Ryan's Working Page: Difference between revisions

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<h2>IMPORTED REVISION FROM WIKISPACES</h2>

This is an imported revision from Wikispaces. The revision metadata is included below for reference:<br>

: This revision was by author [[User:~~mbattaglia1~~|~~mbattaglia1~~]] and made on <tt>2015-08-22 04:32:08 UTC</tt>.<br>

: This revision was by author [[User:Sarzadoce|Sarzadoce]] and made on <tt>2015-08-22 12:57:17 UTC</tt>.<br>

: The original revision id was <tt>~~557156615~~</tt>.<br>

: The original revision id was <tt>557169127</tt>.<br>

: The revision comment was: <tt></tt><br>

The revision contents are below, presented both in the original Wikispaces Wikitext format, and in HTML exactly as Wikispaces rendered it.<br>

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\displaystyle

\left| \text{f} (s) \right|^2 = \zeta(2a) + \sum_{n > d} \left[\frac{\left({\tfrac{n}{d}}\right)^{-bi} + \left({\tfrac{d}{n}}\right)^{-bi}}{n^{2a}}\right]

[[math]]

\displaystyle

\left| \text{f} (s) \right|^2 = \zeta(2a) + \sum_{n > d} \left[ n^{ - \left(2a+bi \right) } d^{ - \left( -bi \right) } + n^{ - \left(2a-bi \right) } d^{ - \left( bi \right) } \right]

[[math]]

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\displaystyle

\left| \text{f} (s) \right|^2 = \zeta(2a) + 2 \text{Re} \left( \sum_{n > d} n^{ - \left(2a+bi \right) } d^{ - \left( -bi \right) } \right)

[[math]]

\displaystyle

\left| \text{f} (s) \right|^2 = \zeta(2a) + 2 \text{Re} \left( \sum \limits_{d = 1}^{\infty} \sum \limits_{n = d+1}^{\infty} n^{ - \left(2a+bi \right) } d^{ - \left( -bi \right) } \right)

[[math]]

\displaystyle

\left| \text{f} (s) \right|^2 = \zeta(2a) + 2 \text{Re} \left( \sum \limits_{d = 1}^{\infty} \left[ d^{ - \left( -bi \right) } \sum \limits_{n = d+1}^{\infty} n^{ - \left(2a+bi \right) } \right] \right)

[[math]]

\displaystyle

\left| \text{f} (s) \right|^2 = \zeta(2a) + 2 \text{Re} \left( \sum \limits_{d = 1}^{\infty} \left[ d^{ - \left( -bi \right) } \zeta(2a+ib,d+1) \right] \right)

[[math]]

\displaystyle

\left| \text{f} (s) \right|^2 = \zeta(2a) + \~~sum_~~{~~n >~~ d} ~~\left[ n~~^{ - \~~left(2a+bi~~ \~~right)~~ } d^{ - \left( -bi \right) } ~~+ n^{ -~~ \~~left~~(2a~~-bi \right) }~~ d~~^{ - \left( bi \right~~) } \right]

\left| \text{f} (s) \right|^2 = \zeta(2a) + 2 \sum \limits_{d = 1}^{\infty} \text{Re} \left[ d^{ - \left( -bi \right) } \zeta(2a+ib,d+1) \right]

[[math]]

~~Note that the RHS of~~ this ~~equation can be broken up into~~ a sum ~~of conjugates. However, we ideally want to break it up into a //product// of conjugates, in order~~ to ~~solve for f(s). This is where I get stuck~~.

At this point I get stuck. Having a generalized zeta function inside a sum isn't exactly the easiest function to manipulate.

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Alright!! I'm going to bed.</pre></div>

<h4>Original HTML content:</h4>

<div style="width:100%; max-height:400pt; overflow:auto; background-color:#f8f9fa; border: 1px solid #eaecf0; padding:0em"><pre style="margin:0px;border:none;background:none;word-wrap:break-word;width:200%;white-space: pre-wrap ! important" class="old-revision-html"><html><head><title>Ryan's Working Page</title></head><body><h1 id="toc0"><a name="Attempt to backwards-engineer a Weil-weighted analog for &quot;Zeta&quot;"></a>Attempt to backwards-engineer a Weil-weighted analog for &quot;Zeta&quot;</h1>

<div style="width:100%; max-height:400pt; overflow:auto; background-color:#f8f9fa; border: 1px solid #eaecf0; padding:0em"><pre style="margin:0px;border:none;background:none;word-wrap:break-word;width:200%;white-space: pre-wrap ! important" class="old-revision-html"><html><head><title>Ryan's Working Page</title></head><body><h1 id="toc0"><a name="Attempt to backwards-engineer a Weil-weighted analog for &quot;Zeta&quot;"></a>Attempt to backwards-engineer a Weil-weighted analog for &quot;Zeta&quot;</h1>

<br />

In Mike's Zeta Function Working Page, we see that zeta can be thought of as a superposition of weighted &quot;cosine accuracy&quot; functions for every unreduced rational:<br />

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<br />

<!-- ws:start:WikiTextMathRule:4:

[[math]]&lt;br/&gt;

\displaystyle&lt;br /&gt;

\left| \text{f} (s) \right|^2 = \zeta(2a) + \sum_{n &gt; d} \left[ n^{ - \left(2a+bi \right) } d^{ - \left( -bi \right) } + n^{ - \left(2a-bi \right) } d^{ - \left( bi \right) } \right]&lt;br/&gt;[[math]]

--><script type="math/tex">\displaystyle

\left| \text{f} (s) \right|^2 = \zeta(2a) + \sum_{n > d} \left[ n^{ - \left(2a+bi \right) } d^{ - \left( -bi \right) } + n^{ - \left(2a-bi \right) } d^{ - \left( bi \right) } \right]</script><br />

<br />

<!-- ws:start:WikiTextMathRule:5:

[[math]]&lt;br/&gt;

\displaystyle&lt;br /&gt;

\left| \text{f} (s) \right|^2 = \zeta(2a) + 2 \text{Re} \left( \sum_{n &gt; d} n^{ - \left(2a+bi \right) } d^{ - \left( -bi \right) } \right)&lt;br/&gt;[[math]]

--><script type="math/tex">\displaystyle

\left| \text{f} (s) \right|^2 = \zeta(2a) + 2 \text{Re} \left( \sum_{n > d} n^{ - \left(2a+bi \right) } d^{ - \left( -bi \right) } \right)</script><br />

\left| \text{f} (s) \right|^2 = \zeta(2a) + 2 \text{Re} \left( \sum_{n > d} n^{ - \left(2a+bi \right) } d^{ - \left( -bi \right) } \right)</script><br />

<br />

<!-- ws:start:WikiTextMathRule:6:

