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    \hbox to 6.28in { {\bf Combinatorial Analysis
                        \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
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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\begin{document}
%FILL IN THE RIGHT INFO.
%\lecture{**LECTURE-NUMBER**}{**DATE**}{**LECTURER**}{**SCRIBE**}
\lecture{11}{September 10}{Neil Calkin}{Shihwei Chao and Honghai Xu}

\section{How to Compute $\left[x^{n}\right]f\left(g\left(x\right)\right)$}
Back to computing $\left[x^{n}\right]e^{f\left(x\right)}$ and more
generally $\left[x^{n}\right]f\left(g\left(x\right)\right)$.
We want to find $\left[x^{n}\right]f\left(g\left(x\right)\right)$,
this is, of course
\[
\left.\frac{1}{n!}\left(\frac{d}{dx}\right)^{n}f\left(g\left(x\right)\right)\right|_{x=0}.\]\\
To put this in context:
recall the rule for diffferentiating a product
\[
\left(\frac{d}{dx}\right)^{n}f\left(x\right)g\left(x\right)=\sum_{k=0}^{n}\binom{n}{k}f^{\left(k\right)}\left(x\right)g^{\left(n-k\right)}\left(x\right).\]
\textbf{
{\color{red}Exercise :  Prove this by induction on $n$.}}\\

Now, how do we differentiate $f\left(g\left(x\right)\right)$?\\\\
Observation:
\begin{eqnarray*}
n=0, \hspace{0.62 in}f(g(x))&=&f^{\left(0\right)}\left(g\left(x\right)\right)g^{\left(1\right)}\left(x\right)^{0},
\hspace{0.1 in}\mbox{silly}\\
n=1, \hspace{0.39 in}\frac{d}{dx}f\left(g\left(x\right)\right)&=&
f^{\left(1\right)}\left(g\left(x\right)\right)g^{\left(1\right)}\left(x\right).\\
n=2, \hspace{0.1 in}\left(\frac{d}{dx}\right)^{2}f\left(g\left(x\right)\right)
&=&\frac{d}{dx}\left(f^{\left(1\right)}\left(g\left(x\right)\right)\right)g^{\left(1\right)}\left(x\right)+f^{\left(1\right)}\left(g\left(x\right)\right)\frac{d}{dx}\left(g^{\left(1\right)}\left(x\right)\right)\\
&=&f^{\left(2\right)}\left(g\left(x\right)\right)g^{\left(1\right)}\left(x\right)^{2}+f^{\left(1\right)}\left(g\left(x\right)\right)g^{\left(2\right)}\left(x\right).\\
n=3, \hspace{0.1 in}
\left(\frac{d}{dx}\right)^{3}f\left(g\left(x\right)\right)
&=&f^{\left(3\right)}\left(g\left(x\right)\right)g^{\left(1\right)}\left(x\right)^{3}+f^{\left(2\right)}\left(g\left(x\right)\right)2g^{\left(1\right)}\left(x\right)g^{\left(2\right)}\left(x\right)+f^{\left(2\right)}\left(g\left(x\right)\right)g^{\left(1\right)}\left(x\right)g^{\left(2\right)}\left(x\right)\\
               &+&f^{\left(1\right)}\left(g\left(x\right)\right)g^{\left(3\right)}\left(x\right)\\\\
&=&f^{\left(3\right)}\left(g\left(x\right)\right)g^{\left(1\right)}\left(x\right)^{3}+3f^{\left(2\right)}\left(g\left(x\right)\right)g^{\left(1\right)}\left(x\right)g^{\left(2\right)}\left(x\right)+f^{\left(1\right)}\left(g\left(x\right)\right)g^{\left(3\right)}\left(x\right).
\end{eqnarray*}
Write $f^{\left(m\right)}$ for $f^{\left(m\right)}\left(g\left(x\right)\right)$
and $g_{m}$ for $g^{\left(m\right)}\left(x\right)$. Then $\left(\frac{d}{dx}\right)^{3}f\left(g\left(x\right)\right)$ can be rewritten as $f^{(3)}g_1^3+3f^{(2)}g_2g_1+f^{(1)}g_3$.\\\\
\underline{General term for $n$:}
For some $p\leqslant n$
\[
f^{\left(p\right)}g_{\lambda_{1}}g_{\lambda_{2}}\cdots g_{\lambda_{n}}\]
with $\lambda_{1}\geqslant\lambda_{2}\geqslant\cdots\geqslant\lambda_{p}\geqslant1$
positive integers and $\lambda_{1}+\lambda_{2}+\cdots+\lambda_{p}=n$.\\\\
\underline{Definition:}
An integer partition $\lambda\vdash n$ of $n$ is a non-increasing
sequence
\[
\lambda_{1}\geqslant\lambda_{2}\geqslant\cdots\geqslant\lambda_{p}\geqslant1\]
of positive integers such that
\[
\lambda_{1}+\lambda_{2}+\cdots+\lambda_{p}=n,\]
where $p$ is the number of parts in the partition, $p=p\left(\lambda\right)$,
and $\lambda$ is the size of the largest part.\\\\
It appears then, that
\[
\left(\frac{d}{dx}\right)^{n}f\left(g\left(x\right)\right)=\sum_{\lambda\vdash n}c_{\lambda}f^{\left(p\left(\lambda\right)\right)}g_{\lambda}\]
where $g_{\lambda}$ denotes $g^{\left(\lambda_{1}\right)}g^{\left(\lambda_{2}\right)}\cdots g^{\left(\lambda_{p}\right)}$.\\\\
\textbf{
{\color{red}Exercise :  Prove this by induction on $n$.}}\\\\
\textbf{
{\color{red}Exercise :  Given $\lambda\vdash n$, write $c_{\lambda}$ as a sum over partitions
of $\lambda'$ of $n-1$ .
The sum will involve the $c_{\lambda'}$'s.}}\\\\
Hint:
($c_{2,1}$ will contribute to $c_{3,1}$, $c_{2,2}$, $c_{2,1,1}$.
$c_{2,2}$ will contribute (twice) to $c_{3,2}$, and (once) to $c_{2,1,1}$.
$c_{3,2}$ comes from $c_{2,2}$.
$c_{3,2}=2c_{2,2}+c_{3,1}$.)\\\\
\textbf{
{\color{red}Exercise :  Compute enough $c_{\lambda}$'s to conjecture and prove a formula.
The formula will be a simple expression form involving factorials
or other functions of $\lambda_{1}$, $\lambda_{2}$, $\cdots$, $\lambda_{p}$,
$p\left(\lambda\right)$ and $i_{1}$, $i_{2}$, $i_{3}$, $\cdots$
where $i_{k}$ is the number of parts of size exactly $k$.}}\\

