APM346-2015F > Final Exam

FE-3

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Emily Deibert:
Thanks Vivian Tan! Great answer, as far as I recall I got the same!  8)

Bruce Wu:
Great work everyone!

Victor Ivrii:
\begin{equation}
B_n = \frac{2}{3 n \pi} \int_0^2 (2x-x^2) \sin( \frac{n \pi x}{2}) dx = -( \frac{2}{3 n \pi} ) \frac{8(\pi \sin(n \pi) + 2\cos(n \pi) - 2)}{\pi^3 n^3} ={\color{red}{-}}( \frac{2}{3 n \pi} ) \frac{16 ((-1)^n - 1)}{\pi^3 n^3}\tag{11}
\end{equation}
which leads to the error in the sign. Correct answer is
\begin{equation*}
u =\sum_{m=0}^\infty \frac{64}{3\pi ^4(2m+1)^4} \sin \bigl(\frac{3\pi (2m+1) t}{2}\bigr) \sin \bigl(\frac{\pi (2m+1) x}{2}\bigr).
\end{equation*}
which coincides with Vivian's except has an opposite sign.

PS To get $$\sum_{n\text{ odd}}$$ type \sum_{n\text{ odd}}  or \sum_{n\ \text{odd}}

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