4 replaced http://mathoverflow.net/ with https://mathoverflow.net/ edited Apr 13 '17 at 12:58 Community♦ 122 silver badges33 bronze badges This question arose from an earlier one and the MO user's useful answers there: http://mathoverflow.net/questions/80710/what-are-the-values-of-the-derivative-of-riemanns-zeta-function-at-the-known-non/80783#80783What are the values of the derivative of Riemann's zeta function at the known non-trivial zeros? (which is not a prerequisite for this version). As usual, let $$\rho=\beta+i\gamma$$ denote the non-trivial 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 mean-value: $$M(x)=\lim_{T\rightarrow\infty}\frac{1}{N(T)}\sum_{0 <\gamma < T}e^{i\phi(\rho)x}.$$ Let $$t>0$$ and $$\theta(t)$$ denote the Riemann-Siegel theta-function. Firstly, by separating the sums over zeros on and off the line $$\beta=1/2$$ and, secondly, by the Bohr-Landau 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/what-are-the-values-of-the-derivative-of-riemanns-zeta-function-at-the-known-non/80783#80783 (which is not a prerequisite for this version). As usual, let $$\rho=\beta+i\gamma$$ denote the non-trivial 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 mean-value: $$M(x)=\lim_{T\rightarrow\infty}\frac{1}{N(T)}\sum_{0 <\gamma < T}e^{i\phi(\rho)x}.$$ Let $$t>0$$ and $$\theta(t)$$ denote the Riemann-Siegel theta-function. Firstly, by separating the sums over zeros on and off the line $$\beta=1/2$$ and, secondly, by the Bohr-Landau 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 non-trivial zeros? (which is not a prerequisite for this version). As usual, let $$\rho=\beta+i\gamma$$ denote the non-trivial 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 mean-value: $$M(x)=\lim_{T\rightarrow\infty}\frac{1}{N(T)}\sum_{0 <\gamma < T}e^{i\phi(\rho)x}.$$ Let $$t>0$$ and $$\theta(t)$$ denote the Riemann-Siegel theta-function. Firstly, by separating the sums over zeros on and off the line $$\beta=1/2$$ and, secondly, by the Bohr-Landau 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 added 34 characters in body edited Nov 13 '11 at 22:18 Kevin Smith 1,32088 silver badges2424 bronze badges This question arose from an earlier one and the MO user's useful answers there: http://mathoverflow.net/questions/80710/what-are-the-values-of-the-derivative-of-riemanns-zeta-function-at-the-known-non/80783#80783 (which is not a prerequisite for this version). As usual, let $$\rho=\beta+i\gamma$$ denote the non-trivial 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 mean-value: $$M(x)=\lim_{T\rightarrow\infty}\frac{1}{N(T)}\sum_{0 <\gamma < T}e^{i\phi(\rho)x}.$$ Let $$t>0$$ and $$\theta(t)$$ denote the Riemann-Siegel theta-function. Firstly, by separating the sums over zeros on and off the line $$\beta=1/2$$ and, secondly, by the Bohr-Landau 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/what-are-the-values-of-the-derivative-of-riemanns-zeta-function-at-the-known-non/80783#80783 (which is not a prerequisite for this version). As usual, let $$\rho=\beta+i\gamma$$ denote the non-trivial 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 mean-value: $$M(x)=\lim_{T\rightarrow\infty}\frac{1}{N(T)}\sum_{0 <\gamma < T}e^{i\phi(\rho)x}.$$ Let $$t>0$$ and $$\theta(t)$$ denote the Riemann-Siegel theta-function. Firstly, by separating the sums over zeros on and off the line $$\beta=1/2$$ and, secondly, by the Bohr-Landau 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/what-are-the-values-of-the-derivative-of-riemanns-zeta-function-at-the-known-non/80783#80783 (which is not a prerequisite for this version). As usual, let $$\rho=\beta+i\gamma$$ denote the non-trivial 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 mean-value: $$M(x)=\lim_{T\rightarrow\infty}\frac{1}{N(T)}\sum_{0 <\gamma < T}e^{i\phi(\rho)x}.$$ Let $$t>0$$ and $$\theta(t)$$ denote the Riemann-Siegel theta-function. Firstly, by separating the sums over zeros on and off the line $$\beta=1/2$$ and, secondly, by the Bohr-Landau 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 added 68 characters in body edited Nov 13 '11 at 20:49 Kevin Smith 1,32088 silver badges2424 bronze badges This question arose from an earlier one and the MO user's useful answers there: http://mathoverflow.net/questions/80710/what-are-the-values-of-the-derivative-of-riemanns-zeta-function-at-the-known-non/80783#80783 (which is not a prerequisite for this version). As usual, let $$\rho=\beta+i\gamma$$ denote the non-trivial 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 mean-value: $$M(x)=\lim_{T\rightarrow\infty}\frac{1}{N(T)}\sum_{0 <\gamma < T}e^{i\phi(\rho)x}.$$ Let $$t>0$$ and $$\theta(t)$$ denote the Riemann-Siegel theta-function. Firstly, by separating the sums over zeros on and off the line $$\beta=1/2$$ and, secondly, by the Bohr-Landau 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/what-are-the-values-of-the-derivative-of-riemanns-zeta-function-at-the-known-non/80783#80783 (which is not a prerequisite for this version). As usual, let $$\rho=\beta+i\gamma$$ denote the non-trivial 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 mean-value: $$M(x)=\lim_{T\rightarrow\infty}\frac{1}{N(T)}\sum_{0 <\gamma < T}e^{i\phi(\rho)x}.$$ Let $$t>0$$ and $$\theta(t)$$ denote the Riemann-Siegel theta-function. Firstly, by separating the sums over zeros on and off the line $$\beta=1/2$$ and, secondly, by the Bohr-Landau 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/what-are-the-values-of-the-derivative-of-riemanns-zeta-function-at-the-known-non/80783#80783 (which is not a prerequisite for this version). As usual, let $$\rho=\beta+i\gamma$$ denote the non-trivial 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 mean-value: $$M(x)=\lim_{T\rightarrow\infty}\frac{1}{N(T)}\sum_{0 <\gamma < T}e^{i\phi(\rho)x}.$$ Let $$t>0$$ and $$\theta(t)$$ denote the Riemann-Siegel theta-function. Firstly, by separating the sums over zeros on and off the line $$\beta=1/2$$ and, secondly, by the Bohr-Landau 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 Kevin Smith 1,32088 silver badges2424 bronze badges