Fourier cosine series for $cos x$












0












$begingroup$


I was trying to find the Fourier cosine series of the function $cos x$ in $[0, pi]$. But I am getting all $a_n$ zero. How to proceed?






fourier-series







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$endgroup$








  • 2




    $begingroup$
    Can you show your development ?
    $endgroup$
    – Yves Daoust
    Mar 9 '16 at 11:10










  • $begingroup$
    You sure the terms are #all# zero? Doesn't the product cos $mx$ cos $nx$ have a different sort of antiderivative when $m=n$ versus otherwise?
    $endgroup$
    – Oscar Lanzi
    Mar 9 '16 at 11:19






  • 1




    $begingroup$
    @Lanzi Sorry, I overlooked the case when $m=n$. Now I got $a_1=1$ and other $a_i$'s are zero.
    $endgroup$
    – user3575652
    Mar 10 '16 at 10:51








  • 1




    $begingroup$
    Note that if you approximate $f(x)$ with some collection of functions that includes $f(x)$ then it perfectly approximates itself.
    $endgroup$
    – CyclotomicField
    Jul 5 '18 at 4:03
















0












$begingroup$


I was trying to find the Fourier cosine series of the function $cos x$ in $[0, pi]$. But I am getting all $a_n$ zero. How to proceed?






fourier-series







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    Can you show your development ?
    $endgroup$
    – Yves Daoust
    Mar 9 '16 at 11:10










  • $begingroup$
    You sure the terms are #all# zero? Doesn't the product cos $mx$ cos $nx$ have a different sort of antiderivative when $m=n$ versus otherwise?
    $endgroup$
    – Oscar Lanzi
    Mar 9 '16 at 11:19






  • 1




    $begingroup$
    @Lanzi Sorry, I overlooked the case when $m=n$. Now I got $a_1=1$ and other $a_i$'s are zero.
    $endgroup$
    – user3575652
    Mar 10 '16 at 10:51








  • 1




    $begingroup$
    Note that if you approximate $f(x)$ with some collection of functions that includes $f(x)$ then it perfectly approximates itself.
    $endgroup$
    – CyclotomicField
    Jul 5 '18 at 4:03














0












0








0





$begingroup$


I was trying to find the Fourier cosine series of the function $cos x$ in $[0, pi]$. But I am getting all $a_n$ zero. How to proceed?






fourier-series







share|cite|improve this question











$endgroup$




I was trying to find the Fourier cosine series of the function $cos x$ in $[0, pi]$. But I am getting all $a_n$ zero. How to proceed?






fourier-series




fourier-series






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Mar 9 '16 at 11:11









H. R.

9,42093262




9,42093262










asked Mar 9 '16 at 11:05









user3575652user3575652

87112




87112








  • 2




    $begingroup$
    Can you show your development ?
    $endgroup$
    – Yves Daoust
    Mar 9 '16 at 11:10










  • $begingroup$
    You sure the terms are #all# zero? Doesn't the product cos $mx$ cos $nx$ have a different sort of antiderivative when $m=n$ versus otherwise?
    $endgroup$
    – Oscar Lanzi
    Mar 9 '16 at 11:19






  • 1




    $begingroup$
    @Lanzi Sorry, I overlooked the case when $m=n$. Now I got $a_1=1$ and other $a_i$'s are zero.
    $endgroup$
    – user3575652
    Mar 10 '16 at 10:51








  • 1




    $begingroup$
    Note that if you approximate $f(x)$ with some collection of functions that includes $f(x)$ then it perfectly approximates itself.
    $endgroup$
    – CyclotomicField
    Jul 5 '18 at 4:03














  • 2




    $begingroup$
    Can you show your development ?
    $endgroup$
    – Yves Daoust
    Mar 9 '16 at 11:10










  • $begingroup$
    You sure the terms are #all# zero? Doesn't the product cos $mx$ cos $nx$ have a different sort of antiderivative when $m=n$ versus otherwise?
    $endgroup$
    – Oscar Lanzi
    Mar 9 '16 at 11:19






  • 1




    $begingroup$
    @Lanzi Sorry, I overlooked the case when $m=n$. Now I got $a_1=1$ and other $a_i$'s are zero.
    $endgroup$
    – user3575652
    Mar 10 '16 at 10:51








  • 1




    $begingroup$
    Note that if you approximate $f(x)$ with some collection of functions that includes $f(x)$ then it perfectly approximates itself.
    $endgroup$
    – CyclotomicField
    Jul 5 '18 at 4:03








2




2




$begingroup$
Can you show your development ?
$endgroup$
– Yves Daoust
Mar 9 '16 at 11:10




