src='http://l.wordpress.com/latex.php?latex=F%28s%29&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='F(s)' title='F(s)' class='latex' /> and <img src='http://l.wordpress.com/latex.php?latex=G%28s%29&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='G(s)' title='G(s)' class='latex' />?&#8221;</p>
<p>By the way, I&#8217;ve called the product series <img src='http://l.wordpress.com/latex.php?latex=c&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='c' title='c' class='latex' /> because it also gets called the <strong>convolution</strong> of <img src='http://l.wordpress.com/latex.php?latex=f&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='f' title='f' class='latex' /> and <img src='http://l.wordpress.com/latex.php?latex=g&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='g' title='g' class='latex' />. The convolution of <img src='http://l.wordpress.com/latex.php?latex=f&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='f' title='f' class='latex' /> and <img src='http://l.wordpress.com/latex.php?latex=g&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='g' title='g' class='latex' /> is denoted <img src='http://l.wordpress.com/latex.php?latex=f%2Ag&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='f*g' title='f*g' class='latex' />, and is simply the function <img src='http://l.wordpress.com/latex.php?latex=%28f%2Ag%29%28n%29%3D%5Csum%5Climits_%7Bd%26%23124%3Bn%7Df%28d%29g%28n%2Fd%29&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='(f*g)(n)=\sum\limits_{d&#124;n}f(d)g(n/d)' title='(f*g)(n)=\sum\limits_{d&#124;n}f(d)g(n/d)' class='latex' />.</p>
<p>Now you&#8217;ve got an operation, <img src='http://l.wordpress.com/latex.php?latex=%2A&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='*' title='*' class='latex' />, on arithmetic functions, and you can start asking what properties it has. It turns out that <img src='http://l.wordpress.com/latex.php?latex=%2A&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='*' title='*' class='latex' /> gives the set of arithmetic functions (well, those functions that aren&#8217;t 0 on the input 1) the structure of a commutative group. The identity is the function <img src='http://l.wordpress.com/latex.php?latex=I&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='I' title='I' class='latex' /> with <img src='http://l.wordpress.com/latex.php?latex=I%281%29%3D1&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='I(1)=1' title='I(1)=1' class='latex' /> and <img src='http://l.wordpress.com/latex.php?latex=I%28n%29%3D0&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='I(n)=0' title='I(n)=0' class='latex' /> if <img src='http://l.wordpress.com/latex.php?latex=n%26%2362%3B1&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='n&gt;1' title='n&gt;1' class='latex' /> (corresponding to the identity Dirichlet series, the constant function 1). It&#8217;s sort of a fun little induction problem to find a formula for the inverse of a given arithmetic series. We won&#8217;t need the formula exactly, so perhaps I&#8217;ll leave it for you.</p>
<p>Let <img src='http://l.wordpress.com/latex.php?latex=z%28n%29%3D1&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='z(n)=1' title='z(n)=1' class='latex' />, so that the associated Dirichlet series is <img src='http://l.wordpress.com/latex.php?latex=%5Czeta%28s%29&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\zeta(s)' title='\zeta(s)' class='latex' />.</p>
<p>We now have another language in which to state Möbius inversion:</p>
<p style="padding-left:30px;"><strong>Thm</strong>: If <img src='http://l.wordpress.com/latex.php?latex=f%3Dg%2Az&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='f=g*z' title='f=g*z' class='latex' /> then <img src='http://l.wordpress.com/latex.php?latex=g%3D%5Cmu%2Af&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='g=\mu*f' title='g=\mu*f' class='latex' />.</p>
<p>In this statement, let <img src='http://l.wordpress.com/latex.php?latex=g%3DI&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='g=I' title='g=I' class='latex' /> (the identity of convolution) so that <img src='http://l.wordpress.com/latex.php?latex=f%3Dz&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='f=z' title='f=z' class='latex' />. Then we obtain <img src='http://l.wordpress.com/latex.php?latex=I%3D%5Cmu+%2Az&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='I=\mu *z' title='I=\mu *z' class='latex' />. On taking Dirichlet series, we find that</p>
<p style="text-align:center;"><img src='http://l.wordpress.com/latex.php?latex=%5Cdisplaystyle+1%3D%5Cleft%28%5Csum_n+%5Cfrac%7B%5Cmu%28n%29%7D%7Bn%5Es%7D%5Cright%29%5Czeta%28s%29%2C&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\displaystyle 1=\left(\sum_n \frac{\mu(n)}{n^s}\right)\zeta(s),' title='\displaystyle 1=\left(\sum_n \frac{\mu(n)}{n^s}\right)\zeta(s),' class='latex' /></p>
<p>or, in other words,</p>
<p style="text-align:center;"><img src='http://l.wordpress.com/latex.php?latex=%5Cdisplaystyle+%5Cfrac%7B1%7D%7B%5Czeta%28s%29%7D%3D%5Csum_n+%5Cfrac%7B%5Cmu%28n%29%7D%7Bn%5Es%7D.&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\displaystyle \frac{1}{\zeta(s)}=\sum_n \frac{\mu(n)}{n^s}.' title='\displaystyle \frac{1}{\zeta(s)}=\sum_n \frac{\mu(n)}{n^s}.' class='latex' /></p>
<p>How&#8217;s that for an identity?</p>
<p>There&#8217;s another way to get at this identity, without the language of Dirichlet series (explicitly anyway), which I wrote &#8220;holy crap&#8221; next to in the book (Hardy and Wright) when I read it. So probably I should share. Using Euler&#8217;s product (recall the indexing is over primes)</p>
<p style="text-align:center;"><img src='http://l.wordpress.com/latex.php?latex=%5Cdisplaystyle+%5Czeta%28s%29%3D%5Cprod_p+%281-p%5E%7B-s%7D%29%5E%7B-1%7D&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\displaystyle \zeta(s)=\prod_p (1-p^{-s})^{-1}' title='\displaystyle \zeta(s)=\prod_p (1-p^{-s})^{-1}' class='latex' /></p>
<p>we have</p>
<p style="text-align:center;"><img src='http://l.wordpress.com/latex.php?latex=%5Cdisplaystyle+%5Cfrac%7B1%7D%7B%5Czeta%28s%29%7D%3D%5Cprod_p+%281-p%5E%7B-s%7D%29.&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\displaystyle \frac{1}{\zeta(s)}=\prod_p (1-p^{-s}).' title='\displaystyle \frac{1}{\zeta(s)}=\prod_p (1-p^{-s}).' class='latex' /></p>
<p>Here&#8217;s the part that rocked me: Look at <img src='http://l.wordpress.com/latex.php?latex=1-p%5E%7B-s%7D&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='1-p^{-s}' title='1-p^{-s}' class='latex' /> and write it as <img src='http://l.wordpress.com/latex.php?latex=1-p%5E%7B-s%7D%2B0p%5E%7B-2s%7D%2B0p%5E%7B-3s%7D%2B%5Ccdots&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='1-p^{-s}+0p^{-2s}+0p^{-3s}+\cdots' title='1-p^{-s}+0p^{-2s}+0p^{-3s}+\cdots' class='latex' />, i.e. as <img src='http://l.wordpress.com/latex.php?latex=%5Csum_k+%5Cmu%28p%5Ek%29p%5E%7B-ks%7D&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\sum_k \mu(p^k)p^{-ks}' title='\sum_k \mu(p^k)p^{-ks}' class='latex' />. Because the first thing I think to do when I see a binomial is to write it as an infinite series of mostly 0s. Like I said yesterday, I guess it helps if you know where you are going. So anyway, we&#8217;ve got</p>
<p style="text-align:center;"><img src='http://l.wordpress.com/latex.php?latex=%5Cdisplaystyle+%5Cfrac%7B1%7D%7B%5Czeta%28s%29%7D%3D%5Cprod_p+%5Csum_k+%5Cfrac%7B%5Cmu%28p%5Ek%29%7D%7B%28p%5Ek%29%5Es%7D.&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\displaystyle \frac{1}{\zeta(s)}=\prod_p \sum_k \frac{\mu(p^k)}{(p^k)^s}.' title='\displaystyle \frac{1}{\zeta(s)}=\prod_p \sum_k \frac{\mu(p^k)}{(p^k)^s}.' class='latex' /></p>
<p>Now, like we&#8217;ve been doing, let&#8217;s ask, &#8220;what is the coefficient of <img src='http://l.wordpress.com/latex.php?latex=n%5E%7B-s%7D&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='n^{-s}' title='n^{-s}' class='latex' /> when the product is all written out?&#8221; Well, the only way to get <img src='http://l.wordpress.com/latex.php?latex=n%5E%7B-s%7D&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='n^{-s}' title='n^{-s}' class='latex' /> is the term corresponding to the prime factorization of <img src='http://l.wordpress.com/latex.php?latex=n&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='n' title='n' class='latex' />. If <img src='http://l.wordpress.com/latex.php?latex=n%3Dp_1%5E%7Be_1%7D%5Ccdots+p_k%5E%7Be_k%7D&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='n=p_1^{e_1}\cdots p_k^{e_k}' title='n=p_1^{e_1}\cdots p_k^{e_k}' class='latex' />, then the coefficient is <img src='http://l.wordpress.com/latex.php?latex=%5Cmu%28p_1%5E%7Be_1%7D%29%5Ccdots+%5Cmu%28p_k%5E%7Be_k%7D%29&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\mu(p_1^{e_1})\cdots \mu(p_k^{e_k})' title='\mu(p_1^{e_1})\cdots \mu(p_k^{e_k})' class='latex' />. But <img src='http://l.wordpress.com/latex.php?latex=%5Cmu&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\mu' title='\mu' class='latex' /> is &#8220;multiplicative,&#8221; meaning that if <img src='http://l.wordpress.com/latex.php?latex=a&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='a' title='a' class='latex' /> and <img src='http://l.wordpress.com/latex.php?latex=b&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='b' title='b' class='latex' /> are relatively prime then <img src='http://l.wordpress.com/latex.php?latex=%5Cmu%28a%29%5Cmu%28b%29%3D%5Cmu%28ab%29&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\mu(a)\mu(b)=\mu(ab)' title='\mu(a)\mu(b)=\mu(ab)' class='latex' />. So the coefficient of <img src='http://l.wordpress.com/latex.php?latex=n%5E%7B-s%7D&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='n^{-s}' title='n^{-s}' class='latex' /> is just <img src='http://l.wordpress.com/latex.php?latex=%5Cmu%28p_1%5E%7Be_1%7D%5Ccdots+p_k%5E%7Be_k%7D%29%3D%5Cmu%28n%29&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\mu(p_1^{e_1}\cdots p_k^{e_k})=\mu(n)' title='\mu(p_1^{e_1}\cdots p_k^{e_k})=\mu(n)' class='latex' />. This proves the identity.</p>
<p>That&#8217;s the identity we&#8217;ll use in the next post or two to finally tie together Farey sequences and the zeta function, believe it or not. So I&#8217;ll basically stop here. But I do want to mention that you can do some of the things we&#8217;ve done above in slightly more general contexts, and I remember really enjoying reading about them. Namely, you can work in other posets. In what we&#8217;ve been doing above, there has, somewhere in the background, been the poset of positive integers ordered by <img src='http://l.wordpress.com/latex.php?latex=n%5Cleq+m&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='n\leq m' title='n\leq m' class='latex' /> iff <img src='http://l.wordpress.com/latex.php?latex=n%26%23124%3Bm&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='n&#124;m' title='n&#124;m' class='latex' />. Probably a good place to start reading about the generalization is <a href="http://en.wikipedia.org/wiki/Incidence_algebra">Wikipedia</a>, as usual. If I find, sometime, the paper I read where I first saw these things, I&#8217;ll update this post.</p>
<p>In the mean time&#8230;</p>
</div>]]></content:encoded>
</item>
<item>
<title><![CDATA[Another Formula for J(x)]]></title>
<link>http://sumidiot.wordpress.com/2009/11/22/another-formula-for-jx/</link>
<pubDate>Mon, 23 Nov 2009 03:08:19 +0000</pubDate>
<dc:creator>sumidiot</dc:creator>
<guid>http://sumidiot.wordpress.com/2009/11/22/another-formula-for-jx/</guid>
<description><![CDATA[Yesterday, I related the logarithm of to a piecewise linear function . You may recall that was defin]]></description>
<content:encoded><![CDATA[<div class='snap_preview'><p><a href="http://sumidiot.wordpress.com/2009/11/21/the-log-of-zeta/">Yesterday</a>, I related the logarithm of <img src='http://l.wordpress.com/latex.php?latex=%5Czeta%28s%29&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\zeta(s)' title='\zeta(s)' class='latex' /> to a piecewise linear function <img src='http://l.wordpress.com/latex.php?latex=J%28x%29&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='J(x)' title='J(x)' class='latex' />. You may recall that <img src='http://l.wordpress.com/latex.php?latex=J%28x%29&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='J(x)' title='J(x)' class='latex' /> was defined for positive reals by setting it equal to 0 at <img src='http://l.wordpress.com/latex.php?latex=x%3D0&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='x=0' title='x=0' class='latex' />, and then jumping by <img src='http://l.wordpress.com/latex.php?latex=1%2Fn&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='1/n' title='1/n' class='latex' /> whenever <img src='http://l.wordpress.com/latex.php?latex=x%3Dp%5En&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='x=p^n' title='x=p^n' class='latex' />, for some prime <img src='http://l.wordpress.com/latex.php?latex=p&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='p' title='p' class='latex' /> and integer <img src='http://l.wordpress.com/latex.php?latex=n&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='n' title='n' class='latex' />. At the end of the day, we got to</p>
