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Dirichlet's Theorem for Turan's Method of Analysis

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2

$begingroup$

This question concerns a special case of Turan's power sum, which looks reasonably straightforward.

Let $S(N,nu)=sum_n=1^N z_n^nu,$ where all the $z_n$ are on the complex unit circle. Then, according to Gonek's paper A Note on Turan's Method, using Dirichlet's theorem on uniform approximation, one can show that there is some $nu in 1,2,ldots,6^N,$ such that
$$
left Vert nu arg left(fracz_n2piright) rightVert leq frac16,quad 1leq nleq N,
$$
where $||x||$ is the nearest distance from $x$ to an integer. This immediately implies that
$$
max_1leq nu leq 6^N |S(v,N)|geq fracS(0,N)2=fracN2.
$$
How can one show this?

complex-analysis fourier-analysis

share|cite|improve this question

edited Mar 14 at 8:33

kodlu

asked Mar 14 at 7:32

kodlukodlu

3,422816

$endgroup$

add a comment | 

2

$begingroup$

This question concerns a special case of Turan's power sum, which looks reasonably straightforward.

Let $S(N,nu)=sum_n=1^N z_n^nu,$ where all the $z_n$ are on the complex unit circle. Then, according to Gonek's paper A Note on Turan's Method, using Dirichlet's theorem on uniform approximation, one can show that there is some $nu in 1,2,ldots,6^N,$ such that
$$
left Vert nu arg left(fracz_n2piright) rightVert leq frac16,quad 1leq nleq N,
$$
where $||x||$ is the nearest distance from $x$ to an integer. This immediately implies that
$$
max_1leq nu leq 6^N |S(v,N)|geq fracS(0,N)2=fracN2.
$$
How can one show this?

complex-analysis fourier-analysis

share|cite|improve this question

edited Mar 14 at 8:33

kodlu

asked Mar 14 at 7:32

kodlukodlu

3,422816

$endgroup$

add a comment | 

$begingroup$

This question concerns a special case of Turan's power sum, which looks reasonably straightforward.

Let $S(N,nu)=sum_n=1^N z_n^nu,$ where all the $z_n$ are on the complex unit circle. Then, according to Gonek's paper A Note on Turan's Method, using Dirichlet's theorem on uniform approximation, one can show that there is some $nu in 1,2,ldots,6^N,$ such that
$$
left Vert nu arg left(fracz_n2piright) rightVert leq frac16,quad 1leq nleq N,
$$
where $||x||$ is the nearest distance from $x$ to an integer. This immediately implies that
$$
max_1leq nu leq 6^N |S(v,N)|geq fracS(0,N)2=fracN2.
$$
How can one show this?

complex-analysis fourier-analysis

share|cite|improve this question

edited Mar 14 at 8:33

kodlu

asked Mar 14 at 7:32

kodlukodlu

3,422816

$endgroup$

share|cite|improve this question

edited Mar 14 at 8:33

kodlu

asked Mar 14 at 7:32

kodlukodlu

3,422816

share|cite|improve this question

edited Mar 14 at 8:33

kodlu

asked Mar 14 at 7:32

kodlukodlu

3,422816

asked Mar 14 at 7:32

kodlukodlu

3,422816

asked Mar 14 at 7:32

kodlukodlu

3,422816

1 Answer
1

active

oldest

votes

1

$begingroup$

The statement you want is this:

Given $𝑁$ reals $𝑎_𝑘$ (here $𝑎_𝑘=arg fracz_k2pi$), a positive integer $q$ (here 6), we can find $1leq nu leq q^N,nu$ integral, s.t $Vert v a_kVertleq frac1q$ and the proof uses the standard $N$ dimensional unit cube, the $q^𝑁+1$ points $(𝑚𝑎_1,...𝑚𝑎_𝑁)$,$0leq 𝑚leq 𝑞^𝑁$ modulo 1, so in the cube (by using the pigeonhole principle when you divide each edge into $𝑞$ parts) you have $𝑞^𝑁$ compartments.

Armed with this result, you use the inequality $|arg(w_k)| leq theta < fracpi2$, where $arg$ is taken in $[-pi, pi]$, implies $Rew_k geq |w_k|cos theta$, so if say $|w_k|=1$, $|Sigmaw_k| geq |ReSigmaw_k| geq Ncos theta $, where $N$ is the number of terms in the sum