$begingroup$
Can you show your development ?
$endgroup$
– Yves Daoust
Mar 9 '16 at 11:10












$begingroup$
You sure the terms are #all# zero? Doesn't the product cos $mx$ cos $nx$ have a different sort of antiderivative when $m=n$ versus otherwise?
$endgroup$
– Oscar Lanzi
Mar 9 '16 at 11:19




$begingroup$
You sure the terms are #all# zero? Doesn't the product cos $mx$ cos $nx$ have a different sort of antiderivative when $m=n$ versus otherwise?
$endgroup$
– Oscar Lanzi
Mar 9 '16 at 11:19




1




1




$begingroup$
@Lanzi Sorry, I overlooked the case when $m=n$. Now I got $a_1=1$ and other $a_i$'s are zero.
$endgroup$
– user3575652
Mar 10 '16 at 10:51






$begingroup$
@Lanzi Sorry, I overlooked the case when $m=n$. Now I got $a_1=1$ and other $a_i$'s are zero.
$endgroup$
– user3575652
Mar 10 '16 at 10:51






1




1




$begingroup$
Note that if you approximate $f(x)$ with some collection of functions that includes $f(x)$ then it perfectly approximates itself.
$endgroup$
– CyclotomicField
Jul 5 '18 at 4:03




$begingroup$
Note that if you approximate $f(x)$ with some collection of functions that includes $f(x)$ then it perfectly approximates itself.
$endgroup$
– CyclotomicField
Jul 5 '18 at 4:03










3 Answers
3






active

oldest

votes


















0












$begingroup$

Don't forget the sine terms:



begin{align*}
f(x+2L) &= f(x) \
f(x) &=frac{a_{0}}{2}+
sum_{n=1}^{infty}
left(
a_{n}cos frac{npi x}{L}+
b_{n}sin frac{npi x}{L}
right) \
a_{n} &= frac{1}{L}int_{0}^{2L} f(t)cos frac{npi t}{L} dt \
&= frac{2}{pi}int_{0}^{pi} cos t cos (2nt) dt \
&= 0 \
b_{n} &= frac{1}{L}int_{0}^{2L} f(t)sin frac{npi t}{L} dt \
&= frac{2}{pi}int_{0}^{pi} cos t sin (2nt) dt \
&= frac{8n}{(4n^{2}-1)pi}
end{align*}






share|cite|improve this answer









$endgroup$













  • $begingroup$
    "Fourier cosine series" implies a different formula, with $cos(nt)$ instead of $cos(t).$ See mathworld.wolfram.com/FourierCosineSeries.html
    $endgroup$
    – David K
    May 3 '18 at 0:28



















0












$begingroup$

This question seems to have been answered in effect in comments, but here the answer is fleshed out in the "answer" section.



Under what seems to be the usual interpretation of a "Fourier cosine series",
you treat the region of integration $[0,pi]$ as just one half of the period of a periodic function, with the other half defined by the assumption that the function is even.
When you have an even function with period $2pi$ such as $cos x,$ its Fourier cosine series on $[0,pi]$ is identical to the Fourier series of the function on $[-pi,pi],$ which in turn can be computed by integrating over just the interval $[0,pi].$ At least that's how I look at it.



The formula for the coefficients of the series (as shown in Wolfram Mathworld, for example), is
$$
a_n = frac2pi int_0^pi f(x)cos(n x) dx,
$$
and setting $f(x) = cos x,$ this formula gives $a_1 = 1.$






share|cite|improve this answer









$endgroup$





















    0












    $begingroup$

    $cosleft(xright)$ is itself a basis vector, so without any calculation you already know there is exactly one non-zero coefficient and its value is $1$... You can find that coefficient with a glance at the basis (and its numbering).






    share|cite|improve this answer









    $endgroup$













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      3 Answers
      3






      active

      oldest

      votes








      3 Answers
      3






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      0












      $begingroup$

      Don't forget the sine terms:



      begin{align*}
      f(x+2L) &= f(x) \
      f(x) &=frac{a_{0}}{2}+
      sum_{n=1}^{infty}
      left(
      a_{n}cos frac{npi x}{L}+
      b_{n}sin frac{npi x}{L}
      right) \
      a_{n} &= frac{1}{L}int_{0}^{2L} f(t)cos frac{npi t}{L} dt \
      &= frac{2}{pi}int_{0}^{pi} cos t cos (2nt) dt \
      &= 0 \
      b_{n} &= frac{1}{L}int_{0}^{2L} f(t)sin frac{npi t}{L} dt \
      &= frac{2}{pi}int_{0}^{pi} cos t sin (2nt) dt \
      &= frac{8n}{(4n^{2}-1)pi}
      end{align*}






      share|cite|improve this answer









      $endgroup$













      • $begingroup$
        "Fourier cosine series" implies a different formula, with $cos(nt)$ instead of $cos(t).$ See mathworld.wolfram.com/FourierCosineSeries.html
        $endgroup$
        – David K
        May 3 '18 at 0:28
















