### Author Topic: Problem 4  (Read 10345 times)

#### Ian Kivlichan

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##### Problem 4
« on: October 31, 2012, 09:32:02 PM »
Hopeful solutions for 4.c)!

edit: Note that sketch is for m=1.
« Last Edit: October 31, 2012, 09:36:18 PM by Ian Kivlichan »

#### Ian Kivlichan

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##### Re: Problem 4
« Reply #1 on: October 31, 2012, 09:44:27 PM »
Additional solution for 4.a) (essentially the same as Aida's post here http://forum.math.toronto.edu/index.php?topic=108.msg552#msg552 , but showing more of the sketch, as well as with details on the odd continuation used for sin Fourier series).

#### Ian Kivlichan

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##### Re: Problem 4
« Reply #2 on: October 31, 2012, 09:53:35 PM »
Hopeful solution for 4.b) attached!

#### Zarak Mahmud

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##### Re: Problem 4
« Reply #3 on: October 31, 2012, 10:53:40 PM »
Part (e):

We use an odd continuation for this function. Integration was done using the angle-sum identity as in Problem 3.

\begin{equation*}

a_n = 0, a_0 = 0.
\end{equation*}

\begin{equation*}
b_n = \frac{2}{\pi} \int_{0}^{\pi} \sin{(m-\frac{1}{2})x}\sin{nx} dx\\
= \frac{1}{\pi}\left[-\frac{\sin{m + n - \frac{1}{2}} x }{m + n - \frac{1}{2}} + \frac{\sin{m - n - \frac{1}{2}} x }{m - n - \frac{1}{2}} \right]_{0}^{\pi}\\
= \frac{1}{\pi}\left(\frac{(-1)^{m+n}}{m + n - \frac{1}{2}} + \frac{(-1)^{m-n}}{m - n - \frac{1}{2}} \right).
\end{equation*}

\begin{equation*}
\sin{((m-\frac{1}{2})x)} = \frac{1}{\pi}(-1)^m \sum_{n=1}^{\infty} (-1)^n \left(\frac{1}{m + n - \frac{1}{2}} + \frac{1}{m - n - \frac{1}{2}} \right) \sin{n x}.
\end{equation*}
« Last Edit: October 31, 2012, 11:30:16 PM by Zarak Mahmud »

#### Djirar

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##### Re: Problem 4
« Reply #4 on: December 15, 2012, 11:06:40 PM »
includes solution to part (d) and what I think is the graph to part (e).

#### Victor Ivrii

Obviously in (d) only terms with odd $n-m$ do not vanish