
4


edited Apr 13 '17 at 12:58

This question arose from an earlier one and the MO user's useful answers there: http://mathoverflow.net/questions/80710/whatarethevaluesofthederivativeofriemannszetafunctionattheknownnon/80783#80783What are the values of the derivative of Riemann's zeta function at the known nontrivial zeros? (which is not a prerequisite for this version).
As usual, let $\rho=\beta+i\gamma$ denote the nontrivial zeros of $\zeta(s)$. Essentially, the question is this:
How are the numbers $e^{i\phi(\rho)}=\lim_{s\rightarrow\rho}\zeta(s)/\overline{\zeta(s)}$ distributed in $\mathbb{T}$?
Of course $\phi\in\mathbb{R}$ and the limit could be replaced by quotients of derivatives if we knew the order of the zeros (in which case $\phi$ is just twice their argument there), but I don't think it is necessary to know that any way. Moreover, I think the answer is independent of RH. My reasoning is as follows.
For $x\in\mathbb{R}$, one may consider the meanvalue:
$$M(x)=\lim_{T\rightarrow\infty}\frac{1}{N(T)}\sum_{0 <\gamma < T}e^{i\phi(\rho)x}.$$
Let $t>0$ and $\theta(t)$ denote the RiemannSiegel thetafunction. Firstly, by separating the sums over zeros on and off the line $\beta=1/2$ and, secondly, by the BohrLandau theorem (that the number of zeros with $\beta>1/2$ is $O(T)$), one gets
$$M(x)=\lim_{T\rightarrow\infty}\frac{1}{N(T)}\left(\sum_{0 <\gamma < T :\beta=1/2}e^{i\phi(\rho)x}+O\left(T\right)\right)$$
$$=\lim_{T\rightarrow\infty}\frac{\pi}{\theta(T)}\sum_{0 <\gamma < T :\beta=1/2}e^{i2\theta(\gamma)x}.$$
If my reasoning is correct, $M(1)=0$ is equivalent to $e^{i\phi(\rho)}$ being dense in $\mathbb{T}$, and $M(n)=0$, $n\in\mathbb{N}$, to them being uniformly distributed. Any advice, insights and corrections would be most welcome.
EDIT 1: the arguments are $2\theta x$ corrected.
EDIT 2: $O(T)$ inside the bracket corrected.
This question arose from an earlier one and the MO user's useful answers there: http://mathoverflow.net/questions/80710/whatarethevaluesofthederivativeofriemannszetafunctionattheknownnon/80783#80783 (which is not a prerequisite for this version).
As usual, let $\rho=\beta+i\gamma$ denote the nontrivial zeros of $\zeta(s)$. Essentially, the question is this:
How are the numbers $e^{i\phi(\rho)}=\lim_{s\rightarrow\rho}\zeta(s)/\overline{\zeta(s)}$ distributed in $\mathbb{T}$?
Of course $\phi\in\mathbb{R}$ and the limit could be replaced by quotients of derivatives if we knew the order of the zeros (in which case $\phi$ is just twice their argument there), but I don't think it is necessary to know that any way. Moreover, I think the answer is independent of RH. My reasoning is as follows.
For $x\in\mathbb{R}$, one may consider the meanvalue:
$$M(x)=\lim_{T\rightarrow\infty}\frac{1}{N(T)}\sum_{0 <\gamma < T}e^{i\phi(\rho)x}.$$
Let $t>0$ and $\theta(t)$ denote the RiemannSiegel thetafunction. Firstly, by separating the sums over zeros on and off the line $\beta=1/2$ and, secondly, by the BohrLandau theorem (that the number of zeros with $\beta>1/2$ is $O(T)$), one gets
$$M(x)=\lim_{T\rightarrow\infty}\frac{1}{N(T)}\left(\sum_{0 <\gamma < T :\beta=1/2}e^{i\phi(\rho)x}+O\left(T\right)\right)$$
$$=\lim_{T\rightarrow\infty}\frac{\pi}{\theta(T)}\sum_{0 <\gamma < T :\beta=1/2}e^{i2\theta(\gamma)x}.$$
If my reasoning is correct, $M(1)=0$ is equivalent to $e^{i\phi(\rho)}$ being dense in $\mathbb{T}$, and $M(n)=0$, $n\in\mathbb{N}$, to them being uniformly distributed. Any advice, insights and corrections would be most welcome.
EDIT 1: the arguments are $2\theta x$ corrected.
EDIT 2: $O(T)$ inside the bracket corrected.
This question arose from an earlier one and the MO user's useful answers there: What are the values of the derivative of Riemann's zeta function at the known nontrivial zeros? (which is not a prerequisite for this version).
As usual, let $\rho=\beta+i\gamma$ denote the nontrivial zeros of $\zeta(s)$. Essentially, the question is this:
How are the numbers $e^{i\phi(\rho)}=\lim_{s\rightarrow\rho}\zeta(s)/\overline{\zeta(s)}$ distributed in $\mathbb{T}$?
Of course $\phi\in\mathbb{R}$ and the limit could be replaced by quotients of derivatives if we knew the order of the zeros (in which case $\phi$ is just twice their argument there), but I don't think it is necessary to know that any way. Moreover, I think the answer is independent of RH. My reasoning is as follows.
For $x\in\mathbb{R}$, one may consider the meanvalue:
$$M(x)=\lim_{T\rightarrow\infty}\frac{1}{N(T)}\sum_{0 <\gamma < T}e^{i\phi(\rho)x}.$$
Let $t>0$ and $\theta(t)$ denote the RiemannSiegel thetafunction. Firstly, by separating the sums over zeros on and off the line $\beta=1/2$ and, secondly, by the BohrLandau theorem (that the number of zeros with $\beta>1/2$ is $O(T)$), one gets
$$M(x)=\lim_{T\rightarrow\infty}\frac{1}{N(T)}\left(\sum_{0 <\gamma < T :\beta=1/2}e^{i\phi(\rho)x}+O\left(T\right)\right)$$
$$=\lim_{T\rightarrow\infty}\frac{\pi}{\theta(T)}\sum_{0 <\gamma < T :\beta=1/2}e^{i2\theta(\gamma)x}.$$
If my reasoning is correct, $M(1)=0$ is equivalent to $e^{i\phi(\rho)}$ being dense in $\mathbb{T}$, and $M(n)=0$, $n\in\mathbb{N}$, to them being uniformly distributed. Any advice, insights and corrections would be most welcome.
EDIT 1: the arguments are $2\theta x$ corrected.
EDIT 2: $O(T)$ inside the bracket corrected.



