Find a sequence $a_f(k)$ so that $sum_k=1^infty a_f(k) operatornamesinc (pi k f) = 0$.

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I am having problems either finding such a sequence or showing that one does not exist. Here is the situation.



Take $f$ to be some positive real number which is not an integer.



Recall that $operatornamesinc(x)$ is $sin(x) / x$ if $x neq 0$, and $operatornamesinc(x) := 1$.



I would like to show that no non-zero series $a_f(k)_k=1^infty subset mathbbR$ exists which satisfies
$$sum_k=1^infty a_f(k) operatornamesinc (pi k f) = 0.$$
(I use non-zero to simply mean that there is at least one non-zero element of the sequence.)



If it is possible to do this, then I would perhaps like to say that for $N$ a large positive integer, there is no sequence of numbers $a_f(k)_k=1^N subset mathbbR$ which satisfies
$$sum_k=1^N a_f(k) operatornamesinc (pi k f) = 0.$$
If this is also possible, then I would maybe add some conditions like $f$ being irrational so that no such sequence would exist.







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    up vote
    1
    down vote

    favorite












    I am having problems either finding such a sequence or showing that one does not exist. Here is the situation.



    Take $f$ to be some positive real number which is not an integer.



    Recall that $operatornamesinc(x)$ is $sin(x) / x$ if $x neq 0$, and $operatornamesinc(x) := 1$.



    I would like to show that no non-zero series $a_f(k)_k=1^infty subset mathbbR$ exists which satisfies
    $$sum_k=1^infty a_f(k) operatornamesinc (pi k f) = 0.$$
    (I use non-zero to simply mean that there is at least one non-zero element of the sequence.)



    If it is possible to do this, then I would perhaps like to say that for $N$ a large positive integer, there is no sequence of numbers $a_f(k)_k=1^N subset mathbbR$ which satisfies
    $$sum_k=1^N a_f(k) operatornamesinc (pi k f) = 0.$$
    If this is also possible, then I would maybe add some conditions like $f$ being irrational so that no such sequence would exist.







    share|cite|improve this question






















      up vote
      1
      down vote

      favorite









      up vote
      1
      down vote

      favorite











      I am having problems either finding such a sequence or showing that one does not exist. Here is the situation.



      Take $f$ to be some positive real number which is not an integer.



      Recall that $operatornamesinc(x)$ is $sin(x) / x$ if $x neq 0$, and $operatornamesinc(x) := 1$.



      I would like to show that no non-zero series $a_f(k)_k=1^infty subset mathbbR$ exists which satisfies
      $$sum_k=1^infty a_f(k) operatornamesinc (pi k f) = 0.$$
      (I use non-zero to simply mean that there is at least one non-zero element of the sequence.)



      If it is possible to do this, then I would perhaps like to say that for $N$ a large positive integer, there is no sequence of numbers $a_f(k)_k=1^N subset mathbbR$ which satisfies
      $$sum_k=1^N a_f(k) operatornamesinc (pi k f) = 0.$$
      If this is also possible, then I would maybe add some conditions like $f$ being irrational so that no such sequence would exist.







      share|cite|improve this question












      I am having problems either finding such a sequence or showing that one does not exist. Here is the situation.



      Take $f$ to be some positive real number which is not an integer.



      Recall that $operatornamesinc(x)$ is $sin(x) / x$ if $x neq 0$, and $operatornamesinc(x) := 1$.



      I would like to show that no non-zero series $a_f(k)_k=1^infty subset mathbbR$ exists which satisfies
      $$sum_k=1^infty a_f(k) operatornamesinc (pi k f) = 0.$$
      (I use non-zero to simply mean that there is at least one non-zero element of the sequence.)



      If it is possible to do this, then I would perhaps like to say that for $N$ a large positive integer, there is no sequence of numbers $a_f(k)_k=1^N subset mathbbR$ which satisfies
      $$sum_k=1^N a_f(k) operatornamesinc (pi k f) = 0.$$
      If this is also possible, then I would maybe add some conditions like $f$ being irrational so that no such sequence would exist.









