Fourier Analysis: Prove $c_n=a_nb_n$












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$begingroup$


So I am stuck in the middle of this proof:




Let $f,g in E[-pi,pi]$ be periodic functions with periods of $2pi$, where



$f(x)$ ~ $sum_{n=-inf}^{inf}a_ne^{inx}$



$g(x)$ ~ $sum_{n=-inf}^{inf}b_ne^{inx}$



$forall xin R: h(x) = frac{1}{2pi}int_{-pi}^{pi}f(x-t)g(t)dt$ | $h(x)$ ~ $sum_{n=-inf}^{inf}c_ne^{inx}$



Prove that $c_n=a_nb_n$



($forall n in Z$)




Here is what I've got so far:



By definition, $c_n=frac{1}{2pi}int_{-pi}^{pi}h(x)e^{-inx}dx=frac{1}{2pi}int_{-pi}^{pi}frac{1}{2pi}int_{-pi}^{pi}f(x-t)g(t)dte^{-inx}dx$



$c_n=frac{1}{(2pi)^2}int_{-pi}^{pi}(int_{-pi}^{pi}f(x-t)e^{-inx}dx)g(t)dt$



However, in order to finish the proof, I need to, presumably using periodicity, say that



$c_n=frac{1}{2pi}int_{-pi}^{pi}f(x)e^{-inx}dx cdot frac{1}{2pi}int_{-pi}^{pi}g(t)dt = a_nb_n$



Any help would be appreciated!






fourier-analysis fourier-series







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    0












    $begingroup$


    So I am stuck in the middle of this proof:




    Let $f,g in E[-pi,pi]$ be periodic functions with periods of $2pi$, where



    $f(x)$ ~ $sum_{n=-inf}^{inf}a_ne^{inx}$



    $g(x)$ ~ $sum_{n=-inf}^{inf}b_ne^{inx}$



    $forall xin R: h(x) = frac{1}{2pi}int_{-pi}^{pi}f(x-t)g(t)dt$ | $h(x)$ ~ $sum_{n=-inf}^{inf}c_ne^{inx}$



    Prove that $c_n=a_nb_n$



    ($forall n in Z$)




    Here is what I've got so far:



    By definition, $c_n=frac{1}{2pi}int_{-pi}^{pi}h(x)e^{-inx}dx=frac{1}{2pi}int_{-pi}^{pi}frac{1}{2pi}int_{-pi}^{pi}f(x-t)g(t)dte^{-inx}dx$



    $c_n=frac{1}{(2pi)^2}int_{-pi}^{pi}(int_{-pi}^{pi}f(x-t)e^{-inx}dx)g(t)dt$



    However, in order to finish the proof, I need to, presumably using periodicity, say that



    $c_n=frac{1}{2pi}int_{-pi}^{pi}f(x)e^{-inx}dx cdot frac{1}{2pi}int_{-pi}^{pi}g(t)dt = a_nb_n$



    Any help would be appreciated!






    fourier-analysis fourier-series







    share|cite|improve this question











    $endgroup$















      0












      0








      0





      $begingroup$


      So I am stuck in the middle of this proof:




      Let $f,g in E[-pi,pi]$ be periodic functions with periods of $2pi$, where



      $f(x)$ ~ $sum_{n=-inf}^{inf}a_ne^{inx}$



      $g(x)$ ~ $sum_{n=-inf}^{inf}b_ne^{inx}$



      $forall xin R: h(x) = frac{1}{2pi}int_{-pi}^{pi}f(x-t)g(t)dt$ | $h(x)$ ~ $sum_{n=-inf}^{inf}c_ne^{inx}$



      Prove that $c_n=a_nb_n$



      ($forall n in Z$)




      Here is what I've got so far:



      By definition, $c_n=frac{1}{2pi}int_{-pi}^{pi}h(x)e^{-inx}dx=frac{1}{2pi}int_{-pi}^{pi}frac{1}{2pi}int_{-pi}^{pi}f(x-t)g(t)dte^{-inx}dx$



      $c_n=frac{1}{(2pi)^2}int_{-pi}^{pi}(int_{-pi}^{pi}f(x-t)e^{-inx}dx)g(t)dt$



      However, in order to finish the proof, I need to, presumably using periodicity, say that



      $c_n=frac{1}{2pi}int_{-pi}^{pi}f(x)e^{-inx}dx cdot frac{1}{2pi}int_{-pi}^{pi}g(t)dt = a_nb_n$



      Any help would be appreciated!






      fourier-analysis fourier-series







      share|cite|improve this question











      $endgroup$




      So I am stuck in the middle of this proof:




      Let $f,g in E[-pi,pi]$ be periodic functions with periods of $2pi$, where



      $f(x)$ ~ $sum_{n=-inf}^{inf}a_ne^{inx}$



      $g(x)$ ~ $sum_{n=-inf}^{inf}b_ne^{inx}$



      $forall xin R: h(x) = frac{1}{2pi}int_{-pi}^{pi}f(x-t)g(t)dt$ | $h(x)$ ~ $sum_{n=-inf}^{inf}c_ne^{inx}$



