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### Messages - Zarak Mahmud

Pages: 1  3
16
##### Home Assignment 6 / Re: Problem 2
« on: November 07, 2012, 05:26:22 PM »
This integral can be computed using residues, but I think the whole point of the Fourier tranform is lost if one uses that approach...

17
##### Home Assignment 5 / Re: Problem 2
« on: November 06, 2012, 06:17:44 PM »
Since $f(x) = x^2$ was already computed in part c, to calculate part a (say $g(x) = x$), we could have also noted that $$g(x) = \frac{1}{2}\frac{d}{dx}f(x)$$ and applied this transformation term by term to the Fourier series of $f(x)$.

18
##### Home Assignment 5 / Re: Problem 3
« on: November 02, 2012, 12:15:11 AM »
Here are the graphs of the $|\sin x|$ function. You can see both the original function and the fourier series of the function overlaid on the same graph.

19
##### Home Assignment 5 / Re: Problem 3
« on: November 01, 2012, 10:18:31 PM »
3(a) Need to mention that function is even so $b_n=0$.

3(b) pending

Sketches?

Thanks, fixed. I was going to write some code in python to produce plots with a few values of $n$, but it might take a bit.

20
##### Home Assignment 5 / Problem 3
« on: October 31, 2012, 11:14:52 PM »
Part (b):
The function is even, so $b_n = 0$.

\begin{equation*}
a_0 = \frac{2}{\pi} \int_{0}^{\pi} \sin x dx\\
= \frac{4}{\pi}.
\end{equation*}

\begin{equation*}
a_n = \frac{2}{\pi} \int_{0}^{\pi} \sin x \cos nx dx
\end{equation*}

Using $2\sin a \cos b = \sin(a+b) + \sin(a-b)$,

\begin{equation*}
=\frac{1}{\pi} \int_{0}^{\pi} \sin ((n+1)x) + \sin ((1-n)x) dx\\
\end{equation*}
Notice that for $n = 1$, the solution is $0$.
\begin{equation*}

=-\frac{1}{\pi}\left[\frac{\cos((n+1)x)}{n+1} - \frac{\cos((n-1)x)}{n-1}  \right]_{0}^{\pi}\\
=-\frac{1}{\pi}\left[\frac{(-1)^{n+1}}{n+1} - \frac{1}{n+1} - \frac{(-1)^{n+1}}{n-1} + \frac{1}{n-1}  \right]_{0}^{\pi}\\
=\frac{1}{\pi}\big((-1)^n + 1 \big) \big(\frac{1}{n+1} + \frac{1}{n-1}  \big)\\
=\frac{-2}{\pi} \big((-1)^n + 1  \big) \big( \frac{1}{n^2 - 1} \big).\\
\end{equation*}

Let $n=2k$. Thus, for $n>1$,
\begin{equation*}
|\sin x| = \frac{2}{\pi} -\frac{4}{\pi} \sum_{k=1}^{\infty} \frac{1}{(2k)^2 - 1}\cos (2kx).
\end{equation*}

Part (a):
Again, the function $|\cos x|$ is even, so $b_n = 0$.
We proceed as above:

\begin{equation*}
a_0 = \frac{2}{\pi} \int_{0}^{\pi} |\cos x| dx \\
=  \frac{2}{\pi} \int_{0}^{\frac{\pi}{2}} \cos x dx -  \frac{2}{\pi} \int_{\frac{\pi}{2}}^{\pi} \cos x dx\\
= \frac{4}{\pi}\\
\end{equation*}

\begin{equation*}
a_n =  \frac{2}{\pi} \int_{0}^{\pi} |\cos x|\cos{nx} dx \\
= \frac{2}{\pi} \int_{0}^{\frac{\pi}{2}} \cos x  \cos{nx} dx -  \frac{2}{\pi} \int_{\frac{\pi}{2}}^{\pi} \cos x \cos{nx} dx\\
= \frac{2}{\pi}\left(-2\frac{\cos{\frac{\pi n}{2}}}{n^2 - 1}\right)\\
\end{equation*}

Therefore,
\begin{equation*}
|\cos x| = \frac{2}{\pi} - \frac{4}{\pi}\sum_{n=1}^{\infty} \frac{\cos{\frac{\pi n}{2}}}{n^2 - 1} \cos{nx}.
\end{equation*}

Both functions are continuous, so the Fourier series will converge to each function at every point.

