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Math Help - Problem with Fourier Series

  1. #1
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    Problem with Fourier Series

    I have a problem with the below fourier series

    -2, -2 < t < -1
    -1, -1 < t < 0
    f(t) = 1, 0 < t < 1
    2, 1 < t < 2

    f(t + 4) = f(t)

    I want to intergrate by parts to go on to solve bn but dont how what U will be.
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  2. #2
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    Quote Originally Posted by hunterage2000 View Post
    I have a problem with the below fourier series

    -2, -2 < t < -1
    -1, -1 < t < 0
    f(t) = 1, 0 < t < 1
    2, 1 < t < 2

    f(t + 4) = f(t)

    I want to intergrate by parts to go on to solve bn but dont how what U will be.
    You want to find the Fourier series of

    f(t)=\begin{cases} -2, -2 < t < -1 \\ -1, -1 < t < 0 \\ 0, 0 < t < 1 \\ 1, 1 < t < 2 \\<br />
\end{cases}

    You should not need integration by part. The function is constant on each interval.

    Here is a PDF with the series.
    Problem with Fourier Series-fourier.pdf
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  3. #3
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    its f(t)=\begin{cases} -2, -2 < t < -1 \\ -1, -1 < t < 0 \\ 1, 0 < t < 1 \\ 2, 1 < t < 2 \\<br />
 \end{cases}
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  4. #4
    Behold, the power of SARDINES!
    TheEmptySet's Avatar
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    That make the problem easier. f(x) is an odd function so all of the a_n must be zero.

    \displaystyle b_n=\frac{1}{2}\left[ \int_{-2}^{-1}-2\sin\left(\frac{n\pi x}{2} \right)dx + \int_{-1}^{0}-\sin\left(\frac{n\pi x}{2} \right)dx+\int_{0}^{1}\sin\left(\frac{n\pi x}{2} \right)dx+\int_{1}^{2}2\sin\left(\frac{n\pi x}{2} \right)dx \right]

    \displaystyle b_n=-\frac{2}{\pi n}\left[ 2\cos(n\pi)-\cos\left( \frac{n \pi}{2}\right)-1\right]
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  5. #5
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    so what is the value of U? is it 2 + 1 or 3?
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    Quote Originally Posted by hunterage2000 View Post
    so what is the value of U? is it 2 + 1 or 3?
    I don't understand what you mean. What is U?? and 2+1=3 so I am really confused.
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  7. #7
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    U as in the intergration by parts U.

    U = ?
    dv/dt = cos (n*pi*t/2)
    du/dt = ?
    v = -2/n*pi sin(n*pi*t/2)
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  8. #8
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    As I said in my first post you do not need integration by parts.

    \displaystyle \int \sin(ax)dx=-\frac{1}{a}\cos(ax)

    Or in your case

    \displaystyle \int \sin\left(\frac{\pi n}{L}x \right)dx=-\frac{L}{\pi n}\cos\left(\frac{\pi n}{L}x \right)
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  9. #9
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    totally new to this. can u show me from start to finish how you got the answer?
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