3


edited Nov 13 '11 at 22:18

This question arose from an earlier one and the MO user's useful answers there: http://mathoverflow.net/questions/80710/whatarethevaluesofthederivativeofriemannszetafunctionattheknownnon/80783#80783 (which is not a prerequisite for this version).
As usual, let $\rho=\beta+i\gamma$ denote the nontrivial zeros of $\zeta(s)$. Essentially, the question is this:
How are the numbers $e^{i\phi(\rho)}=\lim_{s\rightarrow\rho}\zeta(s)/\overline{\zeta(s)}$ distributed in $\mathbb{T}$?
Of course $\phi\in\mathbb{R}$ and the limit could be replaced by quotients of derivatives if we knew the order of the zeros (in which case $\phi$ is just twice their argument there), but I don't think it is necessary to know that any way. Moreover, I think the answer is independent of RH. My reasoning is as follows.
For $x\in\mathbb{R}$, one may consider the meanvalue:
$$M(x)=\lim_{T\rightarrow\infty}\frac{1}{N(T)}\sum_{0 <\gamma < T}e^{i\phi(\rho)x}.$$
Let $t>0$ and $\theta(t)$ denote the RiemannSiegel thetafunction. Firstly, by separating the sums over zeros on and off the line $\beta=1/2$ and, secondly, by the BohrLandau theorem (that the number of zeros with $\beta>1/2$ is $O(T)$), one gets
$$M(x)=\lim_{T\rightarrow\infty}\frac{1}{N(T)}\left(\sum_{0 <\gamma < T :\beta=1/2}e^{i\phi(\rho)x}+O\left(\frac{1}{\log T}\right)\right)$$$$M(x)=\lim_{T\rightarrow\infty}\frac{1}{N(T)}\left(\sum_{0 <\gamma < T :\beta=1/2}e^{i\phi(\rho)x}+O\left(T\right)\right)$$
$$=\lim_{T\rightarrow\infty}\frac{\pi}{\theta(T)}\sum_{0 <\gamma < T :\beta=1/2}e^{i2\theta(\gamma)x}.$$
If my reasoning is correct, $M(1)=0$ is equivalent to $e^{i\phi(\rho)}$ being dense in $\mathbb{T}$, and $M(n)=0$, $n\in\mathbb{N}$, to them being uniformly distributed. Any advice, insights and corrections would be most welcome.
EDIT 1: the arguments are $2\theta x$ corrected.
EDIT 2: $O(T)$ inside the bracket corrected.
This question arose from an earlier one and the MO user's useful answers there: http://mathoverflow.net/questions/80710/whatarethevaluesofthederivativeofriemannszetafunctionattheknownnon/80783#80783 (which is not a prerequisite for this version).
As usual, let $\rho=\beta+i\gamma$ denote the nontrivial zeros of $\zeta(s)$. Essentially, the question is this:
How are the numbers $e^{i\phi(\rho)}=\lim_{s\rightarrow\rho}\zeta(s)/\overline{\zeta(s)}$ distributed in $\mathbb{T}$?
Of course $\phi\in\mathbb{R}$ and the limit could be replaced by quotients of derivatives if we knew the order of the zeros (in which case $\phi$ is just twice their argument there), but I don't think it is necessary to know that any way. Moreover, I think the answer is independent of RH. My reasoning is as follows.