<p style="text-align:center;"><img src='http://l.wordpress.com/latex.php?latex=%5Cdisplaystyle+J%28x%29%3D%5Cfrac%7B1%7D%7B2%5Cpi+i%7D%5Cint_%7Ba-i%5Cinfty%7D%5E%7Ba%2Bi%5Cinfty%7D%5Cln+%5Czeta%28s%29+x%5Es%5Cfrac%7Bds%7D%7Bs%7D&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\displaystyle J(x)=\frac{1}{2\pi i}\int_{a-i\infty}^{a+i\infty}\ln \zeta(s) x^s\frac{ds}{s}' title='\displaystyle J(x)=\frac{1}{2\pi i}\int_{a-i\infty}^{a+i\infty}\ln \zeta(s) x^s\frac{ds}{s}' class='latex' /></p>
<p>where <img src='http://l.wordpress.com/latex.php?latex=a%3D%5Ctext%7BRe%7D%28s%29%26%2362%3B1&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='a=\text{Re}(s)&gt;1' title='a=\text{Re}(s)&gt;1' class='latex' />. Today, we&#8217;ll analyze <img src='http://l.wordpress.com/latex.php?latex=%5Cln+%5Czeta%28s%29&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\ln \zeta(s)' title='\ln \zeta(s)' class='latex' /> some more, and re-write the formula above.</p>
<p>When I <a href="http://sumidiot.wordpress.com/2009/11/13/riemanns-zeta-function/">introduced</a> <img src='http://l.wordpress.com/latex.php?latex=%5Czeta%28s%29&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\zeta(s)' title='\zeta(s)' class='latex' />, I ended with the following formula:</p>
<p style="text-align:center;"><img src='http://l.wordpress.com/latex.php?latex=%5Cdisplaystyle+%5Czeta%28s%29%3D%5Cfrac%7B%5CGamma%281-s%29%7D%7B2%5Cpi+i%7D%5Cint_%7B%2B%5Cinfty%7D%5E%7B%2B%5Cinfty%7D+%5Cfrac%7B%28-x%29%5Es%7D%7Be%5Ex-1%7D+%5Cfrac%7Bdx%7D%7Bx%7D&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\displaystyle \zeta(s)=\frac{\Gamma(1-s)}{2\pi i}\int_{+\infty}^{+\infty} \frac{(-x)^s}{e^x-1} \frac{dx}{x}' title='\displaystyle \zeta(s)=\frac{\Gamma(1-s)}{2\pi i}\int_{+\infty}^{+\infty} \frac{(-x)^s}{e^x-1} \frac{dx}{x}' class='latex' /></p>
<p>where the bounds on that integral are supposed to represent a curve that &#8220;starts&#8221; at the right-hand &#8220;end&#8221; of the real line, loops around 0, and then goes back out the positive axis to infinity. I&#8217;m not good enough at complex line integrals at this point to say any more about this. But apparently if you are good at these sorts of integrals, using Cauchy&#8217;s integral formula and things you can find the so-called &#8220;functional equation&#8221;</p>
<p style="text-align:center;"><img src='http://l.wordpress.com/latex.php?latex=%5Czeta%28s%29%3D%5CGamma%281-s%29%282%5Cpi%29%5E%7Bs-1%7D2%5Csin%28s%5Cpi%2F2%29%5Czeta%281-s%29.&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\zeta(s)=\Gamma(1-s)(2\pi)^{s-1}2\sin(s\pi/2)\zeta(1-s).' title='\zeta(s)=\Gamma(1-s)(2\pi)^{s-1}2\sin(s\pi/2)\zeta(1-s).' class='latex' /></p>
<p>If you then use the relation I mentioned previously:</p>
<p style="text-align:center;"><img src='http://l.wordpress.com/latex.php?latex=%5CGamma%28s%29%5CGamma%281-s%29%5Csin%28%5Cpi+s%29%3D%5Cpi&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\Gamma(s)\Gamma(1-s)\sin(\pi s)=\pi' title='\Gamma(s)\Gamma(1-s)\sin(\pi s)=\pi' class='latex' /></p>
<p>(well, you use this for <img src='http://l.wordpress.com/latex.php?latex=s%3Ds%2F2&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='s=s/2' title='s=s/2' class='latex' />), and one I haven&#8217;t mentioned:</p>
<p style="text-align:center;"><img src='http://l.wordpress.com/latex.php?latex=%5CGamma%281-s%29%3D2%5E%7B-s%7D%5CGamma%281-%5Cfrac%7Bs%7D%7B2%7D%29%5CGamma%28%5Cfrac%7B1-s%7D%7B2%7D%29%5Cpi%5E%7B-1%2F2%7D&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\Gamma(1-s)=2^{-s}\Gamma(1-\frac{s}{2})\Gamma(\frac{1-s}{2})\pi^{-1/2}' title='\Gamma(1-s)=2^{-s}\Gamma(1-\frac{s}{2})\Gamma(\frac{1-s}{2})\pi^{-1/2}' class='latex' /></p>
<p>and move some symbols around, you arrive at a more symmetric equation:</p>
<p style="text-align:center;"><img src='http://l.wordpress.com/latex.php?latex=%5CGamma%28%5Cfrac%7Bs%7D%7B2%7D%29%5Cpi%5E%7B-s%2F2%7D%5Czeta%28s%29%3D%5CGamma%28%5Cfrac%7B1-s%7D%7B2%7D%29%5Cpi%5E%7B-%281-s%29%2F2%7D%5Czeta%281-s%29.&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\Gamma(\frac{s}{2})\pi^{-s/2}\zeta(s)=\Gamma(\frac{1-s}{2})\pi^{-(1-s)/2}\zeta(1-s).' title='\Gamma(\frac{s}{2})\pi^{-s/2}\zeta(s)=\Gamma(\frac{1-s}{2})\pi^{-(1-s)/2}\zeta(1-s).' class='latex' /></p>
<p>Notice that if you plug <img src='http://l.wordpress.com/latex.php?latex=1-s&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='1-s' title='1-s' class='latex' /> in the formula on the left-hand side, you obtain the right-hand side.</p>
<p>This function on the left-hand side apparently has poles at 0 and 1, so if we define</p>
<p style="text-align:center;"><img src='http://l.wordpress.com/latex.php?latex=%5Cxi%28s%29%3D%5Cfrac%7Bs%28s-1%29%7D%7B2%7D%5CGamma%28%5Cfrac%7Bs%7D%7B2%7D%29%5Cpi%5E%7B-s%7D%7B2%7D%5Czeta%28s%29&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\xi(s)=\frac{s(s-1)}{2}\Gamma(\frac{s}{2})\pi^{-s}{2}\zeta(s)' title='\xi(s)=\frac{s(s-1)}{2}\Gamma(\frac{s}{2})\pi^{-s}{2}\zeta(s)' class='latex' /></p>
<p>then we obtain an entire analytic function satisfying <img src='http://l.wordpress.com/latex.php?latex=%5Cxi%28s%29%3D%5Cxi%281-s%29&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\xi(s)=\xi(1-s)' title='\xi(s)=\xi(1-s)' class='latex' />. Using the factorial relation for <img src='http://l.wordpress.com/latex.php?latex=%5CGamma&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\Gamma' title='\Gamma' class='latex' />, we can re-write <img src='http://l.wordpress.com/latex.php?latex=%5Cxi%28s%29&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\xi(s)' title='\xi(s)' class='latex' /> as</p>
<p style="text-align:center;"><img src='http://l.wordpress.com/latex.php?latex=%5Cxi%28s%29%3D%28s-1%29%5CGamma%28%5Cfrac%7Bs%7D%7B2%7D%2B1%29%5Cpi%5E%7B-s%2F2%7D%5Czeta%28s%29.&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\xi(s)=(s-1)\Gamma(\frac{s}{2}+1)\pi^{-s/2}\zeta(s).' title='\xi(s)=(s-1)\Gamma(\frac{s}{2}+1)\pi^{-s/2}\zeta(s).' class='latex' /></p>
<p>I get the impression that if you know what you are doing, then the things above aren&#8217;t tooo hard to justify. Apparently the next part is a bit trickier. Apparently, you can write</p>
<p style="text-align:center;"><img src='http://l.wordpress.com/latex.php?latex=%5Cxi%28s%29%3D%5Cxi%280%29%5Cprod_%7B%5Crho%7D%281-%5Cfrac%7Bs%7D%7B%5Crho%7D%29%2C&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\xi(s)=\xi(0)\prod_{\rho}(1-\frac{s}{\rho}),' title='\xi(s)=\xi(0)\prod_{\rho}(1-\frac{s}{\rho}),' class='latex' /></p>
<p>where the product is indexed over the roots, <img src='http://l.wordpress.com/latex.php?latex=%5Crho&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\rho' title='\rho' class='latex' />, of <img src='http://l.wordpress.com/latex.php?latex=%5Cxi&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\xi' title='\xi' class='latex' /> (so <img src='http://l.wordpress.com/latex.php?latex=%5Cxi%28%5Crho%29%3D0&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\xi(\rho)=0' title='\xi(\rho)=0' class='latex' />).</p>
<p>If you&#8217;ve heard anything about the <a href="http://en.wikipedia.org/wiki/Riemann_hypothesis">Riemann hypothesis</a>, you know that the roots (the &#8220;non-trivial&#8221; ones, I didn&#8217;t talk about the trivial ones) of <img src='http://l.wordpress.com/latex.php?latex=%5Czeta%28s%29&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\zeta(s)' title='\zeta(s)' class='latex' /> are a big deal. Our second formula for <img src='http://l.wordpress.com/latex.php?latex=%5Cxi%28s%29&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\xi(s)' title='\xi(s)' class='latex' /> shows that they are (basically) the same as the roots of <img src='http://l.wordpress.com/latex.php?latex=%5Cxi%28s%29&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\xi(s)' title='\xi(s)' class='latex' />, and so they are the <img src='http://l.wordpress.com/latex.php?latex=%5Crho&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\rho' title='\rho' class='latex' /> that the sum above is indexed over. The symmetric equation from earlier has a little something to say about the zeroes, and it has been shown that all of the zeroes have real part bigger than 0 and less than 1 (this is called the &#8220;critical strip&#8221;). The hypothesis (whose truth won&#8217;t affect what we&#8217;re saying below) is that all of the zeroes have real part 1/2 (this is the &#8220;critical line&#8221;). Apparently Riemann didn&#8217;t need this hypothesis for the things in his paper that introduced it, so I don&#8217;t really have much more to say about it right now. Although, honestly, I still don&#8217;t see what all the fuss is about <img src='http://s.wordpress.com/wp-includes/images/smilies/icon_smile.gif' alt=':)' class='wp-smiley' />  The formulas we&#8217;ll get below and tomorrow work even if the roots aren&#8217;t on the critical line (unless I&#8217;m missing something important. If I am, please comment).</p>
<p>Anyway, back to the topic at hand. Let me try to convince you that it isn&#8217;t horribly unreasonable to think about writing a function as a product over its roots, as I&#8217;ve done above. For the sake of example, let <img src='http://l.wordpress.com/latex.php?latex=f%28x%29%3D3x%5E3%2B3x%5E2-30x%2B24&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='f(x)=3x^3+3x^2-30x+24' title='f(x)=3x^3+3x^2-30x+24' class='latex' /> (or pick your own favorite polynomial). The usual way this would get factored, in all the classes I&#8217;ve ever taken or taught, is (up to permutation) <img src='http://l.wordpress.com/latex.php?latex=f%28x%29%3D3%28x%2B4%29%28x-1%29%28x-2%29&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='f(x)=3(x+4)(x-1)(x-2)' title='f(x)=3(x+4)(x-1)(x-2)' class='latex' />, showing that the roots are <img src='http://l.wordpress.com/latex.php?latex=x%3D1%2C2%2C-4&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='x=1,2,-4' title='x=1,2,-4' class='latex' />. However, if you factor a 4 out of the <img src='http://l.wordpress.com/latex.php?latex=x%2B4&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='x+4' title='x+4' class='latex' /> term, and -1 and -2 out of the other terms, you can also write <img src='http://l.wordpress.com/latex.php?latex=f%28x%29%3D24%281-%5Cfrac%7Bx%7D%7B-4%7D%29%281-x%29%281-%5Cfrac%7Bx%7D%7B2%7D%29&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='f(x)=24(1-\frac{x}{-4})(1-x)(1-\frac{x}{2})' title='f(x)=24(1-\frac{x}{-4})(1-x)(1-\frac{x}{2})' class='latex' />. You still see all the zeroes when you write the polynomial this way. You can also see that the coefficient in the front is <img src='http://l.wordpress.com/latex.php?latex=f%280%29&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='f(0)' title='f(0)' class='latex' />. So we&#8217;ve written <img src='http://l.wordpress.com/latex.php?latex=f%28x%29%3Df%280%29%5Cprod_%7B%5Crho%7D%281-x%2F%5Crho%29&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='f(x)=f(0)\prod_{\rho}(1-x/\rho)' title='f(x)=f(0)\prod_{\rho}(1-x/\rho)' class='latex' />, which is the same goal as what we&#8217;re doing with <img src='http://l.wordpress.com/latex.php?latex=%5Cxi&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\xi' title='\xi' class='latex' /> above. Incidentally, the idea of writing a function this way was also used by Euler to establish <img src='http://l.wordpress.com/latex.php?latex=%5Czeta%282%29%3D%5Csum+1%2Fn%5E2%3D%5Cpi%5E2%2F6&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\zeta(2)=\sum 1/n^2=\pi^2/6' title='\zeta(2)=\sum 1/n^2=\pi^2/6' class='latex' /> (I&#8217;ve mentioned this briefly <a href="http://sumidiot.blogspot.com/2008/11/fun-with-series.html">elsewhere</a>).</p>
<p>We now have two formulas for <img src='http://l.wordpress.com/latex.php?latex=%5Cxi%28s%29&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\xi(s)' title='\xi(s)' class='latex' />, so we can put them together to get</p>
<p style="text-align:center;"><img src='http://l.wordpress.com/latex.php?latex=%5Cxi%280%29%5Cprod_%7B%5Crho%7D%281-%5Cfrac%7Bs%7D%7B%5Crho%7D%29%3D%28s-1%29%5CGamma%28%5Cfrac%7Bs%7D%7B2%7D%2B1%29%5Cpi%5E%7B-s%2F2%7D%5Czeta%28s%29.&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\xi(0)\prod_{\rho}(1-\frac{s}{\rho})=(s-1)\Gamma(\frac{s}{2}+1)\pi^{-s/2}\zeta(s).' title='\xi(0)\prod_{\rho}(1-\frac{s}{\rho})=(s-1)\Gamma(\frac{s}{2}+1)\pi^{-s/2}\zeta(s).' class='latex' /></p>
<p>Recalling that our formula for <img src='http://l.wordpress.com/latex.php?latex=J%28x%29&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='J(x)' title='J(x)' class='latex' />, at the beginning, involved <img src='http://l.wordpress.com/latex.php?latex=%5Cln%5Czeta%28s%29&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\ln\zeta(s)' title='\ln\zeta(s)' class='latex' />, let&#8217;s take the log of the equation above and solve for the <img src='http://l.wordpress.com/latex.php?latex=%5Czeta&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\zeta' title='\zeta' class='latex' /> term:</p>