\section{Solving Some Special Cases with Exponential Generating Functions}
Observation:
Once we've done this in too much generality, we see that it would
be simpler to consider
\[
\left(\frac{d}{dx}\right)^{n}e^{g\left(x\right)}=\sum_{\lambda\vdash n}c_{\lambda}e^{g\left(\lambda\right)}g\left(\lambda\right).\]
This also gives quite a bit of information about $c_{\lambda}$ when
$\lambda$ has restricted part size.
e.g. If $\lambda$ has part size $\leqslant2$, choose $g\left(x\right)$
to be a poly of degree $2$ so that $g^{\left(3\right)}=0$.\\


\underline{Example:}
Observation: If we take $g(x) = e^x-1$, then $g^{(k)}(x) = e^x-1, \forall k \geq 1$. So\\
 \[\left(\frac{d}{dx}\right)^{n}e^{e^x-1}=\sum_{\lambda\vdash n}c_{\lambda}e^{e^x-1}e^{P(\lambda)x}.\]
Setting $x=0$ on both sides, we get\\
\[n![x^n]e^{e^x-1} = [\frac{x^n}{n!}]e^{e^x-1} = \sum_{\lambda\vdash n}c_{\lambda}1.\]
So \[\sum_{\lambda\vdash n}c_{\lambda}=[\frac{x^n}{n!}]e^{e^x-1}.\]
That is, \[e^{e^x-1}=\sum_{n\geq0}b_n[\frac{x^n}{n!}],\] where $\sum_{\lambda\vdash n}c_{\lambda}=b_n.$ This is the exponential function for $b_0, b_1, ...$.\\\\
Now, by using the exponential function for $b_0, b_1, ...$, we have:
\begin{eqnarray*}
\frac{d}{dx}(e^{e^x-1})=e^{e^x-1}e^x.
&\Rightarrow& \sum_{n\geq1}\frac{b_nx^{n-1}}{{(n-1)!}}=(\sum_{n\geq0}\frac{b_nx^{n}}{{n!}})(\sum_{m\geq0}\frac{b_mx^{m}}{{m!}})
=\sum_{n}\sum_{k=0}^n\frac{b_kx^{k}}{k!}\frac{x^{n-k}}{(n-k)!}\\
&\Rightarrow&[x^{n-1}](\sum_{n\geq1}\frac{b_nx^{n-1}}{(n-1)!})=[x^{n-1}](\sum_{n}\sum_{k=0}^n\frac{b_kx^{k}}{k!}\frac{x^{n-k}}{(n-k)!})\\
&\Rightarrow&\frac{b_n}{(n-1)!}=\frac{1}{(n-1)!}\sum_{k=0}^{n-1}{n-1 \choose k} b_k\\
&\Rightarrow&b_n=\sum_{k=0}^{n-1}{n-1 \choose k} b_k.
\end{eqnarray*}



\end{document}