For $x\in\mathbb{R}$, one may consider the meanvalue:
$$M(x)=\lim_{T\rightarrow\infty}\frac{1}{N(T)}\sum_{0 <\gamma < T}e^{i\phi(\rho)x}.$$
Let $t>0$ and $\theta(t)$ denote the RiemannSiegel thetafunction. Firstly, by separating the sums over zeros on and off the line $\beta=1/2$ and, secondly, by the BohrLandau theorem (that the number of zeros with $\beta>1/2$ is $O(T)$), one gets
$$M(x)=\lim_{T\rightarrow\infty}\frac{1}{N(T)}\left(\sum_{0 <\gamma < T :\beta=1/2}e^{i\phi(\rho)x}+O\left(\frac{1}{\log T}\right)\right)$$
$$=\lim_{T\rightarrow\infty}\frac{\pi}{\theta(T)}\sum_{0 <\gamma < T :\beta=1/2}e^{i2\theta(\gamma)x}.$$
If my reasoning is correct, $M(1)=0$ is equivalent to $e^{i\phi(\rho)}$ being dense in $\mathbb{T}$, and $M(n)=0$, $n\in\mathbb{N}$, to them being uniformly distributed. Any advice, insights and corrections would be most welcome.
EDIT 1: the arguments are $2\theta x$ corrected.
This question arose from an earlier one and the MO user's useful answers there: http://mathoverflow.net/questions/80710/whatarethevaluesofthederivativeofriemannszetafunctionattheknownnon/80783#80783 (which is not a prerequisite for this version).
As usual, let $\rho=\beta+i\gamma$ denote the nontrivial zeros of $\zeta(s)$. Essentially, the question is this:
How are the numbers $e^{i\phi(\rho)}=\lim_{s\rightarrow\rho}\zeta(s)/\overline{\zeta(s)}$ distributed in $\mathbb{T}$?
Of course $\phi\in\mathbb{R}$ and the limit could be replaced by quotients of derivatives if we knew the order of the zeros (in which case $\phi$ is just twice their argument there), but I don't think it is necessary to know that any way. Moreover, I think the answer is independent of RH. My reasoning is as follows.
For $x\in\mathbb{R}$, one may consider the meanvalue:
$$M(x)=\lim_{T\rightarrow\infty}\frac{1}{N(T)}\sum_{0 <\gamma < T}e^{i\phi(\rho)x}.$$
Let $t>0$ and $\theta(t)$ denote the RiemannSiegel thetafunction. Firstly, by separating the sums over zeros on and off the line $\beta=1/2$ and, secondly, by the BohrLandau theorem (that the number of zeros with $\beta>1/2$ is $O(T)$), one gets
$$M(x)=\lim_{T\rightarrow\infty}\frac{1}{N(T)}\left(\sum_{0 <\gamma < T :\beta=1/2}e^{i\phi(\rho)x}+O\left(T\right)\right)$$
$$=\lim_{T\rightarrow\infty}\frac{\pi}{\theta(T)}\sum_{0 <\gamma < T :\beta=1/2}e^{i2\theta(\gamma)x}.$$
If my reasoning is correct, $M(1)=0$ is equivalent to $e^{i\phi(\rho)}$ being dense in $\mathbb{T}$, and $M(n)=0$, $n\in\mathbb{N}$, to them being uniformly distributed. Any advice, insights and corrections would be most welcome.
EDIT 1: the arguments are $2\theta x$ corrected.
EDIT 2: $O(T)$ inside the bracket corrected.