[[math]]&lt;br/&gt;

\displaystyle&lt;br /&gt;

\left| \text{f} (s) \right|^2 = \zeta(2a) + 2 \text{Re} \left( \sum \limits_{d = 1}^{\infty} \sum \limits_{n = d+1}^{\infty} n^{ - \left(2a+bi \right) } d^{ - \left( -bi \right) } \right)&lt;br/&gt;[[math]]

--><script type="math/tex">\displaystyle

\left| \text{f} (s) \right|^2 = \zeta(2a) + 2 \text{Re} \left( \sum \limits_{d = 1}^{\infty} \sum \limits_{n = d+1}^{\infty} n^{ - \left(2a+bi \right) } d^{ - \left( -bi \right) } \right)</script><br />

<br />

<!-- ws:start:WikiTextMathRule:7:

[[math]]&lt;br/&gt;

\displaystyle&lt;br /&gt;

\left| \text{f} (s) \right|^2 = \zeta(2a) + 2 \text{Re} \left( \sum \limits_{d = 1}^{\infty} \left[ d^{ - \left( -bi \right) } \sum \limits_{n = d+1}^{\infty} n^{ - \left(2a+bi \right) } \right] \right)&lt;br/&gt;[[math]]

--><script type="math/tex">\displaystyle

\left| \text{f} (s) \right|^2 = \zeta(2a) + 2 \text{Re} \left( \sum \limits_{d = 1}^{\infty} \left[ d^{ - \left( -bi \right) } \sum \limits_{n = d+1}^{\infty} n^{ - \left(2a+bi \right) } \right] \right)</script><br />

<br />

<!-- ws:start:WikiTextMathRule:8:

[[math]]&lt;br/&gt;

\displaystyle&lt;br /&gt;

\left| \text{f} (s) \right|^2 = \zeta(2a) + 2 \text{Re} \left( \sum \limits_{d = 1}^{\infty} \left[ d^{ - \left( -bi \right) } \zeta(2a+ib,d+1) \right] \right)&lt;br/&gt;[[math]]

--><script type="math/tex">\displaystyle

\left| \text{f} (s) \right|^2 = \zeta(2a) + 2 \text{Re} \left( \sum \limits_{d = 1}^{\infty} \left[ d^{ - \left( -bi \right) } \zeta(2a+ib,d+1) \right] \right)</script><br />

<br />

<!-- ws:start:WikiTextMathRule:5:

<!-- ws:start:WikiTextMathRule:9:

[[math]]&lt;br/&gt;

\displaystyle&lt;br /&gt;

\left| \text{f} (s) \right|^2 = \zeta(2a) + \~~sum_~~{~~n &gt;~~ d} ~~\left[ n~~^{ - \~~left(2a+bi~~ \~~right)~~ } d^{ - \left( -bi \right) } ~~+ n^{ -~~ \~~left~~(2a~~-bi \right) }~~ d~~^{ - \left( bi \right~~) } \right]&lt;br/&gt;[[math]]

\left| \text{f} (s) \right|^2 = \zeta(2a) + 2 \sum \limits_{d = 1}^{\infty} \text{Re} \left[ d^{ - \left( -bi \right) } \zeta(2a+ib,d+1) \right]&lt;br/&gt;[[math]]

--><script type="math/tex">\displaystyle

\left| \text{f} (s) \right|^2 = \zeta(2a) + \~~sum_~~{~~n >~~ d} ~~\left[ n~~^{ - \~~left(2a+bi~~ \~~right)~~ } d^{ - \left( -bi \right) } ~~+ n^{ -~~ \~~left~~(2a~~-bi \right) }~~ d~~^{ - \left( bi \right~~) } \right]</script><br />

\left| \text{f} (s) \right|^2 = \zeta(2a) + 2 \sum \limits_{d = 1}^{\infty} \text{Re} \left[ d^{ - \left( -bi \right) } \zeta(2a+ib,d+1) \right]</script><br />

<br />

At this point I get stuck. Having a generalized zeta function inside a sum isn't exactly the easiest function to manipulate.<br />

<br />

Note that the RHS of this equation can be broken up into a sum of conjugates. However, we ideally want to break it up into a <em>product</em> of conjugates, in order to solve for f(s). This is where I get stuck.<br />

<br />

<hr />

<br />

<h1 id="toc1"><a name="Mike's attempt"></a>Mike's attempt</h1>

<h1 id="toc1"><a name="Mike's attempt"></a>Mike's attempt</h1>

<br />

Let's start here<br />

<br />

<!-- ws:start:WikiTextMathRule:6:

<!-- ws:start:WikiTextMathRule:10:

[[math]]&lt;br/&gt;

\displaystyle&lt;br /&gt;

\left| \zeta(s) \right|^2 = \sum_{n,d} \left[ \frac{\cos\left(b \ln\left({\tfrac{n}{d}}\right)\right)}{(nd)^{a}} \right]&lt;br/&gt;[[math]]

--><script type="math/tex">\displaystyle

\left| \zeta(s) \right|^2 = \sum_{n,d} \left[ \frac{\cos\left(b \ln\left({\tfrac{n}{d}}\right)\right)}{(nd)^{a}} \right]</script><br />

\left| \zeta(s) \right|^2 = \sum_{n,d} \left[ \frac{\cos\left(b \ln\left({\tfrac{n}{d}}\right)\right)}{(nd)^{a}} \right]</script><br />

<br />

and change the denominator to max(n,d)^2a<br />

<br />

<!-- ws:start:WikiTextMathRule:7:

<!-- ws:start:WikiTextMathRule:11:

[[math]]&lt;br/&gt;

\displaystyle&lt;br /&gt;

\left| \zeta(s) \right|^2 = \sum_{n,d} \left[ \frac{\cos\left(b \ln\left({\tfrac{n}{d}}\right)\right)}{\max(n,d)^{2a}} \right]&lt;br/&gt;[[math]]

--><script type="math/tex">\displaystyle

\left| \zeta(s) \right|^2 = \sum_{n,d} \left[ \frac{\cos\left(b \ln\left({\tfrac{n}{d}}\right)\right)}{\max(n,d)^{2a}} \right]</script><br />

\left| \zeta(s) \right|^2 = \sum_{n,d} \left[ \frac{\cos\left(b \ln\left({\tfrac{n}{d}}\right)\right)}{\max(n,d)^{2a}} \right]</script><br />

<br />

OK, so now let's split it into three sub-series -- n=d, n&gt;d, n&lt;d<br />

<br />

<!-- ws:start:WikiTextMathRule:8:

<!-- ws:start:WikiTextMathRule:12:

[[math]]&lt;br/&gt;

\displaystyle&lt;br /&gt;

