Find the half-range Fourier series expansion of $f(x) = cos(x)$












2












$begingroup$


I am stuck on the problem of calculating the half-range Fourier series expansion of $$f(x) = cos(x),$$ $$0 < x < frac{pi}{2}$$



I am at the point where I have calculated the definite integral of $b_n$.



$$b_n = frac{4}{pi}int_{0}^{frac{pi}{2}} cos(x)sin(nx) dx = frac{4}{pi} cdot frac{n - sin(frac{pi n}{2})}{(n^2 - 1)}$$




According to this Wikipedia link, the correct answer for $b_n$ is



$$b_n = frac{4}{pi} int_{0}^{frac{pi}{2}} cos(x)sin(nx) dx = frac{4}{pi} cdot frac{n((-1)^n + 1)}{(n^2 - 1)}.$$



Can someone please explain to me how to go from
$$frac{n - sin(frac{pi n}{2})}{(n^2 - 1)}$$
to $$frac{n((-1)^n + 1)}{(n^2 - 1)}.$$



I understand that, for example, $$cos(npi)$$ can be written as $$(-1)^n,$$
because $cos(npi)$ has alternating $1's$ and $0's$ as $n$ increases.

However, for $sin(frac{npi}{2})$ the alteration is $1$, $0$, $-1$, $0$.

I have been stuck on this problem for hours now.
I will greatly appreciate any explanation :)






fourier-series







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  • $begingroup$
    Recheck your work for $b_n$. The formula doesn't look right - the numerator should have a factor of $n$ in the least.
    $endgroup$
    – A.S.
    Oct 17 '15 at 10:25


















2












$begingroup$


I am stuck on the problem of calculating the half-range Fourier series expansion of $$f(x) = cos(x),$$ $$0 < x < frac{pi}{2}$$



I am at the point where I have calculated the definite integral of $b_n$.



$$b_n = frac{4}{pi}int_{0}^{frac{pi}{2}} cos(x)sin(nx) dx = frac{4}{pi} cdot frac{n - sin(frac{pi n}{2})}{(n^2 - 1)}$$




According to this Wikipedia link, the correct answer for $b_n$ is



$$b_n = frac{4}{pi} int_{0}^{frac{pi}{2}} cos(x)sin(nx) dx = frac{4}{pi} cdot frac{n((-1)^n + 1)}{(n^2 - 1)}.$$



Can someone please explain to me how to go from
$$frac{n - sin(frac{pi n}{2})}{(n^2 - 1)}$$
to $$frac{n((-1)^n + 1)}{(n^2 - 1)}.$$



I understand that, for example, $$cos(npi)$$ can be written as $$(-1)^n,$$
because $cos(npi)$ has alternating $1's$ and $0's$ as $n$ increases.

However, for $sin(frac{npi}{2})$ the alteration is $1$, $0$, $-1$, $0$.

I have been stuck on this problem for hours now.
I will greatly appreciate any explanation :)






fourier-series







share|cite|improve this question









$endgroup$












  • $begingroup$
    Recheck your work for $b_n$. The formula doesn't look right - the numerator should have a factor of $n$ in the least.
    $endgroup$
    – A.S.
    Oct 17 '15 at 10:25
















2












2








2


2



$begingroup$


I am stuck on the problem of calculating the half-range Fourier series expansion of $$f(x) = cos(x),$$ $$0 < x < frac{pi}{2}$$



I am at the point where I have calculated the definite integral of $b_n$.



$$b_n = frac{4}{pi}int_{0}^{frac{pi}{2}} cos(x)sin(nx) dx = frac{4}{pi} cdot frac{n - sin(frac{pi n}{2})}{(n^2 - 1)}$$




According to this Wikipedia link, the correct answer for $b_n$ is



$$b_n = frac{4}{pi} int_{0}^{frac{pi}{2}} cos(x)sin(nx) dx = frac{4}{pi} cdot frac{n((-1)^n + 1)}{(n^2 - 1)}.$$



Can someone please explain to me how to go from
$$frac{n - sin(frac{pi n}{2})}{(n^2 - 1)}$$
to $$frac{n((-1)^n + 1)}{(n^2 - 1)}.$$



I understand that, for example, $$cos(npi)$$ can be written as $$(-1)^n,$$
because $cos(npi)$ has alternating $1's$ and $0's$ as $n$ increases.