      0












      $begingroup$

      Don't forget the sine terms:



      begin{align*}
      f(x+2L) &= f(x) \
      f(x) &=frac{a_{0}}{2}+
      sum_{n=1}^{infty}
      left(
      a_{n}cos frac{npi x}{L}+
      b_{n}sin frac{npi x}{L}
      right) \
      a_{n} &= frac{1}{L}int_{0}^{2L} f(t)cos frac{npi t}{L} dt \
      &= frac{2}{pi}int_{0}^{pi} cos t cos (2nt) dt \
      &= 0 \
      b_{n} &= frac{1}{L}int_{0}^{2L} f(t)sin frac{npi t}{L} dt \
      &= frac{2}{pi}int_{0}^{pi} cos t sin (2nt) dt \
      &= frac{8n}{(4n^{2}-1)pi}
      end{align*}






      share|cite|improve this answer









      $endgroup$













      • $begingroup$
        "Fourier cosine series" implies a different formula, with $cos(nt)$ instead of $cos(t).$ See mathworld.wolfram.com/FourierCosineSeries.html
        $endgroup$
        – David K
        May 3 '18 at 0:28














      0












      0








      0





      $begingroup$

      Don't forget the sine terms:



      begin{align*}
      f(x+2L) &= f(x) \
      f(x) &=frac{a_{0}}{2}+
      sum_{n=1}^{infty}
      left(
      a_{n}cos frac{npi x}{L}+
      b_{n}sin frac{npi x}{L}
      right) \
      a_{n} &= frac{1}{L}int_{0}^{2L} f(t)cos frac{npi t}{L} dt \
      &= frac{2}{pi}int_{0}^{pi} cos t cos (2nt) dt \
      &= 0 \
      b_{n} &= frac{1}{L}int_{0}^{2L} f(t)sin frac{npi t}{L} dt \
      &= frac{2}{pi}int_{0}^{pi} cos t sin (2nt) dt \
      &= frac{8n}{(4n^{2}-1)pi}
      end{align*}






      share|cite|improve this answer









      $endgroup$



      Don't forget the sine terms:



      begin{align*}
      f(x+2L) &= f(x) \
      f(x) &=frac{a_{0}}{2}+
      sum_{n=1}^{infty}
      left(
      a_{n}cos frac{npi x}{L}+
      b_{n}sin frac{npi x}{L}
      right) \
      a_{n} &= frac{1}{L}int_{0}^{2L} f(t)cos frac{npi t}{L} dt \
      &= frac{2}{pi}int_{0}^{pi} cos t cos (2nt) dt \
      &= 0 \
      b_{n} &= frac{1}{L}int_{0}^{2L} f(t)sin frac{npi t}{L} dt \
      &= frac{2}{pi}int_{0}^{pi} cos t sin (2nt) dt \
      &= frac{8n}{(4n^{2}-1)pi}
      end{align*}







      share|cite|improve this answer












      share|cite|improve this answer



      share|cite|improve this answer










      answered Mar 9 '16 at 12:45









      Ng Chung TakNg Chung Tak

      14.3k31334




      14.3k31334












      • $begingroup$
        "Fourier cosine series" implies a different formula, with $cos(nt)$ instead of $cos(t).$ See mathworld.wolfram.com/FourierCosineSeries.html
        $endgroup$
        – David K
        May 3 '18 at 0:28


















      • $begingroup$
        "Fourier cosine series" implies a different formula, with $cos(nt)$ instead of $cos(t).$ See mathworld.wolfram.com/FourierCosineSeries.html
        $endgroup$
        – David K
        May 3 '18 at 0:28
















      $begingroup$
      "Fourier cosine series" implies a different formula, with $cos(nt)$ instead of $cos(t).$ See mathworld.wolfram.com/FourierCosineSeries.html
      $endgroup$
      – David K
      May 3 '18 at 0:28




      $begingroup$
      "Fourier cosine series" implies a different formula, with $cos(nt)$ instead of $cos(t).$ See mathworld.wolfram.com/FourierCosineSeries.html
      $endgroup$
      – David K
      May 3 '18 at 0:28











      0












      $begingroup$

      This question seems to have been answered in effect in comments, but here the answer is fleshed out in the "answer" section.



