Prove an equality of Fourier series.












0












$begingroup$


Consider the function:



$f(x)= left{ begin{array}{lcc}
frac{3x}{2} & if & 0leq x leq frac{pi}{3} \
\ frac{pi}{2} & si & frac{pi}3 < x < frac{2pi}3 \
\ frac{3(pi - x)}{2} & si & frac{2pi}{3}leq x leq pi
end{array}
right.$



Prove



$$f(x)=sum_{i=1}^inftyfrac{sin(2n-1)pi}{3(2n-1)^2}times sin(2n-1)x,,,, text{for 0$leq$x$leqpi$ }$$



My attempt:



We need find the representation of fourier series in $sin$ of $f$.



As $f$ is continuous then



$$f(x)=sum_{i=1}^nb_ntimes sin(nx)$ with $xin[0,pi]$$



where,



$$b_n=frac{2}{pi}int_{0}^{pi} f(x)times sin(nx)dx$$



Using Maple I have:



$$b_n=frac{2}{pi}[frac{3}{2},{frac {sin left( frac{pi n}{3} right) -sin left( pi ,n
right) +sin left( frac{2 pi n}{3} right) }{{n}^{2}}}
]$$



then,



$$b_n=frac{1}{pi}[3,{frac {sin left( frac{pi n}{3} right) -sin left( pi ,n
right) +sin left( frac{2 pi n}{3} right) }{{n}^{2}}}]$$



this implies:



$$f(x)=sum_{i=1}^nfrac{1}{pi}[3,{frac {sin left( frac{pi n}{3} right) -sin left( pi ,n
right) +sin left( frac{2 pi n}{3} right) }{{n}^{2}}}]sin(nx)$$
for $0leq xleq pi$



here, I'm stuck. can someone help me?










share|cite|improve this question











$endgroup$

















    0












    $begingroup$


    Consider the function:



    $f(x)= left{ begin{array}{lcc}
    frac{3x}{2} & if & 0leq x leq frac{pi}{3} \
    \ frac{pi}{2} & si & frac{pi}3 < x < frac{2pi}3 \
    \ frac{3(pi - x)}{2} & si & frac{2pi}{3}leq x leq pi
    end{array}
    right.$



    Prove



    $$f(x)=sum_{i=1}^inftyfrac{sin(2n-1)pi}{3(2n-1)^2}times sin(2n-1)x,,,, text{for 0$leq$x$leqpi$ }$$



    My attempt:



    We need find the representation of fourier series in $sin$ of $f$.



    As $f$ is continuous then



    $$f(x)=sum_{i=1}^nb_ntimes sin(nx)$ with $xin[0,pi]$$



    where,



    $$b_n=frac{2}{pi}int_{0}^{pi} f(x)times sin(nx)dx$$



    Using Maple I have:



    $$b_n=frac{2}{pi}[frac{3}{2},{frac {sin left( frac{pi n}{3} right) -sin left( pi ,n
    right) +sin left( frac{2 pi n}{3} right) }{{n}^{2}}}
    ]$$



    then,



    $$b_n=frac{1}{pi}[3,{frac {sin left( frac{pi n}{3} right) -sin left( pi ,n
    right) +sin left( frac{2 pi n}{3} right) }{{n}^{2}}}]$$



    this implies:



    $$f(x)=sum_{i=1}^nfrac{1}{pi}[3,{frac {sin left( frac{pi n}{3} right) -sin left( pi ,n
    right) +sin left( frac{2 pi n}{3} right) }{{n}^{2}}}]sin(nx)$$
    for $0leq xleq pi$



    here, I'm stuck. can someone help me?










    share|cite|improve this question











    $endgroup$















      0












      0








      0





      $begingroup$


      Consider the function:



      $f(x)= left{ begin{array}{lcc}
      frac{3x}{2} & if & 0leq x leq frac{pi}{3} \
      \ frac{pi}{2} & si & frac{pi}3 < x < frac{2pi}3 \
      \ frac{3(pi - x)}{2} & si & frac{2pi}{3}leq x leq pi
      end{array}
      right.$



      Prove



      $$f(x)=sum_{i=1}^inftyfrac{sin(2n-1)pi}{3(2n-1)^2}times sin(2n-1)x,,,, text{for 0$leq$x$leqpi$ }$$



      My attempt:



      We need find the representation of fourier series in $sin$ of $f$.



      As $f$ is continuous then



      $$f(x)=sum_{i=1}^nb_ntimes sin(nx)$ with $xin[0,pi]$$



      where,



      $$b_n=frac{2}{pi}int_{0}^{pi} f(x)times sin(nx)dx$$



      Using Maple I have:



      $$b_n=frac{2}{pi}[frac{3}{2},{frac {sin left( frac{pi n}{3} right) -sin left( pi ,n
      right) +sin left( frac{2 pi n}{3} right) }{{n}^{2}}}
      ]$$



      then,



      $$b_n=frac{1}{pi}[3,{frac {sin left( frac{pi n}{3} right) -sin left( pi ,n
      right) +sin left( frac{2 pi n}{3} right) }{{n}^{2}}}]$$



      this implies:



      $$f(x)=sum_{i=1}^nfrac{1}{pi}[3,{frac {sin left( frac{pi n}{3} right) -sin left( pi ,n
      right) +sin left( frac{2 pi n}{3} right) }{{n}^{2}}}]sin(nx)$$
      for $0leq xleq pi$



      here, I'm stuck. can someone help me?










      share|cite|improve this question











      $endgroup$




      Consider the function:



      $f(x)= left{ begin{array}{lcc}
      frac{3x}{2} & if & 0leq x leq frac{pi}{3} \
      \ frac{pi}{2} & si & frac{pi}3 < x < frac{2pi}3 \
      \ frac{3(pi - x)}{2} & si & frac{2pi}{3}leq x leq pi
      end{array}
      right.$



      Prove



      $$f(x)=sum_{i=1}^inftyfrac{sin(2n-1)pi}{3(2n-1)^2}times sin(2n-1)x,,,, text{for 0$leq$x$leqpi$ }$$



      My attempt:



      We need find the representation of fourier series in $sin$ of $f$.



      As $f$ is continuous then



      $$f(x)=sum_{i=1}^nb_ntimes sin(nx)$ with $xin[0,pi]$$



      where,



      $$b_n=frac{2}{pi}int_{0}^{pi} f(x)times sin(nx)dx$$



      Using Maple I have:



      $$b_n=frac{2}{pi}[frac{3}{2},{frac {sin left( frac{pi n}{3} right) -sin left( pi ,n
      right) +sin left( frac{2 pi n}{3} right) }{{n}^{2}}}
      ]$$



      then,



      $$b_n=frac{1}{pi}[3,{frac {sin left( frac{pi n}{3} right) -sin left( pi ,n
      right) +sin left( frac{2 pi n}{3} right) }{{n}^{2}}}]$$



      this implies:



      $$f(x)=sum_{i=1}^nfrac{1}{pi}[3,{frac {sin left( frac{pi n}{3} right) -sin left( pi ,n
      right) +sin left( frac{2 pi n}{3} right) }{{n}^{2}}}]sin(nx)$$
      for $0leq xleq pi$



      here, I'm stuck. can someone help me?







      fourier-series






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 29 '18 at 16:20









      Bernard

      119k639112




      119k639112










      asked Dec 29 '18 at 16:16









      Bvss12Bvss12

      1,778617




      1,778617






















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