Results 1 to 4 of 4

Math Help - Fourier Series

  1. #1
    Member Maccaman's Avatar
    Joined
    Sep 2008
    Posts
    85

    Fourier Series

    Find the fourier series for

    f(x) = \left\{\begin{array}{c l} 0 & -\pi < x < 0 \\ 1 & 0 < x < \pi \\ f(x + 2\pi) & -\infty < x < \infty \end{array}\right.<br /> <br />

    By considering the Fourier series at x = \frac{\pi}{2} show that \frac{\pi}{4} = \sum_{m=1}^{\infty} \frac{(-1)^{m+1}}{(2m-1)}

    I am not enjoying these at the moment. I get lost. Really lost.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member
    Joined
    Aug 2008
    Posts
    903
    Start with the definitions:

    A_n=\frac{1}{\pi}\int_{-\pi}^{\pi} f(x) \cos(nx)dx,\quad n=0,1,2,\cdots

    B_n=\frac{1}{\pi}\int_{-\pi}^{\pi} f(x) \sin(nx)dx,\quad n=1,2,\cdots

    You can calculate those right?

    A_n=\frac{1}{\pi} \int_0^{\pi} \cos(nx)dx=\frac{1-\cos(n\pi)}{n},\quad n=1,2,\cdots

    B_n=\frac{1}{\pi} \int_0^{\pi} \sin(nx)dx=0

    That's a good start. Then the Fourier series is:

    f(x)=\frac{1}{2}+\frac{1}{\pi} \sum_{n=1}^{\infty} \frac{1-\cos(n\pi)}{n} \sin(n x)

    Now, evaluate that at x=\pi/2 and convert the sines and cosines to minus one raised to powers and adjust the summation expression accordingly.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member Maccaman's Avatar
    Joined
    Sep 2008
    Posts
    85
    I get  a_n = 0

    b_n = 0 if n is even, b_n = \frac{2}{n\pi} if n is odd

    and

    a_0 = \frac{1}{2}

    So fourier series is

     f(x) = \frac{1}{2} + \sum_{n = 1}^{\infty} \frac{2}{n\pi} \sin (nx) .

    Is this correct?

    Im confused with how to "convert the sines and cosines to minus one raised to powers and adjust the summation expression accordingly." Can someone please explain?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    Opalg's Avatar
    Joined
    Aug 2007
    From
    Leeds, UK
    Posts
    4,041
    Thanks
    7
    Quote Originally Posted by shawsend View Post
    ... the Fourier series is:

    f(x)=\frac{1}{2}+\frac{1}{\pi} \sum_{n=1}^{\infty} \frac{1-\cos(n\pi)}{n} \sin(n x)

    Now, evaluate that at x=\pi/2 and convert the sines and cosines to minus one raised to powers and adjust the summation expression accordingly.
    Quote Originally Posted by Maccaman View Post
    Im confused with how to "convert the sines and cosines to minus one raised to powers and adjust the summation expression accordingly." Can someone please explain?
    \cos(0) = 1,\ \cos(\pi)=-1,\ \cos(2\pi)=1,\ \cos(3\pi) = -1,\ldots. Does that suggest to you that \cos(n\pi) has something to do with 1 raised to some power?
    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: 0
    Last Post: May 30th 2009, 06:55 PM
  4. Fourier series
    Posted in the Calculus Forum
    Replies: 2
    Last Post: April 14th 2009, 04:33 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