Finding a series of cosines to represent the absolute value function












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Question: Prove the following identity:



$|x|=frac{pi}{2}+sum_{n=1}^{infty}frac{cos(2n-1)x}{(2n-1)^2}$. Our current topic is Fourier analysis, but I am truly lost here. I tried to compute the Fourier series, but it seemed extremely different than the wanted identity.






fourier-analysis







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  • 1




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    Just a comment. Note that the right hand side is periodic while the left hand side is not (as it is written). What you are doing is in fact the Fourier series of a sawtooth-like function. So the result depends on which interval you pick. Common choices are $[-pi,pi]$ or $[0,2pi]$. Since the right hand side is an even function, your choice is the former. This may be the source of confusion.
    $endgroup$
    – lcv
    Dec 13 '18 at 18:47
















2












$begingroup$


Question: Prove the following identity:



$|x|=frac{pi}{2}+sum_{n=1}^{infty}frac{cos(2n-1)x}{(2n-1)^2}$. Our current topic is Fourier analysis, but I am truly lost here. I tried to compute the Fourier series, but it seemed extremely different than the wanted identity.






fourier-analysis







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    Just a comment. Note that the right hand side is periodic while the left hand side is not (as it is written). What you are doing is in fact the Fourier series of a sawtooth-like function. So the result depends on which interval you pick. Common choices are $[-pi,pi]$ or $[0,2pi]$. Since the right hand side is an even function, your choice is the former. This may be the source of confusion.
    $endgroup$
    – lcv
    Dec 13 '18 at 18:47














2












2








2


1



$begingroup$


Question: Prove the following identity:



$|x|=frac{pi}{2}+sum_{n=1}^{infty}frac{cos(2n-1)x}{(2n-1)^2}$. Our current topic is Fourier analysis, but I am truly lost here. I tried to compute the Fourier series, but it seemed extremely different than the wanted identity.






fourier-analysis







share|cite|improve this question









$endgroup$




Question: Prove the following identity:



$|x|=frac{pi}{2}+sum_{n=1}^{infty}frac{cos(2n-1)x}{(2n-1)^2}$. Our current topic is Fourier analysis, but I am truly lost here. I tried to compute the Fourier series, but it seemed extremely different than the wanted identity.






fourier-analysis




fourier-analysis






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asked Dec 13 '18 at 17:00









Uri George PeterzilUri George Peterzil

929




929








  • 1




    $begingroup$
    Just a comment. Note that the right hand side is periodic while the left hand side is not (as it is written). What you are doing is in fact the Fourier series of a sawtooth-like function. So the result depends on which interval you pick. Common choices are $[-pi,pi]$ or $[0,2pi]$. Since the right hand side is an even function, your choice is the former. This may be the source of confusion.
    $endgroup$
    – lcv
    Dec 13 '18 at 18:47














  • 1




    $begingroup$
    Just a comment. Note that the right hand side is periodic while the left hand side is not (as it is written). What you are doing is in fact the Fourier series of a sawtooth-like function. So the result depends on which interval you pick. Common choices are $[-pi,pi]$ or $[0,2pi]$. Since the right hand side is an even function, your choice is the former. This may be the source of confusion.
    $endgroup$
    – lcv
    Dec 13 '18 at 18:47








1




1




$begingroup$
Just a comment. Note that the right hand side is periodic while the left hand side is not (as it is written). What you are doing is in fact the Fourier series of a sawtooth-like function. So the result depends on which interval you pick. Common choices are $[-pi,pi]$ or $[0,2pi]$. Since the right hand side is an even function, your choice is the former. This may be the source of confusion.
$endgroup$
– lcv
Dec 13 '18 at 18:47




$begingroup$
Just a comment. Note that the right hand side is periodic while the left hand side is not (as it is written). What you are doing is in fact the Fourier series of a sawtooth-like function. So the result depends on which interval you pick. Common choices are $[-pi,pi]$ or $[0,2pi]$. Since the right hand side is an even function, your choice is the former. This may be the source of confusion.
$endgroup$
– lcv
Dec 13 '18 at 18:47










2 Answers
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$begingroup$

You can evaluate it using complex fourier coefficients $c_k$. You get: $frac{1}{2pi}$($int_{0}^{-pi}xe^{-ikx}dx$ + $int_{0}^{pi}xe^{-ikx}dx$) which results after integration by parts in $frac{1}{pi k^2}((-1)^k-1)=c_k$ Now, we use $a_k=c_k+c_{-k}$ and $b_k=i(c_k-c_{-k})$ which is $0$.

