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\begin{document}
\section*{Math 222 Fall 2022, Lecture 5: Introduction}

\textbf{website:}
\texttt{\href{https://www.cip.ifi.lmu.de/~grinberg/t/22fco}{\texttt{https://www.cip.ifi.lmu.de/\symbol{126}%
grinberg/t/22fco}}}

\section{Introduction (cont'd)}

\setcounter{subsection}{1}

\subsection{Sums of powers}

\setcounter{subsubsection}{2}

\subsubsection{Some rules for sums (cont'd)}

\setcounter{theo}{3}

The last summation rule we saw last time says the following:

\begin{itemize}
\item \textbf{Splitting a sum by a value of a function:} Let $S$ and $W$ be
two finite sets. Let $f:S\rightarrow W$ be a map. Let $a_{s}$ be a number for
each $s\in S$. Then,%
\begin{equation}
\sum_{s\in S}a_{s}=\sum_{w\in W}\ \ \sum_{\substack{s\in S;\\f\left(
s\right)  =w}}a_{s} \label{eq.sums.split-by-f}%
\end{equation}
(the right hand side is a sum of sums, which is why you are seeing two $\sum$
signs side by side).
\end{itemize}

This splitting rule can be used to derive a slightly more general sum rule for combinatorics:

\begin{theorem}
[the sum rule, in summation-sign form]\label{thm.sum-rule-in-sum-form}Let $S$
and $W$ be two finite sets. Let $f:S\rightarrow W$ be a map. Then,%
\[
\left\vert S\right\vert =\sum_{w\in W}\left(  \text{\# of }s\in S\text{
satisfying }f\left(  s\right)  =w\right)  .
\]

\end{theorem}

For example, if you have a finite set of socks, each of which is either red or
green or blue, then%
\[
\left(  \text{\# of socks}\right)  =\left(  \text{\# of red socks}\right)
+\left(  \text{\# of green socks}\right)  +\left(  \text{\# of blue
socks}\right)  .
\]
(This follows by applying Theorem \ref{thm.sum-rule-in-sum-form} to
$S=\left\{  \text{socks}\right\}  $, $W=\left\{  \text{colors}\right\}  $ and
$f\left(  s\right)  =\left(  \text{color of the sock }s\right)  $.)

\begin{proof}
[Proof of Theorem \ref{thm.sum-rule-in-sum-form}.]Applying
(\ref{eq.sums.split-by-f}) to $a_{s}=1$, we find%
\[
\sum_{s\in S}1=\sum_{w\in W}\ \ \sum_{\substack{s\in S;\\f\left(  s\right)
=w}}1.
\]
However, in view of the equalities
\begin{align*}
\sum_{s\in S}1  &  =\left\vert S\right\vert \cdot1=\left\vert S\right\vert
\ \ \ \ \ \ \ \ \ \ \text{and}\\
\sum_{\substack{s\in S;\\f\left(  s\right)  =w}}1  &  =\left(  \text{\# of
}s\in S\text{ satisfying }f\left(  s\right)  =w\right)  \cdot1\\
&  =\left(  \text{\# of }s\in S\text{ satisfying }f\left(  s\right)
=w\right)  ,
\end{align*}
this can be rewritten as%
\[
\left\vert S\right\vert =\sum_{w\in W}\left(  \text{\# of }s\in S\text{
satisfying }f\left(  s\right)  =w\right)  ,
\]
and this is precisely the claim of Theorem \ref{thm.sum-rule-in-sum-form}.
\end{proof}

\subsubsection{A few words about finite products}

Finite products are defined in the same way as finite sums, but using the
operation $\cdot$ instead of $+$.

The product of a family of numbers $a_{s}$ is denoted by $\prod\limits_{s\in
S}a_{s}$.