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\left| \zeta(s) \right|^2 = \sum_{n=d} \left[ \frac{\cos\left(b \ln\left({\tfrac{n}{d}}\right)\right)}{(nd)^{a}} \right] +

\sum_{n>d} \left[ \frac{\cos\left(b \ln\left({\tfrac{n}{d}}\right)\right)}{(nn)^{a}} \right] +

\sum_{n< d} \left[ \frac{\cos\left(b \ln\left({\tfrac{n}{d}}\right)\right)}{(dd)^{a}} \right]</script><br />

\sum_{n< d} \left[ \frac{\cos\left(b \ln\left({\tfrac{n}{d}}\right)\right)}{(dd)^{a}} \right]</script><br />

<br />

OK, so for every term in the middle sum, there's a corresponding term on the right where n and d are interchanged. So if n=p and d=q in the middle sum, and p&gt;q, then we get the following two terms (the left from the middle sum, the right from the right sum)<br />

<br />

<!-- ws:start:WikiTextMathRule:9:

<!-- ws:start:WikiTextMathRule:13:

[[math]]&lt;br/&gt;

\displaystyle&lt;br /&gt;

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--><script type="math/tex">\displaystyle

\frac{\cos\left(b \ln\left({\tfrac{p}{q}}\right)\right)}{(pp)^{a}} +

\frac{\cos\left(b \ln\left({\tfrac{q}{p}}\right)\right)}{(pp)^{a}}</script><br />

\frac{\cos\left(b \ln\left({\tfrac{q}{p}}\right)\right)}{(pp)^{a}}</script><br />

<br />

OK, let's throw some extra sin terms in there which cancel out to make this look more like Euler's identity<br />

<br />

<!-- ws:start:WikiTextMathRule:10:

<!-- ws:start:WikiTextMathRule:14:

[[math]]&lt;br/&gt;

\displaystyle&lt;br /&gt;

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--><script type="math/tex">\displaystyle

\frac{\cos\left(b \ln\left({\tfrac{p}{q}}\right)\right) + i\sin\left(b \ln\left({\tfrac{p}{q}}\right)\right)}{(pp)^{a}} +

\frac{\cos\left(b \ln\left({\tfrac{q}{p}}\right)\right) + i\sin\left(b \ln\left({\tfrac{q}{p}}\right)\right)}{(pp)^{a}}</script><br />

\frac{\cos\left(b \ln\left({\tfrac{q}{p}}\right)\right) + i\sin\left(b \ln\left({\tfrac{q}{p}}\right)\right)}{(pp)^{a}}</script><br />

<br />

OK, so we can throw these extra terms back in the original three-part series and get this<br />

<br />

<!-- ws:start:WikiTextMathRule:11:

<!-- ws:start:WikiTextMathRule:15:

[[math]]&lt;br/&gt;

\displaystyle&lt;br /&gt;

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\left| \zeta(s) \right|^2 = \sum_{n=d} \left[ \frac{\cos\left(b \ln\left({\tfrac{n}{d}}\right)\right)}{(nd)^{a}} \right] +

\sum_{n>d} \left[ \frac{\cos\left(b \ln\left({\tfrac{n}{d}}\right)\right) + i\sin\left(b \ln\left({\tfrac{n}{d}}\right)\right)}{(nn)^{a}} \right] +

\sum_{n< d} \left[ \frac{\cos\left(b \ln\left({\tfrac{n}{d}}\right)\right) + i\sin\left(b \ln\left({\tfrac{n}{d}}\right)\right)}{(dd)^{a}} \right]</script><br />

\sum_{n< d} \left[ \frac{\cos\left(b \ln\left({\tfrac{n}{d}}\right)\right) + i\sin\left(b \ln\left({\tfrac{n}{d}}\right)\right)}{(dd)^{a}} \right]</script><br />

<br />

Also, we can add a magical sin term that evaluates to 0 on the left side, yielding<br />

<br />

<!-- ws:start:WikiTextMathRule:12:

<!-- ws:start:WikiTextMathRule:16:

[[math]]&lt;br/&gt;

\displaystyle&lt;br /&gt;

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\left| \zeta(s) \right|^2 = \sum_{n=d} \left[ \frac{\cos\left(b \ln\left({\tfrac{n}{d}}\right)\right) + i\sin\left(b \ln\left({\tfrac{n}{d}}\right)\right)}{(nd)^{a}} \right] +

\sum_{n>d} \left[ \frac{\cos\left(b \ln\left({\tfrac{n}{d}}\right)\right) + i\sin\left(b \ln\left({\tfrac{n}{d}}\right)\right)}{(nn)^{a}} \right] +

\sum_{n< d} \left[ \frac{\cos\left(b \ln\left({\tfrac{n}{d}}\right)\right) + i\sin\left(b \ln\left({\tfrac{n}{d}}\right)\right)}{(dd)^{a}} \right]</script><br />

\sum_{n< d} \left[ \frac{\cos\left(b \ln\left({\tfrac{n}{d}}\right)\right) + i\sin\left(b \ln\left({\tfrac{n}{d}}\right)\right)}{(dd)^{a}} \right]</script><br />

<br />

Euler baby, Euler<br />

<br />

<!-- ws:start:WikiTextMathRule:13:

<!-- ws:start:WikiTextMathRule:17:

[[math]]&lt;br/&gt;

\displaystyle&lt;br /&gt;

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\left| \zeta(s) \right|^2 = \sum_{n=d} \left[ \frac{\left({\tfrac{n}{d}}\right)^{-bi}}{(nd)^{a}} \right] +

\sum_{n>d} \left[ \frac{\left({\tfrac{n}{d}}\right)^{-bi}}{(nn)^{a}} \right] +

\sum_{n< d} \left[ \frac{\left({\tfrac{n}{d}}\right)^{-bi}}{(dd)^{a}} \right]</script><br />

\sum_{n< d} \left[ \frac{\left({\tfrac{n}{d}}\right)^{-bi}}{(dd)^{a}} \right]</script><br />

<br />

OK, let's make it all one sum again<br />

<br />

<!-- ws:start:WikiTextMathRule:14:

<!-- ws:start:WikiTextMathRule:18:

[[math]]&lt;br/&gt;

\displaystyle&lt;br /&gt;

\left| \zeta(s) \right|^2 = \sum_{n,d} \left[ \frac{\left({\tfrac{n}{d}}\right)^{-bi}}{\max(n,d)^{2a}} \right] = \sum_{n,d} \left[\max(n,d)^{-2a} \cdot \left({\tfrac{n}{d}}\right)^{-bi} \right]&lt;br/&gt;[[math]]

