In this post I derived for $s=a + ti$, that assuming the RH, the following should be true:

$$\displaystyle \frac{\xi(\frac12 - a + s)}{\xi(\frac12 - a)} = \prod_{n=1}^\infty \left(1- \frac{s}{\mu_n} \right) \left(1- \frac{s}{\overline{\mu_n}} \right)$$

with $\mu_n = a + \Im(\rho_n) i$ and $\xi(z) = \frac12 z(z-1) \pi^{-\frac{z}{2}} \Gamma(\frac{z}{2}) \zeta(z)$ (i.e. the Riemann xi-function).

$\\\\$

I believe a similar result can be obtained without assuming the RH. With $s=\sigma + ti$ it is:

$$\displaystyle \frac{\xi(s+\sigma)}{\xi(\sigma)} = \prod_{n=1}^\infty \left(1- \frac{s}{\rho_n-\sigma} \right) \left(1- \frac{s}{1-\rho_n -\sigma} \right)$$

I derived it as follows ($t \in \mathbb{R};\ \sigma, s, \rho_n \in \mathbb{C}; \ (\rho_n -\sigma) \ne 0,(1-\rho_n -\sigma)\ne0$):

Start from Hadamard's proof that:

$$\xi(s) = \xi(0) \prod_{n=1}^\infty \left(1- \frac{s}{\rho_n} \right) \left(1- \frac{s}{1-\rho_n} \right) \qquad (1)$$

and firstly work away the $1-$-parts from both factors. Then split the resulting products via multiplying each ot them by: $\dfrac{(\rho_n-\sigma)}{(\rho_n -\sigma)}$ and $\dfrac{(1-\rho_n-\sigma)}{(1-\rho_n -\sigma)}$ respectively. Then re-introduce the $1-$ -parts and this gives:

$$\displaystyle \xi(s)= \xi(0) \prod_{n=1}^\infty \left(1- \frac{\sigma}{\rho_n} \right) \left(1- \frac{\sigma}{1-\rho_n} \right) \prod_{n=1}^\infty \left(1- \frac{ti}{(\rho_n-\sigma)} \right) \left(1- \frac{ti}{(1-\rho_n -\sigma)} \right)$$

Plug $\sigma$ into (1) and the first part becomes $\xi(\sigma)$. Also take $it = s-\sigma$, and the outcome is there:

$$\displaystyle \xi(s+\sigma)= \xi(\sigma) \prod_{n=1}^\infty \left(1- \frac{s}{(\rho_n-\sigma)} \right) \left(1- \frac{s}{(1-\rho_n -\sigma)} \right)$$

Checked it numerically (with the first 100K $\rho$s) and it seems to work fine. Note that it also works when $\sigma$ is exchanged by $ti$.

Addition:

Via the same method, the following can also be derived:

\begin{align*} \xi(2s) &= \xi(s) \prod_{n=1}^\infty \left(1- \frac{s}{(\rho_n-s)} \right) \left(1- \frac{s}{(1-\rho_n -s)} \right) \\ \xi(0) &= \xi(-s) \prod_{n=1}^\infty \left(1- \frac{s}{(\rho_n+s)} \right) \left(1- \frac{s}{(1-\rho_n +s)} \right) \\ \xi(1) &= \xi(1-s) \prod_{n=1}^\infty \left(1- \frac{s}{(\rho_n+s-1)} \right) \left(1- \frac{s}{(1-\rho_n +s-1)} \right) \\ \end{align*}

Using $\xi(s)=\xi(1-s)$ and $\xi(0)=\xi(1)=\frac12$, this relation between $\xi(s)$ and $\xi(-s)$ follows:

$$\frac{\xi(s)}{\xi(-s)} = \prod_{n=1}^\infty \left(1- \frac{1}{(\rho_n+s)} \right) \left(1- \frac{1}{(1-\rho_n +s)} \right)$$

Question:

These outcomes seem pretty trivial, however I am keen to understand intuitively why it is that a linear shift in the $\rho$'s can always be expressed in a division of two $\xi$'s.