However, for $sin(frac{npi}{2})$ the alteration is $1$, $0$, $-1$, $0$.

I have been stuck on this problem for hours now.
I will greatly appreciate any explanation :)






fourier-series







share|cite|improve this question









$endgroup$




I am stuck on the problem of calculating the half-range Fourier series expansion of $$f(x) = cos(x),$$ $$0 < x < frac{pi}{2}$$



I am at the point where I have calculated the definite integral of $b_n$.



$$b_n = frac{4}{pi}int_{0}^{frac{pi}{2}} cos(x)sin(nx) dx = frac{4}{pi} cdot frac{n - sin(frac{pi n}{2})}{(n^2 - 1)}$$




According to this Wikipedia link, the correct answer for $b_n$ is



$$b_n = frac{4}{pi} int_{0}^{frac{pi}{2}} cos(x)sin(nx) dx = frac{4}{pi} cdot frac{n((-1)^n + 1)}{(n^2 - 1)}.$$



Can someone please explain to me how to go from
$$frac{n - sin(frac{pi n}{2})}{(n^2 - 1)}$$
to $$frac{n((-1)^n + 1)}{(n^2 - 1)}.$$



I understand that, for example, $$cos(npi)$$ can be written as $$(-1)^n,$$
because $cos(npi)$ has alternating $1's$ and $0's$ as $n$ increases.

However, for $sin(frac{npi}{2})$ the alteration is $1$, $0$, $-1$, $0$.

I have been stuck on this problem for hours now.
I will greatly appreciate any explanation :)






fourier-series




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asked Oct 17 '15 at 10:17









Joe NiemandJoe Niemand

111




111












  • $begingroup$
    Recheck your work for $b_n$. The formula doesn't look right - the numerator should have a factor of $n$ in the least.
    $endgroup$
    – A.S.
    Oct 17 '15 at 10:25




















  • $begingroup$
    Recheck your work for $b_n$. The formula doesn't look right - the numerator should have a factor of $n$ in the least.
    $endgroup$
    – A.S.
    Oct 17 '15 at 10:25


















$begingroup$
Recheck your work for $b_n$. The formula doesn't look right - the numerator should have a factor of $n$ in the least.
$endgroup$
– A.S.
Oct 17 '15 at 10:25






$begingroup$
Recheck your work for $b_n$. The formula doesn't look right - the numerator should have a factor of $n$ in the least.
$endgroup$
– A.S.
Oct 17 '15 at 10:25












2 Answers
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$begingroup$

for $n=1,5,9,....$ the $sin(frac{npi}{2})=1$
$$frac{n-sin(frac{npi}{2})}{n^2-1}=frac{n-1}{n^2-1}=frac{1}{n+1}$$



for $n=2,4,6,8....$ all terms equal zero



for $n=3,7,11,...$ the $sin(frac{npi}{2})=-1$
$$frac{n-sin(frac{npi}{2})}{n^2-1}=frac{n+1}{n^2-1}=frac{1}{n-1}$$



now add the $frac{1}{n+1}$ with $frac{1}{n-1}$ to get what you want
$$frac{1}{n+1}+frac{1}{n-1}=frac{2n}{n^2-1}=frac{n((-1)^n + 1)}{(n^2 - 1)}.$$






share|cite|improve this answer









$endgroup$





















    0












    $begingroup$

    In that link the integral is from $0$ to $pi$, but your integral is from $0$ to $frac{pi}{2}$






    share|cite|improve this answer









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      2 Answers
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      2 Answers
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      0