      Prove that $c_n=a_nb_n$



      ($forall n in Z$)




      Here is what I've got so far:



      By definition, $c_n=frac{1}{2pi}int_{-pi}^{pi}h(x)e^{-inx}dx=frac{1}{2pi}int_{-pi}^{pi}frac{1}{2pi}int_{-pi}^{pi}f(x-t)g(t)dte^{-inx}dx$



      $c_n=frac{1}{(2pi)^2}int_{-pi}^{pi}(int_{-pi}^{pi}f(x-t)e^{-inx}dx)g(t)dt$



      However, in order to finish the proof, I need to, presumably using periodicity, say that



      $c_n=frac{1}{2pi}int_{-pi}^{pi}f(x)e^{-inx}dx cdot frac{1}{2pi}int_{-pi}^{pi}g(t)dt = a_nb_n$



      Any help would be appreciated!






      fourier-analysis fourier-series




      fourier-analysis fourier-series






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 3 '18 at 19:23







      CluelessButCurious

















      asked Dec 3 '18 at 19:13









      CluelessButCuriousCluelessButCurious

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          $begingroup$

          Almost there



          begin{eqnarray}
          c_n&=&frac{1}{2pi}int_{-pi}^{pi}h(x)e^{-inx}dx=frac{1}{2pi}int_{-pi}^{pi}left{frac{1}{2pi}int_{-pi}^{pi}color{blue}{f(x-t)}color{red}{g(t)}dtright}~e^{-inx}~dx \
          &=& frac{1}{(2pi)^2} int_{-pi}^{pi}int_{-pi}^{pi} color{blue}{sum_{k} a_ke^{ik (x-t)}}color{red}{sum_l b_l e^{il t}}e^{-inx} ~dx dt \
          &=& frac{1}{(2pi)^2} int_{-pi}^{pi}int_{-pi}^{pi} sum_{k,l} a_k b_l e^{ikx}e^{-i(k-l)t}dt dx \
          &=& frac{1}{2pi} int_{-pi}^{pi} sum_{k,l} a_k b_l e^{-i(n - k)x} color{orange}{left{frac{1}{2pi}int_{-pi}^{pi}e^{-i(k-l)t}dtright}} dx \
          &=& frac{1}{2pi} int_{-pi}^{pi} sum_{k,l} a_k b_l e^{-i(n - k)x} color{orange}{delta_{kl}} dx \
          &=& frac{1}{2pi}int_{-pi}^{pi}sum_k a_kb_k e^{-i(n - k)x}dx \
          &=&sum_k a_k b_k color{magenta}{left{frac{1}{2pi}int_{-pi}^{pi}e^{-i(n-k)t}dxright}} \
          &=& sum_k a_k b_k color{magenta}{delta_{nk}}
          = a_n b_n
          end{eqnarray}






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            $begingroup$

            Almost there



            begin{eqnarray}
            c_n&=&frac{1}{2pi}int_{-pi}^{pi}h(x)e^{-inx}dx=frac{1}{2pi}int_{-pi}^{pi}left{frac{1}{2pi}int_{-pi}^{pi}color{blue}{f(x-t)}color{red}{g(t)}dtright}~e^{-inx}~dx \
            &=& frac{1}{(2pi)^2} int_{-pi}^{pi}int_{-pi}^{pi} color{blue}{sum_{k} a_ke^{ik (x-t)}}color{red}{sum_l b_l e^{il t}}e^{-inx} ~dx dt \
            &=& frac{1}{(2pi)^2} int_{-pi}^{pi}int_{-pi}^{pi} sum_{k,l} a_k b_l e^{ikx}e^{-i(k-l)t}dt dx \
            &=& frac{1}{2pi} int_{-pi}^{pi} sum_{k,l} a_k b_l e^{-i(n - k)x} color{orange}{left{frac{1}{2pi}int_{-pi}^{pi}e^{-i(k-l)t}dtright}} dx \
            &=& frac{1}{2pi} int_{-pi}^{pi} sum_{k,l} a_k b_l e^{-i(n - k)x} color{orange}{delta_{kl}} dx \
            &=& frac{1}{2pi}int_{-pi}^{pi}sum_k a_kb_k e^{-i(n - k)x}dx \
            &=&sum_k a_k b_k color{magenta}{left{frac{1}{2pi}int_{-pi}^{pi}e^{-i(n-k)t}dxright}} \
            &=& sum_k a_k b_k color{magenta}{delta_{nk}}
            = a_n b_n
            end{eqnarray}