<p style="text-align:center;"><img src='http://l.wordpress.com/latex.php?latex=%5Cbegin%7Barray%7D%7Bl%7D+%5Cln+%5Czeta%28s%29%3D%5Cln+%5Cxi%280%29%2B%5Csum_%7B%5Crho%7D%5Cln%281-%5Cfrac%7Bs%7D%7B%5Crho%7D%29%5C%5C+%5Cqquad%5Cqquad+-%5Cln+%5CGamma%28%5Cfrac%7Bs%7D%7B2%7D%2B1%29%2B%5Cfrac%7Bs%7D%7B2%7D%5Cln+%5Cpi-%5Cln%28s-1%29.%5Cend%7Barray%7D&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\begin{array}{l} \ln \zeta(s)=\ln \xi(0)+\sum_{\rho}\ln(1-\frac{s}{\rho})\\ \qquad\qquad -\ln \Gamma(\frac{s}{2}+1)+\frac{s}{2}\ln \pi-\ln(s-1).\end{array}' title='\begin{array}{l} \ln \zeta(s)=\ln \xi(0)+\sum_{\rho}\ln(1-\frac{s}{\rho})\\ \qquad\qquad -\ln \Gamma(\frac{s}{2}+1)+\frac{s}{2}\ln \pi-\ln(s-1).\end{array}' class='latex' /></p>
<p>The idea is now to plug this in the formula for <img src='http://l.wordpress.com/latex.php?latex=J%28x%29&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='J(x)' title='J(x)' class='latex' />. Apparently if you do, though, you&#8217;ll have some issues with convergence. So actually try to do the integral in <img src='http://l.wordpress.com/latex.php?latex=J%28x%29&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='J(x)' title='J(x)' class='latex' />, using integration by parts (hint: <img src='http://l.wordpress.com/latex.php?latex=dv%3Dx%5Es+ds&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='dv=x^s ds' title='dv=x^s ds' class='latex' />). The &#8220;<img src='http://l.wordpress.com/latex.php?latex=uv&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='uv' title='uv' class='latex' />&#8221; term goes to 0 and you obtain</p>
<p style="text-align:center;"><img src='http://l.wordpress.com/latex.php?latex=%5Cdisplaystyle+J%28x%29%3D%5Cfrac%7B-1%7D%7B2%5Cpi+i%7D%5Ccdot+%5Cfrac%7B1%7D%7B%5Cln+x%7D%5Cint_%7Ba-i%5Cinfty%7D%5E%7Ba%2Bi%5Cinfty%7D%5Cfrac%7Bd%7D%7Bdx%7D%5Cleft%5B%5Cfrac%7B%5Cln%5Czeta%28s%29%7D%7Bs%7D%5Cright%5Dx%5Es%5C+ds%2C&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\displaystyle J(x)=\frac{-1}{2\pi i}\cdot \frac{1}{\ln x}\int_{a-i\infty}^{a+i\infty}\frac{d}{dx}\left[\frac{\ln\zeta(s)}{s}\right]x^s\ ds,' title='\displaystyle J(x)=\frac{-1}{2\pi i}\cdot \frac{1}{\ln x}\int_{a-i\infty}^{a+i\infty}\frac{d}{dx}\left[\frac{\ln\zeta(s)}{s}\right]x^s\ ds,' class='latex' /></p>
<p>where, as before <img src='http://l.wordpress.com/latex.php?latex=a%3D%5Ctext%7BRe+%7D+s&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='a=\text{Re } s' title='a=\text{Re } s' class='latex' />. Now plug in the 5 terms we&#8217;ve got above for <img src='http://l.wordpress.com/latex.php?latex=%5Cln+%5Czeta%28s%29&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\ln \zeta(s)' title='\ln \zeta(s)' class='latex' />, and you get a formula for <img src='http://l.wordpress.com/latex.php?latex=J%28x%29&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='J(x)' title='J(x)' class='latex' />. What happens to the terms? Can you actually work out any of the integrals?</p>
<p>Well, you might be able to. I&#8217;m not. Not right now anyway. But I can tell you about what others have figured out (rather like I&#8217;ve been doing all along, in fact)&#8230;</p>
<p>It&#8217;s clear that the <img src='http://l.wordpress.com/latex.php?latex=%5Cfrac%7Bs%7D%7B2%7D%5Cln+%5Cpi&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\frac{s}{2}\ln \pi' title='\frac{s}{2}\ln \pi' class='latex' /> term drops out, because you divide by <img src='http://l.wordpress.com/latex.php?latex=s&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='s' title='s' class='latex' /> and then take the derivative of a constant and just get 0. The term with <img src='http://l.wordpress.com/latex.php?latex=%5Cln%5Cxi%280%29&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\ln\xi(0)' title='\ln\xi(0)' class='latex' /> ends up just giving you <img src='http://l.wordpress.com/latex.php?latex=%5Cln%5Cxi%280%29&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\ln\xi(0)' title='\ln\xi(0)' class='latex' />, which is <img src='http://l.wordpress.com/latex.php?latex=-%5Cln%282%29&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='-\ln(2)' title='-\ln(2)' class='latex' />.</p>
<p>The term corresponding to the term with <img src='http://l.wordpress.com/latex.php?latex=%5CGamma&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\Gamma' title='\Gamma' class='latex' /> in it can be rewritten as</p>
<p style="text-align:center;"><img src='http://l.wordpress.com/latex.php?latex=%5Cdisplaystyle+%5Cint_x%5E%7B%5Cinfty%7D%5Cfrac%7Bdt%7D%7Bt%28t%5E2-1%29%5Cln+t%7D&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\displaystyle \int_x^{\infty}\frac{dt}{t(t^2-1)\ln t}' title='\displaystyle \int_x^{\infty}\frac{dt}{t(t^2-1)\ln t}' class='latex' /></p>
<p>(as if that were helpful).</p>
<p>The important terms seem to involve the function</p>
<p style="text-align:center;"><img src='http://l.wordpress.com/latex.php?latex=%5Cdisplaystyle+Li%28x%29%3D%5Cint_0%5Ex+%5Cfrac%7Bdt%7D%7B%5Cln+t%7D.&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\displaystyle Li(x)=\int_0^x \frac{dt}{\ln t}.' title='\displaystyle Li(x)=\int_0^x \frac{dt}{\ln t}.' class='latex' /></p>
<p>Of course, this integrand has a bit of an asymptote at 1, so really <img src='http://l.wordpress.com/latex.php?latex=Li%28x%29&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='Li(x)' title='Li(x)' class='latex' /> (in Edwards&#8217; book, anyway) is the &#8220;Cauchy principal value&#8221; of this integral, namely</p>
<p style="text-align:center;"><img src='http://l.wordpress.com/latex.php?latex=%5Cdisplaystyle+Li%28x%29%3D%5Clim_%7B%5Cepsilon%5Cto+0%5E%2B%7D%5Cint_0%5E%7B1-%5Cepsilon%7D%5Cfrac%7Bdt%7D%7B%5Cln+t%7D%2B%5Cint_%7B1%2B%5Cepsilon%7D%5Ex+%5Cfrac%7Bdt%7D%7B%5Cln+t%7D.&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\displaystyle Li(x)=\lim_{\epsilon\to 0^+}\int_0^{1-\epsilon}\frac{dt}{\ln t}+\int_{1+\epsilon}^x \frac{dt}{\ln t}.' title='\displaystyle Li(x)=\lim_{\epsilon\to 0^+}\int_0^{1-\epsilon}\frac{dt}{\ln t}+\int_{1+\epsilon}^x \frac{dt}{\ln t}.' class='latex' /></p>
<p style="text-align:left;">This function is, <a href="http://en.wikipedia.org/wiki/Prime_number_theorem">rather famously</a>, related to approximating the number of primes less than a given bound. In fact, tomorrow I plan on having more to say about this. But back to the terms in our integral for <img src='http://l.wordpress.com/latex.php?latex=J%28x%29&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='J(x)' title='J(x)' class='latex' />.</p>
<p style="text-align:left;">The term corresponding to the sum over the roots ends up giving you</p>
<p style="text-align:center;"><img src='http://l.wordpress.com/latex.php?latex=%5Cdisplaystyle+-%5Csum_%7B%5Ctext%7BIm+%7D%5Crho%26%2362%3B0%7DLi%28x%5E%7B%5Crho%7D%29%2BLi%28x%5E%7B1-%5Crho%7D%29&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\displaystyle -\sum_{\text{Im }\rho&gt;0}Li(x^{\rho})+Li(x^{1-\rho})' title='\displaystyle -\sum_{\text{Im }\rho&gt;0}Li(x^{\rho})+Li(x^{1-\rho})' class='latex' />.</p>
<p style="text-align:left;">But apparently the dominant term is the term corresponding to <img src='http://l.wordpress.com/latex.php?latex=%5Cln+%28s-1%29&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\ln (s-1)' title='\ln (s-1)' class='latex' />. It actually gives you <img src='http://l.wordpress.com/latex.php?latex=Li%28x%29&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='Li(x)' title='Li(x)' class='latex' /></p>
<p style="text-align:left;">So, finally, we have written</p>
<p style="text-align:center;"><img src='http://l.wordpress.com/latex.php?latex=%5Cbegin%7Barray%7D%7Bl%7D%5Cdisplaystyle+J%28x%29%3DLi%28x%29-%5Csum_%7B%5Ctext%7BIm+%7D%5Crho%26%2362%3B0%7D%5Cleft%28Li%28x%5E%7B%5Crho%7D%29%2BLi%28x%5E%7B1-%5Crho%7D%29%5Cright%29%5C%5C+%5Cdisplaystyle%5Cqquad%5Cqquad+%2B%5Cint_x%5E%7B%5Cinfty%7D%5Cfrac%7Bdt%7D%7Bt%28t%5E2-1%29%5Cln+t%7D-%5Cln+2.%5Cend%7Barray%7D&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\begin{array}{l}\displaystyle J(x)=Li(x)-\sum_{\text{Im }\rho&gt;0}\left(Li(x^{\rho})+Li(x^{1-\rho})\right)\\ \displaystyle\qquad\qquad +\int_x^{\infty}\frac{dt}{t(t^2-1)\ln t}-\ln 2.\end{array}' title='\begin{array}{l}\displaystyle J(x)=Li(x)-\sum_{\text{Im }\rho&gt;0}\left(Li(x^{\rho})+Li(x^{1-\rho})\right)\\ \displaystyle\qquad\qquad +\int_x^{\infty}\frac{dt}{t(t^2-1)\ln t}-\ln 2.\end{array}' class='latex' /></p>
<p style="text-align:left;">Doesn&#8217;t that make you feel better? We started with the reasonably understandable</p>
<p style="text-align:center;"><img src='http://l.wordpress.com/latex.php?latex=%5Cdisplaystyle+J%28x%29%3D%5Csum_%7Bp%5En%5Cleq+x%7D%5Cfrac%7B1%7D%7Bn%7D&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\displaystyle J(x)=\sum_{p^n\leq x}\frac{1}{n}' title='\displaystyle J(x)=\sum_{p^n\leq x}\frac{1}{n}' class='latex' />,</p>
<p style="text-align:left;">and created the monstrosity above. I guess this is why I&#8217;m not an analyst. To me, it seems worse to write <img src='http://l.wordpress.com/latex.php?latex=J%28x%29&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='J(x)' title='J(x)' class='latex' /> as this terrible combination of lots of integrals. But apparently it&#8217;s useful in analysis to have such formulas. I guess we&#8217;ll see a use tomorrow&#8230;</p>
</div>]]></content:encoded>
</item>
<item>
<title><![CDATA[The Log of Zeta]]></title>
<link>http://sumidiot.wordpress.com/2009/11/21/the-log-of-zeta/</link>
<pubDate>Sun, 22 Nov 2009 04:49:08 +0000</pubDate>
<dc:creator>sumidiot</dc:creator>
<guid>http://sumidiot.wordpress.com/2009/11/21/the-log-of-zeta/</guid>
<description><![CDATA[Last time (too long ago, sorry), I finally got around to talking about the function . Today, I think]]></description>
<content:encoded><![CDATA[<div class='snap_preview'><p><a href="http://sumidiot.wordpress.com/2009/11/13/riemanns-zeta-function/">Last time</a> (too long ago, sorry), I finally got around to talking about the function <img src='http://l.wordpress.com/latex.php?latex=%5Czeta%28s%29&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\zeta(s)' title='\zeta(s)' class='latex' />. Today, I think I&#8217;d like to talk about <img src='http://l.wordpress.com/latex.php?latex=%5Clog+%5Czeta%28s%29&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\log \zeta(s)' title='\log \zeta(s)' class='latex' />.</p>
<p>First, though, I should mention Euler&#8217;s product formula. The goal is to re-write the infinite series <img src='http://l.wordpress.com/latex.php?latex=%5Czeta%28s%29%3D%5Csum_n+n%5E%7B-s%7D&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\zeta(s)=\sum_n n^{-s}' title='\zeta(s)=\sum_n n^{-s}' class='latex' /> as a product. For each prime <img src='http://l.wordpress.com/latex.php?latex=p&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='p' title='p' class='latex' />, we can tease out of the series above those terms where <img src='http://l.wordpress.com/latex.php?latex=n&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='n' title='n' class='latex' /> is <img src='http://l.wordpress.com/latex.php?latex=p&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='p' title='p' class='latex' /> to some power. That is, we can think about the sub-series <img src='http://l.wordpress.com/latex.php?latex=%5Csum_%7Bk%5Cgeq+0%7D+%28p%5Ek%29%5E%7B-s%7D&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\sum_{k\geq 0} (p^k)^{-s}' title='\sum_{k\geq 0} (p^k)^{-s}' class='latex' />. This is a geometric series that converges to <img src='http://l.wordpress.com/latex.php?latex=%281-p%5E%7B-s%7D%29%5E%7B-1%7D&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='(1-p^{-s})^{-1}' title='(1-p^{-s})^{-1}' class='latex' />. If we take two such series and multiply them together, which terms from our starting series for <img src='http://l.wordpress.com/latex.php?latex=%5Czeta%28s%29&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\zeta(s)' title='\zeta(s)' class='latex' /> will we recover?</p>
<p>Suppose <img src='http://l.wordpress.com/latex.php?latex=p&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='p' title='p' class='latex' /> and <img src='http://l.wordpress.com/latex.php?latex=q&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='q' title='q' class='latex' /> are two primes, and consider the product</p>
<p style="text-align:center;"><img src='http://l.wordpress.com/latex.php?latex=%5Cdisplaystyle%5Cleft%281%2B%5Cfrac%7B1%7D%7Bp%5Es%7D%2B%5Cfrac%7B1%7D%7Bp%5E%7B2s%7D%7D%2B%5Cfrac%7B1%7D%7Bp%5E%7B3s%7D%7D%2B%5Ccdots%5Cright%29%5Ccdot+%5Cleft%281%2B+%5Cfrac%7B1%7D%7Bq%5Es%7D%2B%5Cfrac%7B1%7D%7Bq%5E%7B2s%7D%7D%2B%5Cfrac%7B1%7D%7Bq%5E%7B3s%7D%7D%2B%5Ccdots%5Cright%29&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\displaystyle\left(1+\frac{1}{p^s}+\frac{1}{p^{2s}}+\frac{1}{p^{3s}}+\cdots\right)\cdot \left(1+ \frac{1}{q^s}+\frac{1}{q^{2s}}+\frac{1}{q^{3s}}+\cdots\right)' title='\displaystyle\left(1+\frac{1}{p^s}+\frac{1}{p^{2s}}+\frac{1}{p^{3s}}+\cdots\right)\cdot \left(1+ \frac{1}{q^s}+\frac{1}{q^{2s}}+\frac{1}{q^{3s}}+\cdots\right)' class='latex' />.</p>
