Ryan's Working Page: Difference between revisions

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:~~Sarzadoce~~|~~Sarzadoce~~]] and made on <tt>2015-08-~~21 20~~:21:08 UTC</tt>.<br>

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

: The original revision id was <tt>~~557147201~~</tt>.<br>

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Line 44:

[[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.</pre></div>

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.

----

=Mike's attempt=

Let's start here

[[math]]

\displaystyle

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

[[math]]

and change the denominator to max(n,d)^2a

[[math]]

\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]

[[math]]

OK, so now let's split it into three sub-series -- n=d, n>d, n<d

[[math]]

\displaystyle

\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]

[[math]]

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>q, then we get the following two terms (the left from the middle sum, the right from the right sum)

[[math]]

\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}}

[[math]]

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

[[math]]

\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}}

[[math]]

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

[[math]]

\displaystyle

\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]

[[math]]

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

[[math]]

\displaystyle

\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]

[[math]]

Euler baby, Euler

[[math]]

\displaystyle

\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]

[[math]]

OK, let's make it all one sum again

[[math]]

\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]

[[math]]

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

[[math]]

\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]

[[math]]

The right terms become

[[math]]

\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]

[[math]]

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 />

Line 96:

Line 191:

\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 />

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.</body></html></pre></div>

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>

<br />

Let's start here<br />

<br />

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

[[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 />

<br />

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

<br />

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

[[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 />

<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:

[[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;

\sum_{n&gt;d} \left[ \frac{\cos\left(b \ln\left({\tfrac{n}{d}}\right)\right)}{(nn)^{a}} \right] +&lt;br /&gt;

\sum_{n&lt; d} \left[ \frac{\cos\left(b \ln\left({\tfrac{n}{d}}\right)\right)}{(dd)^{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] +

\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 />

<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:

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

\displaystyle&lt;br /&gt;

\frac{\cos\left(b \ln\left({\tfrac{p}{q}}\right)\right)}{(pp)^{a}} +&lt;br /&gt;

\frac{\cos\left(b \ln\left({\tfrac{q}{p}}\right)\right)}{(pp)^{a}}&lt;br/&gt;[[math]]

--><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 />

<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:

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

\displaystyle&lt;br /&gt;

\frac{\cos\left(b \ln\left({\tfrac{p}{q}}\right)\right) + i\sin\left(b \ln\left({\tfrac{p}{q}}\right)\right)}{(pp)^{a}} +&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;br/&gt;[[math]]

--><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 />

<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:

[[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;

\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] +&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;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] +

\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 />

<br />

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

<br />

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

[[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) + i\sin\left(b \ln\left({\tfrac{n}{d}}\right)\right)}{(nd)^{a}} \right] +&lt;br /&gt;

\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] +&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;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) + 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 />

<br />

Euler baby, Euler<br />

<br />

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

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

\displaystyle&lt;br /&gt;

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

\sum_{n&gt;d} \left[ \frac{\left({\tfrac{n}{d}}\right)^{-bi}}{(nn)^{a}} \right] +&lt;br /&gt;

\sum_{n&lt; d} \left[ \frac{\left({\tfrac{n}{d}}\right)^{-bi}}{(dd)^{a}} \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}}{(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 />

<br />

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

<br />

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

[[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 />

<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:

[[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 />

<br />

The right terms become<br />

<br />

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

[[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 />

<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:Sarzadoce|Sarzadoce]] and made on <tt>2015-08-21 20:21:08 UTC</tt>.<br>
+: This revision was by author [[User:mbattaglia1|mbattaglia1]] and made on <tt>2015-08-22 04:32:08 UTC</tt>.<br>
-: The original revision id was <tt>557147201</tt>.<br>
+: The original revision id was <tt>557156615</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 44: / Line 44: @@
 [[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.</pre></div>
+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.
+----
+=Mike's attempt=
+Let's start here
+[[math]]
+\displaystyle
+\left| \zeta(s) \right|^2 = \sum_{n,d} \left[ \frac{\cos\left(b \ln\left({\tfrac{n}{d}}\right)\right)}{(nd)^{a}} \right]
+[[math]]
+and change the denominator to max(n,d)^2a
+[[math]]
+\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]
+[[math]]
+OK, so now let's split it into three sub-series -- n=d, n&gt;d, n&lt;d
+[[math]]
+\displaystyle
+\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]
+[[math]]
+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)
+[[math]]
+\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}}
+[[math]]
+OK, let's throw some extra sin terms in there which cancel out to make this look more like Euler's identity
+[[math]]
+\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}}
+[[math]]
+OK, so we can throw these extra terms back in the original three-part series and get this
+[[math]]
+\displaystyle
+\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]
+[[math]]
+Also, we can add a magical sin term that evaluates to 0 on the left side, yielding
+[[math]]
+\displaystyle
+\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]
+[[math]]
+Euler baby, Euler
+[[math]]
+\displaystyle
+\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]
+[[math]]
+OK, let's make it all one sum again
+[[math]]
+\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]
+[[math]]
+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
+[[math]]
+\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]
+[[math]]
+The right terms become
+[[math]]
+\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]
+[[math]]
+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:6:&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:6 --&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: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;
   &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 96: / Line 191: @@
 \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;
 &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;/body&gt;&lt;/html&gt;</pre></div>
+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;br /&gt;
+Let's start here&lt;br /&gt;
+&lt;br /&gt;
+&lt;!-- ws:start:WikiTextMathRule:6:
+[[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;
+&lt;br /&gt;
+and change the denominator to max(n,d)^2a&lt;br /&gt;
+&lt;br /&gt;
+&lt;!-- ws:start:WikiTextMathRule:7:
+[[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;
+&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:
+[[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;
+\sum_{n&amp;gt;d} \left[ \frac{\cos\left(b \ln\left({\tfrac{n}{d}}\right)\right)}{(nn)^{a}} \right] +&amp;lt;br /&amp;gt;
+\sum_{n&amp;lt; d} \left[ \frac{\cos\left(b \ln\left({\tfrac{n}{d}}\right)\right)}{(dd)^{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] +
+\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;
+&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:
+[[math]]&amp;lt;br/&amp;gt;
+\displaystyle&amp;lt;br /&amp;gt;
+\frac{\cos\left(b \ln\left({\tfrac{p}{q}}\right)\right)}{(pp)^{a}} +&amp;lt;br /&amp;gt;
+\frac{\cos\left(b \ln\left({\tfrac{q}{p}}\right)\right)}{(pp)^{a}}&amp;lt;br/&amp;gt;[[math]]
+ --&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;
+&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:
+[[math]]&amp;lt;br/&amp;gt;
+\displaystyle&amp;lt;br /&amp;gt;
+\frac{\cos\left(b \ln\left({\tfrac{p}{q}}\right)\right) + i\sin\left(b \ln\left({\tfrac{p}{q}}\right)\right)}{(pp)^{a}} +&amp;lt;br /&amp;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}}&amp;lt;br/&amp;gt;[[math]]
+ --&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;
+&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:
+[[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;
+\sum_{n&amp;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] +&amp;lt;br /&amp;gt;
+\sum_{n&amp;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]&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] +
+\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;
+&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:
+[[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) + i\sin\left(b \ln\left({\tfrac{n}{d}}\right)\right)}{(nd)^{a}} \right] +&amp;lt;br /&amp;gt;
+\sum_{n&amp;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] +&amp;lt;br /&amp;gt;
+\sum_{n&amp;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]&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) + 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;
+&lt;br /&gt;
+Euler baby, Euler&lt;br /&gt;
+&lt;br /&gt;
+&lt;!-- ws:start:WikiTextMathRule:13:
+[[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}}{(nd)^{a}} \right] +&amp;lt;br /&amp;gt;
+\sum_{n&amp;gt;d} \left[ \frac{\left({\tfrac{n}{d}}\right)^{-bi}}{(nn)^{a}} \right] +&amp;lt;br /&amp;gt;
+\sum_{n&amp;lt; d} \left[ \frac{\left({\tfrac{n}{d}}\right)^{-bi}}{(dd)^{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{\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;
+&lt;br /&gt;
+OK, let's make it all one sum again&lt;br /&gt;
+&lt;br /&gt;
+&lt;!-- ws:start:WikiTextMathRule:14:
+[[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;
+&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:
+[[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;
+&lt;br /&gt;
+The right terms become&lt;br /&gt;
+&lt;br /&gt;
+&lt;!-- ws:start:WikiTextMathRule:16:
+[[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;
+&lt;br /&gt;
+&lt;br /&gt;
+Alright!! I'm going to bed.&lt;/body&gt;&lt;/html&gt;</pre></div>