$a_k$ becomes $frac{-4}{pi k^2}$ where k is odd,i.e. $k=2n-1$ and $a_0$, after evaluation, becomes $pi$



So now, you have : $$f(x)=frac{a_0}{2}+sum_{k=1}^{infty}a_kcos(kx), k=(2n-1)$$ So after plugging in all values, we get $$f(x)=|x|=frac{pi}{2}+frac{-4}{pi} sum_{n=1}^{infty} frac{1}{(2n-1)^2} cos((2n-1)x)$$



Hope this helps !






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    1












    $begingroup$

    It should be
    $$
    |x|=frac{pi}{2}-frac{4}{pi}sum_{n=1}^{infty}frac{cos(2n-1)x}{(2n-1)^2}.
    $$
    Since $|x|$ is even, there are no sine terms, and the coefficient of $cos(n,x)$ is
    $$
    frac{1}{pi}int_{-pi}^pi|x|cos(n,x),dx=frac{2}{pi}int_0^pi xcos(n,x),dx,quad nge0.
    $$
    This integral easily computed using integration by parts.






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      2 Answers
      2






      active

      oldest

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      2 Answers
      2






      active

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      active

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      2












      $begingroup$

      You can evaluate it using complex fourier coefficients $c_k$. You get: $frac{1}{2pi}$($int_{0}^{-pi}xe^{-ikx}dx$ + $int_{0}^{pi}xe^{-ikx}dx$) which results after integration by parts in $frac{1}{pi k^2}((-1)^k-1)=c_k$ Now, we use $a_k=c_k+c_{-k}$ and $b_k=i(c_k-c_{-k})$ which is $0$.

      $a_k$ becomes $frac{-4}{pi k^2}$ where k is odd,i.e. $k=2n-1$ and $a_0$, after evaluation, becomes $pi$



      So now, you have : $$f(x)=frac{a_0}{2}+sum_{k=1}^{infty}a_kcos(kx), k=(2n-1)$$ So after plugging in all values, we get $$f(x)=|x|=frac{pi}{2}+frac{-4}{pi} sum_{n=1}^{infty} frac{1}{(2n-1)^2} cos((2n-1)x)$$



      Hope this helps !






      share|cite|improve this answer









      $endgroup$


















        2












        $begingroup$

        You can evaluate it using complex fourier coefficients $c_k$. You get: $frac{1}{2pi}$($int_{0}^{-pi}xe^{-ikx}dx$ + $int_{0}^{pi}xe^{-ikx}dx$) which results after integration by parts in $frac{1}{pi k^2}((-1)^k-1)=c_k$ Now, we use $a_k=c_k+c_{-k}$ and $b_k=i(c_k-c_{-k})$ which is $0$.

        $a_k$ becomes $frac{-4}{pi k^2}$ where k is odd,i.e. $k=2n-1$ and $a_0$, after evaluation, becomes $pi$



        So now, you have : $$f(x)=frac{a_0}{2}+sum_{k=1}^{infty}a_kcos(kx), k=(2n-1)$$ So after plugging in all values, we get $$f(x)=|x|=frac{pi}{2}+frac{-4}{pi} sum_{n=1}^{infty} frac{1}{(2n-1)^2} cos((2n-1)x)$$



        Hope this helps !






        share|cite|improve this answer









        $endgroup$
















          2












          2








          2





          $begingroup$

          You can evaluate it using complex fourier coefficients $c_k$. You get: $frac{1}{2pi}$($int_{0}^{-pi}xe^{-ikx}dx$ + $int_{0}^{pi}xe^{-ikx}dx$) which results after integration by parts in $frac{1}{pi k^2}((-1)^k-1)=c_k$ Now, we use $a_k=c_k+c_{-k}$ and $b_k=i(c_k-c_{-k})$ which is $0$.

