Author Topic: Problem 5  (Read 9823 times)

Zixin Nie

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Problem 5
« on: October 24, 2012, 06:27:15 PM »
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?

Victor Ivrii

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Re: Problem 5
« Reply #1 on: October 24, 2012, 06:54:33 PM »
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!
« Last Edit: October 24, 2012, 08:23:19 PM by Victor Ivrii »

Ian Kivlichan

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Re: Problem 5
« Reply #2 on: October 31, 2012, 09:32:43 PM »
Hopeful solution for 5.a)!

Zarak Mahmud

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Re: Problem 5
« Reply #3 on: October 31, 2012, 10:31:31 PM »
Part 5(e):

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

\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*}
« Last Edit: October 31, 2012, 11:30:43 PM by Zarak Mahmud »

Fanxun Zeng

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Re: Problem 5
« Reply #4 on: October 31, 2012, 11:50:24 PM »
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!

Victor Ivrii

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Re: Problem 5
« Reply #5 on: November 01, 2012, 01:46:54 AM »
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.

Djirar

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Re: Problem 5
« Reply #6 on: December 15, 2012, 10:40:57 PM »
Solutions to problem 5.

Victor Ivrii

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!