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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 $(0,2l)$. –  Matt Oct 14 '11 at 2:30
    
Thanks for the comment, but my current concern is not its rigor. –  user16859 Oct 14 '11 at 2:49

1 Answer 1

Here's a hint: Forget about the sine expansion of $\phi$ on $(0,l)$; that's not going to get you anywhere. Consider instead (as the text suggests) the sine expansion of $\tilde{\phi}$ on $(0,2l)$. You will get an equality of the form "$\tilde{\phi}(x) = \sum\dots$ for $0 < x < 2l$", which when restricted to $0 < x < l$ gives you the desired result.

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Thank you! In fact, I got the cosine expansion on (0,2l), then I used $\sin(\pi/2-x)=\cosx$ in the hope of getting the sine expansion, but the result is not the desired one. –  user16859 Oct 14 '11 at 7:06

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