$Vert frac12piargz_n^nuVert = Vertnu arg left(fracz_n2piright)Vert$, so we can take $|argz_n^nu| leq fracpi3, cos (argz_n^nu) geq frac12$ and apply the above to deduce the inequality for precisely the $nu$ given by Dirichlet which is bounded as noted

share|cite|improve this answer

edited Mar 15 at 0:25

kodlu

3,422816

answered Mar 14 at 13:02

ConradConrad

1,12345

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  thanks. I am still unsure where exactly $6^N$ is needed?
  $endgroup$
  – kodlu
  Mar 14 at 20:08
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Dirichlet theorem that implies the existence of $nu$ with that strong property about arguments requires it
  $endgroup$
  – Conrad
  Mar 14 at 20:47
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  thanks again for your patience. That the part I am unsure about. Could you provide a reference or a link to a good discussion of this if it's too involved to incorporate into the answer.
  $endgroup$
  – kodlu
  Mar 14 at 21:15
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  There is a good discussion in titchmarsh book on Riemann zeta - a pdf was linked here at some point - in chapter 8 about omega theorems and how dirichlet bound even exponential as it is allows more precise results while kronecker approximation which is needed for the inverse of zeta has no such
  $endgroup$
  – Conrad
  Mar 14 at 21:34
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  The statement you want is this: Given $N$ reals $a_k$ (here $a_k=argfracz_k2pi$), a positive integer $q$ (here $6$), we can find $1 leq nu leq q^N, nu$ integral, s.t $||nu a_k|| leq frac1q$ and the proof uses the standard $N$ dimensional unit cube, the $q^N+1$ points $(ma_1,...ma_N), 0 leq m leq q^N$ modulo $1$, so in the cube, and the pigeonhole principle when you divide each edge in $q$ parts, so you have $q^N$ compartments
  $endgroup$
  – Conrad
  Mar 14 at 23:01
  
  

  
  
  
  

 | 
show 3 more comments

Your Answer

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1

$begingroup$

The statement you want is this:

Given $𝑁$ reals $𝑎_𝑘$ (here $𝑎_𝑘=arg fracz_k2pi$), a positive integer $q$ (here 6), we can find $1leq nu leq q^N,nu$ integral, s.t $Vert v a_kVertleq frac1q$ and the proof uses the standard $N$ dimensional unit cube, the $q^𝑁+1$ points $(𝑚𝑎_1,...𝑚𝑎_𝑁)$,$0leq 𝑚leq 𝑞^𝑁$ modulo 1, so in the cube (by using the pigeonhole principle when you divide each edge into $𝑞$ parts) you have $𝑞^𝑁$ compartments.

Armed with this result, you use the inequality $|arg(w_k)| leq theta < fracpi2$, where $arg$ is taken in $[-pi, pi]$, implies $Rew_k geq |w_k|cos theta$, so if say $|w_k|=1$, $|Sigmaw_k| geq |ReSigmaw_k| geq Ncos theta $, where $N$ is the number of terms in the sum

$Vert frac12piargz_n^nuVert = Vertnu arg left(fracz_n2piright)Vert$, so we can take $|argz_n^nu| leq fracpi3, cos (argz_n^nu) geq frac12$ and apply the above to deduce the inequality for precisely the $nu$ given by Dirichlet which is bounded as noted

share|cite|improve this answer

edited Mar 15 at 0:25

kodlu

3,422816

answered Mar 14 at 13:02

ConradConrad

1,12345

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  thanks. I am still unsure where exactly $6^N$ is needed?
  $endgroup$
  – kodlu
  Mar 14 at 20:08
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Dirichlet theorem that implies the existence of $nu$ with that strong property about arguments requires it
  $endgroup$
  – Conrad
  Mar 14 at 20:47
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  thanks again for your patience. That the part I am unsure about. Could you provide a reference or a link to a good discussion of this if it's too involved to incorporate into the answer.
  $endgroup$
  – kodlu
  Mar 14 at 21:15
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  There is a good discussion in titchmarsh book on Riemann zeta - a pdf was linked here at some point - in chapter 8 about omega theorems and how dirichlet bound even exponential as it is allows more precise results while kronecker approximation which is needed for the inverse of zeta has no such
  $endgroup$
  – Conrad
  Mar 14 at 21:34
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  The statement you want is this: Given $N$ reals $a_k$ (here $a_k=argfracz_k2pi$), a positive integer $q$ (here $6$), we can find $1 leq nu leq q^N, nu$ integral, s.t $||nu a_k|| leq frac1q$ and the proof uses the standard $N$ dimensional unit cube, the $q^N+1$ points $(ma_1,...ma_N), 0 leq m leq q^N$ modulo $1$, so in the cube, and the pigeonhole principle when you divide each edge in $q$ parts, so you have $q^N$ compartments
  $endgroup$
  – Conrad
  Mar 14 at 23:01
  
  

  
  
  
  

 | 
show 3 more comments

1

$begingroup$

The statement you want is this:

Given $𝑁$ reals $𝑎_𝑘$ (here $𝑎_𝑘=arg fracz_k2pi$), a positive integer $q$ (here 6), we can find $1leq nu leq q^N,nu$ integral, s.t $Vert v a_kVertleq frac1q$ and the proof uses the standard $N$ dimensional unit cube, the $q^𝑁+1$ points $(𝑚𝑎_1,...𝑚𝑎_𝑁)$,$0leq 𝑚leq 𝑞^𝑁$ modulo 1, so in the cube (by using the pigeonhole principle when you divide each edge into $𝑞$ parts) you have $𝑞^𝑁$ compartments.