      $begingroup$

      for $n=1,5,9,....$ the $sin(frac{npi}{2})=1$
      $$frac{n-sin(frac{npi}{2})}{n^2-1}=frac{n-1}{n^2-1}=frac{1}{n+1}$$



      for $n=2,4,6,8....$ all terms equal zero



      for $n=3,7,11,...$ the $sin(frac{npi}{2})=-1$
      $$frac{n-sin(frac{npi}{2})}{n^2-1}=frac{n+1}{n^2-1}=frac{1}{n-1}$$



      now add the $frac{1}{n+1}$ with $frac{1}{n-1}$ to get what you want
      $$frac{1}{n+1}+frac{1}{n-1}=frac{2n}{n^2-1}=frac{n((-1)^n + 1)}{(n^2 - 1)}.$$






      share|cite|improve this answer









      $endgroup$


















        0












        $begingroup$

        for $n=1,5,9,....$ the $sin(frac{npi}{2})=1$
        $$frac{n-sin(frac{npi}{2})}{n^2-1}=frac{n-1}{n^2-1}=frac{1}{n+1}$$



        for $n=2,4,6,8....$ all terms equal zero



        for $n=3,7,11,...$ the $sin(frac{npi}{2})=-1$
        $$frac{n-sin(frac{npi}{2})}{n^2-1}=frac{n+1}{n^2-1}=frac{1}{n-1}$$



        now add the $frac{1}{n+1}$ with $frac{1}{n-1}$ to get what you want
        $$frac{1}{n+1}+frac{1}{n-1}=frac{2n}{n^2-1}=frac{n((-1)^n + 1)}{(n^2 - 1)}.$$






        share|cite|improve this answer









        $endgroup$
















          0












          0








          0





          $begingroup$

          for $n=1,5,9,....$ the $sin(frac{npi}{2})=1$
          $$frac{n-sin(frac{npi}{2})}{n^2-1}=frac{n-1}{n^2-1}=frac{1}{n+1}$$



          for $n=2,4,6,8....$ all terms equal zero



          for $n=3,7,11,...$ the $sin(frac{npi}{2})=-1$
          $$frac{n-sin(frac{npi}{2})}{n^2-1}=frac{n+1}{n^2-1}=frac{1}{n-1}$$



          now add the $frac{1}{n+1}$ with $frac{1}{n-1}$ to get what you want
          $$frac{1}{n+1}+frac{1}{n-1}=frac{2n}{n^2-1}=frac{n((-1)^n + 1)}{(n^2 - 1)}.$$






          share|cite|improve this answer









          $endgroup$



          for $n=1,5,9,....$ the $sin(frac{npi}{2})=1$
          $$frac{n-sin(frac{npi}{2})}{n^2-1}=frac{n-1}{n^2-1}=frac{1}{n+1}$$



          for $n=2,4,6,8....$ all terms equal zero



          for $n=3,7,11,...$ the $sin(frac{npi}{2})=-1$
          $$frac{n-sin(frac{npi}{2})}{n^2-1}=frac{n+1}{n^2-1}=frac{1}{n-1}$$



          now add the $frac{1}{n+1}$ with $frac{1}{n-1}$ to get what you want
          $$frac{1}{n+1}+frac{1}{n-1}=frac{2n}{n^2-1}=frac{n((-1)^n + 1)}{(n^2 - 1)}.$$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Oct 17 '15 at 10:41









          E.H.EE.H.E

          16.9k11969




          16.9k11969























              0












              $begingroup$

              In that link the integral is from $0$ to $pi$, but your integral is from $0$ to $frac{pi}{2}$






              share|cite|improve this answer









              $endgroup$


















                0












                $begingroup$

                In that link the integral is from $0$ to $pi$, but your integral is from $0$ to $frac{pi}{2}$






                share|cite|improve this answer









                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  In that link the integral is from $0$ to $pi$, but your integral is from $0$ to $frac{pi}{2}$






                  share|cite|improve this answer









                  $endgroup$



                  In that link the integral is from $0$ to $pi$, but your integral is from $0$ to $frac{pi}{2}$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Oct 17 '15 at 10:59









                  Fikri Irvan HalimFikri Irvan Halim

                  261




                  261






























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