Another form of partial sum of Fourier series












1












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I've been struggling to prove that partial sum of Fourier series can be defined in the following form:



$$ S_{n}(f)(x) = frac{1}{pi} int_{-pi}^{pi} f(t-x)D_{n}(t)dt, $$ where:



$$ D_{n}(t) = frac{1}{2} + sum_{k=1}^n cos(kt)=frac{sin(t(n+frac{1}{2}))}{2sin(frac{t}{2})}.$$ And $f$ is a $2pi$-periodic function. I've managed to show that $$ S_{n}(f)(x) = frac{1}{pi} int_{-pi}^{pi} f(t)D_{n}(t-x)dt $$, but I don't know how to use the fact of $f$ being $2pi$-periodic.



Any ideas?






real-analysis integration fourier-series







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    $begingroup$


    I've been struggling to prove that partial sum of Fourier series can be defined in the following form:



    $$ S_{n}(f)(x) = frac{1}{pi} int_{-pi}^{pi} f(t-x)D_{n}(t)dt, $$ where:



    $$ D_{n}(t) = frac{1}{2} + sum_{k=1}^n cos(kt)=frac{sin(t(n+frac{1}{2}))}{2sin(frac{t}{2})}.$$ And $f$ is a $2pi$-periodic function. I've managed to show that $$ S_{n}(f)(x) = frac{1}{pi} int_{-pi}^{pi} f(t)D_{n}(t-x)dt $$, but I don't know how to use the fact of $f$ being $2pi$-periodic.



    Any ideas?






    real-analysis integration fourier-series







    share|cite|improve this question











    $endgroup$















      1












      1








      1





      $begingroup$


      I've been struggling to prove that partial sum of Fourier series can be defined in the following form:



      $$ S_{n}(f)(x) = frac{1}{pi} int_{-pi}^{pi} f(t-x)D_{n}(t)dt, $$ where:



      $$ D_{n}(t) = frac{1}{2} + sum_{k=1}^n cos(kt)=frac{sin(t(n+frac{1}{2}))}{2sin(frac{t}{2})}.$$ And $f$ is a $2pi$-periodic function. I've managed to show that $$ S_{n}(f)(x) = frac{1}{pi} int_{-pi}^{pi} f(t)D_{n}(t-x)dt $$, but I don't know how to use the fact of $f$ being $2pi$-periodic.



      Any ideas?






      real-analysis integration fourier-series







      share|cite|improve this question











      $endgroup$




      I've been struggling to prove that partial sum of Fourier series can be defined in the following form:



      $$ S_{n}(f)(x) = frac{1}{pi} int_{-pi}^{pi} f(t-x)D_{n}(t)dt, $$ where:



      $$ D_{n}(t) = frac{1}{2} + sum_{k=1}^n cos(kt)=frac{sin(t(n+frac{1}{2}))}{2sin(frac{t}{2})}.$$ And $f$ is a $2pi$-periodic function. I've managed to show that $$ S_{n}(f)(x) = frac{1}{pi} int_{-pi}^{pi} f(t)D_{n}(t-x)dt $$, but I don't know how to use the fact of $f$ being $2pi$-periodic.



      Any ideas?






      real-analysis integration fourier-series




      real-analysis integration fourier-series






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      share|cite|improve this question













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      edited Jan 19 at 15:51









      Scientifica

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      asked Jan 19 at 15:48









      BreadBread

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          $begingroup$

          Note that $D_n$ is even, so we have
          $$S_{n}(f)(x) = frac{1}{pi} int_{-pi}^{pi} f(t)D_{n}(t-x)dt=frac{1}{pi} int_{-pi}^{pi} f(t)D_{n}(x-t)dt$$



          Write $s=x-t$, so that $t=x-s$, and
          $$frac{1}{pi} int_{-pi}^{pi} f(t)D_{n}(t-x)dt=frac{1}{pi}int_{x-pi}^{x+pi}f(x-s)D_n(s)ds=frac{1}{pi}int_{-pi}^{pi}f(x-s)D_n(s)ds.$$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            So we use being $2pi$-periodic to change the integration interval after substitution $s=x-t$? Additionally I can't see how to proceed any further. :(
            $endgroup$
            – Bread
            Jan 21 at 22:30