Armed with this result, you use the inequality $|arg(w_k)| leq theta < fracpi2$, where $arg$ is taken in $[-pi, pi]$, implies $Rew_k geq |w_k|cos theta$, so if say $|w_k|=1$, $|Sigmaw_k| geq |ReSigmaw_k| geq Ncos theta $, where $N$ is the number of terms in the sum

$Vert frac12piargz_n^nuVert = Vertnu arg left(fracz_n2piright)Vert$, so we can take $|argz_n^nu| leq fracpi3, cos (argz_n^nu) geq frac12$ and apply the above to deduce the inequality for precisely the $nu$ given by Dirichlet which is bounded as noted

share|cite|improve this answer

edited Mar 15 at 0:25

kodlu

3,422816

answered Mar 14 at 13:02

ConradConrad

1,12345

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  thanks. I am still unsure where exactly $6^N$ is needed?
  $endgroup$
  – kodlu
  Mar 14 at 20:08
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Dirichlet theorem that implies the existence of $nu$ with that strong property about arguments requires it
  $endgroup$
  – Conrad
  Mar 14 at 20:47
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  thanks again for your patience. That the part I am unsure about. Could you provide a reference or a link to a good discussion of this if it's too involved to incorporate into the answer.
  $endgroup$
  – kodlu
  Mar 14 at 21:15
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  There is a good discussion in titchmarsh book on Riemann zeta - a pdf was linked here at some point - in chapter 8 about omega theorems and how dirichlet bound even exponential as it is allows more precise results while kronecker approximation which is needed for the inverse of zeta has no such
  $endgroup$
  – Conrad
  Mar 14 at 21:34
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  The statement you want is this: Given $N$ reals $a_k$ (here $a_k=argfracz_k2pi$), a positive integer $q$ (here $6$), we can find $1 leq nu leq q^N, nu$ integral, s.t $||nu a_k|| leq frac1q$ and the proof uses the standard $N$ dimensional unit cube, the $q^N+1$ points $(ma_1,...ma_N), 0 leq m leq q^N$ modulo $1$, so in the cube, and the pigeonhole principle when you divide each edge in $q$ parts, so you have $q^N$ compartments
  $endgroup$
  – Conrad
  Mar 14 at 23:01
  
  

  
  
  
  

 | 
show 3 more comments

$begingroup$

The statement you want is this:

Given $𝑁$ reals $𝑎_𝑘$ (here $𝑎_𝑘=arg fracz_k2pi$), a positive integer $q$ (here 6), we can find $1leq nu leq q^N,nu$ integral, s.t $Vert v a_kVertleq frac1q$ and the proof uses the standard $N$ dimensional unit cube, the $q^𝑁+1$ points $(𝑚𝑎_1,...𝑚𝑎_𝑁)$,$0leq 𝑚leq 𝑞^𝑁$ modulo 1, so in the cube (by using the pigeonhole principle when you divide each edge into $𝑞$ parts) you have $𝑞^𝑁$ compartments.

Armed with this result, you use the inequality $|arg(w_k)| leq theta < fracpi2$, where $arg$ is taken in $[-pi, pi]$, implies $Rew_k geq |w_k|cos theta$, so if say $|w_k|=1$, $|Sigmaw_k| geq |ReSigmaw_k| geq Ncos theta $, where $N$ is the number of terms in the sum

$Vert frac12piargz_n^nuVert = Vertnu arg left(fracz_n2piright)Vert$, so we can take $|argz_n^nu| leq fracpi3, cos (argz_n^nu) geq frac12$ and apply the above to deduce the inequality for precisely the $nu$ given by Dirichlet which is bounded as noted

share|cite|improve this answer

edited Mar 15 at 0:25

kodlu

3,422816

answered Mar 14 at 13:02

ConradConrad

1,12345

$endgroup$

share|cite|improve this answer

edited Mar 15 at 0:25

kodlu

3,422816

answered Mar 14 at 13:02

ConradConrad

1,12345

edited Mar 15 at 0:25

kodlu

3,422816

edited Mar 15 at 0:25

kodlu

3,422816

answered Mar 14 at 13:02

ConradConrad

1,12345

answered Mar 14 at 13:02

ConradConrad

1,12345