--><script type="math/tex">\displaystyle

\left| \zeta(s) \right|^2 = \sum_{n,d} \left[ \frac{\left({\tfrac{n}{d}}\right)^{-bi}}{\max(n,d)^{2a}} \right] = \sum_{n,d} \left[\max(n,d)^{-2a} \cdot \left({\tfrac{n}{d}}\right)^{-bi} \right]</script><br />

\left| \zeta(s) \right|^2 = \sum_{n,d} \left[ \frac{\left({\tfrac{n}{d}}\right)^{-bi}}{\max(n,d)^{2a}} \right] = \sum_{n,d} \left[\max(n,d)^{-2a} \cdot \left({\tfrac{n}{d}}\right)^{-bi} \right]</script><br />

<br />

Now, let's note that max(n,d)^2a = (nd)^a * max(n/d,d/n)^a, and that max(n/d,d/n) = exp(|log(n/d)|). So we get<br />

<br />

<!-- ws:start:WikiTextMathRule:15:

<!-- ws:start:WikiTextMathRule:19:

[[math]]&lt;br/&gt;

\displaystyle&lt;br /&gt;

\left| \zeta(s) \right|^2 = \sum_{n,d} \left[e^{-a \left| \log(n/d) \right|} \cdot (nd)^{-a} \cdot \left({\tfrac{n}{d}}\right)^{-bi} \right]&lt;br/&gt;[[math]]

--><script type="math/tex">\displaystyle

\left| \zeta(s) \right|^2 = \sum_{n,d} \left[e^{-a \left| \log(n/d) \right|} \cdot (nd)^{-a} \cdot \left({\tfrac{n}{d}}\right)^{-bi} \right]</script><br />

\left| \zeta(s) \right|^2 = \sum_{n,d} \left[e^{-a \left| \log(n/d) \right|} \cdot (nd)^{-a} \cdot \left({\tfrac{n}{d}}\right)^{-bi} \right]</script><br />

<br />

The right terms become<br />

<br />

<!-- ws:start:WikiTextMathRule:16:

<!-- ws:start:WikiTextMathRule:20:

[[math]]&lt;br/&gt;

\displaystyle&lt;br /&gt;

\left| \zeta(s) \right|^2 = \sum_{n,d} \left[e^{-a \left| \log(n/d) \right|} \cdot n^{-(a+bi)} \cdot d^{-(a-bi)} \right]&lt;br/&gt;[[math]]

--><script type="math/tex">\displaystyle

\left| \zeta(s) \right|^2 = \sum_{n,d} \left[e^{-a \left| \log(n/d) \right|} \cdot n^{-(a+bi)} \cdot d^{-(a-bi)} \right]</script><br />

\left| \zeta(s) \right|^2 = \sum_{n,d} \left[e^{-a \left| \log(n/d) \right|} \cdot n^{-(a+bi)} \cdot d^{-(a-bi)} \right]</script><br />

