# 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:

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

The cosines are weighted by 1/(nd)a. 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).

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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)

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

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

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

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

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

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

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

Euler baby, Euler

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

OK, let's make it all one sum again

$\displaystyle \left| \zeta(s) \right|^2 = \sum_{n,d} \left[ \frac{\left({\tfrac{n}{d}}\right)^{-bi}}{\max(n,d)^{2a}} \right] = \sum_{n,d} \left[\max(n,d)^{-2a} \cdot \left({\tfrac{n}{d}}\right)^{-bi} \right]$

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

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

The right terms become

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

Alright!! I'm going to bed.