Almost all rules for finite sums have analogues for finite
products.\footnote{Why \textquotedblleft almost\textquotedblright? Only
because of one trivial exception: Finite sums satisfy the \textquotedblleft
generalized distributivity law\textquotedblright\ $\sum_{s\in S}\lambda
a_{s}=\lambda\sum_{s\in S}a_{s}$ (for any number $\lambda$ and any numbers
$a_{s}$). The analogue of this law for products takes this form
\[
\prod_{s\in S}a_{s}^{\lambda}=\left(  \prod_{s\in S}a_{s}\right)  ^{\lambda},
\]
but holds only for $\lambda\in\mathbb{N}$ (or for $\lambda\in\mathbb{Z}$ if we
assume that all $a_{s}$ are nonzero). If we try to apply it (say) to
$\lambda=1/2$, we get wrong results such as $\left(  -1\right)  ^{1/2}\left(
-1\right)  ^{1/2}=1^{1/2}$.}

The analogue of an empty sum $\sum_{s\in\varnothing}a_{s}$ is an empty product
$\prod\limits_{s\in\varnothing}a_{s}$. This empty product is defined to be
$1$. In particular, this entails that $x^{0}=1$ for any number $x$, because
the $m$-th power $x^{m}$ (for $m\in\mathbb{N}$) is defined to be the finite
product $\underbrace{xx\cdots x}_{m\text{ many }x\text{'s}}=\prod_{i=1}^{m}x$.

\subsubsection{The sums $1^{k}+2^{k}+...+n^{k}$}

So we know that the sum of the first $n$ positive integers is $1+2+\cdots
+n=\dfrac{n\left(  n+1\right)  }{2}$ for any $n\in\mathbb{N}$. What about
summing higher powers (e.g., squares or cubes) of the first $n$ positive integers?

\begin{theorem}
\label{thm.sum-of-squares}Let $n\in\mathbb{N}$. Then,%
\[
1^{2}+2^{2}+\cdots+n^{2}=\dfrac{n\left(  n+1\right)  \left(  2n+1\right)  }%
{6}.
\]

\end{theorem}

\begin{theorem}
Let $n\in\mathbb{N}$. Then,%
\[
1^{3}+2^{3}+\cdots+n^{3}=\dfrac{n^{2}\left(  n+1\right)  ^{2}}{4}.
\]

\end{theorem}

These theorems can again be proved by induction or in various other ways (see
\S 1.2.5 in the 2019 notes for references).

There are similar formulas for sums of $4$th and $5$th powers. Using the
summation sign:%
\begin{align*}
\sum_{i=1}^{n}i^{0}  &  =n;\\
\sum_{i=1}^{n}i^{1}  &  =\dfrac{n\left(  n+1\right)  }{2};\\
\sum_{i=1}^{n}i^{2}  &  =\dfrac{n\left(  n+1\right)  \left(  2n+1\right)  }%
{6};\\
\sum_{i=1}^{n}i^{3}  &  =\dfrac{n^{2}\left(  n+1\right)  ^{2}}{4};\\
\sum_{i=1}^{n}i^{4}  &  =\dfrac{n\left(  2n+1\right)  \left(  n+1\right)
\left(  3n+3n^{2}-1\right)  }{30};\\
\sum_{i=1}^{n}i^{5}  &  =\dfrac{n^{2}\left(  n+1\right)  ^{2}\left(
2n+2n^{2}-1\right)  }{12}.
\end{align*}
How does this continue (if it does)? It looks like for any given (constant)
$k\in\mathbb{N}$, the sum $\sum\limits_{i=1}^{n}i^{k}$ can be written as a
degree-$\left(  k+1\right)  $ polynomial in $n$ (as long as $n\in\mathbb{N}$).
Why, and how do we find this polynomial?

