hw-example.tex

% HMC Math dept HW template example
% v0.04 by Eric J. Malm, 10 Mar 2005
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\name{Eric Malm}
\class{MA 198}
\assignment{Homework \#1}
\duedate{09/03/2004}

\begin{document}

\problemlist{Rudin 3.5, Saff and Snider 1.5.11, 1.7.5ad,\\Dummit and Foote 1.1.25, Logan 1.8.6}

\begin{problem}[Rudin 3.5]
For any two real sequences $\{a_n\}$, $\{b_n\}$, prove that
\[
  \limsup_{n \to \infty} \, (a_n + b_n)
  \leq \limsup_{n \to \infty} a_n + \limsup_{n \to \infty} b_n,
\]
provided the sum on the right is not of the form $\infty - \infty$.
\end{problem}

\begin{solution}
Suppose that $\limsup_{n \to \infty} a_n + \limsup_{n \to \infty} b_n \neq \infty - \infty$, so that this sum is determinate. Define 
\[
  A_n = \sup_{k \geq n} a_n,
  \quad
  B_n = \sup_{k \geq n} b_n, 
  \quad \text{and} \quad 
  C_n = \sup_{k \geq n} \, (a_n + b_n). 
\]
We first show that $C_n \leq A_n + B_n$ for all $n$. For $k$ and $n$ such that $k \geq n$, we have that $a_k \leq A_n$ and $b_k \leq B_n$. Then $a_k + b_k \leq A_n + B_n$ for all $k \geq n$, so $C_n = \sup_{k \geq n} \, (a_n + b_n) \leq A_n + B_n$. Thus, using the alternate definition of the $\limsup$, we have
\begin{align*}
  \limsup_{n \to \infty} \, (a_n + b_n)
  &= \lim_{n \to \infty} C_n \\
  &\leq \lim_{n \to \infty} \, (A_n + B_n)
  = \lim_{n \to \infty} A_n + \lim_{n \to \infty} B_n
  = \limsup_{n \to \infty} a_n + \limsup_{n \to \infty} b_n.
\end{align*}
\end{solution}

\begin{problem}[SS 1.5.11]
Solve the equation $(z + 1)^5 = z^5$.
\end{problem}

\begin{solution}
Taking fifth roots of the equation yields
\[
  z + 1 = z e^{ik \frac{2\pi}{5}},
\]
where $k \in \mathbb{Z}$. We note that $k = 0$ (and all other multiples of 5) yields $z + 1 = z$, which reduces to $1 = 0$, an inconsistent equation. Isolating $z$, we therefore have the solutions
\[
  \boxed{z = \frac{1}{e^{ik \frac{2\pi}{5}} - 1},}
\] 
with four unique solutions obtained using $k = 1, 2, 3, 4$. We expect 4 unique solutions because $(z + 1)^5 - z^5$ is a fourth-degree polynomial.
\end{solution}

\begin{problem}[SS 1.7.5ad]
Describe the projections on the Riemann sphere of the following sets in the complex plane:
\begin{itemize}
  \item[(\textit{a})] the right half-plane $\{ z \mid \text{Re} ~ z > 0 \}$,
  
  \item[(\textit{d})] the set $\{ z \mid |z| > 3 \}$.
\end{itemize}
\end{problem}

\begin{solution}[(\textit{a})]
The right half-plane corresponds to the right open hemisphere $\{ (x_1, x_2, x_3) \in \mathbb{R}^3 \mid x_1 > 0, x_1^2 + x_2^2 + x_3^2 = 1 \}$.
\end{solution}

\begin{solution}[(\textit{d})]
Since $|z| > 3$, $|z|^2 + 1 > 10$, so
\[
  x_3 = \frac{|z|^2 - 1}{|z|^2 + 1} 
  = 1 - \frac{2}{|z|^2 + 1}
  > 1 - \frac{2}{10} = \frac{4}{5}.
\]
Thus, the set $\{ z \mid |z| > 3 \}$ corresponds to the dome of the Riemann sphere above the plane $x_3 = 4/5$. Hand sketches of these projections are shown below:

\noindent%
\begin{center}
\begin{tabular}{cc}
  (\textit{a}) & (\textit{d}) \\
  \framebox[1.75in]{\rule{0pt}{1.5in}} & \framebox[1.75in]{\rule{0pt}{1.5in}}  \\
  \multicolumn{2}{c}{(to be drawn in later)} \\
\end{tabular}
\end{center}
\end{solution}

\begin{problem}[DF 1.1.25]
Prove that if $x^2 = 1$ for all $x \in G$ then $G$ is abelian.
\end{problem}

\begin{solution}
Suppose $x^2 = 1$ for all $x \in G$. Then $x = x^2 x^{-1} = 1 x^{-1} = x^{-1}$ for all $x \in G$. For two arbitrary elements $a$ and $b$ of $G$,
\[
  ab = (ab)^{-1} = b^{-1} a^{-1} = ba,
\]
so $a$ and $b$ commute. Since $a$ and $b$ are arbitrary, $ab = ba$ for all $a$ and $b \in G$, and $G$ is abelian.
\end{solution}

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\newpage

\begin{problem}[Logan 1.8.6]
This exercise illustrates an important numerical procedure for solving Laplace's equation on a reactangle. Consider Laplace's equation on the rectangle $D: 0 < x < 4, 0 < y < 3$ with boundary conditions given on the bottom and top by $u(x,0) = 0$, $u(x,3) = 0$ for $0 \leq x \leq 4$ and on the sides by $u(0,y) = 2y(3 - y)$, $u(4,y) = 0$ for $0 \leq y \leq 3$. Apply the average value property~(1.45) with $h = 1$ at each of the six lattice points $(1,1), (1,2), (2,1), (2,2), (3,1), (3,2)$ inside $D$ to obtain a system of six equations for the six unknown temperatures on these lattice points. Solve the system to approximate the steady temperature distribution and plot the approximate surface using a software package. 
\end{problem}

\begin{solution}
Applying this average value property with $h = 1$ yields the following linear system:
\begin{align*}
  u(1,1) &= \frac{1}{4} (u(1,0) + u(1,2) + u(0,1) + u(2,1)) = \frac{1}{4} (4 + u(1,2) + u(2,1)), \\
  u(1,2) &= \frac{1}{4} (u(1,1) + u(1,3) + u(0,2) + u(2,2)) = \frac{1}{4} (4 + u(1,1) + u(2,2)), \\
  u(2,1) &= \frac{1}{4} (u(2,0) + u(2,2) + u(1,1) + u(3,1)) = \frac{1}{4} (u(1,1) + u(2,2) + u(3,1)), \\
  u(2,2) &= \frac{1}{4} (u(2,1) + u(2,3) + u(1,2) + u(3,2)) = \frac{1}{4} (u(2,1) + u(2,3) + u(3,2)), \\
  u(3,1) &= \frac{1}{4} (u(3,0) + u(3,2) + u(2,1) + u(4,1)) = \frac{1}{4} (u(3,2) + u(2,1)), \\
  u(3,2) &= \frac{1}{4} (u(3,1) + u(3,3) + u(2,2) + u(4,2)) = \frac{1}{4} (u(3,1) + u(2,2)),
\end{align*}
which we solve in \emph{Mathematica} 5.0 to obtain
\[
 \renewcommand{\arraystretch}{1.25}
  \begin{pmatrix} u(1,1) \\ u(1,2) \\ u(2,1) \\ u(2,2) \\ u(3,1) \\ u(3,2) \end{pmatrix}
  = \begin{pmatrix} \frac{32}{21} \\ \frac{32}{21} \\ \frac{4}{7} \\ \frac{4}{7} \\ \frac{4}{21} \\ \frac{4}{21} \end{pmatrix}.
\]
Plotting these lattice point values yields the following approximate temperature surface:
\begin{center}
  \includegraphics[scale=.8]{logan-1-8-6.pdf}
\end{center}
\end{solution}

\end{document}