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Home Assignment 7 / Re: Problem 3
« on: November 19, 2012, 09:55:51 PM »Calvin: it's a bit simpler to write out ((-1)^(n+1) +1) as 0 or 2!
p.s. I am envious of your skills in LaTeX
Me too. Pretty impressive.
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Me too. Pretty impressive.
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Term Test 2 / Re: TT2 -- posting solutions
« on: November 14, 2012, 01:30:15 PM »
But what about the people in the late sitting? They won't be done by then. 
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Home Assignment Y / Re: Problem 2
« on: November 13, 2012, 08:22:01 PM »
edit: I guess I answered my question by reading the above post again.
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Home Assignment Y / Re: HAY--as preparation for TT2
« on: November 12, 2012, 03:49:15 PM »
By the way, I just wanted to remark - the hyperbola dividing the $(\alpha , \beta)$ plane is a great way to keep everything straight. I was getting mixed up with the signs until I start thinking of it in this way.
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Home Assignment Y / Re: Problem 1
« on: November 12, 2012, 11:45:59 AM »In the last line, shouldn't the power to the second exponential be ikx instead of iky?
No--it is Fourier integral for $k\mapsto y$ (remember FT was by $y$). Calvin, WTH, why are you not defending your solution?
He must be taking a long nap after typing it all up.
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Home Assignment Y / Re: HAY--as preparation for TT2
« on: November 12, 2012, 11:44:04 AM »It is discussed in lecture 13. If you have the Strauss textbook, it is discussed in quite a bit of detail in chapter 4.3
See Appendix C and Appendix B.
You linked to appendix C twice.
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Home Assignment Y / Re: HAY--as preparation for TT2
« on: November 11, 2012, 10:38:16 PM »
It is discussed in lecture 13. If you have the Strauss textbook, it is discussed in quite a bit of detail in chapter 4.3
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Home Assignment Y / Re: HAY--as preparation for TT2
« on: November 11, 2012, 09:51:28 PM »
The only way we can have a negative eigenvalue is if the line $y=-\frac{1}{\alpha}$ intersects $\tanh \beta l$. This can't happen if $\alpha$ is positive. Have you tried drawing the graph?
By the way, for negative eigenvalues the convention is to use $\gamma$ instead of $\beta$.
By the way, for negative eigenvalues the convention is to use $\gamma$ instead of $\beta$.
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Term Test 2 / Re: Scope of Term Test 2
« on: November 10, 2012, 06:04:18 PM »
Oh sorry, I saw the post from the second link but missed the first one. I too have a conflict and must go to the early sitting. I assume we just show up.
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Home Assignment 5 / Re: About HA5
« on: November 10, 2012, 05:07:56 PM »I'm confused over why the sketches for the full Fourier series and the sine Fourier series are different. They have the same formula. I would have expected them both to look like the plot for the sine Fourier series. Thanks.
For the function $f(x) = x$? Yes I think they should look the same.
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Term Test 2 / Re: Scope of Term Test 2
« on: November 10, 2012, 04:04:44 PM »
So then I imagine HA7 is not due on the 14th? Since last time the assignment was replaced by 'HAX' and due date was pushed back. Or am I being a bit presumptuous?
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Home Assignment 6 / Re: Problem 2
« on: November 08, 2012, 12:21:22 AM »Calvin, if you type your solutions in LaTeX or TeX, you could post just the source fixing some discrepancies on the fly
I changed the .pdfs to png's for now, when I have some time later I'll try to massage the tex code into mathjax/forum format.
Calvin - just curious, what program are you in?
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Home Assignment 6 / Re: Problem 2
« on: November 08, 2012, 12:19:45 AM »Aida: How did you do the integral?
He didn't. He used a fourier transform pair from part 1(a).
Yes, That's right!
By the way Zarak, I guess it is clear from my name that I am female!
Hahah oops, sorry!
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Home Assignment 6 / Re: Problem 2
« on: November 07, 2012, 09:39:25 PM »Aida: How did you do the integral?
She didn't. She used a fourier transform pair from part 1(a).
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Home Assignment 6 / Re: Problem 4
« on: November 07, 2012, 09:30:01 PM »
Part (a):
\begin{equation*}
\hat{f}(\omega)=
\frac{1}{2\pi}\int_{-\infty}^\infty f(x)e^{-i\omega x}\,dx\\
= \frac{1}{2\pi}\int_{-a}^{a} f(x)e^{-i\omega x}\,dx\\
= \frac{1}{2\pi} \frac{e^{-\omega x}}{i\omega} \big|_{-a}^{a}\\
= \frac{i}{2\pi \omega}(e^{-i\omega a}-e^{i\omega a})\\
= \frac{e^{i\omega a}-e^{-i\omega a}}{2i\pi \omega}\\
=\frac{sin(\omega a)}{\pi \omega}.