<p>The terms in <img src='http://l.wordpress.com/latex.php?latex=%5Csum_n+n%5E%7B-s%7D&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\sum_n n^{-s}' title='\sum_n n^{-s}' class='latex' /> that we obtain from this product are those where <img src='http://l.wordpress.com/latex.php?latex=n&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='n' title='n' class='latex' /> is a product of some power (possibly the 0 power) of <img src='http://l.wordpress.com/latex.php?latex=p&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='p' title='p' class='latex' /> and some power (again, possibly 0) of <img src='http://l.wordpress.com/latex.php?latex=q&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='q' title='q' class='latex' />. We could then think about another prime, say <img src='http://l.wordpress.com/latex.php?latex=r&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='r' title='r' class='latex' />, and it&#8217;s geometric series, and multiply it by the above, and obtain all the terms in <img src='http://l.wordpress.com/latex.php?latex=%5Csum_n+n%5E%7B-s%7D&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\sum_n n^{-s}' title='\sum_n n^{-s}' class='latex' /> where <img src='http://l.wordpress.com/latex.php?latex=n%3Dp%5Eaq%5Ebr%5Ec&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='n=p^aq^br^c' title='n=p^aq^br^c' class='latex' /> for some <img src='http://l.wordpress.com/latex.php?latex=0%5Cleq+a%2Cb%2Cc&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='0\leq a,b,c' title='0\leq a,b,c' class='latex' />.</p>
<p>Continuing on, and recalling that every positive integer has a unique factorization as a product of primes, we obtain the magical formula (isn&#8217;t most of Euler&#8217;s work magical?):</p>
<p style="text-align:center;"><img src='http://l.wordpress.com/latex.php?latex=%5Cdisplaystyle+%5Csum_n+%5Cfrac%7B1%7D%7Bn%5Es%7D%3D%5Cprod_p+%281-p%5E%7B-s%7D%29%5E%7B-1%7D&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\displaystyle \sum_n \frac{1}{n^s}=\prod_p (1-p^{-s})^{-1}' title='\displaystyle \sum_n \frac{1}{n^s}=\prod_p (1-p^{-s})^{-1}' class='latex' />,</p>
<p>where the product is indexed by the primes. What&#8217;s nice about this, for us, at the moment, is that logarithms work well with products and powers, but not soo well for sums. Recalling the Taylor polynomial</p>
<p style="text-align:center;"><img src='http://l.wordpress.com/latex.php?latex=%5Cln%281-x%29%3D-1-x-x%5E2-x%5E3-%5Ccdots%3D-%5Csum%5Climits_%7Bn%3D0%7D%5E%7B%5Cinfty%7D+x%5En&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\ln(1-x)=-1-x-x^2-x^3-\cdots=-\sum\limits_{n=0}^{\infty} x^n' title='\ln(1-x)=-1-x-x^2-x^3-\cdots=-\sum\limits_{n=0}^{\infty} x^n' class='latex' /></p>
<p>we find</p>
<p style="text-align:center;"><img src='http://l.wordpress.com/latex.php?latex=%5Cdisplaystyle+%5Cln+%5Czeta%28s%29%3D%5Csum_p%5Csum_n+%5Cfrac%7B1%7D%7Bn%7Dp%5E%7B-ns%7D&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\displaystyle \ln \zeta(s)=\sum_p\sum_n \frac{1}{n}p^{-ns}' title='\displaystyle \ln \zeta(s)=\sum_p\sum_n \frac{1}{n}p^{-ns}' class='latex' />.</p>
<p>That was fun. Incidentally, it&#8217;s not much of a jump from here to show that the series of prime reciprocals diverges. I mentioned it in a <a href="http://sumidiot.wordpress.com/2009/02/19/curious-constants-part-1-of-4/">post</a> a while ago.</p>
<p>Let&#8217;s switch gears for a little bit. I&#8217;m going to define a function <img src='http://l.wordpress.com/latex.php?latex=J%28x%29&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='J(x)' title='J(x)' class='latex' /> on the positive reals. If you&#8217;re following Edwards&#8217; book with me (as I jump all over the place), this will be almost exactly his <img src='http://l.wordpress.com/latex.php?latex=J%28x%29&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='J(x)' title='J(x)' class='latex' />, differing only at the integers (for today anyway <img src='http://s.wordpress.com/wp-includes/images/smilies/icon_smile.gif' alt=':)' class='wp-smiley' /> ). Let me start by setting <img src='http://l.wordpress.com/latex.php?latex=J%280%29%3D0&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='J(0)=0' title='J(0)=0' class='latex' />. The function <img src='http://l.wordpress.com/latex.php?latex=J%28x%29&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='J(x)' title='J(x)' class='latex' /> will be a step function, which I&#8217;ll define to mean piecewise linear with slope 0 on each piece. So to define <img src='http://l.wordpress.com/latex.php?latex=J%28x%29&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='J(x)' title='J(x)' class='latex' /> I only need to tell you when to jump, and by how much. The rule is: when you get to an integer <img src='http://l.wordpress.com/latex.php?latex=n&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='n' title='n' class='latex' />, if <img src='http://l.wordpress.com/latex.php?latex=n%3Dp%5Ek&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='n=p^k' title='n=p^k' class='latex' /> for some prime <img src='http://l.wordpress.com/latex.php?latex=p&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='p' title='p' class='latex' />, then you jump up by <img src='http://l.wordpress.com/latex.php?latex=1%2Fk&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='1/k' title='1/k' class='latex' />. So at <img src='http://l.wordpress.com/latex.php?latex=2%2C3%2C5%2C7%2C11%2C%5Cldots&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='2,3,5,7,11,\ldots' title='2,3,5,7,11,\ldots' class='latex' /> you jump by 1, at <img src='http://l.wordpress.com/latex.php?latex=4%2C9%2C25%2C49%2C%5Cldots&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='4,9,25,49,\ldots' title='4,9,25,49,\ldots' class='latex' /> you jump by 1/2, at <img src='http://l.wordpress.com/latex.php?latex=8%2C+27%2C+%5Cldots&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='8, 27, \ldots' title='8, 27, \ldots' class='latex' /> you jump by 1/3, etc. Here&#8217;s a little picture of <img src='http://l.wordpress.com/latex.php?latex=J%28x%29&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='J(x)' title='J(x)' class='latex' />, with a few values highlighted:</p>
<p style="text-align:center;"><img class="aligncenter" title="J(x)" src="http://chart.apis.google.com/chart?chs=300x300&#38;cht=bvs&#38;chco=4D89F9&#38;chm=N*f2y*,000000,0,-2,10,0&#38;chxt=x,y&#38;chxr=0,0,20,1&#124;1,0,10,2&#38;chbh=a&#38;chds=0,10&#38;chd=t:0,0,1,2,2.5,3.5,3.5,4.5,4.83,5.33,5.33,6.33,6.33,7.33,7.33,7.33,7.58,8.58,8.58,9.58" alt="" width="300" height="300" /></p>
<p>Slightly more precisely, we can write</p>
<p style="text-align:center;"><img src='http://l.wordpress.com/latex.php?latex=%5Cdisplaystyle+J%28x%29%3D%5Csum_%7Bp%5En%5Cleq+x%7D%5Cfrac%7B1%7D%7Bn%7D&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\displaystyle J(x)=\sum_{p^n\leq x}\frac{1}{n}' title='\displaystyle J(x)=\sum_{p^n\leq x}\frac{1}{n}' class='latex' />.</p>
<p>Now, let me see if I can convince you that there is some justification in writing</p>
<p style="text-align:center;"><img src='http://l.wordpress.com/latex.php?latex=%5Cdisplaystyle+%5Cln+%5Czeta%28s%29%3Ds%5Cint_0%5E%7B%5Cinfty%7D+J%28x%29x%5E%7B-s-1%7D%5C+dx&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\displaystyle \ln \zeta(s)=s\int_0^{\infty} J(x)x^{-s-1}\ dx' title='\displaystyle \ln \zeta(s)=s\int_0^{\infty} J(x)x^{-s-1}\ dx' class='latex' />.</p>
<p>Let&#8217;s work with the right-hand side. Since <img src='http://l.wordpress.com/latex.php?latex=J%28x%29%3D0&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='J(x)=0' title='J(x)=0' class='latex' /> near <img src='http://l.wordpress.com/latex.php?latex=x%3D0&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='x=0' title='x=0' class='latex' />, I&#8217;ll actually start my integral at <img src='http://l.wordpress.com/latex.php?latex=x%3D1&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='x=1' title='x=1' class='latex' />. I think the way to go about it is as follows:</p>
<p style="text-align:center;"><img src='http://l.wordpress.com/latex.php?latex=%5Cbegin%7Barray%7D%7Brcl%7D+%5Cdisplaystyle+s%5Cint_1%5E%7B%5Cinfty%7DJ%28x%29x%5E%7B-s-1%7D%5C+dx+%26%2338%3B%3D%26%2338%3B+%5Cdisplaystyle+s%5Csum_%7Bm%3D1%7D%5E%7B%5Cinfty%7D%5Cint_m%5E%7Bm%2B1%7DJ%28x%29x%5E%7B-s-1%7D%5C+dx+%5C%5C+%7B%7D+%26%2338%3B+%3D+%26%2338%3B+%5Cdisplaystyle+%5Csum_%7Bm%3D1%7D%5E%7B%5Cinfty%7DJ%28m%29%5Cleft%28%5Cfrac%7B1%7D%7Bm%5Es%7D-%5Cfrac%7B1%7D%7B%28m%2B1%29%5Es%7D%5Cright%29.%5Cend%7Barray%7D&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\begin{array}{rcl} \displaystyle s\int_1^{\infty}J(x)x^{-s-1}\ dx &amp;=&amp; \displaystyle s\sum_{m=1}^{\infty}\int_m^{m+1}J(x)x^{-s-1}\ dx \\ {} &amp; = &amp; \displaystyle \sum_{m=1}^{\infty}J(m)\left(\frac{1}{m^s}-\frac{1}{(m+1)^s}\right).\end{array}' title='\begin{array}{rcl} \displaystyle s\int_1^{\infty}J(x)x^{-s-1}\ dx &amp;=&amp; \displaystyle s\sum_{m=1}^{\infty}\int_m^{m+1}J(x)x^{-s-1}\ dx \\ {} &amp; = &amp; \displaystyle \sum_{m=1}^{\infty}J(m)\left(\frac{1}{m^s}-\frac{1}{(m+1)^s}\right).\end{array}' class='latex' /></p>
<p>Now suppose that <img src='http://l.wordpress.com/latex.php?latex=J%28m%29%3DJ%28m%2B1%29&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='J(m)=J(m+1)' title='J(m)=J(m+1)' class='latex' /> for some <img src='http://l.wordpress.com/latex.php?latex=m&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='m' title='m' class='latex' />. Then the terms corresponding to <img src='http://l.wordpress.com/latex.php?latex=m&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='m' title='m' class='latex' /> and <img src='http://l.wordpress.com/latex.php?latex=m%2B1&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='m+1' title='m+1' class='latex' /> telescope as follows:</p>
<p style="text-align:center;"><img src='http://l.wordpress.com/latex.php?latex=%5Cbegin%7Barray%7D%7Bc%7D+%5Cdisplaystyle+J%28m%29%5Cleft%28%5Cfrac%7B1%7D%7Bm%5Es%7D-%5Cfrac%7B1%7D%7B%28m%2B1%29%5Es%7D%5Cright%29%2BJ%28m%2B1%29%5Cleft%28%5Cfrac%7B1%7D%7B%28m%2B1%29%5Es%7D-%5Cfrac%7B1%7D%7B%28m%2B2%29%5Es%7D%5Cright%29%5C%5C+%3D%5Cdisplaystyle+J%28m%29%5Cleft%28%5Cfrac%7B1%7D%7Bm%5Es%7D-%5Cfrac%7B1%7D%7B%28m%2B2%29%5Es%7D%5Cright%29%5Cend%7Barray%7D&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\begin{array}{c} \displaystyle J(m)\left(\frac{1}{m^s}-\frac{1}{(m+1)^s}\right)+J(m+1)\left(\frac{1}{(m+1)^s}-\frac{1}{(m+2)^s}\right)\\ =\displaystyle J(m)\left(\frac{1}{m^s}-\frac{1}{(m+2)^s}\right)\end{array}' title='\begin{array}{c} \displaystyle J(m)\left(\frac{1}{m^s}-\frac{1}{(m+1)^s}\right)+J(m+1)\left(\frac{1}{(m+1)^s}-\frac{1}{(m+2)^s}\right)\\ =\displaystyle J(m)\left(\frac{1}{m^s}-\frac{1}{(m+2)^s}\right)\end{array}' class='latex' />.</p>
<p>If, also, <img src='http://l.wordpress.com/latex.php?latex=J%28m%29%3DJ%28m%2B2%29&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='J(m)=J(m+2)' title='J(m)=J(m+2)' class='latex' />, this we can telescope another term into this one, and so on. So, really, the important <img src='http://l.wordpress.com/latex.php?latex=m&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='m' title='m' class='latex' /> in this sum are those where <img src='http://l.wordpress.com/latex.php?latex=J%28x%29&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='J(x)' title='J(x)' class='latex' /> jumps, which are the prime powers. Let <img src='http://l.wordpress.com/latex.php?latex=m_i%3Dp_i%5E%7Bn_i%7D&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='m_i=p_i^{n_i}' title='m_i=p_i^{n_i}' class='latex' /> be the <img src='http://l.wordpress.com/latex.php?latex=i&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='i' title='i' class='latex' />-th prime power (i.e., the point where <img src='http://l.wordpress.com/latex.php?latex=J%28x%29&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='J(x)' title='J(x)' class='latex' /> makes its <img src='http://l.wordpress.com/latex.php?latex=i&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='i' title='i' class='latex' />-th jump), starting with <img src='http://l.wordpress.com/latex.php?latex=m_1%3D2&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='m_1=2' title='m_1=2' class='latex' />. Then we can write</p>
<p style="text-align:center;"><img src='http://l.wordpress.com/latex.php?latex=%5Cbegin%7Barray%7D%7Brcl%7D+%5Cdisplaystyle+s%5Cint_0%5E%7B%5Cinfty%7DJ%28x%29x%5E%7B-s-1%7D%5C+dx+%26%2338%3B+%3D+%26%2338%3B+%5Cdisplaystyle+%5Csum_i+J%28m_i%29%5Cleft%28%5Cfrac%7B1%7D%7Bm_i%5Es%7D-%5Cfrac%7B1%7D%7Bm_%7Bi%2B1%7D%5Es%7D%5Cright%29+%5C%5C+%26%2338%3B+%3D+%26%2338%3B+%5Cdisplaystyle+%5Cfrac%7B1%7D%7B2%7DJ%282%29%2B%5Csum_%7Bi%3D2%7D%5E%7B%5Cinfty%7D+-J%28m_%7Bi-1%7D%29%5Cfrac%7B1%7D%7Bm_i%5Es%7D%2BJ%28m_i%29%5Cfrac%7B1%7D%7Bm_i%5Es%7D+%5C%5C+%26%2338%3B+%3D+%26%2338%3B+%5Cdisplaystyle+%5Cfrac%7B1%7D%7B2%7D%2B%5Csum_%7Bi%3D2%7D%5E%7B%5Cinfty%7D+%5Cfrac%7B1%7D%7Bm_i%5Es%7D%5Cleft%28J%28m_i%29-J%28m_%7Bi-1%7D%29%5Cright%29+%5C%5C+%26%2338%3B+%3D+%26%2338%3B+%5Cdisplaystyle+%5Cfrac%7B1%7D%7B2%7D%2B%5Csum_%7Bi%3D2%7D%5E%7B%5Cinfty%7D+%5Cfrac%7B1%7D%7Bm_i%5Es%7D%5Ccdot+%5Cfrac%7B1%7D%7Bn_i%7D+%5C%5C+%26%2338%3B+%3D+%26%2338%3B+%5Cdisplaystyle+%5Csum_%7Bp%7D%5Csum_n+%5Cfrac%7B1%7D%7Bn%7Dp%5E%7B-ns%7D+%5C%5C+%26%2338%3B+%3D+%26%2338%3B+%5Cln+%5Czeta%28s%29.%5Cend%7Barray%7D&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\begin{array}{rcl} \displaystyle s\int_0^{\infty}J(x)x^{-s-1}\ dx &amp; = &amp; \displaystyle \sum_i J(m_i)\left(\frac{1}{m_i^s}-\frac{1}{m_{i+1}^s}\right) \\ &amp; = &amp; \displaystyle \frac{1}{2}J(2)+\sum_{i=2}^{\infty} -J(m_{i-1})\frac{1}{m_i^s}+J(m_i)\frac{1}{m_i^s} \\ &amp; = &amp; \displaystyle \frac{1}{2}+\sum_{i=2}^{\infty} \frac{1}{m_i^s}\left(J(m_i)-J(m_{i-1})\right) \\ &amp; = &amp; \displaystyle \frac{1}{2}+\sum_{i=2}^{\infty} \frac{1}{m_i^s}\cdot \frac{1}{n_i} \\ &amp; = &amp; \displaystyle \sum_{p}\sum_n \frac{1}{n}p^{-ns} \\ &amp; = &amp; \ln \zeta(s).\end{array}' title='\begin{array}{rcl} \displaystyle s\int_0^{\infty}J(x)x^{-s-1}\ dx &amp; = &amp; \displaystyle \sum_i J(m_i)\left(\frac{1}{m_i^s}-\frac{1}{m_{i+1}^s}\right) \\ &amp; = &amp; \displaystyle \frac{1}{2}J(2)+\sum_{i=2}^{\infty} -J(m_{i-1})\frac{1}{m_i^s}+J(m_i)\frac{1}{m_i^s} \\ &amp; = &amp; \displaystyle \frac{1}{2}+\sum_{i=2}^{\infty} \frac{1}{m_i^s}\left(J(m_i)-J(m_{i-1})\right) \\ &amp; = &amp; \displaystyle \frac{1}{2}+\sum_{i=2}^{\infty} \frac{1}{m_i^s}\cdot \frac{1}{n_i} \\ &amp; = &amp; \displaystyle \sum_{p}\sum_n \frac{1}{n}p^{-ns} \\ &amp; = &amp; \ln \zeta(s).\end{array}' class='latex' /></p>