      Under what seems to be the usual interpretation of a "Fourier cosine series",
      you treat the region of integration $[0,pi]$ as just one half of the period of a periodic function, with the other half defined by the assumption that the function is even.
      When you have an even function with period $2pi$ such as $cos x,$ its Fourier cosine series on $[0,pi]$ is identical to the Fourier series of the function on $[-pi,pi],$ which in turn can be computed by integrating over just the interval $[0,pi].$ At least that's how I look at it.



      The formula for the coefficients of the series (as shown in Wolfram Mathworld, for example), is
      $$
      a_n = frac2pi int_0^pi f(x)cos(n x) dx,
      $$
      and setting $f(x) = cos x,$ this formula gives $a_1 = 1.$






      share|cite|improve this answer









      $endgroup$


















        0












        $begingroup$

        This question seems to have been answered in effect in comments, but here the answer is fleshed out in the "answer" section.



        Under what seems to be the usual interpretation of a "Fourier cosine series",
        you treat the region of integration $[0,pi]$ as just one half of the period of a periodic function, with the other half defined by the assumption that the function is even.
        When you have an even function with period $2pi$ such as $cos x,$ its Fourier cosine series on $[0,pi]$ is identical to the Fourier series of the function on $[-pi,pi],$ which in turn can be computed by integrating over just the interval $[0,pi].$ At least that's how I look at it.



        The formula for the coefficients of the series (as shown in Wolfram Mathworld, for example), is
        $$
        a_n = frac2pi int_0^pi f(x)cos(n x) dx,
        $$
        and setting $f(x) = cos x,$ this formula gives $a_1 = 1.$






        share|cite|improve this answer









        $endgroup$
















          0












          0








          0





          $begingroup$

          This question seems to have been answered in effect in comments, but here the answer is fleshed out in the "answer" section.



          Under what seems to be the usual interpretation of a "Fourier cosine series",
          you treat the region of integration $[0,pi]$ as just one half of the period of a periodic function, with the other half defined by the assumption that the function is even.
          When you have an even function with period $2pi$ such as $cos x,$ its Fourier cosine series on $[0,pi]$ is identical to the Fourier series of the function on $[-pi,pi],$ which in turn can be computed by integrating over just the interval $[0,pi].$ At least that's how I look at it.



          The formula for the coefficients of the series (as shown in Wolfram Mathworld, for example), is
          $$
          a_n = frac2pi int_0^pi f(x)cos(n x) dx,
          $$
          and setting $f(x) = cos x,$ this formula gives $a_1 = 1.$






          share|cite|improve this answer









          $endgroup$



          This question seems to have been answered in effect in comments, but here the answer is fleshed out in the "answer" section.



          Under what seems to be the usual interpretation of a "Fourier cosine series",
          you treat the region of integration $[0,pi]$ as just one half of the period of a periodic function, with the other half defined by the assumption that the function is even.
          When you have an even function with period $2pi$ such as $cos x,$ its Fourier cosine series on $[0,pi]$ is identical to the Fourier series of the function on $[-pi,pi],$ which in turn can be computed by integrating over just the interval $[0,pi].$ At least that's how I look at it.



          The formula for the coefficients of the series (as shown in Wolfram Mathworld, for example), is
          $$
          a_n = frac2pi int_0^pi f(x)cos(n x) dx,
          $$
          and setting $f(x) = cos x,$ this formula gives $a_1 = 1.$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered May 3 '18 at 1:23









          David KDavid K

          52.9k340115




          52.9k340115























              0












              $begingroup$

              $cosleft(xright)$ is itself a basis vector, so without any calculation you already know there is exactly one non-zero coefficient and its value is $1$... You can find that coefficient with a glance at the basis (and its numbering).






              share|cite|improve this answer









              $endgroup$


















                0












                $begingroup$

                $cosleft(xright)$ is itself a basis vector, so without any calculation you already know there is exactly one non-zero coefficient and its value is $1$... You can find that coefficient with a glance at the basis (and its numbering).






                share|cite|improve this answer









                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  $cosleft(xright)$ is itself a basis vector, so without any calculation you already know there is exactly one non-zero coefficient and its value is $1$... You can find that coefficient with a glance at the basis (and its numbering).






                  share|cite|improve this answer









                  $endgroup$



                  $cosleft(xright)$ is itself a basis vector, so without any calculation you already know there is exactly one non-zero coefficient and its value is $1$... You can find that coefficient with a glance at the basis (and its numbering).







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered May 3 '18 at 1:35









                  mol3574710n0fN074710nmol3574710n0fN074710n

                  524311




                  524311






























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