Determine Fourier series expansion for $f(theta)=cos^4!theta$












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$begingroup$


Q: The function $f(theta) = cos^4! theta$ is a nice smooth function, so will have a Fourier series expansion. That is, it will have an expansion as a sum of functions $cos j theta$ and $sin j theta$ with real coefficients. Determine what the expansion is.



For reference, the Fourier coefficients are:



begin{align*}
c_n &= frac{1}{2pi} int_0^{2 pi} f(t) e^{-int} , dt = frac{1}{2pi} int_0^{2 pi} f(t) left( cos (nt) - i sin (nt) right) , dt \
end{align*}



The coefficients give the Fourier Series expansion:



begin{align*}
f(t) &= sumlimits_{n=0}^infty c_n e^{int} = sumlimits_{n=0}^infty c_n left( cos (nt) + i sin (nt) right) \
end{align*}



One route is to plug $f(theta)$ into the equation for the Fourier series coefficients:



begin{align*}
c_n &= frac{1}{2pi} int_0^{2 pi} cos^4 t cdot left[ cos (nt) - i sin (nt) right] , dt \
end{align*}



That integral is looking complex. I suspect there is an easier solution to this problem?






real-analysis fourier-series fourier-transform







share|cite|improve this question











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    0












    $begingroup$


    Q: The function $f(theta) = cos^4! theta$ is a nice smooth function, so will have a Fourier series expansion. That is, it will have an expansion as a sum of functions $cos j theta$ and $sin j theta$ with real coefficients. Determine what the expansion is.



    For reference, the Fourier coefficients are:



    begin{align*}
    c_n &= frac{1}{2pi} int_0^{2 pi} f(t) e^{-int} , dt = frac{1}{2pi} int_0^{2 pi} f(t) left( cos (nt) - i sin (nt) right) , dt \
    end{align*}



    The coefficients give the Fourier Series expansion:



    begin{align*}
    f(t) &= sumlimits_{n=0}^infty c_n e^{int} = sumlimits_{n=0}^infty c_n left( cos (nt) + i sin (nt) right) \
    end{align*}



    One route is to plug $f(theta)$ into the equation for the Fourier series coefficients:



    begin{align*}
    c_n &= frac{1}{2pi} int_0^{2 pi} cos^4 t cdot left[ cos (nt) - i sin (nt) right] , dt \
    end{align*}



    That integral is looking complex. I suspect there is an easier solution to this problem?






    real-analysis fourier-series fourier-transform







    share|cite|improve this question











    $endgroup$















      0












      0








      0





      $begingroup$


      Q: The function $f(theta) = cos^4! theta$ is a nice smooth function, so will have a Fourier series expansion. That is, it will have an expansion as a sum of functions $cos j theta$ and $sin j theta$ with real coefficients. Determine what the expansion is.



      For reference, the Fourier coefficients are:



      begin{align*}
      c_n &= frac{1}{2pi} int_0^{2 pi} f(t) e^{-int} , dt = frac{1}{2pi} int_0^{2 pi} f(t) left( cos (nt) - i sin (nt) right) , dt \
      end{align*}



      The coefficients give the Fourier Series expansion:



      begin{align*}
      f(t) &= sumlimits_{n=0}^infty c_n e^{int} = sumlimits_{n=0}^infty c_n left( cos (nt) + i sin (nt) right) \
      end{align*}



      One route is to plug $f(theta)$ into the equation for the Fourier series coefficients:



      begin{align*}
      c_n &= frac{1}{2pi} int_0^{2 pi} cos^4 t cdot left[ cos (nt) - i sin (nt) right] , dt \
      end{align*}



      That integral is looking complex. I suspect there is an easier solution to this problem?






      real-analysis fourier-series fourier-transform







      share|cite|improve this question











      $endgroup$




      Q: The function $f(theta) = cos^4! theta$ is a nice smooth function, so will have a Fourier series expansion. That is, it will have an expansion as a sum of functions $cos j theta$ and $sin j theta$ with real coefficients. Determine what the expansion is.