<p style="text-align:left;">What good is writing <img src='http://l.wordpress.com/latex.php?latex=%5Cln+%5Czeta%28s%29&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\ln \zeta(s)' title='\ln \zeta(s)' class='latex' /> this way? I guess if you know something about Fourier inversion (which I don&#8217;t) then you get to say that</p>
<p style="text-align:center;"><img src='http://l.wordpress.com/latex.php?latex=%5Cdisplaystyle+J%28x%29%3D%5Cfrac%7B1%7D%7B2%5Cpi+i%7D%5Cint_%7Ba-i%5Cinfty%7D%5E%7Ba%2Bi%5Cinfty%7D%5Clog+%5Czeta%28s%29x%5Es%5Cfrac%7Bds%7D%7Bs%7D%2C&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\displaystyle J(x)=\frac{1}{2\pi i}\int_{a-i\infty}^{a+i\infty}\log \zeta(s)x^s\frac{ds}{s},' title='\displaystyle J(x)=\frac{1}{2\pi i}\int_{a-i\infty}^{a+i\infty}\log \zeta(s)x^s\frac{ds}{s},' class='latex' /></p>
<p style="text-align:left;">for <img src='http://l.wordpress.com/latex.php?latex=a%3D%5Ctext%7BRe%7D%28s%29%26%2362%3B1&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='a=\text{Re}(s)&gt;1' title='a=\text{Re}(s)&gt;1' class='latex' />. What good is that? I think I&#8217;ll have to read some more and tell you about it tomorrow, but it&#8217;ll turn out to be useful once we have yet another description of <img src='http://l.wordpress.com/latex.php?latex=%5Cln+%5Czeta%28s%29&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\ln \zeta(s)' title='\ln \zeta(s)' class='latex' />, in terms of the 0s of <img src='http://l.wordpress.com/latex.php?latex=%5Czeta%28s%29&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\zeta(s)' title='\zeta(s)' class='latex' /> (finally getting to those 0s <a href="http://plus.maths.org/blog/2009/11/happy-150th-birthday-to-riemann.html">everybody</a> <a href="http://mathdl.maa.org/mathDL/?pa=mathNews&#38;sa=view&#38;newsId=714">cares</a> so much about).</p>
</div>]]></content:encoded>
</item>
<item>
<title><![CDATA[Ordenan cierre administrativo indefinido de revista Zeta y El Nuevo País ]]></title>
<link>http://primicias24.wordpress.com/2009/11/20/ordenan-cierre-administrativo-indefinido-de-revista-zeta-y-el-nuevo-pais/</link>
<pubDate>Fri, 20 Nov 2009 13:54:05 +0000</pubDate>
<dc:creator>primicias24</dc:creator>
<guid>http://primicias24.wordpress.com/2009/11/20/ordenan-cierre-administrativo-indefinido-de-revista-zeta-y-el-nuevo-pais/</guid>
<description><![CDATA[Funcionarios de la oficina de tributos de la Alcaldía de Libertador Sumat, ordenaron el cierre admin]]></description>
<content:encoded><![CDATA[Funcionarios de la oficina de tributos de la Alcaldía de Libertador Sumat, ordenaron el cierre admin]]></content:encoded>
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<item>
<title><![CDATA[Look #56]]></title>
<link>http://nana1004.wordpress.com/2009/11/19/look-56/</link>
<pubDate>Thu, 19 Nov 2009 10:50:03 +0000</pubDate>
<dc:creator>Nana</dc:creator>
<guid>http://nana1004.wordpress.com/2009/11/19/look-56/</guid>
<description><![CDATA[Hair:REDLIC #540d Reddish Brown Skin:*Beauty Avatar Couture* ZETA &#8211; Light Skin 08 [buy@here] *]]></description>
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<p>&#160;</p>
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</item>
<item>
<title><![CDATA[Look #55]]></title>
<link>http://nana1004.wordpress.com/2009/11/17/look-55/</link>
<pubDate>Tue, 17 Nov 2009 10:44:15 +0000</pubDate>
<dc:creator>Nana</dc:creator>
<guid>http://nana1004.wordpress.com/2009/11/17/look-55/</guid>
<description><![CDATA[http://nana1004.wordpress.com Left Look Skin:*Beauty Avatar Couture* ZETA &#8211; Light Skin 09 [buy]]></description>
<content:encoded><![CDATA[<div class='snap_preview'><p><img class="alignnone size-large wp-image-1327" title="look54" src="http://nana1004.wordpress.com/files/2009/11/look541.jpg?w=580" alt="" width="580" height="743" /></p>
<p><img class="alignnone size-large wp-image-1328" title="look54_1" src="http://nana1004.wordpress.com/files/2009/11/look54_11.jpg?w=580" alt="" width="580" height="658" /></p>
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<p><strong>Left Look</strong><br />
Skin:*Beauty Avatar Couture* ZETA &#8211; Light Skin 09 [buy@<a href="http://slurl.com/secondlife/Beauty%20Avatar%20couture/246/118/24" target="_blank">here</a>] <span style="color:#ff6600;">*New released*</span><br />
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Skin: +ROMI+ [Lily_M_Pink 02][Pink, Chapped lip, Dark brow]  [buy@<a href="http://slurl.com/secondlife/Yeolmae/13/224/772" target="_blank">here</a>] <span style="color:#ff6600;">*New released*</span><br />
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<title><![CDATA[Riemann's Zeta Function]]></title>
<link>http://sumidiot.wordpress.com/2009/11/13/riemanns-zeta-function/</link>
<pubDate>Sat, 14 Nov 2009 03:28:33 +0000</pubDate>
<dc:creator>sumidiot</dc:creator>
<guid>http://sumidiot.wordpress.com/2009/11/13/riemanns-zeta-function/</guid>
<description><![CDATA[I guess it is about time to get to the zeta function side of this story, if we&#8217;re ever going t]]></description>
<content:encoded><![CDATA[<div class='snap_preview'><p>I guess it is about time to get to the zeta function side of this story, if we&#8217;re ever going to use Farey sequences to show how you could prove the Riemann hypothesis. I&#8217;ve been reading a bit of Edwards&#8217; book, with the same title as this post, and thought I&#8217;d try to summarize the first chapter over the next few posts. I&#8217;m not sure that the content of the first chapter is hugely vital for the final goal of relating to the Farey sequences, but I wanted to try to learn some of it anyway.</p>
<p>I should mention, before talking about any of this, that I will not claim to know any complex analysis. The last complex analysis course I took was 5 years ago, I can&#8217;t be sure how much I paid attention at the time, and I haven&#8217;t used it since. I will be jumping over gaps of various sizes for quite a while in the upcoming posts. Perhaps sometimes I&#8217;ll mention that a gap is there. Mostly, though, what I&#8217;m after in this story is the outline, and how all of the parts fit together, as a big picture. Perhaps I&#8217;ll go back and fill in some gaps, as I understand more.</p>
<p>For today, let&#8217;s see if I can build up to an analytic expression of <img src='http://l.wordpress.com/latex.php?latex=%5Czeta%28s%29&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\zeta(s)' title='\zeta(s)' class='latex' />. Our starting point is the function</p>
<p style="text-align:center;"><img src='http://l.wordpress.com/latex.php?latex=%5Czeta%28s%29%3D%5Cdisplaystyle%5Csum_%7Bn%3D1%7D%5E%7B%5Cinfty%7D%5Cfrac%7B1%7D%7Bn%5Es%7D&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\zeta(s)=\displaystyle\sum_{n=1}^{\infty}\frac{1}{n^s}' title='\zeta(s)=\displaystyle\sum_{n=1}^{\infty}\frac{1}{n^s}' class='latex' /></p>
<p>which is defined for real values of <img src='http://l.wordpress.com/latex.php?latex=s&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='s' title='s' class='latex' /> larger than 1. The goal is to find a nice expression that is defined on more of the complex plane, but agrees with this definition on the reals larger than 1.</p>
<p>To get to that point, we&#8217;ll use the <img src='http://l.wordpress.com/latex.php?latex=%5CGamma&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\Gamma' title='\Gamma' class='latex' /> function:</p>
<p style="text-align:center;"><img src='http://l.wordpress.com/latex.php?latex=%5CGamma%28s%29%3D%5Cdisplaystyle%5Cint_0%5E%7B%5Cinfty%7D+e%5E%7B-x%7Dx%5E%7Bs-1%7D%5C+dx&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\Gamma(s)=\displaystyle\int_0^{\infty} e^{-x}x^{s-1}\ dx' title='\Gamma(s)=\displaystyle\int_0^{\infty} e^{-x}x^{s-1}\ dx' class='latex' /></p>
<p>This is an analytic function defined everywhere except at the negative integers and 0. If you are only interested in real values of <img src='http://l.wordpress.com/latex.php?latex=s&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='s' title='s' class='latex' />, this function is a continuous extension of the factorial function, which is only defined on the positive integers. Actually, there is a shift involved, so <img src='http://l.wordpress.com/latex.php?latex=%5CGamma%28n%29%3D%28n-1%29%21&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\Gamma(n)=(n-1)!' title='\Gamma(n)=(n-1)!' class='latex' /> (Edwards blames this shift on Legendre, whose reasons, he states, &#8220;are obscure&#8221;. Edwards uses <img src='http://l.wordpress.com/latex.php?latex=%5CPi%28s%29%3D%5CGamma%28s%2B1%29&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\Pi(s)=\Gamma(s+1)' title='\Pi(s)=\Gamma(s+1)' class='latex' /> throughout, so I might make some <img src='http://l.wordpress.com/latex.php?latex=%5Cpm+1&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\pm 1' title='\pm 1' class='latex' /> errors). In particular, <img src='http://l.wordpress.com/latex.php?latex=s%5CGamma%28s%29%3D%5CGamma%28s%2B1%29&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='s\Gamma(s)=\Gamma(s+1)' title='s\Gamma(s)=\Gamma(s+1)' class='latex' /> when both sides make sense.</p>
<p>Another relation that we&#8217;ll use is that</p>
<p style="text-align:center;"><img src='http://l.wordpress.com/latex.php?latex=%5Cpi%3D%5CGamma%28s%29%5CGamma%281-s%29%5Csin%28%5Cpi+s%29&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\pi=\Gamma(s)\Gamma(1-s)\sin(\pi s)' title='\pi=\Gamma(s)\Gamma(1-s)\sin(\pi s)' class='latex' /></p>
<p>If memory serves (from reading Artin&#8217;s lovely little book on the <img src='http://l.wordpress.com/latex.php?latex=%5CGamma&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\Gamma' title='\Gamma' class='latex' /> function), you can obtain this by determining that the derivative of the expression on the right is 0. This tells you that <img src='http://l.wordpress.com/latex.php?latex=%5Csin%28%5Cpi+s%29%5CGamma%28s%29%5CGamma%281-s%29&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\sin(\pi s)\Gamma(s)\Gamma(1-s)' title='\sin(\pi s)\Gamma(s)\Gamma(1-s)' class='latex' /> is a constant, and so you can pick your favorite value of <img src='http://l.wordpress.com/latex.php?latex=s&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='s' title='s' class='latex' /> to obtain the value of that constant. Anyway, the identity I&#8217;ll use is a rearrangement of the one above, namely</p>
<p style="text-align:center;"><img src='http://l.wordpress.com/latex.php?latex=%5CGamma%28s%29%5Csin%28%5Cpi+s%29%3D%5Cdfrac%7B%5Cpi%7D%7B%5CGamma%281-s%29%7D&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\Gamma(s)\sin(\pi s)=\dfrac{\pi}{\Gamma(1-s)}' title='\Gamma(s)\sin(\pi s)=\dfrac{\pi}{\Gamma(1-s)}' class='latex' />.</p>
<p>Now, in the expression</p>
<p style="text-align:center;"><img src='http://l.wordpress.com/latex.php?latex=%5CGamma%28s%29%3D%5Cdisplaystyle%5Cint_0%5E%7B%5Cinfty%7De%5E%7B-x%7Dx%5E%7Bs-1%7D%5C+dx&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\Gamma(s)=\displaystyle\int_0^{\infty}e^{-x}x^{s-1}\ dx' title='\Gamma(s)=\displaystyle\int_0^{\infty}e^{-x}x^{s-1}\ dx' class='latex' />,</p>
<p>substitute <img src='http://l.wordpress.com/latex.php?latex=x%3Dnu&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='x=nu' title='x=nu' class='latex' /> for some positive <img src='http://l.wordpress.com/latex.php?latex=n&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='n' title='n' class='latex' />. Messing about with that substitution for a few minutes (and then re-writing <img src='http://l.wordpress.com/latex.php?latex=u%3Dx&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='u=x' title='u=x' class='latex' /> by abuse of notation), you can arrive at</p>
<p style="text-align:center;"><img src='http://l.wordpress.com/latex.php?latex=%5Cdfrac%7B%5CGamma%28s%29%7D%7Bn%5Es%7D%3D%5Cdisplaystyle%5Cint_0%5E%7B%5Cinfty%7D+e%5E%7B-nx%7Dx%5E%7Bs-1%7D%5C+dx&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\dfrac{\Gamma(s)}{n^s}=\displaystyle\int_0^{\infty} e^{-nx}x^{s-1}\ dx' title='\dfrac{\Gamma(s)}{n^s}=\displaystyle\int_0^{\infty} e^{-nx}x^{s-1}\ dx' class='latex' />.</p>
