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\usepackage{quantumgravity}
\title{Quantum Gravity Homework 3}
\author{Erin Pearse}

\pagestyle{myheadings}
    \markboth{\hfill \textsc{Quantum Gravity Homework 3} \hfill }
             {\hfill \textsc{Quantum Gravity Homework 3} \hfill }
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\begin{document}
\maketitle

\begin{Prob}
  \item By direct analogy with the number of ways to write down a
    finite number as a sum of 1's and 2's,
    \[p(z) = \sum_{n\geq0}{p_n z^n}
      = \frac{1}{1-z} \cdot \frac{1}{1-z^5} \cdot \frac{1}{1-z^{10}}\]

  \item To find the number of different ways to make change for
    \$10 using only pennies, nickels, and dimes, we use the
    following Mathematica code:
    \begin{align*}
      \textsf{In[6]:=}& \, p[z\_] := \text{Series}\left[\tfrac{1}{1 - z}\tfrac{1}{1 - z^5} \tfrac{1}{1 - z^{10}}, \{z, 0,1000\}\right] \\
      \textsf{In[7]:=}& \, \text{SeriesCoefficient}\left[p[z], 1000\right] \vstr \\
      \textsf{Out[7]=}& \, 10201 \vstr
    \end{align*}

  \item If $p_n$ is the number of ways to write $n$ as a sum of
    positive natural numbers chosen from some set $S\ci \bN^+$,then
    \[p(z) = \sum_{n\geq0}{p_n z^n}
      = \prod_{s \in S} \frac{1}{1-z^s} \]

  \item Let $S = \{s_1,s_2, \ldots , s_k\} \ci \bN^+$.
    A $P$-structure consists of chopping a set $A$ into $k=|S|$
    parts, totally ordering each, then chopping the first into
    blocks of length $s_1$, the second into blocks of length
    $s_2$, etc.
    \par Now we obtain the isomorphism
    \[P \cong \frac{1}{1-Z^{s_1}} \cdot \frac{1}{1-Z^{s_2}} \cdots \frac{1}{1-Z^{s_k}},\]
    because each term
    \(\left(1-Z^{s_j}\right)^{-1}\)
    corresponds to the structure type of totally ordering a set
    and chopping it into blocks of length $s_j$.

  \item If $q_n$ is the number of ways to write $n$ as a sum of
    positive numbers (chosen from $S \ci \bN^+$), where each
    number can be used only once, then
    \[q_n = \sum_{n \geq 0}{q_n z^n} = \prod_{s \in S} {(1+z^s)}.\]
    The justification for this comes by decategorifying 6.

  \item Let $S = \{s_1,s_2, \ldots , s_k\} \ci \bN^+$, where all
    the $s_j$ are distinct. A $Q$-structure on a set $A$ consists
    of chopping $A$ into $k=|S|$ parts. The first part is either
    empty or else it gets totally ordered and endowed with the
    structure of ``being an $s_1$-element set''. The second part
    is treated similarly, etc. This leads to the isomorphism
    \[Q \cong \left(1+Z^{s_1}\right)\left(1+Z^{s_2}\right)\ldots\left(1+Z^{s_k}\right).\]

  \item The number of ways of writing $n$ as a sum of odd numbers is
    \begin{align*}
      \prod_{n\geq 0} {\frac{1}{1-z^{2n+1}}}
      &= \frac{1}{1-z^1} \cdot \frac{1}{1-z^3} \cdot \frac{1}{1-z^5} \ldots \\
      &= \frac{1-z^2}{1-z} \cdot \frac{1-z^4}{1-z^2} \cdot \frac{1-z^6}{1-z^3} \cdot \frac{1-z^8}{1-z^4} \ldots \\
      &= (1+z)\cdot(1+z^2)\cdot(1+z^3)\cdot(1+z^4) \\
      &= \prod_{n\geq 0} {(1+z^n)},
    \end{align*}
    which is the number of ways of writing $n$ as a sum of
    distinct numbers.

\end{Prob}

\end{document}