In this post I derived for $s=a + ti$, that assuming the RH, the following should be true:

$$\displaystyle \frac{\xi(\frac12 - a + s)}{\xi(\frac12 - a)} = \prod_{n=1}^\infty \left(1- \frac{s}{\mu_n} \right) \left(1- \frac{s}{\overline{\mu_n}} \right)$$

with $\mu_n = a + \Im(\rho_n) i$ and $\xi(z) = \frac12 z(z-1) \pi^{-\frac{z}{2}} \Gamma(\frac{z}{2}) \zeta(z)$ (i.e. the Riemann xi-function).

$\\\\$

I believe a similar result can be obtained without assuming the RH. With $s=\sigma + ti$ it is:

$$\displaystyle \frac{\xi(s+\sigma)}{\xi(\sigma)} = \prod_{n=1}^\infty \left(1- \frac{s}{\rho_n-\sigma} \right) \left(1- \frac{s}{1-\rho_n -\sigma} \right)$$

I derived it as follows ($t \in \mathbb{R};\ \sigma, s, \rho_n \in \mathbb{C}; \ (\rho_n -\sigma) \ne 0,(1-\rho_n -\sigma)\ne0$):

Start from Hadamard's proof that:

$$\xi(s) = \xi(0) \prod_{n=1}^\infty \left(1- \frac{s}{\rho_n} \right) \left(1- \frac{s}{1-\rho_n} \right) \qquad (1)$$

and firstly work away the $1-$-parts from both factors. Then split the resulting products via multiplying each ot them by: $\dfrac{(\rho_n-\sigma)}{(\rho_n -\sigma)}$ and $\dfrac{(1-\rho_n-\sigma)}{(1-\rho_n -\sigma)}$ respectively. Then re-introduce the $1-$ -parts and this gives:

$$\displaystyle \xi(s)= \xi(0) \prod_{n=1}^\infty \left(1- \frac{\sigma}{\rho_n} \right) \left(1- \frac{\sigma}{1-\rho_n} \right) \prod_{n=1}^\infty \left(1- \frac{ti}{(\rho_n-\sigma)} \right) \left(1- \frac{ti}{(1-\rho_n -\sigma)} \right)$$

Plug $\sigma$ into (1) and the first part becomes $\xi(\sigma)$. Also take $it = s-\sigma$, and the outcome is there:

$$\displaystyle \xi(s+\sigma)= \xi(\sigma) \prod_{n=1}^\infty \left(1- \frac{s}{(\rho_n-\sigma)} \right) \left(1- \frac{s}{(1-\rho_n -\sigma)} \right)$$

Checked it numerically (with the first 100K $\rho$s) and it seems to work fine. Note that it also works when $\sigma$ is exchanged by $ti$.

Addition:

Via the same method, the following can also be derived:

\begin{align*} \xi(2s) &= \xi(s) \prod_{n=1}^\infty \left(1- \frac{s}{(\rho_n-s)} \right) \left(1- \frac{s}{(1-\rho_n -s)} \right) \\ \xi(0) &= \xi(-s) \prod_{n=1}^\infty \left(1- \frac{s}{(\rho_n+s)} \right) \left(1- \frac{s}{(1-\rho_n +s)} \right) \\ \xi(1) &= \xi(1-s) \prod_{n=1}^\infty \left(1- \frac{s}{(\rho_n+s-1)} \right) \left(1- \frac{s}{(1-\rho_n +s-1)} \right) \\ \end{align*}

Using $\xi(s)=\xi(1-s)$ and $\xi(0)=\xi(1)=\frac12$, this relation between $\xi(s)$ and $\xi(-s)$ follows:

$$\frac{\xi(s)}{\xi(-s)} = \prod_{n=1}^\infty \left(1- \frac{1}{(\rho_n+s)} \right) \left(1- \frac{1}{(1-\rho_n +s)} \right)$$

Question:

These outcomes seem pretty trivial, however I am keen to understand intuitively why it is that a linear shift in the $\rho$'s can always be expressed in a division of two $\xi$'s.