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      asked Aug 9 at 19:29









      Open Season

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          As a consequence of the Poisson summation formula, for any $alphainmathbbRsetminusmathbbZ$ we have
          $$ sum_kgeq 1textsinc(pi k alpha)=frac12left[-1+sum_kinmathbbZtextsinc(pi k alpha)right]=frac12left[-1+sum_sinmathbbZfractextsign(alpha-2s)+textsign(alpha+2s)2alpharight] $$
          such that for any $alphainmathbbR^+setminusmathbbZ$ we get
          $$ sum_kgeq 1textsinc(pi kalpha) = frac12left[-1+frac2lfloor alpha/2rfloor +1alpharight]neq 0. $$
          Similarly you may compute in explicit terms $sum_kgeq 1textsinc^2(pi k alpha)$. This disproves your claim for
          $$ a_f(k) = 1+ C_f,textsinc(pi k f).$$






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            up vote
            2
            down vote



            accepted










            As a consequence of the Poisson summation formula, for any $alphainmathbbRsetminusmathbbZ$ we have
            $$ sum_kgeq 1textsinc(pi k alpha)=frac12left[-1+sum_kinmathbbZtextsinc(pi k alpha)right]=frac12left[-1+sum_sinmathbbZfractextsign(alpha-2s)+textsign(alpha+2s)2alpharight] $$
            such that for any $alphainmathbbR^+setminusmathbbZ$ we get
            $$ sum_kgeq 1textsinc(pi kalpha) = frac12left[-1+frac2lfloor alpha/2rfloor +1alpharight]neq 0. $$
            Similarly you may compute in explicit terms $sum_kgeq 1textsinc^2(pi k alpha)$. This disproves your claim for
            $$ a_f(k) = 1+ C_f,textsinc(pi k f).$$






            share|cite|improve this answer
























              up vote
              2
              down vote



              accepted










              As a consequence of the Poisson summation formula, for any $alphainmathbbRsetminusmathbbZ$ we have
              $$ sum_kgeq 1textsinc(pi k alpha)=frac12left[-1+sum_kinmathbbZtextsinc(pi k alpha)right]=frac12left[-1+sum_sinmathbbZfractextsign(alpha-2s)+textsign(alpha+2s)2alpharight] $$
              such that for any $alphainmathbbR^+setminusmathbbZ$ we get
              $$ sum_kgeq 1textsinc(pi kalpha) = frac12left[-1+frac2lfloor alpha/2rfloor +1alpharight]neq 0. $$
              Similarly you may compute in explicit terms $sum_kgeq 1textsinc^2(pi k alpha)$. This disproves your claim for
              $$ a_f(k) = 1+ C_f,textsinc(pi k f).$$






              share|cite|improve this answer






















                up vote
                2
                down vote



                accepted







                up vote
                2
                down vote



                accepted






                As a consequence of the Poisson summation formula, for any $alphainmathbbRsetminusmathbbZ$ we have
                $$ sum_kgeq 1textsinc(pi k alpha)=frac12left[-1+sum_kinmathbbZtextsinc(pi k alpha)right]=frac12left[-1+sum_sinmathbbZfractextsign(alpha-2s)+textsign(alpha+2s)2alpharight] $$
                such that for any $alphainmathbbR^+setminusmathbbZ$ we get
                $$ sum_kgeq 1textsinc(pi kalpha) = frac12left[-1+frac2lfloor alpha/2rfloor +1alpharight]neq 0. $$
                Similarly you may compute in explicit terms $sum_kgeq 1textsinc^2(pi k alpha)$. This disproves your claim for
                $$ a_f(k) = 1+ C_f,textsinc(pi k f).$$






                share|cite|improve this answer












                As a consequence of the Poisson summation formula, for any $alphainmathbbRsetminusmathbbZ$ we have
                $$ sum_kgeq 1textsinc(pi k alpha)=frac12left[-1+sum_kinmathbbZtextsinc(pi k alpha)right]=frac12left[-1+sum_sinmathbbZfractextsign(alpha-2s)+textsign(alpha+2s)2alpharight] $$
                such that for any $alphainmathbbR^+setminusmathbbZ$ we get
                $$ sum_kgeq 1textsinc(pi kalpha) = frac12left[-1+frac2lfloor alpha/2rfloor +1alpharight]neq 0. $$
                Similarly you may compute in explicit terms $sum_kgeq 1textsinc^2(pi k alpha)$. This disproves your claim for
                $$ a_f(k) = 1+ C_f,textsinc(pi k f).$$







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Aug 9 at 20:28









                Jack D'Aurizio♦

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