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Thread: Fourier Series Problem

  1. #1
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    Fourier Series Problem

    Problem: Find the Fourier Series Representation of -
    $\displaystyle f(x)=sin \pi x +3sin2 \pi x - sin 5 \pi x$
    $\displaystyle g(x) =0$ $\displaystyle c=1, L=1$

    Solution from professor:

    $\displaystyle f(x)=sin \pix cos \pi t+3sin2 \pi xcos2 \pi t-sin5 \pi x cos5 \pi t$

    Now, I just simply do not understand how it gets here. I use the formula for fourier series, and I got into some very difficult integral that is impossible to solve.

    Little help, please?

    K
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  2. #2
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    Quote Originally Posted by tttcomrader View Post
    Problem: Find the Fourier Series Representation of -
    $\displaystyle f(x)=sin \pi x +3sin2 \pi x - sin 5 \pi x$
    $\displaystyle g(x) =0$ $\displaystyle c=1, L=1$
    No! You are not finding a Fourier Series representation you are solving the equation:
    $\displaystyle \frac{\partial^2 u}{\partial t^2} = (1)^2 \cdot \frac{\partial^2 u}{\partial x^2}$.
    With boundary value and initial value problems, respectively,
    $\displaystyle \left\{ \begin{array}{c} u(0,t) = u(1,t) = 0 \\ u(x,0) = \sin \pi x + 3\sin 2\pi x - \sin 5\pi x\\u_t(x,0)=0\end{array} \right.$

    The solution to this heat equation is,
    $\displaystyle u(x,t) = \sum_{n=1}^{\infty} \left( A_n \cos \pi n t + B_n \sin \pi n t \right) \sin \pi n x$

    In this case $\displaystyle B_n = \frac{2}{\pi n}\int_0^1 u_t(x,0)\sin \pi n x dx = 0$ which is the easy part.

    The harder part is,
    $\displaystyle A_n = \frac{2}{1}\int_0^1 u(x,0)\sin \pi n x dx $$\displaystyle = 2\left( \int_0^1 \sin \pi x \sin \pi n x dx + \int_0^1 \sin 3\pi x \sin \pi n xdx - \int_0^1 \sin 5\pi x \sin \pi n x dx\right)$

    Use the fact that the sines are orthogonal series of functions. This means that,
    $\displaystyle \int_{-L}^L \sin \frac{\pi n x}{L}\sin \frac{\pi m x}{L} dx= \left\{ \begin{array}{c} 0\mbox{ if }n\not = m \\ L \mbox{ if }n=m \end{array} \right.$

    Therefore the first integral vanishes for every $\displaystyle n\not = 1$. It does not vanish when $\displaystyle n=1$ in that case the integral is $\displaystyle 1$. So $\displaystyle A_1= 2(1+0+0)=2$. Similarly $\displaystyle A_3 = 2$ and $\displaystyle A_5 = -2$.

    Thus,
    $\displaystyle \sum_{n=1}^{\infty} A_n \cos \pi n t = 2\cos \pi t + 2\cos 3\pi t - 2\cos 5\pi t$

    This means,
    $\displaystyle u(x,t) = \sum_{n=1}^{\infty} A_n \cos \pi n t \sin \pi n x = 2\cos \pi t \sin \pi x + 2\cos 3t \sin 3t - 2\cos 5t \sin 5t$
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  3. #3
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    I understand everything that you have posted, however, I'm stuck on actually taking the integral when n=1.

    Okay, so when n=1, everything else vanish except for $\displaystyle A_{1} = (2) \int^{1}_{0}sin \pi x sin \pi xdx$

    Then I have $\displaystyle (2) \int^{1}_{0}sin^2 \pi xdx$

    Using the table of integrals, I have $\displaystyle (2)( \frac{1}{\pi})(\frac{\pi}{2}) = 1$

    But you have $\displaystyle A_{1}=2$, what am I doing wrong here?

    Thank you for your helps.
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  4. #4
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    Because there is a factor of 2 in front the entire paranthesis.
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  5. #5
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    So there is a 2 in front of the one that I just calculated? But I thought I have already include it into my calculation.
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