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### Topics - Zarak Mahmud

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1
##### 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.

2
##### 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.

3
##### Home Assignment 2 / Problem 2
« on: October 01, 2012, 09:01:03 PM »
Part (a): Make the variable change and use product rule to find $u_{rr}$, $u_r$, and $u_{tt}$.
$$\begin{equation*} u = vr \\ \frac{\partial u}{\partial t} = \frac{1}{r}\frac{\partial v}{\partial t} \\ \frac{\partial^2 u}{\partial t^2} = \frac{1}{r} \frac{\partial^2 v}{\partial t^2} \\ u_{tt} = \frac{1}{r}v_{tt} \\ \frac{\partial u}{\partial r} = \frac{\partial}{\partial r}\frac{v}{r} = \frac{1}{r}\frac{\partial v}{\partial r} - \frac{v}{r^2}\\ \frac{\partial^2 u}{\partial r^2} = \frac{\partial}{\partial r}\frac{\partial u}{\partial v} \\ = \frac{\partial}{\partial r} \left(\frac{1}{r}\frac{\partial v}{\partial r} - \frac{v}{r^2} \right) \\ = -\frac{1}{r^2}\frac{\partial v}{\partial r} + \frac{1}{r}\frac{\partial^2 v}{\partial r^2} - \frac{1}{r^2} + \frac{2v}{r^3} \\ u_{rr} + \frac{2}{r}u_r = \frac{v_{rr}}{r} \\ u_{tt} = \frac{v_{tt}}{r} \\ \frac{v_{tt}}{r} = c^2\frac{v_{rr}}{r} \\ v_{tt} = c^2v_{rr} \end{equation*}$$

Part (b):
$$\begin{equation*} v_{tt} = c^2v_{rr} \\ u(r, t) = \frac{f(r + ct) + g(r - ct)}{r} \\ \end{equation*}$$

Part (c): Take initial conditions with change of variable and then plug into D'Alambert formula.
$$\begin{equation*} \phi (r) = v(r, 0) = ru(r, 0) = r \Phi (r) \\ \psi (r) = v_t(r, 0) = ru_t(r, 0) = r \Psi (r) \\ u(r, t) = \frac{1}{2r} \left[(r+ct) \Phi (r+ct) + (r - ct) \Phi (r - ct) \right] + \frac{1}{2cr} \int_{r-ct}^{r+ct}s \Psi (s) ds \\ \end{equation*}$$

Part (d): For function to be continuous at $r = 0$, we must have
$$\begin{equation*} \lim\limits_{r \to 0} \left[(r+ct) \Phi (r+ct) + (r - ct) \Phi (r - ct) \right] = 0 \\ ct \lim\limits_{r \to 0} \left[ \Phi (r+ct) - \Phi (r - ct) \right] = 0 \\ \Phi (ct) - \Phi (-ct) = 0 \\ \implies \Phi (ct) = \Phi(-ct) \end{equation*}$$
Thus, $\Phi$ is an even function. Similarly, $\Psi$ must be odd since the integral of an odd function between symmetric bounds, i.e., $[-ct, ct]$ is equal to $0$.

4
##### Misc Math / Classification criteria for PDEs
« on: September 29, 2012, 05:26:38 PM »
I read somewhere (and I think it was mentioned in class) that all linear PDEs can be categorized into either parabolic, hyperbolic, or elliptic types according to: $B^2 - 4AC$. For example, if we have
$$$$u_{t} = u_{xx}$$$$
How do we determine what the values of $A$, $B$ and $C$ are?
And does this only apply to second order and smaller PDEs?

5
##### Misc Math / 1-D Wave equation derivation
« on: September 28, 2012, 01:27:42 PM »
Starting with

$$$$\frac{\partial}{\partial x} \left[ T(x,t) \sin{\theta (x,t)} \right] = \rho (x) u_{tt}$$$$

where $\rho$ is the density and $T(x,t)$ is the tension force, we made the assumption that the vibrations are small, which gave us a linearized wave equation. I can see why some of the other assumptions (i.e. full flexibility, and no horizontal tension component) make sense, but I don't think I understand the insight behind this one.

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