Ryan's Working Page: Difference between revisions
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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: | 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 | |||
\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> | ||
\left| \zeta(s) \right|^2 = \zeta(2a) + 2 \sum_{n | |||
The cosines are weighted by 1/(nd) | 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 norm, Tenney–Weil norm, and TWp interval and tuning space|Weil]]-weighted analog of the Zeta function. I will denote this function by f(s). | ||
<math>\displaystyle | |||
\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> | ||
\left| \text{f} (s) \right|^2 = \zeta(2a) + 2 \sum_{n | |||
Let's do a few manipulations, to try to work our way backwards to f(s): | Let's do a few manipulations, to try to work our way backwards to f(s): | ||
<math>\displaystyle | |||
\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> | ||
\left| \text{f} (s) \right|^2 = \zeta(2a) + 2 \sum_{n | |||
<math>\displaystyle | |||
\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> | ||
\left| \text{f} (s) \right|^2 = \zeta(2a) + \sum_{n | |||
<math>\displaystyle | |||
\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> | ||
\left| \text{f} (s) \right|^2 = \zeta(2a) + | |||
<math>\displaystyle | |||
\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> | ||
\left| \text{f} (s) \right|^2 = \zeta(2a) + \sum_{n | |||
<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) + 2 \sum \limits_{d = 1}^{\infty} \text{Re} \left[ d^{ - \left( -bi \right) } \zeta(2a+ib,d+1) \right]</math> | |||
\left| \zeta(s) \right|^2 = \ | |||
At this point I get stuck. Having a generalized zeta function inside a sum isn't exactly the easiest function to manipulate. | |||
----- | |||
=Mike's attempt= | |||
\ | |||
Let's start here | |||
\left | <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> | ||
\left | |||
OK, so now let's split it into three sub-series -- n=d, n>d, n<d | |||
\left | |||
<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) | |||
\left | |||
<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 | ||
\left | |||
<math>\displaystyle | |||
\left| \ | \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. | |||
[[Category:Math]] | |||
[[Category:Zeta]] | |||
{{todo|move to userspace}} | |||