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Math Help - diff eqns help plz!

  1. #1
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    diff eqns help plz!

    Find the Fourier series on the interval [-pi, pi] of the EVEN function:

    f(x) = { -x for x < 0
    { 0 for x = 0
    { x for x > 0


    So we are given the hint, that its an even function... which I believe means we only have to solve for the a sub n's..

    However, we just went over this in class on friday... can you show me how to proceed? Thanks in advance!
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  2. #2
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    It's been a while since I've read about this, but I'll give it a try.

    Note that the function f(x)=x\,{\rm{sgn}}(x), where <br />
{\rm{sgn}}(x)=\begin{cases}<br />
-1, & x < 0 \\<br />
0, & x = 0 \\<br />
1, & x > 0<br />
\end{cases}.

    You are correct to note that for an even function, you only need to solve for a_n since b_n is 0 for every n.

    Now, <br />
a_n=\frac1L \int_{-L}^L f(x) \cos(\frac{n \pi x}L)\,dx\\<br />
=\frac1\pi \int_{-\pi}^\pi x\,{\rm{sgn}}(x) \cos(n x)\,dx\\<br />
=\frac2\pi \int_0^\pi x\,\cos(n x)\,dx\\<br />
.

    Finally, after solving the integral above (note that the \sin(n \pi) term always is 0 and \cos(n \pi)=(-1)^n) you'll get the fourier series given by f(x)=\frac{a_0}2+\sum_{n=1}^\infty a_n\, \cos(nx).
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  3. #3
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    fantastic explanation, thank you very much!
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  4. #4
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    diff. eqns.

    Hi Guys,
    I posted this once before, and cyph gave me a nice answer, but my professor told me to take it back home and give it another look...


    Find the Fourier series on the interval [-pi, pi] of the EVEN function:

    f(x) =
    { -x for x < 0
    { 0 for x = 0
    { x for x > 0


    Here's my work:




    Assuming all the above is correct (which it probably isn't) this is where I am stuck, can you help me finish the problem? Thanks very much in advance!
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  5. #5
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    Quote Originally Posted by boousaf View Post
    Hi Guys,
    I posted this once before, and cyph gave me a nice answer, but my professor told me to take it back home and give it another look...


    Find the Fourier series on the interval [-pi, pi] of the EVEN function:

    f(x) =
    { -x for x < 0
    { 0 for x = 0
    { x for x > 0


    Here's my work:




    Assuming all the above is correct (which it probably isn't) this is where I am stuck, can you help me finish the problem? Thanks very much in advance!

    You are trying to find the fourier series of

    f(x)=|x|, x\in [-\pi,\pi]

    a_0=\frac{1}{\pi}\int_{-\pi}^{\pi}|x|dx=2\int_{0}^{\pi}xdx=\pi

    a_n=\frac{1}{\pi}\int_{-\pi}^{\pi}|x|\cos(nx)dx=\frac{2}{\pi}\int_{0}^{\pi  }x\cos(nx)dx=\frac{2}{\pi}\left( \frac{\cos(\pi n)+n\sin(\pi n)-1}{n^2}\right)=
    \frac{2}{\pi}\left( \frac{(-1)^n-1}{n^2} \right)

    b_n=\frac{1}{\pi}\int_{-\pi}^{\pi}|x|\sin(nx)dx=0, \forall n

    |x|=f(x)=\frac{\pi}{2}+\frac{2}{\pi}\sum_{n=1}^{\i  nfty}\left( \frac{(-1)^n-1}{n^2} \right)\cos(nx)

    we can further simplify this becuase whenever n is even the term will be zero so

    f(x)=\frac{\pi}{2}-\frac{4}{\pi}\sum_{n=1}^{\infty}\frac{1}{(2n-1)^2}\cos([2n-1]x)
    Last edited by TheEmptySet; July 14th 2008 at 01:24 PM. Reason: left out Pi
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