Results 1 to 4 of 4
Like Tree2Thanks
  • 1 Post By HallsofIvy
  • 1 Post By zzephod

Math Help - I am stuck in one step of a Fourier sine series

  1. #1
    Junior Member
    Joined
    Sep 2010
    Posts
    64

    I am stuck in one step of a Fourier sine series

    Hi, I had solved this problem in a very complex and lengthy way and got the same result as what our lecturer says we should get. However, his answer (which skips several steps) is much simpler than what I did, but I don't know how he does the last step.

    Here is the problem:

    Find the Fourier sine series of f(t)=2+t on (0, \pi).

    Solution:
    b_k=\frac{2}{\pi}\int_{0}^{\pi}f(t)sin(kt) dt=\frac{2}{\pi}\int_{0}^{\pi}e^{t}sin(kt) dt

    We know that e^{ikt}=cos(kt)+i sin(kt), so we will only use the imaginary part of e^{ikt}

    b_k=\frac{2}{\pi}\Im\int_{0}^{\pi}e^{t} e^{ikt} dt = \frac{2}{\pi}\Im\int_{0}^{\pi}e^{(1+ik)t} dt

    b_k=\frac{2}{\pi}\Im[\frac{e^{(1+ik)t}}{1+ik} ]_{0}^{\pi}

    b_k=\Im (\frac{2}{\pi(1+ik)}\{e^{(1+ik)\pi}-1\})

    And what I don't understand is how it goes from the result above to the final result below:

    =\frac{2k}{\pi(1+k^2)}\{1-(-1)^ke^{\pi}\}

    I would really appreciate it if someone could tell me how this is done! I have no clue where the k in " 2k" and " k^2" come from... Thanks a lot!
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    Apr 2005
    Posts
    15,306
    Thanks
    1282

    Re: I am stuck in one step of a Fourier sine series

    Quote Originally Posted by juanma101285 View Post
    Hi, I had solved this problem in a very complex and lengthy way and got the same result as what our lecturer says we should get. However, his answer (which skips several steps) is much simpler than what I did, but I don't know how he does the last step.

    Here is the problem:

    Find the Fourier sine series of f(t)=2+t on (0, \pi).

    Solution:
    b_k=\frac{2}{\pi}\int_{0}^{\pi}f(t)sin(kt) dt=\frac{2}{\pi}\int_{0}^{\pi}e^{t}sin(kt) dt
    What? You said one line above this that "f(t)= 2+ t". How did it become e^t???
    We know that e^{ikt}=cos(kt)+i sin(kt), so we will only use the imaginary part of e^{ikt}

    b_k=\frac{2}{\pi}\Im\int_{0}^{\pi}e^{t} e^{ikt} dt = \frac{2}{\pi}\Im\int_{0}^{\pi}e^{(1+ik)t} dt

    b_k=\frac{2}{\pi}\Im[\frac{e^{(1+ik)t}}{1+ik} ]_{0}^{\pi}

    b_k=\Im (\frac{2}{\pi(1+ik)}\{e^{(1+ik)\pi}-1\})

    And what I don't understand is how it goes from the result above to the final result below:

    =\frac{2k}{\pi(1+k^2)}\{1-(-1)^ke^{\pi}\}

    I would really appreciate it if someone could tell me how this is done! I have no clue where the k in " 2k" and " k^2" come from... Thanks a lot!
    Thanks from topsquark
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Sep 2010
    Posts
    64

    Re: I am stuck in one step of a Fourier sine series

    Thanks for your reply! Sorry, I wrote it wrong... The problem is:

    Find the Fourier sine series of f(t) = e^t on (0, \pi).
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Member
    Joined
    Apr 2012
    From
    Erewhon
    Posts
    164
    Thanks
    107

    Re: I am stuck in one step of a Fourier sine series

    Quote Originally Posted by juanma101285 View Post
    ..

    b_k=\Im (\frac{2}{\pi(1+ik)}\{e^{(1+ik)\pi}-1\})

    And what I don't understand is how it goes from the result above to the final result below:

    =\frac{2k}{\pi(1+k^2)}\{1-(-1)^ke^{\pi}\}

    I would really appreciate it if someone could tell me how this is done! I have no clue where the k in " 2k" and " k^2" come from... Thanks a lot!
    First look at what is inside the braces:

    \{e^{(1+ik)\pi}-1\} = e^{\pi}e^{ik\pi}-1

    Now observe that e^{ik\pi}=\cos(k\pi)+i\sin(k\pi) and k\in \mathbb{N}, so \cos(k\pi)=(-1)^{k} and \sin(k\pi)=0.

    Hence: e^{ik\pi}=\cos(k\pi)+i\sin(k\pi) = (-1)^k which is real.

    So now we need the imaginary part of what is outside the braces, and you get that by first multiplying top and bottom by (1-ik) ...

    .
    Thanks from topsquark
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Fourier series of step function
    Posted in the Calculus Forum
    Replies: 3
    Last Post: January 17th 2010, 01:57 PM
  2. Fourier sine and cosine series of x(Pi-x)
    Posted in the Advanced Applied Math Forum
    Replies: 3
    Last Post: January 2nd 2010, 12:42 PM
  3. PDE-fourier sine series (help!!!)
    Posted in the Differential Equations Forum
    Replies: 1
    Last Post: May 14th 2009, 05:17 AM
  4. is this a sine or cosine fourier series...
    Posted in the Advanced Math Topics Forum
    Replies: 1
    Last Post: February 14th 2009, 04:04 PM
  5. Fourier sine series...
    Posted in the Calculus Forum
    Replies: 5
    Last Post: February 10th 2008, 07:53 AM

Search Tags


/mathhelpforum @mathhelpforum