21
##### Home Assignment 5 / Re: Problem 4
« 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*}

22
##### Home Assignment 5 / Re: Problem 5
« 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*}

23
##### Home Assignment 5 / Re: Problem 1
« on: October 31, 2012, 09:30:10 PM »

$f(x) = e^{zx}$ for some $z \in \mathbb{C}$

Part (a):
\begin{equation*}

a_0 = \frac{1}{l} \int_{-l}^{l} e^{zx}dx\\
= \frac{1}{zl}  e^{zx}\big|_{-l}^{l}\\
= \frac{1}{zl}\big(e^{zl} - e^{-zl}  \big)\\
\end{equation*}

\begin{equation*}
a_n = \frac{1}{2l} \int_{-l}^{l} e^{zx}\big( \exp{(\frac{in\pi x}{l})} +  \exp{(-\frac{in\pi x}{l})} \big) dx \\
= \frac{1}{2l} \int_{-l}^{l} \exp{((z + \frac{in\pi }{l})x)} +  \frac{1}{2l} \int_{-l}^{l} \exp{((z -\frac{in\pi }{l})x)}  dx \\
= \frac{1}{2l} \frac{\exp{((z +\frac{in\pi}{l})x)}}{z +\frac{in\pi }{l}} \big|_{-l}^{l} + \frac{1}{2l} \frac{\exp{((z -\frac{in\pi  }{l})x)}}{z -\frac{in\pi }{l}} \big|_{-l}^{l}\\
= \frac{1}{2l} \frac{e^{zl + in\pi }}{z +\frac{in\pi }{l}} -  \frac{1}{2l} \frac{e^{-zl - in\pi }}{z +\frac{in\pi }{l}} + \frac{1}{2l} \frac{e^{zl - in\pi }}{z -\frac{in\pi }{l}} - \frac{1}{2l} \frac{e^{-zl + in\pi }}{z -\frac{in\pi }{l}}\\
=\frac{1}{2l}(-1)^n \big[e^{zl} - e^{-zl} \big]\big(\frac{1}{z +\frac{in\pi }{l}} + \frac{1}{z -\frac{in\pi }{l}}  \big)\\
=  \frac{(-1)^n}{l} \big(e^{zl} - e^{-zl} \big) \frac{z}{z^2 + (\frac{n \pi}{l})^2}\\
= \frac{(-1)^nzl \big(e^{zl} - e^{-zl} \big)}{(zl)^2 + (n \pi)^2 }
\end{equation*}

Similarly, for $b_n$:

\begin{equation*}

b_n  = {\color{magenta}- }\frac{(-1)^n {\color{magenta}\pi }{\color{magenta}n } \big(e^{zl} - e^{-zl} \big)}{(zl)^2 + (n \pi)^2 }
\end{equation*}

Note the difference in magenta. Exceptional values are $lz = in\pi$ or $lz = -in\pi$.

\begin{equation}

e^{zx} = (e^{zl} - e^{-zl})\left[\frac{1}{2l} + \sum_{n=1}^{\infty} \frac{(-1)^nzl }{(zl)^2 + (n \pi)^2 } \cos{\frac{n\pi x}{l}} - \sum_{n=1}^{\infty} \frac{(-1)^n n\pi }{(zl)^2 + (n \pi)^2 } \sin{\frac{n\pi x}{l}} \right].
\end{equation}

Part (b):

\begin{equation*}
\cos{\omega x} = \frac{(e^{i\omega} - e^{-i\omega})}{2}
\end{equation*}

Let $z = i\omega$ or $z=-i\omega$. Using the result obtained in (1),
\begin{equation*}

\frac{1}{2}\left[ (e^{i\omega l} - e^{-i\omega l})\left(\frac{1}{2i \omega l} + \sum_{n=1}^{\infty} \frac{(-1)^n i\omega l }{-(\omega l)^2 + (n \pi)^2 } \cos{\frac{n\pi x}{l}} - \sum_{n=1}^{\infty} \frac{(-1)^n n\pi }{-(\omega l)^2 + (n \pi)^2 } \sin{\frac{n\pi x}{l}} \right)  - \\(e^{i\omega l} - e^{-i\omega l})\left(\frac{1}{-2i \omega l} + -\sum_{n=1}^{\infty} \frac{(-1)^n i\omega l }{-(\omega l)^2 + (n \pi)^2 } \cos{\frac{n\pi x}{l}} - \sum_{n=1}^{\infty} \frac{(-1)^n n\pi }{-(\omega l)^2 + (n \pi)^2 } \sin{\frac{n\pi x}{l}} \right) \right]\\
= \sin{\omega l}\left(\frac{1}{\omega l} -  \sum_{n=1}^{\infty} 2(-1)^n  \frac{\omega l \cos{\frac{n\pi x}{l}}}{(n\pi)^2 -( \omega l)^2}  \right).

\end{equation*}

Exceptional values here appear to be $(\omega l)^2 = (n \pi)^2$. At $\omega = 0$, we have an indeterminate form which is defined in the limit as $\omega$ approaches $0$.
Similarly, for $\sin{\omega x}$, we have

\begin{equation*}
\sin{\omega x} = -2\sin{\omega l} \sum_{n=1}^{\infty} 2(-1)^n  \frac{n \pi \sin{\frac{n\pi x}{l}}}{(n\pi)^2 -( \omega l)^2}.

