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Thread: Integration term by term and orthogonality

  1. #1
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    Integration term by term and orthogonality

    $\displaystyle \displaystyle\int_0^L\varphi_m(x)\varphi_n(x) \ dx=0 \ \ \ \text{if} \ m\neq n$

    $\displaystyle \displaystyle\int_0^L\varphi^2_n(x) \ dx>0$

    $\displaystyle \displaystyle\int_0^L f(x)\sin\left(\frac{m\pi x}{L}\right) \ dx=\sum_{n=1}^{\infty}b_n\int_0^L\sin\left(\frac{n \pi x}{L}\right)\sin\left(\frac{m\pi x}{L}\right) \ dx$

    Every term is 0 except for when m = n.

    $\displaystyle \displaystyle\int_0^L f(x)\sin\left(\frac{m\pi x}{L}\right) \ dx=b_m\int_0^L\left[\sin\left(\frac{m\pi x}{L}\right)\right]^2 \ dx$

    $\displaystyle \displaystyle f(x)=\sin\left(\frac{3\pi x}{L}\right)$

    $\displaystyle \displaystyle\int_0^L \sin\left(\frac{3\pi x}{L}\right)\sin\left(\frac{m\pi x}{L}\right) \ dx=b_m\int_0^L\left[\sin\left(\frac{m\pi x}{L}\right)\right]^2 \ dx$

    Looking at the LHS:

    $\displaystyle \displaystyle b_m\int_0^L\left[\sin\left(\frac{m\pi x}{L}\right)\right]^2 \ dx=b_m\frac{1}{2}\int_0^L \left[1-\cos\left(\frac{2m\pi x}{L}\right)\right] \ dx$

    $\displaystyle \displaystyle b_m\frac{1}{2}\left[x-\frac{L\sin\left(\frac{2m\pi x}{L}\right)}{2m\pi x}\right]_0^L\Rightarrow\frac{L}{2} b_m$

    $\displaystyle \displaystyle b_n=\frac{2}{L}\int_0^L\sin\left(\frac{3\pi x}{L}\right)\sin\left(\frac{m\pi x}{L}\right) \ dx$

    What do I do now? My solution I am obtain is way wrong.
    Last edited by dwsmith; Feb 23rd 2011 at 03:35 PM.
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  2. #2
    Senior Member
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    In Fourier series the integral is

    $\displaystyle
    \displaystyle
    \int_{-L}^L
    $

    If

    $\displaystyle
    \displaystyle
    \int_{0}^L
    $

    other orthogonality conditions may be.
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  3. #3
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    Is this the solution:

    $\displaystyle \displaystyle b_n=\begin{cases}0 & \text{if} \ m\neq 3\\ 1 & \text{if} \ m=3\end{cases}$
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