In this post I derived for $s=a + ti$, that assuming the RH, the following should be true:

$$\displaystyle \frac{\xi(\frac12 - a + s)}{\xi(\frac12 - a)} = \prod_{n=1}^\infty \left(1- \frac{s}{\mu_n} \right) \left(1- \frac{s}{\overline{\mu_n}} \right)$$

with $\mu_n = a + \Im(\rho_n) i$ and $\xi(z) = \frac12 z(z-1) \pi^{-\frac{z}{2}} \Gamma(\frac{z}{2}) \zeta(z)$ (i.e. the Riemann xi-function).

$\\\\$

I believe a similar result can be obtained without assuming the RH. With $s=\sigma + ti$ it is:

$$\displaystyle \frac{\xi(s+\sigma)}{\xi(\sigma)} = \prod_{n=1}^\infty \left(1- \frac{s}{\rho_n-\sigma} \right) \left(1- \frac{s}{1-\rho_n -\sigma} \right)$$

I derived it as follows ($t \in \mathbb{R};\ \sigma, s, \rho_n \in \mathbb{C}; \ (\rho_n -\sigma) \ne 0,(1-\rho_n -\sigma)\ne0$):

Start from Hadamard's proof that:

$$\xi(s) = \xi(0) \prod_{n=1}^\infty \left(1- \frac{s}{\rho_n} \right) \left(1- \frac{s}{1-\rho_n} \right) \qquad (1)$$

and firstly work away the $1-$-parts from both factors. Then split the resulting products via multiplying each ot them by: $\dfrac{(\rho_n-\sigma)}{(\rho_n -\sigma)}$ and $\dfrac{(1-\rho_n-\sigma)}{(1-\rho_n -\sigma)}$ respectively. Then re-introduce the $1-$ -parts and this gives:

$$\displaystyle \xi(s)= \xi(0) \prod_{n=1}^\infty \left(1- \frac{\sigma}{\rho_n} \right) \left(1- \frac{\sigma}{1-\rho_n} \right) \prod_{n=1}^\infty \left(1- \frac{ti}{(\rho_n-\sigma)} \right) \left(1- \frac{ti}{(1-\rho_n -\sigma)} \right)$$

Plug $\sigma$ into (1) and the first part becomes $\xi(\sigma)$. Also take $it = s-\sigma$, and the outcome is there:

$$\displaystyle \xi(s+\sigma)= \xi(\sigma) \prod_{n=1}^\infty \left(1- \frac{s}{(\rho_n-\sigma)} \right) \left(1- \frac{s}{(1-\rho_n -\sigma)} \right)$$

Checked it numerically (with the first 100K $\rho$s) and it seems to work fine. Note that it also works when $\sigma$ is exchanged by $ti$.

Questions:

1. The outcome is surprisingly simple and I guess it is pretty trivial. Can it be explained intuitively?

2. Could the second factor be simplified any further?

In this post I derived for $s=a + ti$, that assuming the RH, the following should be true:

$$\displaystyle \frac{\xi(\frac12 - a + s)}{\xi(\frac12 - a)} = \prod_{n=1}^\infty \left(1- \frac{s}{\mu_n} \right) \left(1- \frac{s}{\overline{\mu_n}} \right)$$

with $\mu_n = a + \Im(\rho_n) i$ and $\xi(z) = \frac12 z(z-1) \pi^{-\frac{z}{2}} \Gamma(\frac{z}{2}) \zeta(z)$ (i.e. the Riemann xi-function).

$\\\\$

I believe a similar result can be obtained without assuming the RH. With $s=\sigma + ti$ it is:

$$\displaystyle \frac{\xi(s+\sigma)}{\xi(\sigma)} = \prod_{n=1}^\infty \left(1- \frac{s}{\rho_n-\sigma} \right) \left(1- \frac{s}{1-\rho_n -\sigma} \right)$$

I derived it as follows ($t \in \mathbb{R};\ \sigma, s, \rho_n \in \mathbb{C}; \ (\rho_n -\sigma) \ne 0,(1-\rho_n -\sigma)\ne0$):