\end{equation*}

Part (c):
Just as in (b), we have $\cosh{\eta x} = \frac{e^{\eta x} + e^{-\eta x}}{2}$ and can make the substitution $z = \eta$ or $z = -\eta$. Then

\begin{equation*}
\cosh{\eta x} = \sin{\eta l}\left(\frac{1}{\eta l} -  \sum_{n=1}^{\infty} 2(-1)^n  \frac{\eta l \cos{\frac{n\pi x}{l}}}{(n\pi)^2 +( \eta l)^2}  \right)
\end{equation*}
And for $\sinh{\eta x}$, we have
\begin{equation*}
\sinh(\eta x) = \sinh{\eta l}\left(- 2 \sum_{n=1}^{\infty} (-1)^n  \frac{n\pi  \sin{\frac{n\pi x}{l}}}{(n\pi)^2 +( \eta l)^2}  \right).
\end{equation*}

24
##### Technical Questions / Re: QED symbol?
« on: October 17, 2012, 12:47:32 PM »
Here's a nice searchable list of mathjax symbols I've been using:

http://www.onemathematicalcat.org/MathJaxDocumentation/TeXSyntax.htm

Yup I've been using that resource too. I didn't think to check \square or $\blacklozenge$.

25
##### Term Test 1 / Re: TT1 = Problem 3
« on: October 16, 2012, 10:31:29 PM »
Here is a rather brief typed version.

\begin{equation} E(t)= \frac{1}{2}\int_0^L \big( u_t^2 + K u_{xx}^2 + \omega^2 u^2\big)\,dx \end{equation}

Take the derivative with respect to $t$ and then integrate the middle term by parts.

\begin{equation*} E'(t)= \frac{1}{2}\int_0^L \bigl( u_{tt}u_t + K u_{xx}u_{xxt} + \omega^2 uu_t\big)\,dx \\

K \left( u_{xt}u_{xx}\big|_{0}^{L} - \int_0^L u_{xt}u_{xxx}dx \right) \\
\implies K \left( u_{xt}u_{xx}\big|_{0}^{L} - u_{t}u_{xxx}\big|_{0}^{L} + \int_0^L u_tu_{xxxx} \right)\\

E'(t)= \int_0^L u_t \bigl( u_{tt}+K u_{xxxx} + \omega^2 u \big) dx +  K \big[ u_{xt}u_{xx} - u_{t}u_{xxx} \big]_{0}^{L}\\

=K \big[ u_{xt}u_{xx} - u_{t}u_{xxx} \big]_{0}^{L}

\end{equation*}

After plugging in the bounds and checking boundary conditions, we see that $E'(t) = 0$ and thus $E(t)$ is a constant function, independent of $t$.

26
##### Technical Questions / QED symbol?
« on: October 16, 2012, 09:58:48 PM »
How do you render that square symbol to indicate the end of a proof? I wasn't able to find it doing a quick google search.

27
##### Term Test 1 / Re: TT1 = Problem 1
« on: October 16, 2012, 08:41:46 PM »
My solution. Please check there might be mistakes. Sorry for quality of print and hand writing.

Djirar, I posted my solution 10 seconds after you I was scribbling my solutions as fast as I could Edit: I forgot to write down the axes to part a.  The vertical is x-axis and the horizontal is y-axis. I hope I didn't forget this on the test Do you mean the $t$ axis?

28
##### Home Assignment X / Re: problem 3
« on: October 15, 2012, 12:53:35 PM »
So, after some more time spent on this problem as a result of integration I get
$$\Re(\alpha(|u_t|^2))(x=0) + \Re(\beta(|u_t|^2))(x=L).$$
Does this imply that the answers should be

a) $\Re(\alpha)=0, \Re(\beta)=0$
b) $\Re(\alpha)>0, \Re(\beta)<0$

P.S. I am sorry for my weird math-writing.

After some simple edits Click quote to see what I changed.

29
##### Home Assignment 3 / Re: Problem 1
« on: October 14, 2012, 08:15:42 AM »
Zarak,

You don't need to surround
Code: [Select]
 \begin{equation} ....\end{equation}by double dollars as equation, align, gather (and their * versions) are LaTeX environments, and multline  (and its * version) is AMS-LaTeX environment, all recognizable by MJ. Basically DD are  deprecated in LaTeX.

Oh I see. I edited them out of the code for this post and will dispense with them in future posts. Now that I think about it, the double dollars do seem redundant when using equation, align, gather etc.

30
##### Misc Math / Re: Bonus Web Problem
« on: October 13, 2012, 11:38:31 AM »
Are there any resources or examples of this method apart from what is described in the lecture notes?

I'm a bit confused. In the lecture notes we have

\begin{equation*}
u_{\alpha}(x,t)=\alpha u(\alpha x, \alpha^2 t).
\label{eq-4}
\end{equation*}

but in the question there is no $t$ term

\begin{equation*}
u_{\lambda}:= \lambda u(\lambda x, \lambda^\gamma )=u(x,\lambda)\qquad \forall \lambda>0
\end{equation*}

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