<br />

Alright!! I'm going to bed.</body></html></pre></div>

@@ Line 1: / Line 1: @@
 <h2>IMPORTED REVISION FROM WIKISPACES</h2>
 This is an imported revision from Wikispaces. The revision metadata is included below for reference:<br>
-: This revision was by author [[User:mbattaglia1|mbattaglia1]] and made on <tt>2015-08-22 04:32:08 UTC</tt>.<br>
+: This revision was by author [[User:Sarzadoce|Sarzadoce]] and made on <tt>2015-08-22 12:57:17 UTC</tt>.<br>
-: The original revision id was <tt>557156615</tt>.<br>
+: The original revision id was <tt>557169127</tt>.<br>
 : The revision comment was: <tt></tt><br>
 The revision contents are below, presented both in the original Wikispaces Wikitext format, and in HTML exactly as Wikispaces rendered it.<br>
@@ Line 32: / Line 32: @@
 \displaystyle
 \left| \text{f} (s) \right|^2 = \zeta(2a) + \sum_{n &gt; d} \left[\frac{\left({\tfrac{n}{d}}\right)^{-bi} + \left({\tfrac{d}{n}}\right)^{-bi}}{n^{2a}}\right]
+[[math]]
+[[math]]
+\displaystyle
+\left| \text{f} (s) \right|^2 = \zeta(2a) + \sum_{n &gt; d} \left[ n^{ - \left(2a+bi \right) } d^{ - \left( -bi \right) } + n^{ - \left(2a-bi \right) } d^{ - \left( bi \right) } \right]
 [[math]]
@@ Line 37: / Line 42: @@
 \displaystyle
 \left| \text{f} (s) \right|^2 = \zeta(2a) + 2 \text{Re} \left( \sum_{n &gt; d} n^{ - \left(2a+bi \right) } d^{ - \left( -bi \right) } \right)
+[[math]]
+[[math]]
+\displaystyle
+\left| \text{f} (s) \right|^2 = \zeta(2a) + 2 \text{Re} \left( \sum \limits_{d = 1}^{\infty} \sum \limits_{n = d+1}^{\infty} n^{ - \left(2a+bi \right) } d^{ - \left( -bi \right) } \right)
+[[math]]
+[[math]]
+\displaystyle
+\left| \text{f} (s) \right|^2 = \zeta(2a) + 2 \text{Re} \left( \sum \limits_{d = 1}^{\infty} \left[ d^{ - \left( -bi \right) } \sum \limits_{n = d+1}^{\infty} n^{ - \left(2a+bi \right) } \right] \right)
+[[math]]
+[[math]]
+\displaystyle
+\left| \text{f} (s) \right|^2 = \zeta(2a) + 2 \text{Re} \left( \sum \limits_{d = 1}^{\infty} \left[ d^{ - \left( -bi \right) } \zeta(2a+ib,d+1) \right] \right)
 [[math]]
 [[math]]
 \displaystyle
-\left| \text{f} (s) \right|^2 = \zeta(2a) + \sum_{n &gt; d} \left[ n^{ - \left(2a+bi \right) } d^{ - \left( -bi \right) } + n^{ - \left(2a-bi \right) } d^{ - \left( bi \right) } \right]
+\left| \text{f} (s) \right|^2 = \zeta(2a) + 2 \sum \limits_{d = 1}^{\infty} \text{Re} \left[ d^{ - \left( -bi \right) } \zeta(2a+ib,d+1) \right]
 [[math]]
-Note that the RHS of this equation can be broken up into a sum of conjugates. However, we ideally want to break it up into a //product// of conjugates, in order to solve for f(s). This is where I get stuck.
+At this point I get stuck. Having a generalized zeta function inside a sum isn't exactly the easiest function to manipulate.
@@ Line 141: / Line 163: @@
 Alright!! I'm going to bed.</pre></div>
 <h4>Original HTML content:</h4>
-<div style="width:100%; max-height:400pt; overflow:auto; background-color:#f8f9fa; border: 1px solid #eaecf0; padding:0em"><pre style="margin:0px;border:none;background:none;word-wrap:break-word;width:200%;white-space: pre-wrap ! important" class="old-revision-html">&lt;html&gt;&lt;head&gt;&lt;title&gt;Ryan's Working Page&lt;/title&gt;&lt;/head&gt;&lt;body&gt;&lt;!-- ws:start:WikiTextHeadingRule:17:&amp;lt;h1&amp;gt; --&gt;&lt;h1 id="toc0"&gt;&lt;a name="Attempt to backwards-engineer a Weil-weighted analog for &amp;quot;Zeta&amp;quot;"&gt;&lt;/a&gt;&lt;!-- ws:end:WikiTextHeadingRule:17 --&gt;Attempt to backwards-engineer a Weil-weighted analog for &amp;quot;Zeta&amp;quot;&lt;/h1&gt;
+<div style="width:100%; max-height:400pt; overflow:auto; background-color:#f8f9fa; border: 1px solid #eaecf0; padding:0em"><pre style="margin:0px;border:none;background:none;word-wrap:break-word;width:200%;white-space: pre-wrap ! important" class="old-revision-html">&lt;html&gt;&lt;head&gt;&lt;title&gt;Ryan's Working Page&lt;/title&gt;&lt;/head&gt;&lt;body&gt;&lt;!-- ws:start:WikiTextHeadingRule:21:&amp;lt;h1&amp;gt; --&gt;&lt;h1 id="toc0"&gt;&lt;a name="Attempt to backwards-engineer a Weil-weighted analog for &amp;quot;Zeta&amp;quot;"&gt;&lt;/a&gt;&lt;!-- ws:end:WikiTextHeadingRule:21 --&gt;Attempt to backwards-engineer a Weil-weighted analog for &amp;quot;Zeta&amp;quot;&lt;/h1&gt;
   &lt;br /&gt;
 In Mike's Zeta Function Working Page, we see that zeta can be thought of as a superposition of weighted &amp;quot;cosine accuracy&amp;quot; functions for every unreduced rational:&lt;br /&gt;
@@ Line 178: / Line 200: @@
 &lt;br /&gt;
 &lt;!-- ws:start:WikiTextMathRule:4:
+[[math]]&amp;lt;br/&amp;gt;
+\displaystyle&amp;lt;br /&amp;gt;
+\left| \text{f} (s) \right|^2 = \zeta(2a) + \sum_{n &amp;gt; d} \left[ n^{ - \left(2a+bi \right) } d^{ - \left( -bi \right) } + n^{ - \left(2a-bi \right) } d^{ - \left( bi \right) } \right]&amp;lt;br/&amp;gt;[[math]]
+ --&gt;&lt;script type="math/tex"&gt;\displaystyle
