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   {\bf Disclaimer}: {\it These notes are intended for students in the class
listed above: they are not guaranteed to be complete or even necessarily
correct.  They may only be redistributed with permission, which you may
expect will be liberally granted.  Ask first, please.} \vspace*{4mm}
}

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%\lecture{**LECTURE-NUMBER**}{**DATE**}{**LECTURER**}{**SCRIBE**}
\lecture{7}{September 1}{Neil Calkin}{Jack Cooper/Shihwei Chao}

\begin{flushleft}
Recall: $c_{n}$ is the number of ways of parenthesizing $a_{1}a_{2}\cdots a_{n}$.
We have \[
c_{n}=\frac{1}{4n-2}\binom{2n}{n}.\]
Thus
\[
c_{1}=1\]
\[
c_{2}=1\]
\[
c_{3}=2\]
\[
c_{4}=5\]
\[
c_{5}=14\]
\[
c_{6}=42\]
\[
c_{7}=132\]
\[
c_{8}=429\]
\[
c_{9}=1430\]
\[
c_{10}=4862\]
Note: $\frac{1}{2\left(2n-1\right)}\binom{2n}{n}=\frac{\left(2n\right)!}{2\left(2n-1\right)n!n!}=\frac{\left(2n-2\right)!}{n\left(n-1\right)!\left(n-1\right)!}=\frac{1}{n}\binom{2\left(n-1\right)}{\left(n-1\right)}$
gives a slightly nicer representation: fewer factors.
A useful website, The On-Line Encyclopedia of Integer Sequences, http://www.research.att.com/njas/sequences/
Observe that
\[
c_{1}=1\]
\[
c_{2}=1\]
\[
c_{3}=2\]
\[
c_{4}=5\]
\[
c_{5}=14=2\times7\]
\[
c_{6}=42=2\times3\times7\]
\[
c_{7}=132=2^{2}\times3\times11\]
\[
c_{8}=429=3\times11\times13\]
Then,
\[
\frac{c_{2}}{c_{1}}=1=\frac{2}{2}=\frac{2\times1}{2}\]
\[
\frac{c_{3}}{c_{2}}=2=\frac{6}{3}=\frac{2\times3}{3}\]
\[
\frac{c_{4}}{c_{3}}=\frac{5}{2}=\frac{10}{4}=\frac{2\times5}{4}\]
\[
\frac{c_{5}}{c_{4}}=\frac{2\times7}{5}=\frac{14}{5}=\frac{2\times7}{5}\]
\[
\frac{c_{6}}{c_{5}}=3=\frac{18}{6}=\frac{2\times9}{6}\]
\[
\frac{c_{7}}{c_{6}}=\frac{2\times11}{7}=\frac{22}{7}=\frac{2\times11}{7}\]
\[
\frac{c_{8}}{c_{7}}=\frac{13}{4}=\frac{26}{8}=\frac{2\times13}{8}\]
Therefore, we have for $n\geqslant2$,
\[
\frac{c_{n}}{c_{n-1}}=\frac{2\left(2n-3\right)}{n}.\]
Now, 
$c_{n}=\frac{c_{n}}{c_{n-1}}\times\frac{c_{n-1}}{c_{n-2}}\times\frac{c_{n-2}}{c_{n-3}}\times\cdots\times\frac{c_{2}}{c_{1}}\times c_{1}$
$=\frac{2^{n-1}\left(2n-3\right)\left(2n-5\right)\cdots3\cdot1}{n\left(n-1\right)\cdots3\cdot2\cdot1}$
$=\frac{1}{n}\cdot\frac{2^{n-1}\left(2n-3\right)\left(2n-5\right)\cdots3\cdot1\cdot\left(2n-2\right)\left(2n-4\right)\cdots4\cdot2}{\left(n-1\right)!2^{n-1}\left(n-1\right)!}$
$=\frac{1}{n}\cdot\frac{\left(2n-2\right)!}{\left(n-1\right)!\left(n-1\right)!}$
$=\frac{1}{n}\binom{2n-2}{n-1}$.

\indent

Formal power series $\leftrightarrow$ oneway infinite sequences i.e. $\sum_{n\geq0}a_{n}x^{n}$ $\leftrightarrow$ $\left(a_{0},a_{1},a_{2},\cdots\right)$

\indent

Equip the space of formal power series ( or oneway infinite sequences
) with coefficients in a ring ( say $\mathbb{Z}$, $\mathbb{Q}$,
$\mathbb{R}$, $\mathbb{C}$, $\frac{\mathbb{Z}}{2\mathbb{Z}}$,
etc. ) with a metric $\left|\quad\right|_{u}$, it is actually an
ultrametric via the following.