Here is an answer, which we will prove in the next chapter:

\begin{theorem}
\label{thm.sum-of-k-th-powers-teaser}Let $n\in\mathbb{N}$, and let $k$ be a
positive integer. Then,%
\[
\sum_{i=1}^{n}i^{k}=\sum_{i=0}^{k}\operatorname*{sur}\left(  k,i\right)
\cdot\dbinom{n+1}{i+1},
\]
where

\begin{itemize}
\item $\operatorname*{sur}\left(  k,i\right)  $ denotes the \# of surjective
(= onto) maps from $\left[  k\right]  $ to $\left[  i\right]  $;

\item $\dbinom{n+1}{i+1}$ is an instance of a binomial coefficient. The
binomial coefficient $\dbinom{m}{k}$ is defined by%
\[
\dbinom{m}{k}=\dfrac{m\left(  m-1\right)  \left(  m-2\right)  \cdots\left(
m-k+1\right)  }{k\left(  k-1\right)  \left(  k-2\right)  \cdots1}%
\]
for any number $m$ and any $k\in\mathbb{N}$ (we will soon extend this
definition to all $k$); for fixed $k$, this is a degree-$k$ polynomial in $m$.
\end{itemize}
\end{theorem}

This theorem is actually useful in computing $\sum\limits_{i=1}^{n}i^{k}$:
Even if the right hand side is given as a finite sum, this sum has only $k+1$
addends, so it does not get more complicated with increasing $n$.

For example, we can rederive our above formula for $\sum\limits_{i=1}^{n}%
i^{2}$ from Theorem \ref{thm.sum-of-k-th-powers-teaser}: Namely, plugging
$k=2$ into Theorem \ref{thm.sum-of-k-th-powers-teaser}, we get%
\begin{align*}
&  \sum_{i=1}^{n}i^{2}\\
&  =\sum_{i=0}^{2}\operatorname*{sur}\left(  2,i\right)  \cdot\dbinom
{n+1}{i+1}\\
&  =\underbrace{\operatorname*{sur}\left(  2,0\right)  }%
_{\substack{=0\\\text{(since there are no}\\\text{surjective maps }\left[
2\right]  \rightarrow\left[  0\right]  \text{,}\\\text{or any maps }\left[
2\right]  \rightarrow\left[  0\right]  \\\text{for that matter)}}}\cdot
\dbinom{n+1}{0+1}+\underbrace{\operatorname*{sur}\left(  2,1\right)
}_{\substack{=1\\\text{(since there is exactly}\\\text{one surjective}%
\\\text{map }\left[  2\right]  \rightarrow\left[  1\right]  \text{)}}%
}\cdot\dbinom{n+1}{1+1}+\underbrace{\operatorname*{sur}\left(  2,2\right)
}_{\substack{=2\\\text{(since there are}\\\text{two surjective}\\\text{maps
}\left[  2\right]  \rightarrow\left[  2\right]  \text{)}}}\cdot\dbinom
{n+1}{2+1}\\
&  =0\cdot\dbinom{n+1}{0+1}+1\cdot\dbinom{n+1}{1+1}+2\cdot\dbinom{n+1}{2+1}\\
&  =\dbinom{n+1}{1+1}+2\cdot\dbinom{n+1}{2+1}=\dbinom{n+1}{2}+2\cdot
\dbinom{n+1}{3}\\
&  =\dfrac{\left(  n+1\right)  n}{2\cdot1}+2\cdot\dfrac{\left(  n+1\right)
n\left(  n-1\right)  }{3\cdot2\cdot1}=\dfrac{n\left(  n+1\right)  \left(
2n+1\right)  }{6}.
\end{align*}
So we recover our formula for the sum of the first $n$ squares (Theorem
\ref{thm.sum-of-squares}).

We won't prove Theorem \ref{thm.sum-of-k-th-powers-teaser} right now, but we
will take a closer look at the expressions that appear in it: first, the
factorials and the binomial coefficients.