+\left| \text{f} (s) \right|^2 = \zeta(2a) + \sum_{n &gt; d} \left[ n^{ - \left(2a+bi \right) } d^{ - \left( -bi \right) } + n^{ - \left(2a-bi \right) } d^{ - \left( bi \right) } \right]&lt;/script&gt;&lt;!-- ws:end:WikiTextMathRule:4 --&gt;&lt;br /&gt;
+&lt;br /&gt;
+&lt;!-- ws:start:WikiTextMathRule:5:
 [[math]]&amp;lt;br/&amp;gt;
 \displaystyle&amp;lt;br /&amp;gt;
 \left| \text{f} (s) \right|^2 = \zeta(2a) + 2 \text{Re} \left( \sum_{n &amp;gt; d} n^{ - \left(2a+bi \right) } d^{ - \left( -bi \right) } \right)&amp;lt;br/&amp;gt;[[math]]
   --&gt;&lt;script type="math/tex"&gt;\displaystyle
-\left| \text{f} (s) \right|^2 = \zeta(2a) + 2 \text{Re} \left( \sum_{n &gt; d} n^{ - \left(2a+bi \right) } d^{ - \left( -bi \right) } \right)&lt;/script&gt;&lt;!-- ws:end:WikiTextMathRule:4 --&gt;&lt;br /&gt;
+\left| \text{f} (s) \right|^2 = \zeta(2a) + 2 \text{Re} \left( \sum_{n &gt; d} n^{ - \left(2a+bi \right) } d^{ - \left( -bi \right) } \right)&lt;/script&gt;&lt;!-- ws:end:WikiTextMathRule:5 --&gt;&lt;br /&gt;
+&lt;br /&gt;
+&lt;br /&gt;
+&lt;!-- ws:start:WikiTextMathRule:6:
+[[math]]&amp;lt;br/&amp;gt;
+\displaystyle&amp;lt;br /&amp;gt;
+\left| \text{f} (s) \right|^2 = \zeta(2a) + 2 \text{Re} \left( \sum \limits_{d = 1}^{\infty} \sum \limits_{n = d+1}^{\infty} n^{ - \left(2a+bi \right) } d^{ - \left( -bi \right) } \right)&amp;lt;br/&amp;gt;[[math]]
+ --&gt;&lt;script type="math/tex"&gt;\displaystyle
+\left| \text{f} (s) \right|^2 = \zeta(2a) + 2 \text{Re} \left( \sum \limits_{d = 1}^{\infty} \sum \limits_{n = d+1}^{\infty} n^{ - \left(2a+bi \right) } d^{ - \left( -bi \right) } \right)&lt;/script&gt;&lt;!-- ws:end:WikiTextMathRule:6 --&gt;&lt;br /&gt;
+&lt;br /&gt;
+&lt;!-- ws:start:WikiTextMathRule:7:
+[[math]]&amp;lt;br/&amp;gt;
+\displaystyle&amp;lt;br /&amp;gt;
+\left| \text{f} (s) \right|^2 = \zeta(2a) + 2 \text{Re} \left( \sum \limits_{d = 1}^{\infty} \left[ d^{ - \left( -bi \right) } \sum \limits_{n = d+1}^{\infty} n^{ - \left(2a+bi \right) } \right] \right)&amp;lt;br/&amp;gt;[[math]]
+ --&gt;&lt;script type="math/tex"&gt;\displaystyle
+\left| \text{f} (s) \right|^2 = \zeta(2a) + 2 \text{Re} \left( \sum \limits_{d = 1}^{\infty} \left[ d^{ - \left( -bi \right) } \sum \limits_{n = d+1}^{\infty} n^{ - \left(2a+bi \right) } \right] \right)&lt;/script&gt;&lt;!-- ws:end:WikiTextMathRule:7 --&gt;&lt;br /&gt;
+&lt;br /&gt;
+&lt;!-- ws:start:WikiTextMathRule:8:
+[[math]]&amp;lt;br/&amp;gt;
+\displaystyle&amp;lt;br /&amp;gt;
+\left| \text{f} (s) \right|^2 = \zeta(2a) + 2 \text{Re} \left( \sum \limits_{d = 1}^{\infty} \left[ d^{ - \left( -bi \right) } \zeta(2a+ib,d+1) \right] \right)&amp;lt;br/&amp;gt;[[math]]
+ --&gt;&lt;script type="math/tex"&gt;\displaystyle
+\left| \text{f} (s) \right|^2 = \zeta(2a) + 2 \text{Re} \left( \sum \limits_{d = 1}^{\infty} \left[ d^{ - \left( -bi \right) } \zeta(2a+ib,d+1) \right] \right)&lt;/script&gt;&lt;!-- ws:end:WikiTextMathRule:8 --&gt;&lt;br /&gt;
 &lt;br /&gt;
-&lt;!-- ws:start:WikiTextMathRule:5:
+&lt;!-- ws:start:WikiTextMathRule:9:
 [[math]]&amp;lt;br/&amp;gt;
 \displaystyle&amp;lt;br /&amp;gt;
-\left| \text{f} (s) \right|^2 = \zeta(2a) + \sum_{n &amp;gt; d} \left[ n^{ - \left(2a+bi \right) } d^{ - \left( -bi \right) } + n^{ - \left(2a-bi \right) } d^{ - \left( bi \right) } \right]&amp;lt;br/&amp;gt;[[math]]
+\left| \text{f} (s) \right|^2 = \zeta(2a) + 2 \sum \limits_{d = 1}^{\infty} \text{Re} \left[ d^{ - \left( -bi \right) } \zeta(2a+ib,d+1) \right]&amp;lt;br/&amp;gt;[[math]]
   --&gt;&lt;script type="math/tex"&gt;\displaystyle
-\left| \text{f} (s) \right|^2 = \zeta(2a) + \sum_{n &gt; d} \left[ n^{ - \left(2a+bi \right) } d^{ - \left( -bi \right) } + n^{ - \left(2a-bi \right) } d^{ - \left( bi \right) } \right]&lt;/script&gt;&lt;!-- ws:end:WikiTextMathRule:5 --&gt;&lt;br /&gt;
+\left| \text{f} (s) \right|^2 = \zeta(2a) + 2 \sum \limits_{d = 1}^{\infty} \text{Re} \left[ d^{ - \left( -bi \right) } \zeta(2a+ib,d+1) \right]&lt;/script&gt;&lt;!-- ws:end:WikiTextMathRule:9 --&gt;&lt;br /&gt;
+&lt;br /&gt;
+At this point I get stuck. Having a generalized zeta function inside a sum isn't exactly the easiest function to manipulate.&lt;br /&gt;
 &lt;br /&gt;
-Note that the RHS of this equation can be broken up into a sum of conjugates. However, we ideally want to break it up into a &lt;em&gt;product&lt;/em&gt; of conjugates, in order to solve for f(s). This is where I get stuck.&lt;br /&gt;
 &lt;br /&gt;
 &lt;br /&gt;
 &lt;hr /&gt;
 &lt;br /&gt;
-&lt;!-- ws:start:WikiTextHeadingRule:19:&amp;lt;h1&amp;gt; --&gt;&lt;h1 id="toc1"&gt;&lt;a name="Mike's attempt"&gt;&lt;/a&gt;&lt;!-- ws:end:WikiTextHeadingRule:19 --&gt;Mike's attempt&lt;/h1&gt;
+&lt;!-- ws:start:WikiTextHeadingRule:23:&amp;lt;h1&amp;gt; --&gt;&lt;h1 id="toc1"&gt;&lt;a name="Mike's attempt"&gt;&lt;/a&gt;&lt;!-- ws:end:WikiTextHeadingRule:23 --&gt;Mike's attempt&lt;/h1&gt;
   &lt;br /&gt;
 Let's start here&lt;br /&gt;
 &lt;br /&gt;