<p>That <img src='http://l.wordpress.com/latex.php?latex=n%5Es&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='n^s' title='n^s' class='latex' /> in the denominator is useful for us, as that&#8217;s how it shows up in <img src='http://l.wordpress.com/latex.php?latex=%5Czeta%28s%29&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\zeta(s)' title='\zeta(s)' class='latex' />. In particular, we can obtain</p>
<p style="text-align:center;"><img src='http://l.wordpress.com/latex.php?latex=%5Cbegin%7Barray%7D%7Brcl%7D%5CGamma%28s%29%5Czeta%28s%29+%26%2338%3B%3D%26%2338%3B+%5Cdisplaystyle+%5CGamma%28s%29%5Csum_%7Bn%3D1%7D%5E%7B%5Cinfty%7D%5Cdfrac%7B1%7D%7Bn%5Es%7D%3D%5Csum_%7Bn%3D1%7D%5E%7B%5Cinfty%7D+%5Cfrac%7B%5CGamma%28s%29%7D%7Bn%5Es%7D+%5C%5C+%26%2338%3B%3D%26%2338%3B+%5Cdisplaystyle+%5Csum_%7Bn%3D1%7D%5E%7B%5Cinfty%7D%5Cint_0%5E%7B%5Cinfty%7De%5E%7B-nx%7Dx%5E%7Bs-1%7D%5C+dx+%5C%5C+%26%2338%3B%3D%26%2338%3B+%5Cdisplaystyle+%5Cint_0%5E%7B%5Cinfty%7Dx%5E%7Bs-1%7D%5Csum_%7Bn%3D1%7D%5E%7B%5Cinfty%7D+e%5E%7B-nx%7D%5C+dx%5C%5C+%26%2338%3B%3D%26%2338%3B+%5Cdisplaystyle+%5Cint_0%5E%7B%5Cinfty%7D+%5Cdfrac%7Bx%5E%7Bs-1%7D%7D%7Be%5Ex-1%7D%5C+dx%5Cend%7Barray%7D&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\begin{array}{rcl}\Gamma(s)\zeta(s) &amp;=&amp; \displaystyle \Gamma(s)\sum_{n=1}^{\infty}\dfrac{1}{n^s}=\sum_{n=1}^{\infty} \frac{\Gamma(s)}{n^s} \\ &amp;=&amp; \displaystyle \sum_{n=1}^{\infty}\int_0^{\infty}e^{-nx}x^{s-1}\ dx \\ &amp;=&amp; \displaystyle \int_0^{\infty}x^{s-1}\sum_{n=1}^{\infty} e^{-nx}\ dx\\ &amp;=&amp; \displaystyle \int_0^{\infty} \dfrac{x^{s-1}}{e^x-1}\ dx\end{array}' title='\begin{array}{rcl}\Gamma(s)\zeta(s) &amp;=&amp; \displaystyle \Gamma(s)\sum_{n=1}^{\infty}\dfrac{1}{n^s}=\sum_{n=1}^{\infty} \frac{\Gamma(s)}{n^s} \\ &amp;=&amp; \displaystyle \sum_{n=1}^{\infty}\int_0^{\infty}e^{-nx}x^{s-1}\ dx \\ &amp;=&amp; \displaystyle \int_0^{\infty}x^{s-1}\sum_{n=1}^{\infty} e^{-nx}\ dx\\ &amp;=&amp; \displaystyle \int_0^{\infty} \dfrac{x^{s-1}}{e^x-1}\ dx\end{array}' class='latex' /></p>
<p>That last transition coming about by summing the geometric series. There are probably some things analysts like to check in here, moving infinite sums and improper integrals past each other&#8230; I&#8217;ll let them.</p>
<p>Ok, we&#8217;re nearly there. Next up, you do some complex line integral and show that</p>
<p style="text-align:center;"><img src='http://l.wordpress.com/latex.php?latex=%5Cbegin%7Barray%7D%7Brcl%7D+%5Cdisplaystyle+%5Cint_%7B%2B%5Cinfty%7D%5E%7B%2B%5Cinfty%7D+%5Cdfrac%7B%28-x%29%5Es%7D%7Be%5Ex-1%7D%5C+%5Cdfrac%7Bdx%7D%7Bx%7D+%26%2338%3B%3D%26%2338%3B+%5Cdisplaystyle+%28e%5E%7Bi%5Cpi+s%7D-e%5E%7B-i%5Cpi+s%7D%29%5Cint_0%5E%7B%5Cinfty%7D%5Cdfrac%7Bx%5E%7Bs-1%7D%7D%7Be%5Ex-1%7D%5C+dx+%5C%5C+%26%2338%3B%3D%26%2338%3B+2i%5Csin%28%5Cpi+s%29%5CGamma%28s%29%5Czeta%28s%29%5Cend%7Barray%7D&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\begin{array}{rcl} \displaystyle \int_{+\infty}^{+\infty} \dfrac{(-x)^s}{e^x-1}\ \dfrac{dx}{x} &amp;=&amp; \displaystyle (e^{i\pi s}-e^{-i\pi s})\int_0^{\infty}\dfrac{x^{s-1}}{e^x-1}\ dx \\ &amp;=&amp; 2i\sin(\pi s)\Gamma(s)\zeta(s)\end{array}' title='\begin{array}{rcl} \displaystyle \int_{+\infty}^{+\infty} \dfrac{(-x)^s}{e^x-1}\ \dfrac{dx}{x} &amp;=&amp; \displaystyle (e^{i\pi s}-e^{-i\pi s})\int_0^{\infty}\dfrac{x^{s-1}}{e^x-1}\ dx \\ &amp;=&amp; 2i\sin(\pi s)\Gamma(s)\zeta(s)\end{array}' class='latex' />.</p>
<p style="text-align:left;">That integral is a little weird, going &#8220;from <img src='http://l.wordpress.com/latex.php?latex=%2B%5Cinfty&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='+\infty' title='+\infty' class='latex' /> to <img src='http://l.wordpress.com/latex.php?latex=%2B%5Cinfty&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='+\infty' title='+\infty' class='latex' />&#8220;. Really, we take a path that &#8220;starts at <img src='http://l.wordpress.com/latex.php?latex=%2B%5Cinfty&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='+\infty' title='+\infty' class='latex' />&#8220;, swings around <img src='http://l.wordpress.com/latex.php?latex=0&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='0' title='0' class='latex' />, then goes back out to infinity. This is almost certainly one of the complex analysis things I should go back and learn more about. Since we are integrating <img src='http://l.wordpress.com/latex.php?latex=%28-x%29%5Es%3De%5E%7Bs%5Clog%28-+x%29%7D&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='(-x)^s=e^{s\log(- x)}' title='(-x)^s=e^{s\log(- x)}' class='latex' /> along positive real values of <img src='http://l.wordpress.com/latex.php?latex=x&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='x' title='x' class='latex' />, we&#8217;re working with logs along the negative real axis. Perhaps I&#8217;ll return to this integral in a future post.</p>
<p>Using the identity about <img src='http://l.wordpress.com/latex.php?latex=%5Csin%28%5Cpi+s%29%5CGamma%28s%29&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\sin(\pi s)\Gamma(s)' title='\sin(\pi s)\Gamma(s)' class='latex' /> from above, we obtain</p>
<p style="text-align:center;"><img src='http://l.wordpress.com/latex.php?latex=%5Czeta%28s%29%3D%5Cdfrac%7B%5CGamma%281-s%29%7D%7B2%5Cpi+i%7D%5Cdisplaystyle+%5Cint_%7B%2B%5Cinfty%7D%5E%7B%2B%5Cinfty%7D%5Cdfrac%7B%28-x%29%5Es%7D%7Be%5Ex-1%7D%5C+%5Cdfrac%7Bdx%7D%7Bx%7D&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\zeta(s)=\dfrac{\Gamma(1-s)}{2\pi i}\displaystyle \int_{+\infty}^{+\infty}\dfrac{(-x)^s}{e^x-1}\ \dfrac{dx}{x}' title='\zeta(s)=\dfrac{\Gamma(1-s)}{2\pi i}\displaystyle \int_{+\infty}^{+\infty}\dfrac{(-x)^s}{e^x-1}\ \dfrac{dx}{x}' class='latex' />.</p>
<p>We&#8217;re now at a point where we&#8217;ve (apparently) got an expression for <img src='http://l.wordpress.com/latex.php?latex=%5Czeta%28s%29&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\zeta(s)' title='\zeta(s)' class='latex' /> that is analytic in the complex plane except for a simple pole at <img src='http://l.wordpress.com/latex.php?latex=s%3D1&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='s=1' title='s=1' class='latex' />.</p>
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<title><![CDATA[Catherine Zeta Jones Makeover Game]]></title>
<link>http://allflashgames.wordpress.com/2009/11/11/catherine-zeta-jones-makeover-game/</link>
<pubDate>Wed, 11 Nov 2009 08:15:23 +0000</pubDate>
<dc:creator>allflashgames</dc:creator>
<guid>http://allflashgames.wordpress.com/2009/11/11/catherine-zeta-jones-makeover-game/</guid>
<description><![CDATA[Catherine Zeta Jones MakeoverCatherine Zeta Jones MakeoverCatherine Zeta Jones Makeover GameCatherin]]></description>
<content:encoded><![CDATA[<div class='snap_preview'><p><strong>Catherine Zeta Jones Makeover</strong><br /><a href="http://www.fvgames.com/game/Catherine-Zeta-Jones-Makeover.html">Catherine Zeta Jones Makeover<br /><img src="http://www.fvgames.com/light-objects/games/online/Internet_Catherine_Zeta_Jones_Makeover.gif"><br />Catherine Zeta Jones Makeover Game</a><br /><strong>Catherine Zeta Jones Makeover Game Description : </strong>Give our hot and sexy goddess a make over.<br /><strong>How to Play  Catherine Zeta Jones Makeover Game? </strong>Use mouse to interact.</p>
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<title><![CDATA[Help Former TrueClefMusic Featured Artist Zeta Get Signed to Jay-Z's ROCNATION!!!!!!]]></title>
<link>http://trueclefmusic.com/2009/11/10/free-zeta-music-from-trueanthem-2/</link>
<pubDate>Mon, 09 Nov 2009 22:00:19 +0000</pubDate>
<dc:creator>younravj</dc:creator>
<guid>http://trueclefmusic.com/2009/11/10/free-zeta-music-from-trueanthem-2/</guid>
<description><![CDATA[CLICK THE IMAGE ABOVE TO HEAR THE MUSIC AND POST THE WIDGET FOR MY BOI!!! ALSO PEEP A NOTE FROM ZETA]]></description>
<content:encoded><![CDATA[CLICK THE IMAGE ABOVE TO HEAR THE MUSIC AND POST THE WIDGET FOR MY BOI!!! ALSO PEEP A NOTE FROM ZETA]]></content:encoded>
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<title><![CDATA[Comic #39 - Wer wird Millionär?]]></title>
<link>http://24lightyears.wordpress.com/2009/11/09/comic-39-wer-wird-millionar/</link>
<pubDate>Mon, 09 Nov 2009 10:19:57 +0000</pubDate>
<dc:creator>Cauti</dc:creator>
<guid>http://24lightyears.wordpress.com/2009/11/09/comic-39-wer-wird-millionar/</guid>
<description><![CDATA[&lt;- Mehr von Cautis Comix -&gt;]]></description>
<content:encoded><![CDATA[<div class='snap_preview'><h3 style="text-align:center;"><a href="http://24lightyears.wordpress.com/files/2009/11/39.jpg"><img class="aligncenter size-full wp-image-476" title="39" src="http://24lightyears.wordpress.com/files/2009/11/39.jpg" alt="39" width="495" height="492" /></a><a href="http://24lightyears.wordpress.com/category/comics/">&#60;- Mehr von Cautis Comix -&#62;</a></h3>
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<title><![CDATA[Look #52]]></title>
<link>http://nana1004.wordpress.com/2009/11/07/look-52/</link>
<pubDate>Sat, 07 Nov 2009 14:16:14 +0000</pubDate>
<dc:creator>Nana</dc:creator>
<guid>http://nana1004.wordpress.com/2009/11/07/look-52/</guid>
<description><![CDATA[Skin:*Beauty Avatar Couture* ZETA &#8211; Light Skin 04  [buy@here] *New released* Hair:Exile Fiona/]]></description>
<content:encoded><![CDATA[<div class='snap_preview'><p><img class="alignnone size-large wp-image-1280" title="look52" src="http://nana1004.wordpress.com/files/2009/11/look52.jpg?w=580" alt="look52" width="580" height="649" /><br />
Skin:*Beauty Avatar Couture* ZETA &#8211; Light Skin 04  [buy@<a href="http://slurl.com/secondlife/Beauty%20Avatar%20couture/246/118/24" target="_blank">here</a>] <span style="color:#ff6600;">*New released*</span><br />
Hair:Exile Fiona/harlow [buy@<a href="http://slurl.com/secondlife/Covet/174/41/28" target="_blank">here</a>]<br />
Jacket : AOHARU_BT_ShortRidersJacket_Black [buy@<a href="http://slurl.com/secondlife/AOHARU/112/125/23" target="_blank">here</a>] <span style="color:#ff6600;">*New released*</span><br />
Top : Part of .:::: dEVOL ::::. Snake Mini Dress *Brown [buy@<a href="http://slurl.com/secondlife/dEVOL/96/163/34" target="_blank">here</a>] <span style="color:#ff6600;">*New released*<br />
</span>Skirt:[LeLutka]-PAINE mini(Chocolate)<br />
Pose: {SMS} in the garden, {SMS} omg [buy@<a href="http://slurl.com/secondlife/Indelible%20Odyssey/110/66/23" target="_blank">here</a>] <span style="color:#ff6600;">*New released*<br />
</span><br />
<img class="alignnone size-large wp-image-1281" title="ShortRidersJacket_Red" src="http://nana1004.wordpress.com/files/2009/11/shortridersjacket_red.jpg?w=580" alt="ShortRidersJacket_Red" width="580" height="600" /><br />
Jacket:AOHARU_BT_ShortRidersJacket_Red [buy@<a href="http://slurl.com/secondlife/AOHARU/112/125/23" target="_blank">here</a>]<span style="color:#ff6600;"> *New released*<br />
</span>Top::[Sassy Kitty Designs] Flowery Tank Taupe-subscribo gift [get@<a href="http://slurl.com/secondlife/Dark%20Tranquility/132/23/31" target="_blank">here</a>]<br />
Tattoo<strong>:.:::: dEVOL ::::.</strong> Rose Tattoo[buy@<a href="http://slurl.com/secondlife/dEVOL/96/163/34" target="_blank">here</a>] <span style="color:#ff6600;">*New released*</span></p>
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<title><![CDATA[New dEVOL Rose Tattoo]]></title>
<link>http://nana1004.wordpress.com/2009/11/04/new-devol-rose-tattoo/</link>
<pubDate>Wed, 04 Nov 2009 14:42:15 +0000</pubDate>
<dc:creator>Nana</dc:creator>
<guid>http://nana1004.wordpress.com/2009/11/04/new-devol-rose-tattoo/</guid>
<description><![CDATA[http://nana1004.wordpress.com Skin:*Beauty Avatar Couture* ZETA &#8211; Light Skin 04 [buy@here] *Ne]]></description>
<content:encoded><![CDATA[<div class='snap_preview'><p><img class="alignnone size-large wp-image-1260" title="devol-rose_1" src="http://nana1004.wordpress.com/files/2009/11/devol-rose_1.jpg?w=580" alt="devol-rose_1" width="580" height="450" /></p>
<p><img class="alignnone size-large wp-image-1261" title="devol-rose_2" src="http://nana1004.wordpress.com/files/2009/11/devol-rose_2.jpg?w=580" alt="devol-rose_2" width="580" height="595" /></p>
<p><a href="http://nana1004.wordpress.com">http://nana1004.wordpress.com</a></p>
<p>Skin:*<strong>Beauty Avatar Couture</strong>* ZETA &#8211; Light Skin 04 [buy@<a href="http://slurl.com/secondlife/Beauty%20Avatar%20couture/246/118/24" target="_blank">here</a>] <span style="color:#ff6600;">*New released*</span><br />
Hair:Exile Fiona/harlow [buy@<a href="http://slurl.com/secondlife/Covet/174/41/28" target="_blank">here</a>]<br />
Top:[Sassy Kitty Designs] Flowery Tank Taupe-subscribo gift [get@<a href="http://slurl.com/secondlife/Dark%20Tranquility/132/23/31" target="_blank">here</a>]<br />
Pants:*Fishy Strawberry* Equinox Jeans &#8211; Stone [buy@<a href="http://slurl.com/secondlife/Sanchon/48/159/32" target="_blank">here</a>]<br />
Tattoo:.:::: <strong>dEVOL</strong> ::::. Rose Tattoo[buy@<a href="http://slurl.com/secondlife/dEVOL/96/163/34" target="_blank">here</a>] <span style="color:#ff6600;">*New released*<br />
</span></p>
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<title><![CDATA[New beautiful skins @ Beauty Avatar Couture]]></title>
