Series of $sin(nx)$ terms that sum up to $0$.












1












$begingroup$


Working in $mathbb{R}$, what sequences $a_n$ satisfy $sum_{n=0}^{infty} a_n sin(nx)=0$ for all $x$, pointwise ?



I've never thought about this and I'm not sure whether I'm not missing something trivial - I would have expected that $a_n = 0$ for all $n$ is the only solution, but I'm not sure I've seen this stated anywhere.






real-analysis fourier-analysis







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Fourier series. They're linearly independent, even with infinitely many terms.
    $endgroup$
    – Matt Samuel
    Jan 5 at 23:07










  • $begingroup$
    But what justifies exchanging the integral and sum order?
    $endgroup$
    – John P
    Jan 6 at 14:31
















1












$begingroup$


Working in $mathbb{R}$, what sequences $a_n$ satisfy $sum_{n=0}^{infty} a_n sin(nx)=0$ for all $x$, pointwise ?



I've never thought about this and I'm not sure whether I'm not missing something trivial - I would have expected that $a_n = 0$ for all $n$ is the only solution, but I'm not sure I've seen this stated anywhere.






real-analysis fourier-analysis







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Fourier series. They're linearly independent, even with infinitely many terms.
    $endgroup$
    – Matt Samuel
    Jan 5 at 23:07










  • $begingroup$
    But what justifies exchanging the integral and sum order?
    $endgroup$
    – John P
    Jan 6 at 14:31














1












1








1


1



$begingroup$


Working in $mathbb{R}$, what sequences $a_n$ satisfy $sum_{n=0}^{infty} a_n sin(nx)=0$ for all $x$, pointwise ?



I've never thought about this and I'm not sure whether I'm not missing something trivial - I would have expected that $a_n = 0$ for all $n$ is the only solution, but I'm not sure I've seen this stated anywhere.






real-analysis fourier-analysis







share|cite|improve this question











$endgroup$




Working in $mathbb{R}$, what sequences $a_n$ satisfy $sum_{n=0}^{infty} a_n sin(nx)=0$ for all $x$, pointwise ?



I've never thought about this and I'm not sure whether I'm not missing something trivial - I would have expected that $a_n = 0$ for all $n$ is the only solution, but I'm not sure I've seen this stated anywhere.






real-analysis fourier-analysis




real-analysis fourier-analysis






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 5 at 22:52









Bernard

124k741117




124k741117










asked Jan 5 at 22:42









John PJohn P

55727




55727








  • 1




    $begingroup$
    Fourier series. They're linearly independent, even with infinitely many terms.
    $endgroup$
    – Matt Samuel
    Jan 5 at 23:07










  • $begingroup$
    But what justifies exchanging the integral and sum order?
    $endgroup$
    – John P
    Jan 6 at 14:31














  • 1




    $begingroup$
    Fourier series. They're linearly independent, even with infinitely many terms.
    $endgroup$
    – Matt Samuel
    Jan 5 at 23:07










  • $begingroup$
    But what justifies exchanging the integral and sum order?
    $endgroup$
    – John P
    Jan 6 at 14:31








1




1




$begingroup$
Fourier series. They're linearly independent, even with infinitely many terms.
$endgroup$
– Matt Samuel
Jan 5 at 23:07




$begingroup$
Fourier series. They're linearly independent, even with infinitely many terms.
$endgroup$
– Matt Samuel
Jan 5 at 23:07












$begingroup$
But what justifies exchanging the integral and sum order?
$endgroup$
– John P
Jan 6 at 14:31




$begingroup$
But what justifies exchanging the integral and sum order?
$endgroup$
– John P
Jan 6 at 14:31










2 Answers
2






active

oldest

votes


















1












$begingroup$

Riemann's uniqueness theorem says that if a trigonometric series converges pointwise to $0$, then its coefficients are all equal to $0$. This was later generalized by Cantor, leading him to the development of set theory.






share|cite|improve this answer









$endgroup$





















    2












    $begingroup$

    It may depend on the meaning of the summation (and, as a consequence, the speed of the decay of $a$).



    To solve all convergence problems, we assume that the series converges normally on $mathbb{R}$, ie $a in ell^1$.



    Let then $f(x)=sum_n{a_nsin(nx)}$. $f$ is continuous.



    I suggest that you compute $int_0^{2pi}{f(x)sin(nx),dx}$ as a function of $a_n$ (using the normal convergence of the series).






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      I'm sorry but I have no idea what sort of approach to take here - exchanging the integral and sum signs would of course solve the problem, but I don't see how the use of this is justified. Uniform convergence would be enough, but I don't see how $sum a_n sin(nx)=0$ pointwise implies that this convergence is necessarily uniform. The general Lebesgue theorem I know concerning this requires some sort of bound on absolute value of the partial sums, or for example that $sum_n int |f_n| < infty$.
      $endgroup$
      – John P
      Jan 6 at 14:29












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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    1












    $begingroup$

    Riemann's uniqueness theorem says that if a trigonometric series converges pointwise to $0$, then its coefficients are all equal to $0$. This was later generalized by Cantor, leading him to the development of set theory.






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      Riemann's uniqueness theorem says that if a trigonometric series converges pointwise to $0$, then its coefficients are all equal to $0$. This was later generalized by Cantor, leading him to the development of set theory.






