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I know how to do (a). I know the sine expansion of $\phi(x)$ on $(0,l)$: $\phi(x)=\sum_{n=1}^\infty B_n \sin \frac{n\pi x}{l}$, but could not get the desired form. Through the formula I mentioned above, we can write $\tilde{\phi}(x)=\phi(2l-x)=-\sum_{n=1}^\infty B_n \sin \frac{n\pi x}{l}$ for $x\in (l,2l)$. I guess there might be something wrong here.

If (b) is solved, the rest should not be too hard. Thank you!

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    By "function" I guess this exercise means something like piecewise continuous function with piecewise continuous derivative ... Surely we can't expect *any* function on $(0,l)$ to have such an expansion. Also, it should say that we extend it to be $2l$-periodic to the rest of $\mathbb{R}$ instead of just a function on $(0,2l)$.2011-10-14
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    Thanks for the comment, but my current concern is not its rigor.2011-10-14

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