\end{equation*}
Part (b):
Using the result from part (a) along with Theorem 3d:
\begin{equation*}
g = xf(x)\implies
\hat{g}(\omega) = i\hat{f}(\omega)\\
=i\frac{d}{d\omega}\big(\frac{sin(\omega a)}{\pi \omega} \big)\\
=i \frac{a\omega \cos{\omega a} - \sin{\omega a}}{\pi \omega^2}\\
=\frac{ia\cos{\omega a}}{\pi \omega} - \frac{i\sin{}\omega a}{\pi \omega^2}.
\end{equation*}
Part (c):
Let $$ f(x)=\left\{\begin{aligned} & 1&& |x|\le a,\\ & 0 && |x|> a;\end{aligned}\right.$$
Then using the result obtained from part (a), we have a fourier transform pair:
\begin{equation*}
f(x) = \int_{-\infty}^\infty \hat{f}(\omega)e^{i\omega x}\,d\omega\\
\implies f(x) = \int_{-\infty}^\infty \frac{\sin{\omega a}}{\pi \omega}e^{i\omega x}\,d\omega\\
\end{equation*}
Switch $\omega$ with $x$.
\begin{equation*}
\implies \left\{\begin{aligned} & \pi&& |\omega|\le a,\\ & 0 && |\omega|> a;\end{aligned}\right.= \int_{-\infty}^\infty \frac{\sin{x a}}{x}e^{i\omega x}\,dx\\
\end{equation*}
Now let $a = 1$, and $\omega = 0$. For these values the function gives us $\pi$.
Thus,
\begin{equation*}
\int_{-\infty}^\infty \frac{\sin{x}}{x}\,dx = \pi.
\end{equation*}
\begin{equation*}
\hat{f}(\omega)=
\frac{1}{2\pi}\int_{-\infty}^\infty f(x)e^{-i\omega x}\,dx\\
= \frac{1}{2\pi}\int_{-a}^{a} f(x)e^{-i\omega x}\,dx\\
= \frac{1}{2\pi} \frac{e^{-\omega x}}{i\omega} \big|_{-a}^{a}\\
= \frac{i}{2\pi \omega}(e^{-i\omega a}-e^{i\omega a})\\
= \frac{e^{i\omega a}-e^{-i\omega a}}{2i\pi \omega}\\
=\frac{sin(\omega a)}{\pi \omega}.
\end{equation*}
Part (b):
Using the result from part (a) along with Theorem 3d:
\begin{equation*}
g = xf(x)\implies
\hat{g}(\omega) = i\hat{f}(\omega)\\
=i\frac{d}{d\omega}\big(\frac{sin(\omega a)}{\pi \omega} \big)\\
=i \frac{a\omega \cos{\omega a} - \sin{\omega a}}{\pi \omega^2}\\
=\frac{ia\cos{\omega a}}{\pi \omega} - \frac{i\sin{}\omega a}{\pi \omega^2}.
\end{equation*}
Part (c):
Let $$ f(x)=\left\{\begin{aligned} & 1&& |x|\le a,\\ & 0 && |x|> a;\end{aligned}\right.$$
Then using the result obtained from part (a), we have a fourier transform pair:
\begin{equation*}
f(x) = \int_{-\infty}^\infty \hat{f}(\omega)e^{i\omega x}\,d\omega\\
\implies f(x) = \int_{-\infty}^\infty \frac{\sin{\omega a}}{\pi \omega}e^{i\omega x}\,d\omega\\
\end{equation*}
Switch $\omega$ with $x$.
\begin{equation*}
\implies \left\{\begin{aligned} & \pi&& |\omega|\le a,\\ & 0 && |\omega|> a;\end{aligned}\right.= \int_{-\infty}^\infty \frac{\sin{x a}}{x}e^{i\omega x}\,dx\\
\end{equation*}
Now let $a = 1$, and $\omega = 0$. For these values the function gives us $\pi$.
Thus,
\begin{equation*}
\int_{-\infty}^\infty \frac{\sin{x}}{x}\,dx = \pi.
\end{equation*}