Finding a series of cosines to represent the absolute value function












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












$begingroup$


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






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 13 '18 at 17:00









Uri George PeterzilUri George Peterzil

949




949








  • 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
2






active

oldest

votes


















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









$endgroup$





















    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









    $endgroup$













      Your Answer





      StackExchange.ifUsing("editor", function () {
      return StackExchange.using("mathjaxEditing", function () {
      StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
      StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
      });
      });
      }, "mathjax-editing");

      StackExchange.ready(function() {
      var channelOptions = {
      tags: "".split(" "),
      id: "69"
      };
      initTagRenderer("".split(" "), "".split(" "), channelOptions);

      StackExchange.using("externalEditor", function() {
      // Have to fire editor after snippets, if snippets enabled
      if (StackExchange.settings.snippets.snippetsEnabled) {
      StackExchange.using("snippets", function() {
      createEditor();
      });
      }
      else {
      createEditor();
      }
      });

      function createEditor() {
      StackExchange.prepareEditor({
      heartbeatType: 'answer',
      autoActivateHeartbeat: false,
      convertImagesToLinks: true,
      noModals: true,
      showLowRepImageUploadWarning: true,
      reputationToPostImages: 10,
      bindNavPrevention: true,
      postfix: "",
      imageUploader: {
      brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
      contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
      allowUrls: true
      },
      noCode: true, onDemand: true,
      discardSelector: ".discard-answer"
      ,immediatelyShowMarkdownHelp:true
      });


      }
      });














      draft saved

      draft discarded


















      StackExchange.ready(
      function () {
      StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3038288%2ffinding-a-series-of-cosines-to-represent-the-absolute-value-function%23new-answer', 'question_page');
      }
      );

      Post as a guest















      Required, but never shown

























      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      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

          616516




          616516























              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

                  69.1k24096




                  69.1k24096






























                      draft saved

                      draft discarded




















































                      Thanks for contributing an answer to Mathematics Stack Exchange!


                      • Please be sure to answer the question. Provide details and share your research!

                      But avoid …



                      • Asking for help, clarification, or responding to other answers.

                      • Making statements based on opinion; back them up with references or personal experience.


                      Use MathJax to format equations. MathJax reference.


                      To learn more, see our tips on writing great answers.




                      draft saved


                      draft discarded














                      StackExchange.ready(
                      function () {
                      StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3038288%2ffinding-a-series-of-cosines-to-represent-the-absolute-value-function%23new-answer', 'question_page');
                      }
                      );

                      Post as a guest















                      Required, but never shown





















































                      Required, but never shown














                      Required, but never shown












                      Required, but never shown







                      Required, but never shown

































                      Required, but never shown














                      Required, but never shown












                      Required, but never shown







                      Required, but never shown







                      -

                      Popular posts from this blog

                      Plaza Victoria

                      Image
                      Plaza Victoria Plaza Victoria en 2017 Otros nombres Plaza de La Victoria Localización El Almendral, Valparaíso, Chile Vías adyacentes Chacabuco, Molina, Plaza Victoria y Edwards Creación 1841 Metro Bellavista Ubicación 33°02′46″S 71°37′11″O  /  -33.04624167, -71.61980833 Coordenadas: 33°02′46″S 71°37′11″O  /  -33.04624167, -71.61980833 [editar datos en Wikidata] La Plaza Victoria , también llamada Plaza de la Victoria , [ 1 ] ​ es una plaza ubicada en el sector El Almendral del plan de la ciudad de Valparaíso, Chile, de importancia patrimonial, que incluye valiosas estatuas y especies vegetales. [ 2 ] ​ La plaza se creó como tal en la primera mitad del siglo XIX, tomando previamente los nombres de Plaza Nueva y Plaza de Orrego . Si bien desde sus inicios tuvo relevancia histórica, no fue sino a partir de fines de los años 1860 cuando comenzó a cobrar mayor importancia y se convirtió en uno de los principales centros sociales de l
                      Read more

                      Puebla de Zaragoza

                      Image
                      Para otros usos de este término, véase Puebla (desambiguación). Heroica Puebla de Zaragoza Localidad De arriba a abajo de izquierda a derecha: Estrella de Puebla, Volcán Popocatépetl, Catedral de Puebla, Monumento a Ignacio Zaragoza, Estadio Cuauhtémoc, Angelópolis y Vista panorámica de la ciudad. Escudo Otros nombres: El relicario de América/ La ciudad de los Ángeles/ La Angelópolis Lema: Muy noble y muy leal ciudad de Puebla de los Ángeles Heroica Puebla de Zaragoza Localización de Heroica Puebla de Zaragoza en México Heroica Puebla de Zaragoza Localización de Heroica Puebla de Zaragoza en Puebla Mapa interactivo Coordenadas 19°03′05″N 98°13′04″O  /  19.051388888889, -98.217777777778 Coordenadas: 19°03′05″N 98°13′04″O  /  19.051388888889, -98.217777777778 Idioma oficial Español Entidad Localidad  • País   México  • Estado Puebla  • Municipio Puebla Presidente municipal Claudia Rivera Vivanc
                      Read more

                      Musa

                      Image
                      Para el género de plantas, véase Musa (planta). Apolo y la Nueve Musas En la mitología griega, las musas (en griego antiguo μοῦσαι « mousai» ) son, según los escritores más antiguos, las divinidades inspiradoras de las Artes: cada una de ellas está relacionada con ramas artísticas y del conocimiento. Son hijas de Zeus y de Mnemósine, compañeras del séquito de Apolo, dios olímpico de la música y patrón de las bellas artes, quien tuvo romances con cada una de ellas, dejando descendientes. Bajaban a la tierra a susurrar ideas e inspirar a aquellos mortales que las invocaran. En la época más arcaica eran las ninfas inspiradoras de las fuentes, en las cuales eran adoradas. Finalmente, alrededor de los siglos VIII-VII a. C. [ 1 ] ​ prevaleció en todo el territorio de la Hélade la adoración de las nueve Musas, que son Calíope, Clío, Erato, Euterpe, Melpómene, Polimnia, Talía, Terpsícore y Urania. El culto a las musas era originalmente de Tracia y Beocia, y fueron de vital impo
                      Read more