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Math Help - Fourier series of step function

  1. #1
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    Fourier series of step function

    Hiya,

    I'm having a bit of trouble figuing out the fourier series for a function f(x) defined as 1 if 0 < x < pi, -1 if -pi < x < 0.

    The problem I'm having is that \int f(x) dx from -pi to pi equals 0, therefore all of the fourier coefficients are 0. Am I doing something wrong?

    Thanks
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by CorruptioN View Post
    Hiya,

    I'm having a bit of trouble figuing out the fourier series for a function f(x) defined as 1 if 0 < x < pi, -1 if -pi < x < 0.

    The problem I'm having is that \int f(x) dx from -pi to pi equals 0, therefore all of the fourier coefficients are 0. Am I doing something wrong?

    Thanks
    The "therefore" is wrong the conclusion does not follow from your premiss. You have shown that a_0 is zero, and as the function is odd all of the a's are zero, but the b 's are non-zero. Write out the integral for b_1 and see.

    CB
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  3. #3
    Senior Member DeMath's Avatar
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    Quote Originally Posted by CorruptioN View Post
    Hiya,

    I'm having a bit of trouble figuing out the fourier series for a function f(x) defined as 1 if 0 < x < pi, -1 if -pi < x < 0.

    The problem I'm having is that \int f(x) dx from -pi to pi equals 0, therefore all of the fourier coefficients are 0. Am I doing something wrong?

    Thanks
    f(x) = \begin{cases}<br />
  -1,  & \mbox{if}~~ -\pi < x < 0 \\<br />
  1, & \mbox{if}~~~~~~ 0 < x < \pi<br />
\end{cases}

    You have an odd function, so you need to find only b_n.

    b_n = -\frac{1}{\pi}\int\limits_{-\pi}^0 \sin nx\,dx  + \frac{1}{\pi}\int\limits_0^\pi \sin nx\,dx = \frac{2}{\pi}\int\limits_0^\pi \sin nx\,dx =

    = \left. {-\frac{2}{\pi n}\cos nx} \right|_0^\pi  = \frac{2}{\pi}\frac{1-(-1)^n}{n} = \begin{cases}<br />
  0,  & \mbox{if} \quad n~~\text{is}~~\text{even} \\<br />
  \dfrac{4}{\pi}\dfrac{1}{2n-1}, & \mbox{if} \quad n~~\text{is}~~\text{odd}\end{cases}

    Finally you have

    f(x) = \frac{4}{\pi}\sum\limits_{n=1}^\infty \frac{\sin\bigl[(2n-1) x\bigl]}{2n-1},~~ - \pi < x < \pi .
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  4. #4
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    Thank you DeMath! I had just reached the same answer myself, but attempts to plot the function were failing, so I thought I had done something wrong I was about to scrap what I had and start again before you came along!
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