Start from Hadamard's proof that:

$$\xi(s) = \xi(0) \prod_{n=1}^\infty \left(1- \frac{s}{\rho_n} \right) \left(1- \frac{s}{1-\rho_n} \right) \qquad (1)$$

and firstly work away the $1-$-parts from both factors. Then split the resulting products via multiplying each ot them by: $\dfrac{(\rho_n-\sigma)}{(\rho_n -\sigma)}$ and $\dfrac{(1-\rho_n-\sigma)}{(1-\rho_n -\sigma)}$ respectively. Then re-introduce the $1-$ -parts and this gives:

$$\displaystyle \xi(s)= \xi(0) \prod_{n=1}^\infty \left(1- \frac{\sigma}{\rho_n} \right) \left(1- \frac{\sigma}{1-\rho_n} \right) \prod_{n=1}^\infty \left(1- \frac{ti}{(\rho_n-\sigma)} \right) \left(1- \frac{ti}{(1-\rho_n -\sigma)} \right)$$

Plug $\sigma$ into (1) and the first part becomes $\xi(\sigma)$. Also take $it = s-\sigma$, and the outcome is there:

$$\displaystyle \xi(s+\sigma)= \xi(\sigma) \prod_{n=1}^\infty \left(1- \frac{s}{(\rho_n-\sigma)} \right) \left(1- \frac{s}{(1-\rho_n -\sigma)} \right)$$

Checked it numerically (with the first 100K $\rho$s) and it seems to work fine. Note that it also works when $\sigma$ is exchanged by $ti$.

Addition:

Via the same method, the following can also be derived:

\begin{align*} \xi(2s) &= \xi(s) \prod_{n=1}^\infty \left(1- \frac{s}{(\rho_n-s)} \right) \left(1- \frac{s}{(1-\rho_n -s)} \right) \\ \xi(0) &= \xi(-s) \prod_{n=1}^\infty \left(1- \frac{s}{(\rho_n+s)} \right) \left(1- \frac{s}{(1-\rho_n +s)} \right) \\ \xi(1) &= \xi(1-s) \prod_{n=1}^\infty \left(1- \frac{s}{(\rho_n+s-1)} \right) \left(1- \frac{s}{(1-\rho_n +s-1)} \right) \\ \end{align*}

Using $\xi(s)=\xi(1-s)$ and $\xi(0)=\xi(1)=\frac12$, this relation between $\xi(s)$ and $\xi(-s)$ follows:

$$\frac{\xi(s)}{\xi(-s)} = \prod_{n=1}^\infty \left(1- \frac{1}{(\rho_n+s)} \right) \left(1- \frac{1}{(1-\rho_n +s)} \right)$$

Question:

These outcomes seem pretty trivial, however I am keen to understand intuitively why it is that a linear shift in the $\rho$'s can always be expressed in a division of two $\xi$'s.

In this post I derived, that assuming the RH, the following should be true:

$$\displaystyle \frac{\xi(\frac12 - a + s)}{\xi(\frac12 - a)} = \prod_{n=1}^\infty \left(1- \frac{s}{\mu_n} \right) \left(1- \frac{s}{\overline{\mu_n}} \right)$$

with $\mu_n = a + \Im(\rho_n) i$ and $\xi(z) = \frac12 z(z-1) \pi^{-\frac{z}{2}} \Gamma(\frac{z}{2}) \zeta(z)$ (i.e. the Riemann xi-function).

$\\\\$

I believe a similar result can be obtained without assuming the RH. With $s=\sigma + ti$ it is:

$$\displaystyle \frac{\xi(s+\sigma)}{\xi(\sigma)} = \prod_{n=1}^\infty \left(1- \frac{s}{\rho_n-\sigma} \right) \left(1- \frac{s}{1-\rho_n -\sigma} \right)$$

I derived it as follows ($t \in \mathbb{R};\ \sigma, s, \rho_n \in \mathbb{C}; \ (\rho_n -\sigma) \ne 0,(1-\rho_n -\sigma)\ne0$):

