Let's write \(\rho \in \mathbb{C}\) were \(0 < \Re \rho < 1\). And let's say that we have a countable list of values \(\rho \in P\); where \(P\) is all values which satisfy:
\[
\zeta_G(\rho) = 0\\
\]
If we can find the values of \(\rho\); or guess and understand them. We'll be able to understand the function \(\zeta_G\). I'm too drunk right now; but this is what we need to look at!
Regards, James
I'm going to talk about the Hadamard product formula. Where "if the Riemann Hypothesis is true" we have a product formula for \(\zeta\). This is typically written as:
\[
\zeta(s) = H(s) \prod_{\rho} \left( 1-\frac{s}{\rho}\right)e^{\frac{s}{\rho}}\\
\]
Where \(H(s)\) is an entire function; and \(\rho\) is a list of a zeroes of \(\zeta\).
We have to be a bit more careful with \(\zeta_G\). But if we write this for \(\Re(s) >0\); we are given the exact same formula:
\[
\zeta_G(s) = H_G(s) \prod_{\rho \in P_G} \left( 1-\frac{s}{\rho}\right)e^{\frac{s}{\rho}}\\
\]
.......
The function \(H_G\) converges for \(\Re(s) >0\) and is non zero.... And as soon as we find the zeroes of \(\zeta_G\), we have found \(\zeta_G\) for \(\Re(s) > 0\). The function \(H_G\) is pretty tame too. Especially because its wall of singularities is at \(\Re(s) = -1\)...
This might look super stupid. And I'm pretty drunk. But the function \(\zeta_G(s)\) looks like a Hadamard product of its zeroes.... The function:
\[
Z_G(s) = \prod_{\rho \in P_G} \left( 1-\frac{s}{\rho}\right)e^{\frac{s}{\rho}}\\
\]
Is the fucking function we care about; where \(\zeta_G\); is just a normalized version...
Lmao, I'm too drunk to give a proper analysis. But it looks something like this. My Old Riemann Hypothesis Training, is finally fucking paying off, lmao. I can write this clearer and more straight. But this is standard analytic number theory work!
If you take a function \(Z_G(s)\) which has zeroes exactly at the points which \(\zeta_G(s)\) has zeroes. Where additionally they are the exact same types of zeroes. This means that \(\zeta_G(s) = H_G(s) Z_G(s)\)... where \(H_G(s)\) is just a standard non zero function that maps \(\Re(s) >0 \to \mathbb{C}/0\)... Where the statement:
\[
\zeta_G(s) = H_G(s) Z_G(s) = H_G(s)\prod_{\rho \in P_G} \left( 1-\frac{s}{\rho}\right)e^{\frac{s}{\rho}}\\
\]
I'm super drunk right now. And just happy to talk math. I apologize if I'm being a dumbass
\[
\zeta_G(\rho) = 0\\
\]
If we can find the values of \(\rho\); or guess and understand them. We'll be able to understand the function \(\zeta_G\). I'm too drunk right now; but this is what we need to look at!
Regards, James
I'm going to talk about the Hadamard product formula. Where "if the Riemann Hypothesis is true" we have a product formula for \(\zeta\). This is typically written as:
\[
\zeta(s) = H(s) \prod_{\rho} \left( 1-\frac{s}{\rho}\right)e^{\frac{s}{\rho}}\\
\]
Where \(H(s)\) is an entire function; and \(\rho\) is a list of a zeroes of \(\zeta\).
We have to be a bit more careful with \(\zeta_G\). But if we write this for \(\Re(s) >0\); we are given the exact same formula:
\[
\zeta_G(s) = H_G(s) \prod_{\rho \in P_G} \left( 1-\frac{s}{\rho}\right)e^{\frac{s}{\rho}}\\
\]
.......
The function \(H_G\) converges for \(\Re(s) >0\) and is non zero.... And as soon as we find the zeroes of \(\zeta_G\), we have found \(\zeta_G\) for \(\Re(s) > 0\). The function \(H_G\) is pretty tame too. Especially because its wall of singularities is at \(\Re(s) = -1\)...
This might look super stupid. And I'm pretty drunk. But the function \(\zeta_G(s)\) looks like a Hadamard product of its zeroes.... The function:
\[
Z_G(s) = \prod_{\rho \in P_G} \left( 1-\frac{s}{\rho}\right)e^{\frac{s}{\rho}}\\
\]
Is the fucking function we care about; where \(\zeta_G\); is just a normalized version...
Lmao, I'm too drunk to give a proper analysis. But it looks something like this. My Old Riemann Hypothesis Training, is finally fucking paying off, lmao. I can write this clearer and more straight. But this is standard analytic number theory work!
If you take a function \(Z_G(s)\) which has zeroes exactly at the points which \(\zeta_G(s)\) has zeroes. Where additionally they are the exact same types of zeroes. This means that \(\zeta_G(s) = H_G(s) Z_G(s)\)... where \(H_G(s)\) is just a standard non zero function that maps \(\Re(s) >0 \to \mathbb{C}/0\)... Where the statement:
\[
\zeta_G(s) = H_G(s) Z_G(s) = H_G(s)\prod_{\rho \in P_G} \left( 1-\frac{s}{\rho}\right)e^{\frac{s}{\rho}}\\
\]
I'm super drunk right now. And just happy to talk math. I apologize if I'm being a dumbass