-&lt;!-- ws:start:WikiTextMathRule:6:
+&lt;!-- ws:start:WikiTextMathRule:10:
 [[math]]&amp;lt;br/&amp;gt;
 \displaystyle&amp;lt;br /&amp;gt;
 \left| \zeta(s) \right|^2 = \sum_{n,d} \left[ \frac{\cos\left(b \ln\left({\tfrac{n}{d}}\right)\right)}{(nd)^{a}} \right]&amp;lt;br/&amp;gt;[[math]]
   --&gt;&lt;script type="math/tex"&gt;\displaystyle
-\left| \zeta(s) \right|^2 = \sum_{n,d} \left[ \frac{\cos\left(b \ln\left({\tfrac{n}{d}}\right)\right)}{(nd)^{a}} \right]&lt;/script&gt;&lt;!-- ws:end:WikiTextMathRule:6 --&gt;&lt;br /&gt;
+\left| \zeta(s) \right|^2 = \sum_{n,d} \left[ \frac{\cos\left(b \ln\left({\tfrac{n}{d}}\right)\right)}{(nd)^{a}} \right]&lt;/script&gt;&lt;!-- ws:end:WikiTextMathRule:10 --&gt;&lt;br /&gt;
 &lt;br /&gt;
 and change the denominator to max(n,d)^2a&lt;br /&gt;
 &lt;br /&gt;
-&lt;!-- ws:start:WikiTextMathRule:7:
+&lt;!-- ws:start:WikiTextMathRule:11:
 [[math]]&amp;lt;br/&amp;gt;
 \displaystyle&amp;lt;br /&amp;gt;
 \left| \zeta(s) \right|^2 = \sum_{n,d} \left[ \frac{\cos\left(b \ln\left({\tfrac{n}{d}}\right)\right)}{\max(n,d)^{2a}} \right]&amp;lt;br/&amp;gt;[[math]]
   --&gt;&lt;script type="math/tex"&gt;\displaystyle
-\left| \zeta(s) \right|^2 = \sum_{n,d} \left[ \frac{\cos\left(b \ln\left({\tfrac{n}{d}}\right)\right)}{\max(n,d)^{2a}} \right]&lt;/script&gt;&lt;!-- ws:end:WikiTextMathRule:7 --&gt;&lt;br /&gt;
+\left| \zeta(s) \right|^2 = \sum_{n,d} \left[ \frac{\cos\left(b \ln\left({\tfrac{n}{d}}\right)\right)}{\max(n,d)^{2a}} \right]&lt;/script&gt;&lt;!-- ws:end:WikiTextMathRule:11 --&gt;&lt;br /&gt;
 &lt;br /&gt;
 OK, so now let's split it into three sub-series -- n=d, n&amp;gt;d, n&amp;lt;d&lt;br /&gt;
 &lt;br /&gt;
-&lt;!-- ws:start:WikiTextMathRule:8:
+&lt;!-- ws:start:WikiTextMathRule:12:
 [[math]]&amp;lt;br/&amp;gt;
 \displaystyle&amp;lt;br /&amp;gt;
@@ Line 227: / Line 279: @@
 \left| \zeta(s) \right|^2 = \sum_{n=d} \left[ \frac{\cos\left(b \ln\left({\tfrac{n}{d}}\right)\right)}{(nd)^{a}} \right] +
 \sum_{n&gt;d} \left[ \frac{\cos\left(b \ln\left({\tfrac{n}{d}}\right)\right)}{(nn)^{a}} \right] +
-\sum_{n&lt; d} \left[ \frac{\cos\left(b \ln\left({\tfrac{n}{d}}\right)\right)}{(dd)^{a}} \right]&lt;/script&gt;&lt;!-- ws:end:WikiTextMathRule:8 --&gt;&lt;br /&gt;
+\sum_{n&lt; d} \left[ \frac{\cos\left(b \ln\left({\tfrac{n}{d}}\right)\right)}{(dd)^{a}} \right]&lt;/script&gt;&lt;!-- ws:end:WikiTextMathRule:12 --&gt;&lt;br /&gt;
 &lt;br /&gt;
 OK, so for every term in the middle sum, there's a corresponding term on the right where n and d are interchanged. So if n=p and d=q in the middle sum, and p&amp;gt;q, then we get the following two terms (the left from the middle sum, the right from the right sum)&lt;br /&gt;
 &lt;br /&gt;
-&lt;!-- ws:start:WikiTextMathRule:9:
+&lt;!-- ws:start:WikiTextMathRule:13:
 [[math]]&amp;lt;br/&amp;gt;
 \displaystyle&amp;lt;br /&amp;gt;
@@ Line 238: / Line 290: @@
   --&gt;&lt;script type="math/tex"&gt;\displaystyle
 \frac{\cos\left(b \ln\left({\tfrac{p}{q}}\right)\right)}{(pp)^{a}} +
-\frac{\cos\left(b \ln\left({\tfrac{q}{p}}\right)\right)}{(pp)^{a}}&lt;/script&gt;&lt;!-- ws:end:WikiTextMathRule:9 --&gt;&lt;br /&gt;
+\frac{\cos\left(b \ln\left({\tfrac{q}{p}}\right)\right)}{(pp)^{a}}&lt;/script&gt;&lt;!-- ws:end:WikiTextMathRule:13 --&gt;&lt;br /&gt;
 &lt;br /&gt;
 OK, let's throw some extra sin terms in there which cancel out to make this look more like Euler's identity&lt;br /&gt;
 &lt;br /&gt;
-&lt;!-- ws:start:WikiTextMathRule:10:
+&lt;!-- ws:start:WikiTextMathRule:14:
 [[math]]&amp;lt;br/&amp;gt;
 \displaystyle&amp;lt;br /&amp;gt;
@@ Line 249: / Line 301: @@
   --&gt;&lt;script type="math/tex"&gt;\displaystyle
 \frac{\cos\left(b \ln\left({\tfrac{p}{q}}\right)\right) + i\sin\left(b \ln\left({\tfrac{p}{q}}\right)\right)}{(pp)^{a}} +
-\frac{\cos\left(b \ln\left({\tfrac{q}{p}}\right)\right) + i\sin\left(b \ln\left({\tfrac{q}{p}}\right)\right)}{(pp)^{a}}&lt;/script&gt;&lt;!-- ws:end:WikiTextMathRule:10 --&gt;&lt;br /&gt;
+\frac{\cos\left(b \ln\left({\tfrac{q}{p}}\right)\right) + i\sin\left(b \ln\left({\tfrac{q}{p}}\right)\right)}{(pp)^{a}}&lt;/script&gt;&lt;!-- ws:end:WikiTextMathRule:14 --&gt;&lt;br /&gt;
 &lt;br /&gt;
 OK, so we can throw these extra terms back in the original three-part series and get this&lt;br /&gt;
 &lt;br /&gt;
-&lt;!-- ws:start:WikiTextMathRule:11:
+&lt;!-- ws:start:WikiTextMathRule:15:
 [[math]]&amp;lt;br/&amp;gt;
 \displaystyle&amp;lt;br /&amp;gt;
@@ Line 262: / Line 314: @@
 \left| \zeta(s) \right|^2 = \sum_{n=d} \left[ \frac{\cos\left(b \ln\left({\tfrac{n}{d}}\right)\right)}{(nd)^{a}} \right] +
 \sum_{n&gt;d} \left[ \frac{\cos\left(b \ln\left({\tfrac{n}{d}}\right)\right) + i\sin\left(b \ln\left({\tfrac{n}{d}}\right)\right)}{(nn)^{a}} \right] +
-\sum_{n&lt; d} \left[ \frac{\cos\left(b \ln\left({\tfrac{n}{d}}\right)\right) + i\sin\left(b \ln\left({\tfrac{n}{d}}\right)\right)}{(dd)^{a}} \right]&lt;/script&gt;&lt;!-- ws:end:WikiTextMathRule:11 --&gt;&lt;br /&gt;