Start from Hadamard's proof that:

$$\xi(s) = \xi(0) \prod_{n=1}^\infty \left(1- \frac{s}{\rho_n} \right) \left(1- \frac{s}{1-\rho_n} \right) \qquad (1)$$

and firstly work away the $1-$-parts from both factors. Then split the resulting products via multiplying each ot them by: $\dfrac{(\rho_n-\sigma)}{(\rho_n -\sigma)}$ and $\dfrac{(1-\rho_n-\sigma)}{(1-\rho_n -\sigma)}$ respectively. Then re-introduce the $1-$ -parts and this gives:

$$\displaystyle \xi(s)= \xi(0) \prod_{n=1}^\infty \left(1- \frac{\sigma}{\rho_n} \right) \left(1- \frac{\sigma}{1-\rho_n} \right) \prod_{n=1}^\infty \left(1- \frac{ti}{(\rho_n-\sigma)} \right) \left(1- \frac{ti}{(1-\rho_n -\sigma)} \right)$$

Plug $\sigma$ into (1) and the first part becomes $\xi(\sigma)$. Also take $it = s-\sigma$, and the outcome is there:

$$\displaystyle \xi(s+\sigma)= \xi(\sigma) \prod_{n=1}^\infty \left(1- \frac{s}{(\rho_n-\sigma)} \right) \left(1- \frac{s}{(1-\rho_n -\sigma)} \right)$$

Checked it numerically (with the first 100K $\rho$s) and it seems to work fine. Note that it also works when $\sigma$ is exchanged by $ti$.

Questions:

1. The outcome is surprisingly simple and I guess it is pretty trivial. Can it be explained intuitively?

2. Could the second factor be simplified any further?

In this post I derived for $s=a + ti$, that assuming the RH, the following should be true:

$$\displaystyle \frac{\xi(\frac12 - a + s)}{\xi(\frac12 - a)} = \prod_{n=1}^\infty \left(1- \frac{s}{\mu_n} \right) \left(1- \frac{s}{\overline{\mu_n}} \right)$$

with $\mu_n = a + \Im(\rho_n) i$ and $\xi(z) = \frac12 z(z-1) \pi^{-\frac{z}{2}} \Gamma(\frac{z}{2}) \zeta(z)$ (i.e. the Riemann xi-function).

$\\\\$

I believe a similar result can be obtained without assuming the RH. With $s=\sigma + ti$ it is:

$$\displaystyle \frac{\xi(s+\sigma)}{\xi(\sigma)} = \prod_{n=1}^\infty \left(1- \frac{s}{\rho_n-\sigma} \right) \left(1- \frac{s}{1-\rho_n -\sigma} \right)$$

I derived it as follows ($t \in \mathbb{R};\ \sigma, s, \rho_n \in \mathbb{C}; \ (\rho_n -\sigma) \ne 0,(1-\rho_n -\sigma)\ne0$):

Start from Hadamard's proof that:

$$\xi(s) = \xi(0) \prod_{n=1}^\infty \left(1- \frac{s}{\rho_n} \right) \left(1- \frac{s}{1-\rho_n} \right) \qquad (1)$$

and firstly work away the $1-$-parts from both factors. Then split the resulting products via multiplying each ot them by: $\dfrac{(\rho_n-\sigma)}{(\rho_n -\sigma)}$ and $\dfrac{(1-\rho_n-\sigma)}{(1-\rho_n -\sigma)}$ respectively. Then re-introduce the $1-$ -parts and this gives:

$$\displaystyle \xi(s)= \xi(0) \prod_{n=1}^\infty \left(1- \frac{\sigma}{\rho_n} \right) \left(1- \frac{\sigma}{1-\rho_n} \right) \prod_{n=1}^\infty \left(1- \frac{ti}{(\rho_n-\sigma)} \right) \left(1- \frac{ti}{(1-\rho_n -\sigma)} \right)$$

Plug $\sigma$ into (1) and the first part becomes $\xi(\sigma)$. Also take $it = s-\sigma$, and the outcome is there:

$$\displaystyle \xi(s+\sigma)= \xi(\sigma) \prod_{n=1}^\infty \left(1- \frac{s}{(\rho_n-\sigma)} \right) \left(1- \frac{s}{(1-\rho_n -\sigma)} \right)$$

Checked it numerically (with the first 100K $\rho$s) and it seems to work fine. Note that it also works when $\sigma$ is exchanged by $ti$.

Questions:

1. The outcome is surprisingly simple and I guess it is pretty trivial. Can it be explained intuitively?

2. Could the second factor be simplified any further?