          $a_k$ becomes $frac{-4}{pi k^2}$ where k is odd,i.e. $k=2n-1$ and $a_0$, after evaluation, becomes $pi$



          So now, you have : $$f(x)=frac{a_0}{2}+sum_{k=1}^{infty}a_kcos(kx), k=(2n-1)$$ So after plugging in all values, we get $$f(x)=|x|=frac{pi}{2}+frac{-4}{pi} sum_{n=1}^{infty} frac{1}{(2n-1)^2} cos((2n-1)x)$$



          Hope this helps !






          share|cite|improve this answer









          $endgroup$



          You can evaluate it using complex fourier coefficients $c_k$. You get: $frac{1}{2pi}$($int_{0}^{-pi}xe^{-ikx}dx$ + $int_{0}^{pi}xe^{-ikx}dx$) which results after integration by parts in $frac{1}{pi k^2}((-1)^k-1)=c_k$ Now, we use $a_k=c_k+c_{-k}$ and $b_k=i(c_k-c_{-k})$ which is $0$.

          $a_k$ becomes $frac{-4}{pi k^2}$ where k is odd,i.e. $k=2n-1$ and $a_0$, after evaluation, becomes $pi$



          So now, you have : $$f(x)=frac{a_0}{2}+sum_{k=1}^{infty}a_kcos(kx), k=(2n-1)$$ So after plugging in all values, we get $$f(x)=|x|=frac{pi}{2}+frac{-4}{pi} sum_{n=1}^{infty} frac{1}{(2n-1)^2} cos((2n-1)x)$$



          Hope this helps !







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 13 '18 at 19:29









          PoujhPoujh

          609516




          609516























              1












              $begingroup$

              It should be
              $$
              |x|=frac{pi}{2}-frac{4}{pi}sum_{n=1}^{infty}frac{cos(2n-1)x}{(2n-1)^2}.
              $$
              Since $|x|$ is even, there are no sine terms, and the coefficient of $cos(n,x)$ is
              $$
              frac{1}{pi}int_{-pi}^pi|x|cos(n,x),dx=frac{2}{pi}int_0^pi xcos(n,x),dx,quad nge0.
              $$
              This integral easily computed using integration by parts.






              share|cite|improve this answer









              $endgroup$


















                1












                $begingroup$

                It should be
                $$
                |x|=frac{pi}{2}-frac{4}{pi}sum_{n=1}^{infty}frac{cos(2n-1)x}{(2n-1)^2}.
                $$
                Since $|x|$ is even, there are no sine terms, and the coefficient of $cos(n,x)$ is
                $$
                frac{1}{pi}int_{-pi}^pi|x|cos(n,x),dx=frac{2}{pi}int_0^pi xcos(n,x),dx,quad nge0.
                $$
                This integral easily computed using integration by parts.






                share|cite|improve this answer









                $endgroup$
















                  1












                  1








                  1





                  $begingroup$

                  It should be
                  $$
                  |x|=frac{pi}{2}-frac{4}{pi}sum_{n=1}^{infty}frac{cos(2n-1)x}{(2n-1)^2}.
                  $$
                  Since $|x|$ is even, there are no sine terms, and the coefficient of $cos(n,x)$ is
                  $$
                  frac{1}{pi}int_{-pi}^pi|x|cos(n,x),dx=frac{2}{pi}int_0^pi xcos(n,x),dx,quad nge0.
                  $$
                  This integral easily computed using integration by parts.






                  share|cite|improve this answer









                  $endgroup$



                  It should be
                  $$
                  |x|=frac{pi}{2}-frac{4}{pi}sum_{n=1}^{infty}frac{cos(2n-1)x}{(2n-1)^2}.
                  $$
                  Since $|x|$ is even, there are no sine terms, and the coefficient of $cos(n,x)$ is
                  $$
                  frac{1}{pi}int_{-pi}^pi|x|cos(n,x),dx=frac{2}{pi}int_0^pi xcos(n,x),dx,quad nge0.
                  $$
                  This integral easily computed using integration by parts.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 13 '18 at 18:41









                  Julián AguirreJulián Aguirre

                  68.7k24096




                  68.7k24096






























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