How to prove that $left|sum_{k=0}^{n-1}sin(2k+1)xright|$ is bounded












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I'm currently studying Fourier series using a textbook which unfortunately provides only partial solutions to its explanations. To show the uniform convergence of a Fourier series I need to prove the assumption from the title above. Any help is very much appreciated.






calculus functional-analysis fourier-analysis







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  • 2




    $begingroup$
    $|sum_{k=0}^{n-1}sin(2k+1)x| le sum_{k=0}^{n-1}|sin(2k+1)x|$
    $endgroup$
    – N74
    Jan 2 at 11:50


















2












$begingroup$


I'm currently studying Fourier series using a textbook which unfortunately provides only partial solutions to its explanations. To show the uniform convergence of a Fourier series I need to prove the assumption from the title above. Any help is very much appreciated.






calculus functional-analysis fourier-analysis







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    $|sum_{k=0}^{n-1}sin(2k+1)x| le sum_{k=0}^{n-1}|sin(2k+1)x|$
    $endgroup$
    – N74
    Jan 2 at 11:50
















2












2








2





$begingroup$


I'm currently studying Fourier series using a textbook which unfortunately provides only partial solutions to its explanations. To show the uniform convergence of a Fourier series I need to prove the assumption from the title above. Any help is very much appreciated.






calculus functional-analysis fourier-analysis







share|cite|improve this question











$endgroup$




I'm currently studying Fourier series using a textbook which unfortunately provides only partial solutions to its explanations. To show the uniform convergence of a Fourier series I need to prove the assumption from the title above. Any help is very much appreciated.






calculus functional-analysis fourier-analysis




calculus functional-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 2 at 12:33









mechanodroid

28.9k62748




28.9k62748










asked Jan 2 at 11:25









GuitarmanGuitarman

152




152








  • 2




    $begingroup$
    $|sum_{k=0}^{n-1}sin(2k+1)x| le sum_{k=0}^{n-1}|sin(2k+1)x|$
    $endgroup$
    – N74
    Jan 2 at 11:50
















  • 2




    $begingroup$
    $|sum_{k=0}^{n-1}sin(2k+1)x| le sum_{k=0}^{n-1}|sin(2k+1)x|$
    $endgroup$
    – N74
    Jan 2 at 11:50










2




2




$begingroup$
$|sum_{k=0}^{n-1}sin(2k+1)x| le sum_{k=0}^{n-1}|sin(2k+1)x|$
$endgroup$
– N74
Jan 2 at 11:50






$begingroup$
$|sum_{k=0}^{n-1}sin(2k+1)x| le sum_{k=0}^{n-1}|sin(2k+1)x|$
$endgroup$
– N74
Jan 2 at 11:50












1 Answer
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$begingroup$

You can calculate this sum explicitly:



begin{align}
sum_{k=0}^{n-1}sin(2k+1)x &= operatorname{Im}left(sum_{k=0}^{n-1}e^{(2k+1)ix}right)\
&= operatorname{Im}left(e^{ix}sum_{k=0}^{n-1}left(e^{2ix}right)^{k}right)\
&= operatorname{Im}left(e^{ix}frac{e^{2nix}-1}{e^{2ix}-1}right)\
&= operatorname{Im}left(frac{e^{2nix}-1}{e^{ix}-e^{-ix}}right)\
&= operatorname{Im}left(frac{cos(2nix)-1+isin(2pi ix)}{2isin x}right)\
&= frac{sin(2nx)}{2sin x}
end{align}



Now it should be clear that it is bounded.






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    1 Answer
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    1 Answer
    1






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    active

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    1












    $begingroup$

    You can calculate this sum explicitly:



    begin{align}
    sum_{k=0}^{n-1}sin(2k+1)x &= operatorname{Im}left(sum_{k=0}^{n-1}e^{(2k+1)ix}right)\
    &= operatorname{Im}left(e^{ix}sum_{k=0}^{n-1}left(e^{2ix}right)^{k}right)\
    &= operatorname{Im}left(e^{ix}frac{e^{2nix}-1}{e^{2ix}-1}right)\
    &= operatorname{Im}left(frac{e^{2nix}-1}{e^{ix}-e^{-ix}}right)\
    &= operatorname{Im}left(frac{cos(2nix)-1+isin(2pi ix)}{2isin x}right)\
    &= frac{sin(2nx)}{2sin x}
    end{align}



    Now it should be clear that it is bounded.






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      You can calculate this sum explicitly:



      begin{align}
      sum_{k=0}^{n-1}sin(2k+1)x &= operatorname{Im}left(sum_{k=0}^{n-1}e^{(2k+1)ix}right)\
      &= operatorname{Im}left(e^{ix}sum_{k=0}^{n-1}left(e^{2ix}right)^{k}right)\
      &= operatorname{Im}left(e^{ix}frac{e^{2nix}-1}{e^{2ix}-1}right)\
      &= operatorname{Im}left(frac{e^{2nix}-1}{e^{ix}-e^{-ix}}right)\
      &= operatorname{Im}left(frac{cos(2nix)-1+isin(2pi ix)}{2isin x}right)\
      &= frac{sin(2nx)}{2sin x}
      end{align}



      Now it should be clear that it is bounded.






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        You can calculate this sum explicitly:



        begin{align}
        sum_{k=0}^{n-1}sin(2k+1)x &= operatorname{Im}left(sum_{k=0}^{n-1}e^{(2k+1)ix}right)\
        &= operatorname{Im}left(e^{ix}sum_{k=0}^{n-1}left(e^{2ix}right)^{k}right)\
        &= operatorname{Im}left(e^{ix}frac{e^{2nix}-1}{e^{2ix}-1}right)\
        &= operatorname{Im}left(frac{e^{2nix}-1}{e^{ix}-e^{-ix}}right)\
        &= operatorname{Im}left(frac{cos(2nix)-1+isin(2pi ix)}{2isin x}right)\
        &= frac{sin(2nx)}{2sin x}
        end{align}



        Now it should be clear that it is bounded.






        share|cite|improve this answer









        $endgroup$



        You can calculate this sum explicitly:



        begin{align}
        sum_{k=0}^{n-1}sin(2k+1)x &= operatorname{Im}left(sum_{k=0}^{n-1}e^{(2k+1)ix}right)\
        &= operatorname{Im}left(e^{ix}sum_{k=0}^{n-1}left(e^{2ix}right)^{k}right)\
        &= operatorname{Im}left(e^{ix}frac{e^{2nix}-1}{e^{2ix}-1}right)\
        &= operatorname{Im}left(frac{e^{2nix}-1}{e^{ix}-e^{-ix}}right)\
        &= operatorname{Im}left(frac{cos(2nix)-1+isin(2pi ix)}{2isin x}right)\
        &= frac{sin(2nx)}{2sin x}
        end{align}



        Now it should be clear that it is bounded.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 2 at 12:33









        mechanodroidmechanodroid

        28.9k62748




        28.9k62748






























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