Toronto Math Forum
APM346-2012 => APM346 Math => Home Assignment 5 => Topic started by: Zixin Nie on October 24, 2012, 06:27:15 PM
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On problem 5, it says to decompose into sine series again. I'm pretty sure that's a typo, since that's what question 4 asked us to do. Do we decomplose into cosine series instead?
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On problem 5, it says to decompose into sine series again. I'm pretty sure that's a typo, since that's what question 4 asked us to do. Do we decomplose into cosine series instead?
Yes, corrected. Thanks!
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Hopeful solution for 5.a)!
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Part 5(e):
We use an even continuation for this function. Integration was done using the angle-sum identity as in Problem 3 (http://forum.math.toronto.edu/index.php?topic=110.0).
\begin{equation*}
b_n = 0.
\end{equation*}
\begin{equation*}
a_0 = \frac{2}{\pi} \int_{0}^{\pi} \sin{((m-\frac{1}{2})x)}dx \\
= \frac{-2}{\pi (m - \frac{1}{2}} \cos{(m-\frac{1}{2})x} \big|_{0}^{\pi}\\
=\frac{-2}{\pi (m- \frac{1}{2})}.
\end{equation*}
\begin{equation*}
a_n = \frac{2}{\pi} \int_{0}^{\pi} \sin{((m-\frac{1}{2})x)}\cos{nx} dx\\
= -\frac{1}{\pi}\left[\frac{\cos{m + n - \frac{1}{2}} x }{m + n - \frac{1}{2}} + \frac{\cos{m - n - \frac{1}{2}} x }{m - n - \frac{1}{2}} \right]_{0}^{\pi}\\
= \frac{1}{\pi}\left(\frac{1}{m + n - \frac{1}{2}} + \frac{1}{m - n - \frac{1}{2}} \right).
\end{equation*}
\begin{equation*}
\sin{((m-\frac{1}{2})x)} = \frac{1}{\pi (m- \frac{1}{2})} + \frac{1}{\pi}\sum_{n=1}^{\infty} \left(\frac{1}{m + n - \frac{1}{2}} + \frac{1}{m - n - \frac{1}{2}} \right) \cos{n x}.
\end{equation*}
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In fact, PDF version of Problem 5 as of the due date still continues to state sin Fourier series instead of cos Fourier series, as homework 5 PDF proof attached downloaded today. We appreciate if the two versions can be consistent in future.
http://www.math.toronto.edu/courses/apm346h1/20129/HA5.pdf
On problem 5, it says to decompose into sine series again. I'm pretty sure that's a typo, since that's what question 4 asked us to do. Do we decomplose into cosine series instead?
Yes, corrected. Thanks!
-
In fact, PDF version of Problem 5 as of the due date still continues to state sin Fourier series instead of cos Fourier series, as homework 5 PDF proof attached downloaded today. We appreciate if the two versions can be consistent in future.
http://www.math.toronto.edu/courses/apm346h1/20129/HA5.pdf
On problem 5, it says to decompose into sine series again. I'm pretty sure that's a typo, since that's what question 4 asked us to do. Do we decomplose into cosine series instead?
Yes, corrected. Thanks!
I thought I fixed. Sorry.
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Solutions to problem 5.
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Some questions require no calculations: f.e. decomposition of $1$ into cos-F.s. is $1=1$ as $1$ is one of the base functions!