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Dirichlet's Theorem for Turan's Method of Analysis


Variations on the Stirling's formula for $Gamma(z)$Real elliptic curves in the fundamental domain of $Gamma(2)$Using Cauchy's Inequality to prove a function's second derivative is zeroProving a sufficient condition for complex differentiabilityProving that the line integral $int_gamma_2 e^ix^2:mathrmdx$ tends to zeroRelation between Besov and Sobolev spaces (Littlewood-Paley-theory)Verifying my proof of Liouville's theorem?show that a complex function vanishes everywhereConvergence of the multidimensional stationary phase asymptotic expansionBehavior on the Circle of Convergence using Dirichlet's Test













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?










share|cite|improve this question











$endgroup$
















    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?










    share|cite|improve this question











    $endgroup$














      2












      2








      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?










      share|cite|improve this question











      $endgroup$




      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















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Mar 14 at 8:33







      kodlu

















      asked Mar 14 at 7:32









      kodlukodlu

      3,422816




      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











          $endgroup$












          • $begingroup$
            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











          Your Answer





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          1 Answer
          1






          active

          oldest

          votes








          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          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











          $endgroup$












          • $begingroup$
            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
















          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











          $endgroup$












          • $begingroup$
            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














          1












          1








          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











          $endgroup$



          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














          share|cite|improve this answer



          share|cite|improve this answer








          edited Mar 15 at 0:25









          kodlu

          3,422816




          3,422816










          answered Mar 14 at 13:02









          ConradConrad

          1,12345




          1,12345











          • $begingroup$
            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

















          • $begingroup$
            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
















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




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




          1




          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




          $begingroup$
          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




          $begingroup$
          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




          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




          $begingroup$
          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




          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





          $begingroup$
          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


















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