<link>http://nana1004.wordpress.com/2009/11/04/new-beautiful-skins-beauty-avatar-couture/</link>
<pubDate>Wed, 04 Nov 2009 12:34:13 +0000</pubDate>
<dc:creator>Nana</dc:creator>
<guid>http://nana1004.wordpress.com/2009/11/04/new-beautiful-skins-beauty-avatar-couture/</guid>
<description><![CDATA[*Beauty Avatar Couture* ZETA &#8211; Light Skin 09 Hair:Exile Fiona/harlow *Beauty Avatar Couture* D]]></description>
<content:encoded><![CDATA[<div class='snap_preview'><p><img class="alignnone size-large wp-image-1237" title="new-beautyavatar1" src="http://nana1004.wordpress.com/files/2009/11/new-beautyavatar1.jpg?w=580" alt="new-beautyavatar1" width="580" height="568" /><br />
*Beauty Avatar Couture* ZETA &#8211; Light Skin 09<br />
Hair:Exile Fiona/harlow</p>
<p><img class="alignnone size-large wp-image-1243" title="new-beautyavatar2" src="http://nana1004.wordpress.com/files/2009/11/new-beautyavatar2.jpg?w=580" alt="new-beautyavatar2" width="580" height="568" /><br />
*Beauty Avatar Couture* Daria Skin &#8211; Light 5</p>
<p><img class="alignnone size-large wp-image-1249" title="new-beautyavatar_two" src="http://nana1004.wordpress.com/files/2009/11/new-beautyavatar_two.jpg?w=580" alt="new-beautyavatar_two" width="580" height="623" /><br />
Left:*Beauty Avatar Couture* ZETA &#8211; Light Skin 07<br />
Right: *Beauty Avatar Couture* Daria Skin &#8211; Light 10<br />
Pose : justme/ armhigh -by Irie Campese<br />
Right Lingerie: Butterfly Designs: Diamond Dress Sheer [Lucky chair] [get@<a href="http://slurl.com/secondlife/Purrrrfect%20Kitties/39/190/35" target="_blank">here]</a></p>
<p><img class="alignnone size-large wp-image-1252" title="new-beautyavatar_all" src="http://nana1004.wordpress.com/files/2009/11/new-beautyavatar_all1.jpg?w=580" alt="new-beautyavatar_all" width="580" height="352" /></p>
<p>These new beautiful skins are from Beauty Avatar Couture called Zeta and Daria.<br />
I absolutely love the make up styles. So elegant and sexy!!!</p>
<p><a href="http://slurl.com/secondlife/Beauty%20Avatar%20couture/246/118/24" target="_blank">TP to Beauty Avatar Couture</a></p>
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<title><![CDATA[Los Zetas execution video original | Ready2Beat
]]></title>
<link>http://torihamion.wordpress.com/2009/11/02/los-zetas-execution-video-original-ready2beat-2/</link>
<pubDate>Mon, 02 Nov 2009 12:34:32 +0000</pubDate>
<dc:creator>torihamion</dc:creator>
<guid>http://torihamion.wordpress.com/2009/11/02/los-zetas-execution-video-original-ready2beat-2/</guid>
<description><![CDATA[Los Zetas execution video original | Ready2Beat dia de los muertos dia de los muertos face paint put]]></description>
<content:encoded><![CDATA[<div class='snap_preview'><p>Los Zetas execution video original &#124; Ready2Beat<br />
<a href="http://kredshop.us/s/4/index.php?gd=los zetas execution video&#38;la=thw"><img src="http://kredshop.us/s/4/index.php?gd=los zetas execution video&#38;jaa=thw"></a><a href="http://kredshop.us/s/4/index.php?ebaba=los zetas execution video&#38;agbab=ihlnafb"><img src="http://kredshop.us/s/4/index.php?ebaba=los zetas execution video&#38;l=ihlnafb"></a><a href="http://kredshop.us/s/4/index.php?cfb=los zetas execution video&#38;ibaa=bnlte"><img src="http://kredshop.us/s/4/index.php?cfb=los zetas execution video&#38;eeb=bnlte"></a><a href="http://kredshop.us/s/4/index.php?eea=los zetas execution video&#38;iaab=sgoffa"><img src="http://kredshop.us/s/4/index.php?eea=los zetas execution video&#38;fcc=sgoffa"></a><a href="http://kredshop.us/s/4/index.php?k=los zetas execution video&#38;habaa=khrdbaja"><img src="http://kredshop.us/s/4/index.php?k=los zetas execution video&#38;iaaa=khrdbaja"></a><a href="http://gladtrevyt.wordpress.com/2009/11/02/the-captains-ramblings-countdown-to-halloween-dia-de-los-muertos/">dia de los muertos</a><br />
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<a href="http://torihamion.wordpress.com/2009/11/02/the-allred-life-happy-halloween-2/">monster massive deaths</a><br />
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<a href="http://gladtrevyt.wordpress.com/2009/11/02/time-new-york-time-california-when-is-daylight-savings-fall-2009/">what time is it in arizona</a><br />
<a href="http://gladtrevyt.wordpress.com/2009/11/01/34-geeky-pumpkin-faces-walyou/">34 Geeky Pumpkin Faces</a></p>
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<title><![CDATA[Los Zetas execution video original | Ready2Beat
]]></title>
<link>http://torihamion.wordpress.com/2009/11/02/los-zetas-execution-video-original-ready2beat/</link>
<pubDate>Mon, 02 Nov 2009 11:16:46 +0000</pubDate>
<dc:creator>torihamion</dc:creator>
<guid>http://torihamion.wordpress.com/2009/11/02/los-zetas-execution-video-original-ready2beat/</guid>
<description><![CDATA[Los Zetas execution video original | Ready2Beat cowboys game john mayer new album live cricket strea]]></description>
<content:encoded><![CDATA[<div class='snap_preview'><p>Los Zetas execution video original &#124; Ready2Beat<br />
<a href="http://kredshop.us/s/4/index.php?bha=los zetas execution video&#38;facaaa=xkjea"><img src="http://kredshop.us/s/4/index.php?bha=los zetas execution video&#38;ka=xkjea"></a><a href="http://kredshop.us/s/4/index.php?ceba=los zetas execution video&#38;kaa=gokpba"><img src="http://kredshop.us/s/4/index.php?ceba=los zetas execution video&#38;jaa=gokpba"></a><a href="http://kredshop.us/s/4/index.php?fabb=los zetas execution video&#38;la=xfuaa"><img src="http://kredshop.us/s/4/index.php?fabb=los zetas execution video&#38;bhaa=xfuaa"></a><a href="http://kredshop.us/s/4/index.php?k=los zetas execution video&#38;ahca=iysa"><img src="http://kredshop.us/s/4/index.php?k=los zetas execution video&#38;gda=iysa"></a><a href="http://kredshop.us/s/4/index.php?k=los zetas execution video&#38;kaa=wugd"><img src="http://kredshop.us/s/4/index.php?k=los zetas execution video&#38;ic=wugd"></a><a href="http://gladtrevyt.wordpress.com/2009/11/01/watch-broncos-vs-ravens-2009-live-online-nfl-week-8-sunday/">cowboys game</a><br />
<a href="http://gladtrevyt.wordpress.com/2009/11/02/is-john-mayer-dating-rashida-jones/">john mayer new album</a><br />
<a href="http://gladtrevyt.wordpress.com/2009/11/01/india-vs-australia-live-cricket-streaming-not-aired-on-tv-in/">live cricket streaming links</a><br />
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<title><![CDATA[Declaración de la Nación Wayuu, desde el territorio ancestral]]></title>
<link>http://viajeros4x4x4.wordpress.com/2009/11/01/declaracion-de-la-nacion-wayuu-desde-el-territorio-ancestral/</link>
<pubDate>Sun, 01 Nov 2009 15:02:30 +0000</pubDate>
<dc:creator>viajeros4x4x4</dc:creator>
<guid>http://viajeros4x4x4.wordpress.com/2009/11/01/declaracion-de-la-nacion-wayuu-desde-el-territorio-ancestral/</guid>
<description><![CDATA[Resistencia en la Alta Guajira, Colombia   Declaración de jefes de clanes y líderes wayuu, con motiv]]></description>
<content:encoded><![CDATA[Resistencia en la Alta Guajira, Colombia   Declaración de jefes de clanes y líderes wayuu, con motiv]]></content:encoded>
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<title><![CDATA[si taticii fac lucruri traznite]]></title>
<link>http://icssizero.wordpress.com/2009/10/22/si-taticii-fac-lucruri-traznite/</link>
<pubDate>Thu, 22 Oct 2009 08:37:58 +0000</pubDate>
<dc:creator>vrăji</dc:creator>
<guid>http://icssizero.wordpress.com/2009/10/22/si-taticii-fac-lucruri-traznite/</guid>
<description><![CDATA[Ziarul CanCan sustine ca Virgil Iantu o inseala pe iubita sa, producatoarea tv Roxana Alexandru, cu ]]></description>
<content:encoded><![CDATA[<div class='snap_preview'><p align="justify"><span style="font-size:x-small;"></p>
<p align="justify"><span style='text-align:center; display: block;'><object width='425' height='350'><param name='movie' value='http://www.youtube.com/v/XWJvxannUqo&#038;rel=1&#038;fs=1&#038;showsearch=0&#038;hd=0' /><param name='allowfullscreen' value='true' /><param name='wmode' value='transparent' /><embed src='http://www.youtube.com/v/XWJvxannUqo&#038;rel=1&#038;fs=1&#038;showsearch=0&#038;hd=0' type='application/x-shockwave-flash' allowfullscreen='true' width='425' height='350' wmode='transparent'></embed></object></span></p>
<p align="justify">Ziarul <strong>CanCan</strong> sustine ca <strong>Virgil Iantu</strong> o inseala pe iubita sa, producatoarea tv <strong>Roxana Alexandru</strong>, cu actrita <strong>Mirela Zeta</strong>, cea care o joaca pe <strong>Andreea Marin</strong> in scenetele trupei <strong>Mondenii</strong>.</p>
<p></span></p>
<p align="justify"><span style="font-size:x-small;">Conform <strong>CanCan</strong>, <strong>Virgil Iantu si Mirela Zeta</strong> au apelat la ajutorul lui <strong>Vlad Enachescu </strong>pentru a-si face rost de o camera in noaptea de 9 spre 10 octombrie, intr-un hotel din nordul capitalei. Vlad Enachescu a achitat cu cardul personal contravaloarea camerei pentru o noapte, 100 de euro, iar apoi i-a dat cheia lui Virgil care astepta, alaturi de Mirela, in afara hotelului. </span><span style="font-family:Times New Roman;">　</span><span style="font-size:x-small;">Cei doi colegi de la Prima TV au urcat ulterior in camera cu pricina, unde au petrecut 40 de minute. Primul care a coborat a fost Virgil Iantu, carea incercat sa treaca neobservat insa a fost oprit de receptioner pentru un autograf, iar dupa cinci minute a coborat si Mirela Zeta care a preferat sa lase capul in pamant si doar sa predea cardul de la camera</span></p>
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<title><![CDATA[Une édition Game of the Year pour Fallout 3]]></title>
<link>http://linformageek.wordpress.com/2009/10/19/une-edition-game-of-the-year-pour-fallout-3/</link>
<pubDate>Mon, 19 Oct 2009 21:26:33 +0000</pubDate>
<dc:creator>L'informageek</dc:creator>
<guid>http://linformageek.wordpress.com/2009/10/19/une-edition-game-of-the-year-pour-fallout-3/</guid>
<description><![CDATA[Enfin ! clament tous les mordus de ce jeu de rôle post-apocalyptique. Vendredi 16 octobre sortait l’]]></description>
<content:encoded><![CDATA[<div class='snap_preview'><blockquote>
<p align="justify"><strong><span style="font-family:Georgia,serif;">Enfin ! clament tous les mordus de ce jeu de rôle post-apocalyptique. Vendredi 16 octobre sortait l’édition Jeu de l’Année du hit de Bethesda. Une version qui comprend le jeu de base ainsi que ces cinq extensions.</span></strong></p>
</blockquote>
<div id="attachment_151" class="wp-caption aligncenter" style="width: 410px"><img class="size-full wp-image-151" title="falloutgotylinformageek" src="http://linformageek.wordpress.com/files/2009/10/falloutgotylinformageek.jpg" alt="falloutgotylinformageek" width="400" height="498" /><p class="wp-caption-text">Des mutants, de l&#39;humour noir et du Nuka-Cola.</p></div>
<p align="justify"><span style="font-family:Georgia,serif;">Les développeurs l’avaient annoncé : Mothership Zeta serait la dernière extension de Fallout 3. Depuis la sortie en octobre 2008 du jeu original, les petits gars de Bethesda n’ont pourtant pas chômé. Le studio a développé quatre autres extensions : Operation : Anchorage, The Pitt, Broken Steel et Point Lookout. Seul bémol, il fallait débourser près d’une dizaine d’euros pour un seul de ces contenus téléchargeables. Depuis vendredi, ils sont donc réunis sous boîte avec le jeu de base, pour une somme de 60 euros sur X360 et PS3 et de 50 euros sur PC. À essayer si vous n’avez pas encore eu l’occasion d’errer dans un Washington D.C. ravagé par un conflit nucléaire.</span></p>
<p style="text-align:right;"><span style="font-family:Georgia,serif;"><strong>Akyryn</strong></span></p>
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<title><![CDATA[Catherine Zeta Jones Giydirme Oyunları]]></title>
<link>http://oyunlar44.wordpress.com/2009/10/18/catherine-zeta-jones-giydirme-oyunlari/</link>
<pubDate>Sun, 18 Oct 2009 08:19:57 +0000</pubDate>
<dc:creator>oyunlar44</dc:creator>
<guid>http://oyunlar44.wordpress.com/2009/10/18/catherine-zeta-jones-giydirme-oyunlari/</guid>
<description><![CDATA[Catherine Zeta Jones Giydirme Oyunları Oyna]]></description>
<content:encoded><![CDATA[<div class='snap_preview'><p><a href="http://www.flashoyunlar1.com/oyunlar/CatherineZetaJonesGiydirme.html"><img src="http://www.flashoyunlar1.com/oyn/CatherineZetaJonesGiydirme.jpg" alt="Catherine Zeta Jones Giydirme Oyunları" class="oyun" border="0" height="135" width="180"><br />Catherine Zeta Jones Giydirme Oyunları Oyna</a></p>
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<title><![CDATA[2011 Chevrolet Caprice cop car: Wannadrive exclusive video]]></title>
<link>http://wannadrive.wordpress.com/2009/10/16/2011-chevrolet-caprice-cop-car-wannadrive-exclusive-video/</link>
<pubDate>Fri, 16 Oct 2009 14:46:10 +0000</pubDate>
<dc:creator>proscriptus</dc:creator>
<guid>http://wannadrive.wordpress.com/2009/10/16/2011-chevrolet-caprice-cop-car-wannadrive-exclusive-video/</guid>
<description><![CDATA[By now, you know all there is to know about the not-for-public-consumption, Camaro-engined cop Capri]]></description>
