Results 1 to 3 of 3

Math Help - Fourier series no.2 :)

  1. #1
    Member
    Joined
    May 2008
    Posts
    171

    Fourier series no.2 :)

    Hey guys.
    So I have this function f(x) between -pi and pi.
    I found the fourier series for it.
    Now I need to find the fourier series for g(x). The problem is, I'm not sure about g(x), is it correct what I did?
    Thanks in advance.
    Attached Thumbnails Attached Thumbnails Fourier series no.2 :)-1.jpg  
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    9
    Quote Originally Posted by asi123 View Post
    Hey guys.
    So I have this function f(x) between -pi and pi.
    I found the fourier series for it.
    Now I need to find the fourier series for g(x). The problem is, I'm not sure about g(x), is it correct what I did?
    Thanks in advance.
    Say that, f(x) = \frac{a_0}{2} + \sum_{n=1}^{\infty} a_n \cos nx + b_n \sin nx (we can think of f as extended periodically).

    This means,
    f(x+y) = \frac{a_0}{2} + \sum_{n=1}^{\infty} a_n \cos [n(x+y)] + b_n \sin [n(x+y)]

    But,
    \cos [n(x+y)] = \cos [nx + ny] = \cos (nx) \cos (ny) - \sin (nx) \sin (ny)
    \sin [n(x+y)] = \sin [nx + ny] = \sin (nx) \cos (ny) + \cos (nx) \sin (ny)

    Therefore,
    g(x) = \frac{a_0}{2} + \sum_{n=1}^{\infty} A_n \cos (nx) + B_n \sin (nx)

    Where, A_n = a_n \cos (ny) + b_n\sin (ny) now find B_n.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    May 2008
    Posts
    171
    Quote Originally Posted by ThePerfectHacker View Post
    Say that, f(x) = \frac{a_0}{2} + \sum_{n=1}^{\infty} a_n \cos nx + b_n \sin nx (we can think of f as extended periodically).

    This means,
    f(x+y) = \frac{a_0}{2} + \sum_{n=1}^{\infty} a_n \cos [n(x+y)] + b_n \sin [n(x+y)]

    But,
    \cos [n(x+y)] = \cos [nx + ny] = \cos (nx) \cos (ny) - \sin (nx) \sin (ny)
    \sin [n(x+y)] = \sin [nx + ny] = \sin (nx) \cos (ny) + \cos (nx) \sin (ny)

    Therefore,
    g(x) = \frac{a_0}{2} + \sum_{n=1}^{\infty} A_n \cos (nx) + B_n \sin (nx)

    Where, A_n = a_n \cos (ny) + b_n\sin (ny) now find B_n.
    Thanks a lot.
    Another thing, what about the complex "version".
    Is it correct what I did in the pic?
    I'm not sure about the last part, can I say that f and g are actually the same?

    And another thing (last one).
    a_n and b_n also depends on x (cos(nx), sin(nx)). is changing only the outside cos(nx) and sin(nx) is enough? shouldn't I also place x+y inside of a_n and b_n instead of the x or maybe to recalculate them using the c_n?
    Attached Thumbnails Attached Thumbnails Fourier series no.2 :)-1.jpg  
    Last edited by asi123; January 13th 2009 at 01:22 PM.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. [SOLVED] Fourier series to calculate an infinite series
    Posted in the Differential Geometry Forum
    Replies: 1
    Last Post: August 4th 2010, 01:49 PM
  2. Complex Fourier Series & Full Fourier Series
    Posted in the Calculus Forum
    Replies: 1
    Last Post: December 9th 2009, 05:39 AM
  3. Fourier Series
    Posted in the Calculus Forum
    Replies: 4
    Last Post: March 15th 2009, 09:30 AM
  4. Fourier series
    Posted in the Calculus Forum
    Replies: 7
    Last Post: December 9th 2008, 09:51 PM
  5. from fourier transform to fourier series
    Posted in the Calculus Forum
    Replies: 1
    Last Post: February 1st 2008, 06:35 AM

Search Tags


/mathhelpforum @mathhelpforum