            share|cite|improve this answer









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              1












              $begingroup$

              Almost there



              begin{eqnarray}
              c_n&=&frac{1}{2pi}int_{-pi}^{pi}h(x)e^{-inx}dx=frac{1}{2pi}int_{-pi}^{pi}left{frac{1}{2pi}int_{-pi}^{pi}color{blue}{f(x-t)}color{red}{g(t)}dtright}~e^{-inx}~dx \
              &=& frac{1}{(2pi)^2} int_{-pi}^{pi}int_{-pi}^{pi} color{blue}{sum_{k} a_ke^{ik (x-t)}}color{red}{sum_l b_l e^{il t}}e^{-inx} ~dx dt \
              &=& frac{1}{(2pi)^2} int_{-pi}^{pi}int_{-pi}^{pi} sum_{k,l} a_k b_l e^{ikx}e^{-i(k-l)t}dt dx \
              &=& frac{1}{2pi} int_{-pi}^{pi} sum_{k,l} a_k b_l e^{-i(n - k)x} color{orange}{left{frac{1}{2pi}int_{-pi}^{pi}e^{-i(k-l)t}dtright}} dx \
              &=& frac{1}{2pi} int_{-pi}^{pi} sum_{k,l} a_k b_l e^{-i(n - k)x} color{orange}{delta_{kl}} dx \
              &=& frac{1}{2pi}int_{-pi}^{pi}sum_k a_kb_k e^{-i(n - k)x}dx \
              &=&sum_k a_k b_k color{magenta}{left{frac{1}{2pi}int_{-pi}^{pi}e^{-i(n-k)t}dxright}} \
              &=& sum_k a_k b_k color{magenta}{delta_{nk}}
              = a_n b_n
              end{eqnarray}






              share|cite|improve this answer









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                1












                1








                1





                $begingroup$

                Almost there



                begin{eqnarray}
                c_n&=&frac{1}{2pi}int_{-pi}^{pi}h(x)e^{-inx}dx=frac{1}{2pi}int_{-pi}^{pi}left{frac{1}{2pi}int_{-pi}^{pi}color{blue}{f(x-t)}color{red}{g(t)}dtright}~e^{-inx}~dx \
                &=& frac{1}{(2pi)^2} int_{-pi}^{pi}int_{-pi}^{pi} color{blue}{sum_{k} a_ke^{ik (x-t)}}color{red}{sum_l b_l e^{il t}}e^{-inx} ~dx dt \
                &=& frac{1}{(2pi)^2} int_{-pi}^{pi}int_{-pi}^{pi} sum_{k,l} a_k b_l e^{ikx}e^{-i(k-l)t}dt dx \
                &=& frac{1}{2pi} int_{-pi}^{pi} sum_{k,l} a_k b_l e^{-i(n - k)x} color{orange}{left{frac{1}{2pi}int_{-pi}^{pi}e^{-i(k-l)t}dtright}} dx \
                &=& frac{1}{2pi} int_{-pi}^{pi} sum_{k,l} a_k b_l e^{-i(n - k)x} color{orange}{delta_{kl}} dx \
                &=& frac{1}{2pi}int_{-pi}^{pi}sum_k a_kb_k e^{-i(n - k)x}dx \
                &=&sum_k a_k b_k color{magenta}{left{frac{1}{2pi}int_{-pi}^{pi}e^{-i(n-k)t}dxright}} \
                &=& sum_k a_k b_k color{magenta}{delta_{nk}}
                = a_n b_n
                end{eqnarray}






                share|cite|improve this answer









                $endgroup$



                Almost there



                begin{eqnarray}
                c_n&=&frac{1}{2pi}int_{-pi}^{pi}h(x)e^{-inx}dx=frac{1}{2pi}int_{-pi}^{pi}left{frac{1}{2pi}int_{-pi}^{pi}color{blue}{f(x-t)}color{red}{g(t)}dtright}~e^{-inx}~dx \
                &=& frac{1}{(2pi)^2} int_{-pi}^{pi}int_{-pi}^{pi} color{blue}{sum_{k} a_ke^{ik (x-t)}}color{red}{sum_l b_l e^{il t}}e^{-inx} ~dx dt \
                &=& frac{1}{(2pi)^2} int_{-pi}^{pi}int_{-pi}^{pi} sum_{k,l} a_k b_l e^{ikx}e^{-i(k-l)t}dt dx \
                &=& frac{1}{2pi} int_{-pi}^{pi} sum_{k,l} a_k b_l e^{-i(n - k)x} color{orange}{left{frac{1}{2pi}int_{-pi}^{pi}e^{-i(k-l)t}dtright}} dx \
                &=& frac{1}{2pi} int_{-pi}^{pi} sum_{k,l} a_k b_l e^{-i(n - k)x} color{orange}{delta_{kl}} dx \
                &=& frac{1}{2pi}int_{-pi}^{pi}sum_k a_kb_k e^{-i(n - k)x}dx \
                &=&sum_k a_k b_k color{magenta}{left{frac{1}{2pi}int_{-pi}^{pi}e^{-i(n-k)t}dxright}} \
                &=& sum_k a_k b_k color{magenta}{delta_{nk}}
                = a_n b_n
                end{eqnarray}







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                share|cite|improve this answer



                share|cite|improve this answer










                answered Dec 3 '18 at 19:47









                caveraccaverac

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