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    \hbox to 6.28in { {\bf Combinatorial Analyis
                        \hfill Fall 2010} }
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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
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expect will be liberally granted.  Ask first, please.} \vspace*{4mm}
}

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\begin{document}
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%\lecture{**LECTURE-NUMBER**}{**DATE**}{**LECTURER**}{**SCRIBE**}
\lecture{6}{August 30}{Neil Calkin}{Sarah Anderson and Honghai Xu}

\begin{flushleft}

\section{The Binomial Theorem for the unusual exponents:}

We know
\[
(1 - x)^{-1/2} = \sum_{n \geq 0} \binom{-1/2}{n}(-1)^n x^n.
\]
where 
\begin{align*}
(-1)^n \binom{-1/2}{n} &= \frac{(-1/2)(-2/3) \ldots (-1/2 - n + 1)}{n!}(-1)^n \\
&= \frac{(1)(3)(5) \ldots (2n - 1)}{2^n n!} \\
&= \frac{(1)(2)(3)(4)(5) \ldots (2n - 1)(2n)}{2^n n! ((2)(4) \ldots (2n))} \\
&= \frac{(2n)!}{2^n(n!)2^n(n!)} \\
&= \binom{2n}{n} \frac{1}{2^{2n}}.
\end{align*}
Hence,
\[
(1 - 4y)^{-1/2} = \sum_{n \geq 0} \binom{2n}{n} y^n.
\]

Exercise: Thus, 
\[
(1 - 4y)^{-1/2}(1 - 4y)^{-1/2} = (1 - 4y)^{-1} = \sum_{n \geq 0} 4^n y^n
\]
and hence
\[
\sum_{k = 0} \binom{2k}{k} \binom{2n - 2k}{n - k} = 4^n.
\]
For example, when $n = 4$, we get 
\[
\binom{0}{0}\binom{8}{4} +\binom{2}{1}\binom{6}{3} +  \binom{4}{2}\binom{4}{2} + \binom{6}{3}\binom{2}{1} + \binom{8}{4}\binom{0}{0} = 256 
=4^4.
\]
Find a combinatorial proof of this. 

Exercise: Find a relation between 
\[
\binom{-1/3}{n} \;\; \text{and} \;\; \binom{-2/3}{n}
\]
and
\[
\binom{-1/4}{n} \;\; \text{and} \;\; \binom{-3/4}{n}.
\]
Simplify 
\[
\binom{-1/6}{n}. 
\]
using potentially, $\binom{-2/6}{n}, \binom{-3/6}{n}, \binom{-4/6}{n}, \binom{-2/6}{n}$. How many do we need?

Parenthesis: Product with $n$ factors has $c_n$ different interpretations as iterated binary products. For example, $a_1a_2a_3$ is $(a_1a_2)a_3$ or $a_1(a_2a_3)$. Then we showed
\[
c_n = \sum_{k = 1}^{n - 1} c_kc_{k - 1}, \;\; n > 1
\] 
where $c = 1$. Hence, we know $c_0 = 0, c_1 = 1, c_2 = 1$, and $c_3 = 2$. Let $C(x) = \sum_{n \geq 1} c_nx^n$. Then consider
\begin{align*}
C(x)^2 &= \sum_{k \geq 1} c_kx^k \sum_{l \geq 1} c_lx^l \\
&= \sum_{n \geq 2} x^n \sum_{k = 1}^{n - 1} c_k c_{n - k} \\
&= C(x) - x.
\end{align*}

So $C(x)$ satisfies $C(x)^2 - C(x) + x = 0$. Applying the quadratic formula, we obtain
\begin{eqnarray*}
C[x]=\frac{1\pm\sqrt{1-4x}}{2}.
\end{eqnarray*}

Now
\begin{eqnarray*}
(1-4x)^{\frac{1}{2}}=\sum\limits_{n\geq0}\left(\begin{array}{c}\frac{1}{2} \\ n \end{array} \right){(-4)}^nx^n
\end{eqnarray*}
Since $\left(\begin{array}{c}\frac{1}{2} \\ 0 \end{array} \right)=1, \left(\begin{array}{c}\frac{1}{2} \\ 1 \end{array} \right)=\frac{1}{2},$ and
\begin{eqnarray*}
\left(\begin{array}{c}\frac{1}{2} \\ n \end{array} \right)
&=&\frac{\frac{1}{2}(\frac{1}{2}-1)(\frac{1}{2}-2)...(\frac{1}{2}-n+1)}{n!}\\
&=&\frac{1\cdot1\cdot3\cdot...\cdot(2n-3)}{2^nn!}(-1)^{n-1}\\
&=&\frac{1\cdot1\cdot3\cdot...\cdot(2n-3)\cdot(2n-1)\cdot2\cdot4\cdot...\cdot(2n-2)\cdot(2n)}{2^nn!\cdot(2n-1)\cdot2\cdot4\cdot...\cdot(2n-2)\cdot(2n)}(-1)^{n-1}\\
&=&\frac{(2n)!}{2^nn!\cdot(2n-1)2^nn!}(-1)^{n-1}\\
&=&\frac{(2n)!}{4^n\cdot(2n-1)n!n!}(-1)^{n-1}\\
&=&\frac{(-1)^{n-1}}{4^n\cdot(2n-1)}\left(\begin{array}{c}2n \\ n \end{array} \right) \hspace{0.1 in}  \mbox{for }n\geq2,
\end{eqnarray*}

it follows that
\begin{eqnarray*}
(1-4x)^{\frac{1}{2}}&=&1-2x-\sum\limits_{n\geq2}\frac{(-1)^{n-1}}{4^n\cdot(2n-1)}\left(\begin{array}{c}2n \\ n \end{array} \right)(-4)^nx^n\\
&=&1-2x-2x^2-4x^3-10x^4-....
\end{eqnarray*}

Thus
\begin{eqnarray*}
C[x]=\frac{1\pm {(1-2x-2x^2-4x^3-10x^4-...)}}{2}.\\\\
\end{eqnarray*}

\indent

Note that here we need to choose the sign to ensure that $C[0]=0$, so we obtain
\begin{eqnarray*}
C[x]&=&\frac{1-(1-2x-2x^2-4x^3-10x^4-...)}{2}\\
&=&x+x^2+2x^3+5x^4+...
\end{eqnarray*}

So for $n\geq1$, 
\begin{eqnarray*}
C_n=\frac{1}{2}\cdot\frac{1}{2n-1}\left(\begin{array}{c}2n \\ n \end{array} \right)=\frac{1}{4n-2}\left(\begin{array}{c}2n \\ n \end{array} \right).\\
\end{eqnarray*}

\textbf{Question:} How can we get $C_n$ without using the generating function?


\end{flushleft}

\end{document}