      For reference, the Fourier coefficients are:



      begin{align*}
      c_n &= frac{1}{2pi} int_0^{2 pi} f(t) e^{-int} , dt = frac{1}{2pi} int_0^{2 pi} f(t) left( cos (nt) - i sin (nt) right) , dt \
      end{align*}



      The coefficients give the Fourier Series expansion:



      begin{align*}
      f(t) &= sumlimits_{n=0}^infty c_n e^{int} = sumlimits_{n=0}^infty c_n left( cos (nt) + i sin (nt) right) \
      end{align*}



      One route is to plug $f(theta)$ into the equation for the Fourier series coefficients:



      begin{align*}
      c_n &= frac{1}{2pi} int_0^{2 pi} cos^4 t cdot left[ cos (nt) - i sin (nt) right] , dt \
      end{align*}



      That integral is looking complex. I suspect there is an easier solution to this problem?






      real-analysis fourier-series fourier-transform




      real-analysis fourier-series fourier-transform






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Jan 21 at 18:41









      Daniele Tampieri

      2,3172922




      2,3172922










      asked Jan 21 at 16:49









      clayclay

      767415




      767415






















          2 Answers
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          $begingroup$

          Using $$cos^2(theta )=frac{1+cos(2theta )}{2},$$ one get $$cos^4(theta )=frac{1+2cos(2theta )+cos^2(2theta )}{4}=frac{1+2cos(2theta )+frac{1+cos(4theta )}{2}}{4}$$
          $$=frac{3}{8}+frac{1}{2}cos(2theta )+frac{1}{8}cos(4theta ).$$






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            $cos^4 theta = frac{3 + 4cos(2theta) + cos (4theta)}{8}$. That's from en.wikipedia.org/wiki/…. You are close. That does it! Thanks!
            $endgroup$
            – clay
            Jan 21 at 17:00










          • $begingroup$
            It's exactly what it is written ;) you are welcome @clay
            $endgroup$
            – Dylan
            Jan 21 at 17:02








          • 2




            $begingroup$
            You are missing $frac{1}{8}$ multiplying $cos(4theta)$
            $endgroup$
            – user289143
            Jan 21 at 17:04










          • $begingroup$
            @user289143: Ouuppss, thank you, typo corrected :)
            $endgroup$
            – Dylan
            Jan 21 at 17:05



















          0












          $begingroup$

          Hint: $$cos(t) = frac{e^{it} + e^{-it}}{2}$$






          share|cite|improve this answer











          $endgroup$













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            2 Answers
            2






            active

            oldest

            votes








            2 Answers
            2






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            2












            $begingroup$

            Using $$cos^2(theta )=frac{1+cos(2theta )}{2},$$ one get $$cos^4(theta )=frac{1+2cos(2theta )+cos^2(2theta )}{4}=frac{1+2cos(2theta )+frac{1+cos(4theta )}{2}}{4}$$
            $$=frac{3}{8}+frac{1}{2}cos(2theta )+frac{1}{8}cos(4theta ).$$






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              $cos^4 theta = frac{3 + 4cos(2theta) + cos (4theta)}{8}$. That's from en.wikipedia.org/wiki/…. You are close. That does it! Thanks!
              $endgroup$
              – clay
              Jan 21 at 17:00










            • $begingroup$
              It's exactly what it is written ;) you are welcome @clay
              $endgroup$
              – Dylan
              Jan 21 at 17:02








            • 2




              $begingroup$
              You are missing $frac{1}{8}$ multiplying $cos(4theta)$
              $endgroup$
              – user289143
              Jan 21 at 17:04










            • $begingroup$
              @user289143: Ouuppss, thank you, typo corrected :)
              $endgroup$
              – Dylan
              Jan 21 at 17:05
















            2












            $begingroup$

            Using $$cos^2(theta )=frac{1+cos(2theta )}{2},$$ one get $$cos^4(theta )=frac{1+2cos(2theta )+cos^2(2theta )}{4}=frac{1+2cos(2theta )+frac{1+cos(4theta )}{2}}{4}$$
            $$=frac{3}{8}+frac{1}{2}cos(2theta )+frac{1}{8}cos(4theta ).$$






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              $cos^4 theta = frac{3 + 4cos(2theta) + cos (4theta)}{8}$. That's from en.wikipedia.org/wiki/…. You are close. That does it! Thanks!
              $endgroup$
              – clay
              Jan 21 at 17:00










            • $begingroup$
              It's exactly what it is written ;) you are welcome @clay
              $endgroup$
              – Dylan
              Jan 21 at 17:02








            • 2




              $begingroup$
              You are missing $frac{1}{8}$ multiplying $cos(4theta)$
              $endgroup$
              – user289143
              Jan 21 at 17:04










            • $begingroup$
              @user289143: Ouuppss, thank you, typo corrected :)
              $endgroup$
              – Dylan
              Jan 21 at 17:05