          • $begingroup$
            Yes, and this is about as far as you'll get. the idea is to show that $S_n(f)$ is the convolution of $f$ and $D_n$, and that is exactly what the last integral in my answer is.
            $endgroup$
            – Aweygan
            Jan 21 at 22:40










          • $begingroup$
            Isn't convolution defined on interval $[-inf, inf]$?
            $endgroup$
            – Bread
            Jan 23 at 21:27










          • $begingroup$
            Convolution on the circle (or equivalently, of $2pi$-periodic functions) is different from convolution on the reals. See here, for instance. Also, type infty for $infty$.
            $endgroup$
            – Aweygan
            Jan 24 at 2:37













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          $begingroup$

          Note that $D_n$ is even, so we have
          $$S_{n}(f)(x) = frac{1}{pi} int_{-pi}^{pi} f(t)D_{n}(t-x)dt=frac{1}{pi} int_{-pi}^{pi} f(t)D_{n}(x-t)dt$$



          Write $s=x-t$, so that $t=x-s$, and
          $$frac{1}{pi} int_{-pi}^{pi} f(t)D_{n}(t-x)dt=frac{1}{pi}int_{x-pi}^{x+pi}f(x-s)D_n(s)ds=frac{1}{pi}int_{-pi}^{pi}f(x-s)D_n(s)ds.$$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            So we use being $2pi$-periodic to change the integration interval after substitution $s=x-t$? Additionally I can't see how to proceed any further. :(
            $endgroup$
            – Bread
            Jan 21 at 22:30










          • $begingroup$
            Yes, and this is about as far as you'll get. the idea is to show that $S_n(f)$ is the convolution of $f$ and $D_n$, and that is exactly what the last integral in my answer is.
            $endgroup$
            – Aweygan
            Jan 21 at 22:40










          • $begingroup$
            Isn't convolution defined on interval $[-inf, inf]$?
            $endgroup$
            – Bread
            Jan 23 at 21:27










          • $begingroup$
            Convolution on the circle (or equivalently, of $2pi$-periodic functions) is different from convolution on the reals. See here, for instance. Also, type infty for $infty$.
            $endgroup$
            – Aweygan
            Jan 24 at 2:37


















          1












          $begingroup$

          Note that $D_n$ is even, so we have
          $$S_{n}(f)(x) = frac{1}{pi} int_{-pi}^{pi} f(t)D_{n}(t-x)dt=frac{1}{pi} int_{-pi}^{pi} f(t)D_{n}(x-t)dt$$



          Write $s=x-t$, so that $t=x-s$, and
          $$frac{1}{pi} int_{-pi}^{pi} f(t)D_{n}(t-x)dt=frac{1}{pi}int_{x-pi}^{x+pi}f(x-s)D_n(s)ds=frac{1}{pi}int_{-pi}^{pi}f(x-s)D_n(s)ds.$$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            So we use being $2pi$-periodic to change the integration interval after substitution $s=x-t$? Additionally I can't see how to proceed any further. :(
            $endgroup$
            – Bread
            Jan 21 at 22:30










          • $begingroup$
            Yes, and this is about as far as you'll get. the idea is to show that $S_n(f)$ is the convolution of $f$ and $D_n$, and that is exactly what the last integral in my answer is.
            $endgroup$
            – Aweygan
            Jan 21 at 22:40










          • $begingroup$
            Isn't convolution defined on interval $[-inf, inf]$?
            $endgroup$
            – Bread
            Jan 23 at 21:27










          • $begingroup$
            Convolution on the circle (or equivalently, of $2pi$-periodic functions) is different from convolution on the reals. See here, for instance. Also, type infty for $infty$.
            $endgroup$
            – Aweygan
            Jan 24 at 2:37
















          1












          1








          1





          $begingroup$

          Note that $D_n$ is even, so we have
          $$S_{n}(f)(x) = frac{1}{pi} int_{-pi}^{pi} f(t)D_{n}(t-x)dt=frac{1}{pi} int_{-pi}^{pi} f(t)D_{n}(x-t)dt$$



          Write $s=x-t$, so that $t=x-s$, and
          $$frac{1}{pi} int_{-pi}^{pi} f(t)D_{n}(t-x)dt=frac{1}{pi}int_{x-pi}^{x+pi}f(x-s)D_n(s)ds=frac{1}{pi}int_{-pi}^{pi}f(x-s)D_n(s)ds.$$






          share|cite|improve this answer









          $endgroup$



          Note that $D_n$ is even, so we have
          $$S_{n}(f)(x) = frac{1}{pi} int_{-pi}^{pi} f(t)D_{n}(t-x)dt=frac{1}{pi} int_{-pi}^{pi} f(t)D_{n}(x-t)dt$$