2


edited Nov 13 '11 at 20:49

This question arose from an earlier one and the MO user's useful answers there: http://mathoverflow.net/questions/80710/whatarethevaluesofthederivativeofriemannszetafunctionattheknownnon/80783#80783 (which is not a prerequisite for this version).
As usual, let $\rho=\beta+i\gamma$ denote the nontrivial zeros of $\zeta(s)$. Essentially, the question is this:
How are the numbers $e^{i\phi(\rho)}=\lim_{s\rightarrow\rho}\zeta(s)/\overline{\zeta(s)}$ distributed in $\mathbb{T}$?
Of course $\phi\in\mathbb{R}$ and the limit could be replaced by quotients of derivatives if we knew the order of the zeros (in which case $\phi$ is just twice their argument there), but I don't think it is necessary to know that any way. Moreover, I think the answer is independent of RH. My reasoning is as follows.
For $x\in\mathbb{R}$, one may consider the meanvalue:
$$M(x)=\lim_{T\rightarrow\infty}\frac{1}{N(T)}\sum_{0 <\gamma < T}e^{i\phi(\rho)x}.$$
Let $t>0$ and $\theta(t)$ denote the RiemannSiegel thetafunction. Firstly, by separating the sums over zeros on and off the line $\beta=1/2$ and, secondly, by the BohrLandau theorem (that the number of zeros with $\beta>1/2$ is $O(T)$), one gets
$$M(x)=\lim_{T\rightarrow\infty}\frac{1}{N(T)}\left(\sum_{0 <\gamma < T :\beta=1/2}e^{i\phi(\rho)x}+O\left(\frac{1}{\log T}\right)\right)$$
$$=\lim_{T\rightarrow\infty}\frac{\pi}{\theta(T)}\sum_{0 <\gamma < T :\beta=1/2}e^{i\theta(\gamma)x}.$$$$=\lim_{T\rightarrow\infty}\frac{\pi}{\theta(T)}\sum_{0 <\gamma < T :\beta=1/2}e^{i2\theta(\gamma)x}.$$
If my reasoning is correct, $M(1)=0$ is equivalent to $e^{i\phi(\rho)}$ being dense in $\mathbb{T}$, and $M(n)=0$, $n\in\mathbb{N}$, to them being uniformly distributed. Any advice, insights and corrections would be most welcome.
EDIT 1: the arguments are $2\theta x$ corrected.
This question arose from an earlier one and the MO user's useful answers there: http://mathoverflow.net/questions/80710/whatarethevaluesofthederivativeofriemannszetafunctionattheknownnon/80783#80783 (which is not a prerequisite for this version).
As usual, let $\rho=\beta+i\gamma$ denote the nontrivial zeros of $\zeta(s)$. Essentially, the question is this:
How are the numbers $e^{i\phi(\rho)}=\lim_{s\rightarrow\rho}\zeta(s)/\overline{\zeta(s)}$ distributed in $\mathbb{T}$?
Of course $\phi\in\mathbb{R}$ and the limit could be replaced by quotients of derivatives if we knew the order of the zeros (in which case $\phi$ is just their argument), but I don't think it is necessary to know that any way. Moreover, I think the answer is independent of RH. My reasoning is as follows.