            2












            2








            2





            $begingroup$

            Using $$cos^2(theta )=frac{1+cos(2theta )}{2},$$ one get $$cos^4(theta )=frac{1+2cos(2theta )+cos^2(2theta )}{4}=frac{1+2cos(2theta )+frac{1+cos(4theta )}{2}}{4}$$
            $$=frac{3}{8}+frac{1}{2}cos(2theta )+frac{1}{8}cos(4theta ).$$






            share|cite|improve this answer











            $endgroup$



            Using $$cos^2(theta )=frac{1+cos(2theta )}{2},$$ one get $$cos^4(theta )=frac{1+2cos(2theta )+cos^2(2theta )}{4}=frac{1+2cos(2theta )+frac{1+cos(4theta )}{2}}{4}$$
            $$=frac{3}{8}+frac{1}{2}cos(2theta )+frac{1}{8}cos(4theta ).$$







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Jan 21 at 17:05

























            answered Jan 21 at 16:54









            DylanDylan

            1968




            1968












            • $begingroup$
              $cos^4 theta = frac{3 + 4cos(2theta) + cos (4theta)}{8}$. That's from en.wikipedia.org/wiki/…. You are close. That does it! Thanks!
              $endgroup$
              – clay
              Jan 21 at 17:00










            • $begingroup$
              It's exactly what it is written ;) you are welcome @clay
              $endgroup$
              – Dylan
              Jan 21 at 17:02








            • 2




              $begingroup$
              You are missing $frac{1}{8}$ multiplying $cos(4theta)$
              $endgroup$
              – user289143
              Jan 21 at 17:04










            • $begingroup$
              @user289143: Ouuppss, thank you, typo corrected :)
              $endgroup$
              – Dylan
              Jan 21 at 17:05


















            • $begingroup$
              $cos^4 theta = frac{3 + 4cos(2theta) + cos (4theta)}{8}$. That's from en.wikipedia.org/wiki/…. You are close. That does it! Thanks!
              $endgroup$
              – clay
              Jan 21 at 17:00










            • $begingroup$
              It's exactly what it is written ;) you are welcome @clay
              $endgroup$
              – Dylan
              Jan 21 at 17:02








            • 2




              $begingroup$
              You are missing $frac{1}{8}$ multiplying $cos(4theta)$
              $endgroup$
              – user289143
              Jan 21 at 17:04










            • $begingroup$
              @user289143: Ouuppss, thank you, typo corrected :)
              $endgroup$
              – Dylan
              Jan 21 at 17:05
















            $begingroup$
            $cos^4 theta = frac{3 + 4cos(2theta) + cos (4theta)}{8}$. That's from en.wikipedia.org/wiki/…. You are close. That does it! Thanks!
            $endgroup$
            – clay
            Jan 21 at 17:00




            $begingroup$
            $cos^4 theta = frac{3 + 4cos(2theta) + cos (4theta)}{8}$. That's from en.wikipedia.org/wiki/…. You are close. That does it! Thanks!
            $endgroup$
            – clay
            Jan 21 at 17:00












            $begingroup$
            It's exactly what it is written ;) you are welcome @clay
            $endgroup$
            – Dylan
            Jan 21 at 17:02






            $begingroup$
            It's exactly what it is written ;) you are welcome @clay
            $endgroup$
            – Dylan
            Jan 21 at 17:02






            2




            2




            $begingroup$
            You are missing $frac{1}{8}$ multiplying $cos(4theta)$
            $endgroup$
            – user289143
            Jan 21 at 17:04




            $begingroup$
            You are missing $frac{1}{8}$ multiplying $cos(4theta)$
            $endgroup$
            – user289143
            Jan 21 at 17:04












            $begingroup$
            @user289143: Ouuppss, thank you, typo corrected :)
            $endgroup$
            – Dylan
            Jan 21 at 17:05




            $begingroup$
            @user289143: Ouuppss, thank you, typo corrected :)
            $endgroup$
            – Dylan
            Jan 21 at 17:05











            0












            $begingroup$

            Hint: $$cos(t) = frac{e^{it} + e^{-it}}{2}$$






            share|cite|improve this answer











            $endgroup$


















              0












              $begingroup$

              Hint: $$cos(t) = frac{e^{it} + e^{-it}}{2}$$






              share|cite|improve this answer











              $endgroup$
















                0












                0








                0





                $begingroup$

                Hint: $$cos(t) = frac{e^{it} + e^{-it}}{2}$$






                share|cite|improve this answer











                $endgroup$



                Hint: $$cos(t) = frac{e^{it} + e^{-it}}{2}$$







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited Jan 21 at 18:11









                Daniele Tampieri

                2,3172922




                2,3172922










                answered Jan 21 at 17:03









                VectorizerVectorizer

                399413




                399413






























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