          Write $s=x-t$, so that $t=x-s$, and
          $$frac{1}{pi} int_{-pi}^{pi} f(t)D_{n}(t-x)dt=frac{1}{pi}int_{x-pi}^{x+pi}f(x-s)D_n(s)ds=frac{1}{pi}int_{-pi}^{pi}f(x-s)D_n(s)ds.$$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 19 at 16:04









          AweyganAweygan

          14.4k21441




          14.4k21441












          • $begingroup$
            So we use being $2pi$-periodic to change the integration interval after substitution $s=x-t$? Additionally I can't see how to proceed any further. :(
            $endgroup$
            – Bread
            Jan 21 at 22:30










          • $begingroup$
            Yes, and this is about as far as you'll get. the idea is to show that $S_n(f)$ is the convolution of $f$ and $D_n$, and that is exactly what the last integral in my answer is.
            $endgroup$
            – Aweygan
            Jan 21 at 22:40










          • $begingroup$
            Isn't convolution defined on interval $[-inf, inf]$?
            $endgroup$
            – Bread
            Jan 23 at 21:27










          • $begingroup$
            Convolution on the circle (or equivalently, of $2pi$-periodic functions) is different from convolution on the reals. See here, for instance. Also, type infty for $infty$.
            $endgroup$
            – Aweygan
            Jan 24 at 2:37




















          • $begingroup$
            So we use being $2pi$-periodic to change the integration interval after substitution $s=x-t$? Additionally I can't see how to proceed any further. :(
            $endgroup$
            – Bread
            Jan 21 at 22:30










          • $begingroup$
            Yes, and this is about as far as you'll get. the idea is to show that $S_n(f)$ is the convolution of $f$ and $D_n$, and that is exactly what the last integral in my answer is.
            $endgroup$
            – Aweygan
            Jan 21 at 22:40










          • $begingroup$
            Isn't convolution defined on interval $[-inf, inf]$?
            $endgroup$
            – Bread
            Jan 23 at 21:27










          • $begingroup$
            Convolution on the circle (or equivalently, of $2pi$-periodic functions) is different from convolution on the reals. See here, for instance. Also, type infty for $infty$.
            $endgroup$
            – Aweygan
            Jan 24 at 2:37


















          $begingroup$
          So we use being $2pi$-periodic to change the integration interval after substitution $s=x-t$? Additionally I can't see how to proceed any further. :(
          $endgroup$
          – Bread
          Jan 21 at 22:30




          $begingroup$
          So we use being $2pi$-periodic to change the integration interval after substitution $s=x-t$? Additionally I can't see how to proceed any further. :(
          $endgroup$
          – Bread
          Jan 21 at 22:30












          $begingroup$
          Yes, and this is about as far as you'll get. the idea is to show that $S_n(f)$ is the convolution of $f$ and $D_n$, and that is exactly what the last integral in my answer is.
          $endgroup$
          – Aweygan
          Jan 21 at 22:40




          $begingroup$
          Yes, and this is about as far as you'll get. the idea is to show that $S_n(f)$ is the convolution of $f$ and $D_n$, and that is exactly what the last integral in my answer is.
          $endgroup$
          – Aweygan
          Jan 21 at 22:40












          $begingroup$
          Isn't convolution defined on interval $[-inf, inf]$?
          $endgroup$
          – Bread
          Jan 23 at 21:27




          $begingroup$
          Isn't convolution defined on interval $[-inf, inf]$?
          $endgroup$
          – Bread
          Jan 23 at 21:27












          $begingroup$
          Convolution on the circle (or equivalently, of $2pi$-periodic functions) is different from convolution on the reals. See here, for instance. Also, type infty for $infty$.
          $endgroup$
          – Aweygan
          Jan 24 at 2:37






          $begingroup$
          Convolution on the circle (or equivalently, of $2pi$-periodic functions) is different from convolution on the reals. See here, for instance. Also, type infty for $infty$.
          $endgroup$
          – Aweygan
          Jan 24 at 2:37




















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