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Math Help - Fourier Series - Wave Eqn

  1. #1
    Rhymes with Orange Chris L T521's Avatar
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    Fourier Series - Wave Eqn

    I just have a quick question:

    I'm evaluating a Fourier coefficient for the following series:

    y(x,t)=\sum_{n=1}^{\infty}B_n\sin\left(\frac{n\pi a}{L}t\right)\sin\left(\frac{n\pi}{L}x\right)

    Where B_n=\frac{2}{n\pi a}\int_0^L \psi(x)\sin\left(\frac{n\pi}{L}x\right)\,dx and \psi \left( x \right) = \left\{ \begin{array}{cl}<br />
  0 &0 \leqslant x \leqslant \frac{{L - a}}<br />
{2} \\<br />
  1 & \frac{{L - a}}<br />
{2} < x < \frac{{L + a}}<br />
{2}  \\<br />
  0 & \,\,\frac{{L + a}}<br />
{2} < x < L  \\ <br />
\end{array}  \right.<br />

    I'm going to show my work:

    I said that

    \begin{aligned}<br />
B_n&=\frac{2}{n\pi a} \left[\int_0^{\frac{L-a}{2}}(0)\sin\left(\frac{n\pi}{L}x\right)\,dx+\int  _{\frac{L-a}{2}}^{\frac{L+a}{2}}\sin\left(\frac{n\pi}{L}x\ri  ght)\,dx+ \int_{\frac{L+a}{2}}^L(0)\sin\left(\frac{n\pi}{L}x  \right)\,dx\right] \\<br />
&=\frac{2}{n\pi a}\int_{\frac{L-a}{2}}^{\frac{L+a}{2}}\sin\left(\frac{n\pi}{L}x\ri  ght)\,dx<br />
\end{aligned}

    \begin{aligned}<br />
B_n&=<br />
\frac{2}{n\pi a}\left.\left[-\frac{L}{n\pi}\cos\left(\frac{n\pi}{L}x\right)\rig  ht]\right|_{\frac{L-a}{2}}^{\frac{L+a}{2}} \\<br />
 &=\frac{-2L}{n^2\pi^2a}\left[\cos\left(\frac{n\pi}{L}\cdot\frac{L+a}{2}\right)-\cos\left(\frac{n\pi}{L}\cdot\frac{L-a}{2}\right)\right] <br />
\end{aligned}<br />

    This is what I did to simplify what was in the brackets:

    \cos\left(\frac{n\pi}{2}+\frac{n\pi a}{2L}\right)-\cos\left(\frac{n\pi}{2}-\frac{n\pi a}{2L}\right)

    \implies \cos\frac{n\pi}{2}\cos\frac{n\pi a}{2L}-\sin\frac{n\pi}{2}\sin\frac{n\pi a}{2L}-\cos\frac{n\pi}{2}\cos\frac{n\pi a}{2L}-\sin\frac{n\pi}{2}\sin\frac{n\pi a}{2L}

    \implies -2\sin\frac{n\pi}{2}\sin\frac{n\pi a}{2L}

    Multiplying by -\frac{2L}{n^2\pi^2a}, I get:

    \frac{4L}{n^2\pi^2a}\sin\frac{n\pi}{2}\sin\frac{n\  pi a}{2L}

    Now this is where I'm somewhat in the dark. Since I'm plugging in values of n, where n\in \mathbb{N}, into \sin\frac{n\pi}{2}, I notice when n is odd, I get either 1 or -1, and when n is even, I get zero. Thus, I rewrote my answer as:

    \frac{4L}{n^2\pi^2a}\sin\frac{(2n-1)\pi}{2}\sin\frac{n\pi a}{2L}

    I noticed that \sin\frac{(2n-1)\pi}{2} could be written as (-1)^{n+1}; \text{ } n\in\mathbb{N}

    So then I saw that B_n=\frac{4(-1)^{n+1}L}{n^2\pi^2a}\sin\frac{n\pi a}{2L}

    Thus, my series solution would be

    y(x,t)=\frac{4L}{\pi^2a}\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n^2}\sin\left(\frac{n\pi a}{2L}\right)\sin\left(\frac{n\pi a}{L}t\right)\sin\left(\frac{n\pi}{L}x\right)

    Is this correct??

    However, my question is this: when I rewrote \sin\frac{n\pi}{2} as \sin\frac{(2n-1)\pi}{2}, do I need to replace the other n's in my equation with 2n-1? meaning, would my answer for B_n be \frac{4L}{(2n-1)^2\pi^2a}\sin\frac{(2n-1)\pi}{2}\sin\frac{(2n-1)\pi a}{2L}????

    I'd appreciate any help.
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  2. #2
    MHF Contributor Reckoner's Avatar
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    Quote Originally Posted by Chris L T521 View Post
    However, my question is this: when I rewrote \sin\frac{n\pi}{2} as \sin\frac{(2n-1)\pi}{2}, do I need to replace the other n's in my equation with 2n-1? meaning, would my answer for B_n be \frac{4L}{(2n-1)^2\pi^2a}\sin\frac{(2n-1)\pi}{2}\sin\frac{(2n-1)\pi a}{2L}????
    Yes. Think about it: with

    B_n = \frac{4L}{n^2\pi^2a}\sin\frac{n\pi}{2}\sin\frac{n\  pi a}{2L}

    any time n is even, this term evaluates to zero, as you observed. The only remaining terms will be those for which n is odd. The values of \frac{4L}{n^2\pi^2a}\text{ and }\sin\frac{n\pi a}{2L} do not matter for even n, because they're being multiplied by zero. However, with

    B_n = \frac{4L}{n^2\pi^2a}\sin\frac{(2n-1)\pi}{2}\sin\frac{n\pi a}{2L}

    this changes. Now, the values of \frac{4L}{n^2\pi^2a}\text{ and }\sin\frac{n\pi a}{2L} will affect the result regardless of whether n is even or odd.

    The rest of your work looks good to me, but I admit that I only gave a quick look through it.
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