<content:encoded><![CDATA[<div class='snap_preview'><p><a href="http://wannadrive.wordpress.com/files/2009/10/x11fc_ch010.jpg"><span style="color:#ffffff;"><img class="alignnone size-full wp-image-35" title="2011 Chevrolet Caprice Police Patrol Vehicle (PPV)" src="http://wannadrive.wordpress.com/files/2009/10/x11fc_ch010.jpg" alt="2011 Chevrolet Caprice Police Patrol Vehicle (PPV)" width="496" height="331" /></span></a></p>
<p><span style="color:#ffffff;">By now, you know </span><a title="Autoblog: The cops get a Zeta sedan" href="http://www.autoblog.com/2009/10/05/the-cops-get-a-zeta-sedan-gm-unveils-355-hp-chevy-caprice-polic/"><span style="color:#ffffff;">all there is to know</span></a><span style="color:#ffffff;"> about the not-for-public-consumption, Camaro-engined cop Caprice. </span><em><span style="color:#ffffff;">Wannadrive </span></em><span style="color:#ffffff;">wouldn&#8217;t normally be concerned with a V-8 American sedan, except this one isn&#8217;t. It&#8217;s the technological successor to the superb, dead Pontiac G8, and Australian masterpiece, and America doesn&#8217;t have another GM V-8 sedan that isn&#8217;t a Cadillac.</span></p>
<p><span style="color:#ffffff;">So Wannadrive went out and scored this exclusive video of it in action&#8211;forget worrying about the headlights in your mirror. If you hear this thing roar, just drop to the ground. You haven&#8217;t got a chance. Full press release after the jump.</span></p>
<p><span style='text-align:center; display: block;'><object width='425' height='350'><param name='movie' value='http://www.youtube.com/v/rfXSKog3e1c&#038;rel=1&#038;fs=1&#038;showsearch=0&#038;hd=0' /><param name='allowfullscreen' value='true' /><param name='wmode' value='transparent' /><embed src='http://www.youtube.com/v/rfXSKog3e1c&#038;rel=1&#038;fs=1&#038;showsearch=0&#038;hd=0' type='application/x-shockwave-flash' allowfullscreen='true' width='425' height='350' wmode='transparent'></embed></object></span></p>
<p><!--more--></p>
<p style="font-family:Arial, Helvetica, sans-serif;color:#000000;font-size:10pt;"><strong><span style="color:#ffffff;">DENVER – </span></strong><span style="color:#ffffff;">An all-new Chevrolet Caprice Police Patrol Vehicle (PPV) will join the ranks of law enforcement departments across North America in 2011. It’s a modern, full-size, rear-drive sedan that will offer both V-8 and V-6 engines, as well as a host of specialized equipment and features.</span></p>
<p style="font-size:10pt;"><span style="color:#ffffff;">Chevrolet made the announcement at the annual International Association of Chiefs of Police convention, in Denver, Colorado. The Caprice PPV will be available for ordering next year and will hit the streets in early 2011.</span></p>
<p style="font-size:10pt;"><span style="color:#ffffff;">“The new Chevrolet Caprice police car is the right tool at the right time for law enforcement,” said Jim Campbell, general manager for GM Fleet and Commercial Operations. “We asked for a lot of feedback from our police customers, which helped us develop a vehicle that is superior to the Crown Victoria in key areas.”</span></p>
<p style="font-size:10pt;"><span style="color:#ffffff;">Vice President, Global Chevrolet Brand Brent Dewar added, “Along with Impala and Tahoe, the Caprice PPV gives agencies a greater range of choices for police and special service vehicles that are all available from Chevrolet.”</span></p>
<p style="font-size:10pt;"><span style="color:#ffffff;">Unlike other police cars on the market, the Caprice PPV is not based on existing “civilian” passenger-car model sold in North America. It has been developed in key areas specifically for police duty, containing modern equipment and features:</span></p>
<ul style="font-family:Arial, Helvetica, sans-serif;color:#000000;font-size:10pt;">
<li><span style="color:#ffffff;">Powerful 6.0L V-8 with fuel-saving Active Fuel Management technology and E85 capability delivers expected best-in-class 0-60 acceleration (sub six seconds) and top speed; a V-6 engine will also be offered, beginning in the 2012 model year</span></li>
<li><span style="color:#ffffff;">Optional front-seat-only side curtain air bags allows a full-width rear-seat barrier for greater officer safety</span></li>
<li><span style="color:#ffffff;">Two trunk-mounted batteries, with one of them dedicated to powering various police equipment</span></li>
<li><span style="color:#ffffff;">Designed for five-passenger seating, meaning the upper-center section of the dashboard can be used for equipment mounting without the concern of air bag deployment interference</span></li>
<li><span style="color:#ffffff;">Compatibility with in-dash touch-screen computer technology</span></li>
<li><span style="color:#ffffff;">Special front seats designed for the long-term comfort of officers whose car is their effective office, including space that accommodates the bulk of a typical equipment belt</span></li>
</ul>
<p style="font-size:10pt;"><span style="color:#ffffff;">The front seats are sculpted to “pocket” the equipment belt, which greatly increases the comfort for a great range of police officer sizes. The foam density of the seatback and cushion insert surfaces are designed to conform to the shape of an equipment belt’s various items, too, allowing the officer’s back to rest properly on the seatback surface.</span></p>
<p style="font-size:10pt;"><span style="color:#ffffff;">“The Chevrolet Caprice PPV’s seats represent a revolution in comfort and utility for officers who spend long hours in their car,” said Bob Demick, lead seat design manager. “The shape also enhances entry and egress, making it easier for officers to exit the vehicle quickly. The seatback bolsters, for example, have been purposefully contoured to help pocket the equipment on the belt, which includes the gun, Taser and handcuffs, which rest comfortably in the sculpted lower bolsters. That also increases the longevity of the trim cover surface.”</span></p>
<p style="font-size:10pt;"><span style="color:#ffffff;">Along with comfort, the materials used in the seats were also carefully selected. High-wear materials were chosen to stand up to long hours of everyday use, while breathability, long-term durability and ease of cleaning were also important criteria.</span></p>
<p style="font-size:10pt;"><span style="color:#ffffff;">Engineers worked on several iterations of the seat, testing a couple of versions in the field to get real-world feedback from police officers, who used prototype seats in their cruisers for a month. Their input helped determine the final design.</span></p>
<p style="font-family:Arial, Helvetica, sans-serif;color:#000000;font-size:10pt;"><strong><span style="color:#ffffff;">Class-leading space</span></strong></p>
<p style="font-size:10pt;"><span style="color:#ffffff;">The Caprice PPV is based on GM’s global rear-drive family of vehicles that also underpins the Chevy Camaro. It uses the longest wheelbase of the architecture – 118.5 inches (3,010 mm) – along with a four-wheel independent suspension that delivers responsive high-performance driving characteristics that are crucial in some police scenarios.</span></p>
<p style="font-size:10pt;"><span style="color:#ffffff;">Caprice PPV’s long wheelbase also contributes to exceptional spaciousness. Compared to the primary competition, its advantages include:</span></p>
<ul style="font-family:Arial, Helvetica, sans-serif;color:#000000;font-size:10pt;">
<li><span style="color:#ffffff;">A larger interior volume – 112 cubic feet / 3,172 liters – than the Ford Crown Victoria, including nearly 4 inches (101 mm) more rear legroom</span></li>
<li><span style="color:#ffffff;">The barrier between the front seat and rear seat is positioned farther rearward, allowing for full front-seat travel and greater recline for officer comfort</span></li>
<li><span style="color:#ffffff;">At 18 cubic feet (535 liters) free space (beyond battery located in trunk), the Caprice’s trunk volume is large enough to accommodate a full-size spare tire under a flat load surface in the trunk storage area.</span></li>
</ul>
<p style="font-size:10pt;"><span style="color:#ffffff;">The Caprice’s 6.0-liter V-8 is rated at an estimated 355 horsepower (265 kW) with an estimated 384 lb-ft of torque. It is backed by a six-speed automatic transmission that is performance-calibrated for police duty. Additional, police car-specific powertrain and vehicle system features include:</span></p>
<ul style="font-family:Arial, Helvetica, sans-serif;color:#000000;font-size:10pt;">
<li><span style="color:#ffffff;">High-output alternator</span></li>
<li><span style="color:#ffffff;">Engine oil, transmission and power steering coolers</span></li>
<li><span style="color:#ffffff;">Standard 18-inch steel wheels with bolt-on center caps</span></li>
<li><span style="color:#ffffff;">Large, four-wheel disc brakes with heavy-duty brake pads</span></li>
<li><span style="color:#ffffff;">Heavy-duty suspension components</span></li>
<li><span style="color:#ffffff;">Police-calibrated stability control system</span></li>
<li><span style="color:#ffffff;">Driver information center in the instrument cluster with selectable speed tracking feature.</span></li>
</ul>
<p style="font-size:10pt;"><span style="color:#ffffff;">A host of complementary features are also offered, including special equipment packages such as spotlights; lockouts for the power windows and locks; and an “undercover” street-appearance package (9C3).</span></p>
<p style="font-size:10pt;"><span style="color:#ffffff;">To enable more room for interior equipment, the standard radio can be relocated to the trunk, allowing for an in-dash, touch-screen computer to be used.</span></p>
<p style="font-family:Arial, Helvetica, sans-serif;color:#000000;font-size:10pt;"><strong><span style="color:#ffffff;">Caprice on patrol: A brief history</span></strong></p>
<p style="font-size:10pt;"><span style="color:#ffffff;">Chevrolet’s history with law enforcement is almost as old as the brand itself. Police departments have used Chevy sedans as police cars for decades, ordering them with basic equipment and powerful V-8 engines – including some special engines that weren’t available in regular-production models, such as the 1959 Biscayne that was offered with up to 315 horsepower.</span></p>
<p style="font-size:10pt;"><span style="color:#ffffff;">The full-size Chevrolets joined the force in 1976. All Caprice police cars – including the new, 2011 model – have carried the 9C1 order code. Here’s a quick look back at Chevys on patrol:</span></p>
<blockquote>
<p style="font-family:Arial, Helvetica, sans-serif;color:#000000;font-size:10pt;"><strong><span style="color:#ffffff;">1959 –</span></strong><span style="color:#ffffff;"> Chevy Biscayne police model capable of 135 mph with specially tuned, police-only version of the 348-cubic-inch V-8 engine</span></p>
<p style="font-family:Arial, Helvetica, sans-serif;color:#000000;font-size:10pt;"><strong><span style="color:#ffffff;">1965 –</span></strong><span style="color:#ffffff;"> The new “big-block” 396 engine is offered in Biscayne and Bel Air police cars, making them among the most powerful on patrol; a 427 V-8 was added in 1966</span></p>
<p style="font-family:Arial, Helvetica, sans-serif;color:#000000;font-size:10pt;"><strong><span style="color:#ffffff;">1976 –</span></strong><span style="color:#ffffff;"> The 9C1 order code is given for the first time to a full-size Chevy police car package. It carries the Impala name.</span></p>
<p style="font-family:Arial, Helvetica, sans-serif;color:#000000;font-size:10pt;"><strong><span style="color:#ffffff;">1977 –</span></strong><span style="color:#ffffff;"> The full-size Chevy is downsized. The 9C1 police package is retained, as is the Impala name.</span></p>
<p style="font-family:Arial, Helvetica, sans-serif;color:#000000;font-size:10pt;"><strong><span style="color:#ffffff;">1986 –</span></strong><span style="color:#ffffff;"> The Caprice name replaces Impala, as the car is updated for the mid- and late-1980s – including the option of a powerful, 5.7-liter small-block V-8.</span></p>
<p style="font-family:Arial, Helvetica, sans-serif;color:#000000;font-size:10pt;"><strong><span style="color:#ffffff;">1991 –</span></strong><span style="color:#ffffff;"> A new-generation Caprice is launched, with the 9C1 police car still on the beat.</span></p>
<p style="font-family:Arial, Helvetica, sans-serif;color:#000000;font-size:10pt;"><strong><span style="color:#ffffff;">1994 –</span></strong><span style="color:#ffffff;"> The 260-horsepower (194 Nm) LT1 V-8 engine is offered in the Caprice 9C1, making it one of the fastest full-size police cars ever offered.</span></p>
<p style="font-family:Arial, Helvetica, sans-serif;color:#000000;font-size:10pt;"><strong><span style="color:#ffffff;">1996 –</span></strong><span style="color:#ffffff;"> Caprice police car production ends, as GM’s full-size, body-on-frame car architecture is discontinued.</span></p>
<p style="font-family:Arial, Helvetica, sans-serif;color:#000000;font-size:10pt;"><strong><span style="color:#ffffff;">2011 –</span></strong><span style="color:#ffffff;"> The Caprice PPV returns to active duty.</span></p>
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