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        Riemann's uniqueness theorem says that if a trigonometric series converges pointwise to $0$, then its coefficients are all equal to $0$. This was later generalized by Cantor, leading him to the development of set theory.






        share|cite|improve this answer









        $endgroup$



        Riemann's uniqueness theorem says that if a trigonometric series converges pointwise to $0$, then its coefficients are all equal to $0$. This was later generalized by Cantor, leading him to the development of set theory.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 6 at 17:54









        Julián AguirreJulián Aguirre

        69.5k24297




        69.5k24297























            2












            $begingroup$

            It may depend on the meaning of the summation (and, as a consequence, the speed of the decay of $a$).



            To solve all convergence problems, we assume that the series converges normally on $mathbb{R}$, ie $a in ell^1$.



            Let then $f(x)=sum_n{a_nsin(nx)}$. $f$ is continuous.



            I suggest that you compute $int_0^{2pi}{f(x)sin(nx),dx}$ as a function of $a_n$ (using the normal convergence of the series).






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              I'm sorry but I have no idea what sort of approach to take here - exchanging the integral and sum signs would of course solve the problem, but I don't see how the use of this is justified. Uniform convergence would be enough, but I don't see how $sum a_n sin(nx)=0$ pointwise implies that this convergence is necessarily uniform. The general Lebesgue theorem I know concerning this requires some sort of bound on absolute value of the partial sums, or for example that $sum_n int |f_n| < infty$.
              $endgroup$
              – John P
              Jan 6 at 14:29
















            2












            $begingroup$

            It may depend on the meaning of the summation (and, as a consequence, the speed of the decay of $a$).



            To solve all convergence problems, we assume that the series converges normally on $mathbb{R}$, ie $a in ell^1$.



            Let then $f(x)=sum_n{a_nsin(nx)}$. $f$ is continuous.



            I suggest that you compute $int_0^{2pi}{f(x)sin(nx),dx}$ as a function of $a_n$ (using the normal convergence of the series).






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              I'm sorry but I have no idea what sort of approach to take here - exchanging the integral and sum signs would of course solve the problem, but I don't see how the use of this is justified. Uniform convergence would be enough, but I don't see how $sum a_n sin(nx)=0$ pointwise implies that this convergence is necessarily uniform. The general Lebesgue theorem I know concerning this requires some sort of bound on absolute value of the partial sums, or for example that $sum_n int |f_n| < infty$.
              $endgroup$
              – John P
              Jan 6 at 14:29














            2












            2








            2





            $begingroup$

            It may depend on the meaning of the summation (and, as a consequence, the speed of the decay of $a$).



            To solve all convergence problems, we assume that the series converges normally on $mathbb{R}$, ie $a in ell^1$.



            Let then $f(x)=sum_n{a_nsin(nx)}$. $f$ is continuous.



            I suggest that you compute $int_0^{2pi}{f(x)sin(nx),dx}$ as a function of $a_n$ (using the normal convergence of the series).






            share|cite|improve this answer









            $endgroup$



            It may depend on the meaning of the summation (and, as a consequence, the speed of the decay of $a$).



            To solve all convergence problems, we assume that the series converges normally on $mathbb{R}$, ie $a in ell^1$.



            Let then $f(x)=sum_n{a_nsin(nx)}$. $f$ is continuous.



            I suggest that you compute $int_0^{2pi}{f(x)sin(nx),dx}$ as a function of $a_n$ (using the normal convergence of the series).







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Jan 5 at 23:08









            MindlackMindlack

            4,900211




            4,900211












            • $begingroup$
              I'm sorry but I have no idea what sort of approach to take here - exchanging the integral and sum signs would of course solve the problem, but I don't see how the use of this is justified. Uniform convergence would be enough, but I don't see how $sum a_n sin(nx)=0$ pointwise implies that this convergence is necessarily uniform. The general Lebesgue theorem I know concerning this requires some sort of bound on absolute value of the partial sums, or for example that $sum_n int |f_n| < infty$.
              $endgroup$
              – John P
              Jan 6 at 14:29


















            • $begingroup$
              I'm sorry but I have no idea what sort of approach to take here - exchanging the integral and sum signs would of course solve the problem, but I don't see how the use of this is justified. Uniform convergence would be enough, but I don't see how $sum a_n sin(nx)=0$ pointwise implies that this convergence is necessarily uniform. The general Lebesgue theorem I know concerning this requires some sort of bound on absolute value of the partial sums, or for example that $sum_n int |f_n| < infty$.
              $endgroup$
              – John P
              Jan 6 at 14:29
















            $begingroup$
            I'm sorry but I have no idea what sort of approach to take here - exchanging the integral and sum signs would of course solve the problem, but I don't see how the use of this is justified. Uniform convergence would be enough, but I don't see how $sum a_n sin(nx)=0$ pointwise implies that this convergence is necessarily uniform. The general Lebesgue theorem I know concerning this requires some sort of bound on absolute value of the partial sums, or for example that $sum_n int |f_n| < infty$.
            $endgroup$
            – John P
            Jan 6 at 14:29




            $begingroup$
            I'm sorry but I have no idea what sort of approach to take here - exchanging the integral and sum signs would of course solve the problem, but I don't see how the use of this is justified. Uniform convergence would be enough, but I don't see how $sum a_n sin(nx)=0$ pointwise implies that this convergence is necessarily uniform. The general Lebesgue theorem I know concerning this requires some sort of bound on absolute value of the partial sums, or for example that $sum_n int |f_n| < infty$.
            $endgroup$
            – John P
            Jan 6 at 14:29


















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