Ryan's Working Page

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=Attempt to backwards-engineer a Weil-weighted analog for "Zeta"= 

In Mike's Zeta Function Working Page, we see that zeta can be thought of as a superposition of weighted "cosine accuracy" functions for every unreduced rational:

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

The cosines are weighted by 1/(nd)<span style="font-size: 11.6999998092651px; vertical-align: super;">a</span>. However, it is of interest to replace n*d with max(n,d)^2 to see if we can derive a Weil-weighted analog of the Zeta function. I will denote this function by f(s).

[[math]]
\displaystyle
\left| \text{f} (s) \right|^2 = \zeta(2a) + 2 \sum_{n > d} \left[\frac{\cos\left(b\ln\left(\tfrac{n}{d}\right)\right)}{\text{max}(n,d)^{2a}}\right]
[[math]]

Let's do a few manipulations, to try to work our way backwards to f(s):

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

[[math]]
\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]]

[[math]]
\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]]

[[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]]

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.

Original HTML content:

<html><head><title>Ryan's Working Page</title></head><body><!-- ws:start:WikiTextHeadingRule:17:&lt;h1&gt; --><h1 id="toc0"><a name="Attempt to backwards-engineer a Weil-weighted analog for &quot;Zeta&quot;"></a><!-- ws:end:WikiTextHeadingRule:17 -->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 />
<br />
<!-- ws:start:WikiTextMathRule:0:
[[math]]&lt;br/&gt;
\displaystyle&lt;br /&gt;
\left| \zeta(s) \right|^2 = \zeta(2a) + 2 \sum_{n &gt; 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 = \zeta(2a) + 2 \sum_{n > d} \left[\frac{\cos\left(b\ln\left(\tfrac{n}{d}\right)\right)}{(nd)^{a}}\right]</script><!-- ws:end:WikiTextMathRule:0 --><br />
<br />
The cosines are weighted by 1/(nd)<span style="font-size: 11.6999998092651px; vertical-align: super;">a</span>. However, it is of interest to replace n*d with max(n,d)^2 to see if we can derive a Weil-weighted analog of the Zeta function. I will denote this function by f(s).<br />
<br />
<!-- ws:start:WikiTextMathRule:1:
[[math]]&lt;br/&gt;
\displaystyle&lt;br /&gt;
\left| \text{f} (s) \right|^2 = \zeta(2a) + 2 \sum_{n &gt; d} \left[\frac{\cos\left(b\ln\left(\tfrac{n}{d}\right)\right)}{\text{max}(n,d)^{2a}}\right]&lt;br/&gt;[[math]]
 --><script type="math/tex">\displaystyle
\left| \text{f} (s) \right|^2 = \zeta(2a) + 2 \sum_{n > d} \left[\frac{\cos\left(b\ln\left(\tfrac{n}{d}\right)\right)}{\text{max}(n,d)^{2a}}\right]</script><!-- ws:end:WikiTextMathRule:1 --><br />
<br />
Let's do a few manipulations, to try to work our way backwards to f(s):<br />
<br />
<!-- ws:start:WikiTextMathRule:2:
[[math]]&lt;br/&gt;
\displaystyle&lt;br /&gt;
\left| \text{f} (s) \right|^2 = \zeta(2a) + 2 \sum_{n &gt; d} \left[\frac{\cos\left(b\ln\left(\tfrac{n}{d}\right)\right)}{n^{2a}}\right]&lt;br/&gt;[[math]]
 --><script type="math/tex">\displaystyle
\left| \text{f} (s) \right|^2 = \zeta(2a) + 2 \sum_{n > d} \left[\frac{\cos\left(b\ln\left(\tfrac{n}{d}\right)\right)}{n^{2a}}\right]</script><!-- ws:end:WikiTextMathRule:2 --><br />
<br />
<!-- ws:start:WikiTextMathRule:3:
[[math]]&lt;br/&gt;
\displaystyle&lt;br /&gt;
\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]&lt;br/&gt;[[math]]
 --><script type="math/tex">\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]</script><!-- ws:end:WikiTextMathRule:3 --><br />
<br />
<!-- ws:start:WikiTextMathRule:4:
[[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><!-- ws:end:WikiTextMathRule:4 --><br />
<br />
<!-- ws:start:WikiTextMathRule:5:
[[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><!-- ws:end:WikiTextMathRule:5 --><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 />
<br />
<hr />
<br />
<!-- ws:start:WikiTextHeadingRule:19:&lt;h1&gt; --><h1 id="toc1"><a name="Mike's attempt"></a><!-- ws:end:WikiTextHeadingRule:19 -->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><!-- ws:end:WikiTextMathRule:6 --><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><!-- ws:end:WikiTextMathRule:7 --><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><!-- ws:end:WikiTextMathRule:8 --><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><!-- ws:end:WikiTextMathRule:9 --><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><!-- ws:end:WikiTextMathRule:10 --><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><!-- ws:end:WikiTextMathRule:11 --><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><!-- ws:end:WikiTextMathRule:12 --><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><!-- ws:end:WikiTextMathRule:13 --><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><!-- ws:end:WikiTextMathRule:14 --><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><!-- ws:end:WikiTextMathRule:15 --><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><!-- ws:end:WikiTextMathRule:16 --><br />
<br />
<br />
Alright!! I'm going to bed.</body></html>