\indent

Fix $0<\gamma<1$.
If $f\left(x\right)=\sum_{n\geq0}a_{n}x^{n}$, set $\left|f\left(x\right)\right|_{u}=\gamma^{k}$
where $k=$ least $n$ so that $a_{n}\neq0$,
that is $f\left(x\right)=a_{k}x^{t}\left(1+\frac{a_{k+1}}{a_{k}}x+\frac{a_{k+2}}{a_{k}}x^{2}+\cdots\right)$
and $a_{k}\neq0$.  Define $\left|0\right|_{u}=0$.

\indent

This metric induces a topology on the set of power series:
Let $f_{n}\left(x\right)$, $n\geqslant0$, be a sequence of power
series, then $f_{n}\left(x\right)\rightarrow0$ as $n\rightarrow\infty$
means $\left|f_{n}\left(x\right)-0\right|_{u}\rightarrow0$, with
respect to the reals, as $n\rightarrow\infty$ .
$\forall\varepsilon>0$, $\exists n_{0}$ such that if $n>n_{0}$
then $\left|f_{n}\left(x\right)\right|_{u}<\varepsilon$ $\equiv$
setting $N=\frac{log\varepsilon}{log\gamma}$, this means degree of
leading term of $f_{n}\left(x\right)>N$.

\indent

Hence $f_{n}\left(x\right)\rightarrow0$ is this topology$\Leftrightarrow$
minimum degrees of $f_{n}\left(x\right)\rightarrow\infty$ as $n\rightarrow\infty$.

\indent

$f_{n}\left(x\right)\rightarrow g\left(x\right)$ if for every $N>0$,
$\exists n_{0}$ so that if $n>n_{0}$ then $f_{n}\left(x\right)$
and $g\left(x\right)$ agree up to terms in $x^{N}$.

\indent

\textbf{Exercises}
\begin{enumerate}
\item Show that $\left|\quad\right|_{u}$ is a metric.
\item Show that $\left|f\left(x\right)g\left(x\right)\right|_{u}=\left|f\left(x\right)\right|_{u}\left|g\left(x\right)\right|_{u}$.
\item Show that $\left|\quad\right|_{u}$ satisfies the ultrametric
inequality: \[
\left|f\left(x\right)+g\left(x\right)\right|_{u}\leqslant max\left\{ \left|f\left(x\right)\right|_{u},\left|g\left(x\right)\right|_{u}\right\} .\]
If $\left|f\left(x\right)\right|_{u}\neq\left|g\left(x\right)\right|_{u}$,
we have equality in the previous inequality.
\end{enumerate}

\textbf{Example:}

Let $f_{n}\left(x\right)=\sum_{k=0}^{n}x^{k}$. Then $f_{n}\left(x\right)\rightarrow\sum_{k\geqslant0}x^{k}$.
Let $f_{n}\left(x\right)=\sum_{k=0}^{n}a_{k}x^{k}$. Then $f_{n}\left(x\right)\rightarrow\sum_{k\geqslant0}a_{k}x^{k}$.
Ultrametric convergence is easy!

\indent

\textbf{Exercise}
Prove:

\indent

$\sum_{n\geqslant0}f_{n}\left(x\right)$ converges with respect to
$\left|\quad\right|_{u}$ $\Leftrightarrow$ $\left|f_{n}\left(x\right)\right|_{u}\rightarrow0$.\\
And, $\prod_{n\geqslant0}h_{n}\left(x\right)$ converges $\Leftrightarrow$
$h_{n}\left(x\right)\rightarrow1$ $\Leftrightarrow$ $h_{n}\left(x\right)-1\rightarrow0$.

\indent


Things can still get strange:
e.g. $f_{n}\left(x\right)=\sum_{n\geqslant0}n!x^{n}$ is a perfectly
valid power series.

\indent

$h\left(x\right)=\left(1+x\right)\left(1+x^{2}\right)\left(1+x^{4}\right)\left(1+x^{8}\right)\cdots=\prod_{i=0}^{\infty}\left(1+x^{2^{i}}\right)$
converges to $1+x+x^{2}+x^{3}+\cdots$.

\indent

\textbf{Exercise}

\indent 

Prove this by
\begin{enumerate}
\item uniqueness of binary representations.
\item showing $\left(1-x\right)h\left(x\right)=1$ and $\left(1-x\right)\left(1+x+x^{2}+x^{3}+\cdots\right)=1$,
and hence $h\left(x\right)=1+x+x^{2}+x^{3}+\cdots$.
\end{enumerate}
\end{flushleft}

\end{document}
