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.Original HTML content:
<html><head><title>Ryan's Working Page</title></head><body><!-- ws:start:WikiTextHeadingRule:6:<h1> --><h1 id="toc0"><a name="Attempt to backwards-engineer a Weil-weighted analog for "Zeta""></a><!-- ws:end:WikiTextHeadingRule:6 -->Attempt to backwards-engineer a Weil-weighted analog for "Zeta"</h1>
<br />
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:<br />
<br />
<!-- ws:start:WikiTextMathRule:0:
[[math]]<br/>
\displaystyle<br />
\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]<br/>[[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]]<br/>
\displaystyle<br />
\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]<br/>[[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]]<br/>
\displaystyle<br />
\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]<br/>[[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]]<br/>
\displaystyle<br />
\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]<br/>[[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]]<br/>
\displaystyle<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)<br/>[[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]]<br/>
\displaystyle<br />
\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]<br/>[[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.</body></html>