+\sum_{n&lt; d} \left[ \frac{\cos\left(b \ln\left({\tfrac{n}{d}}\right)\right) + i\sin\left(b \ln\left({\tfrac{n}{d}}\right)\right)}{(dd)^{a}} \right]&lt;/script&gt;&lt;!-- ws:end:WikiTextMathRule:15 --&gt;&lt;br /&gt;
 &lt;br /&gt;
 Also, we can add a magical sin term that evaluates to 0 on the left side, yielding&lt;br /&gt;
 &lt;br /&gt;
-&lt;!-- ws:start:WikiTextMathRule:12:
+&lt;!-- ws:start:WikiTextMathRule:16:
 [[math]]&amp;lt;br/&amp;gt;
 \displaystyle&amp;lt;br /&amp;gt;
@@ Line 275: / Line 327: @@
 \left| \zeta(s) \right|^2 = \sum_{n=d} \left[ \frac{\cos\left(b \ln\left({\tfrac{n}{d}}\right)\right) + i\sin\left(b \ln\left({\tfrac{n}{d}}\right)\right)}{(nd)^{a}} \right] +
 \sum_{n&gt;d} \left[ \frac{\cos\left(b \ln\left({\tfrac{n}{d}}\right)\right) + i\sin\left(b \ln\left({\tfrac{n}{d}}\right)\right)}{(nn)^{a}} \right] +
-\sum_{n&lt; d} \left[ \frac{\cos\left(b \ln\left({\tfrac{n}{d}}\right)\right) + i\sin\left(b \ln\left({\tfrac{n}{d}}\right)\right)}{(dd)^{a}} \right]&lt;/script&gt;&lt;!-- ws:end:WikiTextMathRule:12 --&gt;&lt;br /&gt;
+\sum_{n&lt; d} \left[ \frac{\cos\left(b \ln\left({\tfrac{n}{d}}\right)\right) + i\sin\left(b \ln\left({\tfrac{n}{d}}\right)\right)}{(dd)^{a}} \right]&lt;/script&gt;&lt;!-- ws:end:WikiTextMathRule:16 --&gt;&lt;br /&gt;
 &lt;br /&gt;
 Euler baby, Euler&lt;br /&gt;
 &lt;br /&gt;
-&lt;!-- ws:start:WikiTextMathRule:13:
+&lt;!-- ws:start:WikiTextMathRule:17:
 [[math]]&amp;lt;br/&amp;gt;
 \displaystyle&amp;lt;br /&amp;gt;
@@ Line 288: / Line 340: @@
 \left| \zeta(s) \right|^2 = \sum_{n=d} \left[ \frac{\left({\tfrac{n}{d}}\right)^{-bi}}{(nd)^{a}} \right] +
 \sum_{n&gt;d} \left[ \frac{\left({\tfrac{n}{d}}\right)^{-bi}}{(nn)^{a}} \right] +
-\sum_{n&lt; d} \left[ \frac{\left({\tfrac{n}{d}}\right)^{-bi}}{(dd)^{a}} \right]&lt;/script&gt;&lt;!-- ws:end:WikiTextMathRule:13 --&gt;&lt;br /&gt;
+\sum_{n&lt; d} \left[ \frac{\left({\tfrac{n}{d}}\right)^{-bi}}{(dd)^{a}} \right]&lt;/script&gt;&lt;!-- ws:end:WikiTextMathRule:17 --&gt;&lt;br /&gt;
 &lt;br /&gt;
 OK, let's make it all one sum again&lt;br /&gt;
 &lt;br /&gt;
-&lt;!-- ws:start:WikiTextMathRule:14:
+&lt;!-- ws:start:WikiTextMathRule:18:
 [[math]]&amp;lt;br/&amp;gt;
 \displaystyle&amp;lt;br /&amp;gt;
 \left| \zeta(s) \right|^2 = \sum_{n,d} \left[ \frac{\left({\tfrac{n}{d}}\right)^{-bi}}{\max(n,d)^{2a}} \right] = \sum_{n,d} \left[\max(n,d)^{-2a} \cdot \left({\tfrac{n}{d}}\right)^{-bi} \right]&amp;lt;br/&amp;gt;[[math]]
   --&gt;&lt;script type="math/tex"&gt;\displaystyle
-\left| \zeta(s) \right|^2 = \sum_{n,d} \left[ \frac{\left({\tfrac{n}{d}}\right)^{-bi}}{\max(n,d)^{2a}} \right] = \sum_{n,d} \left[\max(n,d)^{-2a} \cdot \left({\tfrac{n}{d}}\right)^{-bi} \right]&lt;/script&gt;&lt;!-- ws:end:WikiTextMathRule:14 --&gt;&lt;br /&gt;
+\left| \zeta(s) \right|^2 = \sum_{n,d} \left[ \frac{\left({\tfrac{n}{d}}\right)^{-bi}}{\max(n,d)^{2a}} \right] = \sum_{n,d} \left[\max(n,d)^{-2a} \cdot \left({\tfrac{n}{d}}\right)^{-bi} \right]&lt;/script&gt;&lt;!-- ws:end:WikiTextMathRule:18 --&gt;&lt;br /&gt;
 &lt;br /&gt;
 Now, let's note that max(n,d)^2a = (nd)^a * max(n/d,d/n)^a, and that max(n/d,d/n) = exp(|log(n/d)|). So we get&lt;br /&gt;
 &lt;br /&gt;
-&lt;!-- ws:start:WikiTextMathRule:15:
+&lt;!-- ws:start:WikiTextMathRule:19:
 [[math]]&amp;lt;br/&amp;gt;
 \displaystyle&amp;lt;br /&amp;gt;
 \left| \zeta(s) \right|^2 = \sum_{n,d} \left[e^{-a \left| \log(n/d) \right|} \cdot (nd)^{-a} \cdot \left({\tfrac{n}{d}}\right)^{-bi} \right]&amp;lt;br/&amp;gt;[[math]]
   --&gt;&lt;script type="math/tex"&gt;\displaystyle
-\left| \zeta(s) \right|^2 = \sum_{n,d} \left[e^{-a \left| \log(n/d) \right|} \cdot (nd)^{-a} \cdot \left({\tfrac{n}{d}}\right)^{-bi} \right]&lt;/script&gt;&lt;!-- ws:end:WikiTextMathRule:15 --&gt;&lt;br /&gt;
+\left| \zeta(s) \right|^2 = \sum_{n,d} \left[e^{-a \left| \log(n/d) \right|} \cdot (nd)^{-a} \cdot \left({\tfrac{n}{d}}\right)^{-bi} \right]&lt;/script&gt;&lt;!-- ws:end:WikiTextMathRule:19 --&gt;&lt;br /&gt;
 &lt;br /&gt;
 The right terms become&lt;br /&gt;
 &lt;br /&gt;
-&lt;!-- ws:start:WikiTextMathRule:16:
+&lt;!-- ws:start:WikiTextMathRule:20:
 [[math]]&amp;lt;br/&amp;gt;
 \displaystyle&amp;lt;br /&amp;gt;
 \left| \zeta(s) \right|^2 = \sum_{n,d} \left[e^{-a \left| \log(n/d) \right|} \cdot n^{-(a+bi)} \cdot d^{-(a-bi)} \right]&amp;lt;br/&amp;gt;[[math]]
   --&gt;&lt;script type="math/tex"&gt;\displaystyle
-\left| \zeta(s) \right|^2 = \sum_{n,d} \left[e^{-a \left| \log(n/d) \right|} \cdot n^{-(a+bi)} \cdot d^{-(a-bi)} \right]&lt;/script&gt;&lt;!-- ws:end:WikiTextMathRule:16 --&gt;&lt;br /&gt;
+\left| \zeta(s) \right|^2 = \sum_{n,d} \left[e^{-a \left| \log(n/d) \right|} \cdot n^{-(a+bi)} \cdot d^{-(a-bi)} \right]&lt;/script&gt;&lt;!-- ws:end:WikiTextMathRule:20 --&gt;&lt;br /&gt;
 &lt;br /&gt;
 &lt;br /&gt;
 Alright!! I'm going to bed.&lt;/body&gt;&lt;/html&gt;</pre></div>