For $x\in\mathbb{R}$, one may consider the meanvalue:
$$M(x)=\lim_{T\rightarrow\infty}\frac{1}{N(T)}\sum_{0 <\gamma < T}e^{i\phi(\rho)x}.$$
Let $t>0$ and $\theta(t)$ denote the RiemannSiegel thetafunction. Firstly, by separating the sums over zeros on and off the line $\beta=1/2$ and, secondly, by the BohrLandau theorem (that the number of zeros with $\beta>1/2$ is $O(T)$), one gets
$$M(x)=\lim_{T\rightarrow\infty}\frac{1}{N(T)}\left(\sum_{0 <\gamma < T :\beta=1/2}e^{i\phi(\rho)x}+O\left(\frac{1}{\log T}\right)\right)$$
$$=\lim_{T\rightarrow\infty}\frac{\pi}{\theta(T)}\sum_{0 <\gamma < T :\beta=1/2}e^{i\theta(\gamma)x}.$$
If my reasoning is correct, $M(1)=0$ is equivalent to $e^{i\phi(\rho)}$ being dense in $\mathbb{T}$, and $M(n)=0$, $n\in\mathbb{N}$, to them being uniformly distributed. Any advice, insights and corrections would be most welcome.
This question arose from an earlier one and the MO user's useful answers there: http://mathoverflow.net/questions/80710/whatarethevaluesofthederivativeofriemannszetafunctionattheknownnon/80783#80783 (which is not a prerequisite for this version).
As usual, let $\rho=\beta+i\gamma$ denote the nontrivial zeros of $\zeta(s)$. Essentially, the question is this:
How are the numbers $e^{i\phi(\rho)}=\lim_{s\rightarrow\rho}\zeta(s)/\overline{\zeta(s)}$ distributed in $\mathbb{T}$?
Of course $\phi\in\mathbb{R}$ and the limit could be replaced by quotients of derivatives if we knew the order of the zeros (in which case $\phi$ is just twice their argument there), but I don't think it is necessary to know that any way. Moreover, I think the answer is independent of RH. My reasoning is as follows.
For $x\in\mathbb{R}$, one may consider the meanvalue:
$$M(x)=\lim_{T\rightarrow\infty}\frac{1}{N(T)}\sum_{0 <\gamma < T}e^{i\phi(\rho)x}.$$
Let $t>0$ and $\theta(t)$ denote the RiemannSiegel thetafunction. Firstly, by separating the sums over zeros on and off the line $\beta=1/2$ and, secondly, by the BohrLandau theorem (that the number of zeros with $\beta>1/2$ is $O(T)$), one gets
$$M(x)=\lim_{T\rightarrow\infty}\frac{1}{N(T)}\left(\sum_{0 <\gamma < T :\beta=1/2}e^{i\phi(\rho)x}+O\left(\frac{1}{\log T}\right)\right)$$
$$=\lim_{T\rightarrow\infty}\frac{\pi}{\theta(T)}\sum_{0 <\gamma < T :\beta=1/2}e^{i2\theta(\gamma)x}.$$
If my reasoning is correct, $M(1)=0$ is equivalent to $e^{i\phi(\rho)}$ being dense in $\mathbb{T}$, and $M(n)=0$, $n\in\mathbb{N}$, to them being uniformly distributed. Any advice, insights and corrections would be most welcome.
EDIT 1: the arguments are $2\theta x$ corrected.



1


asked Nov 13 '11 at 19:32