\subsection{Factorials and binomial coefficients}

\subsubsection{Factorials}

\begin{definition}
For any $n\in\mathbb{N}$, we define a positive integer $n!$ by%
\[
n!=1\cdot2\cdot3\cdot\cdots\cdot n=n\cdot\left(  n-1\right)  \cdot\left(
n-2\right)  \cdot\cdots\cdot1=\prod_{i=1}^{n}i.
\]
This integer $n!$ is called \textquotedblleft$n$ \textbf{factorial}%
\textquotedblright.
\end{definition}

So we have%
\begin{align*}
0!  &  =1\cdot2\cdot\cdots\cdot0=\prod_{i=1}^{0}i=\left(  \text{empty
product}\right)  =1;\\
1!  &  =1=1;\\
2!  &  =1\cdot2=2;\\
3!  &  =1\cdot2\cdot3=6;\\
4!  &  =1\cdot2\cdot3\cdot4=24;\\
5!  &  =1\cdot2\cdot3\cdot4\cdot5=120;\\
6!  &  =720;\\
7!  &  =5\ 040;\\
8!  &  =40\ 320.
\end{align*}
(Yes, these numbers grow fast! See
\href{https://en.wikipedia.org/wiki/Stirlings_approximation}{Stirling's
approximation} for about how fast they grow.)

Factorials can be computed recursively:

\begin{proposition}
If $n$ is a positive integer, then $n!=\left(  n-1\right)  !\cdot n$.
\end{proposition}

\begin{proof}
Just split off the factor $n$ from $n!=1\cdot2\cdot\cdots\cdot n$. The
remaining factors form a product that equals $\left(  n-1\right)  !$.
\end{proof}

Factorials can be used to describe quite a few other things; the following fun
exercise (Exercise 1.3.1 in the 2019 notes) is a simple example:

\begin{exercise}
Let $n\in\mathbb{N}$. Prove that the product of the first $n$ \textbf{odd}
positive integers is%
\[
1\cdot3\cdot5\cdot\cdots\cdot\left(  2n-1\right)  =\dfrac{\left(  2n\right)
!}{2^{n}n!}.
\]

\end{exercise}

\subsubsection{Definition of the binomial coefficients}

We can now define binomial coefficients in the generality I want to define
them in:

\begin{definition}
Let $n$ and $k$ be any two numbers (i.e., real or complex numbers). We define
a number $\dbinom{n}{k}$ as follows:

\begin{itemize}
\item If $k\in\mathbb{N}$ (recall that $0\in\mathbb{N}$), then%
\[
\dbinom{n}{k}=\dfrac{n\left(  n-1\right)  \left(  n-2\right)  \cdots\left(
n-k+1\right)  }{k!}.
\]


\item If $k\notin\mathbb{N}$, then%
\[
\dbinom{n}{k}=0.
\]

\end{itemize}

This number $\dbinom{n}{k}$ is pronounced \textquotedblleft$n$ \textbf{choose}
$k$\textquotedblright\ and is called a \textbf{binomial coefficient}.
\end{definition}

This definition is standard across much of the literature, but some authors
nevertheless disagree with it, mostly in situations when $n$ is negative or
$k$ is a non-integer. So be careful with some texts and papers.

Some authors use the notations $C_{k}^{n}$ or $^{n}C_{k}$ or $_{n}C_{k}$ for
$\dbinom{n}{k}$.

Do not mistake the binomial coefficient $\dbinom{n}{k}$ for the size-$2$
vector $\left(
\begin{array}
[c]{c}%
n\\
k
\end{array}
\right)  $. They look very similar (the only real difference is the spacing);
fortunately they rarely appear in the same context.

The above definition of $\dbinom{n}{k}$ is worth memorizing. The product
\[
n\left(  n-1\right)  \left(  n-2\right)  \cdots\left(  n-k+1\right)
\]
in the numerator is a product of $k$ factors, the first of which is $n$ while
each following factor equals the previous factor minus $1$. Let's apply the
definition to compute some examples:

\begin{example}
\textbf{(a)} For any number $n$, we have%
\[
\dbinom{n}{0}=\dfrac{\left(  \text{empty product}\right)  }{0!}=\dfrac{1}%
{1}=1.
\]


\textbf{(b)} For any number $n$, we have%
\[
\dbinom{n}{1}=\dfrac{n}{1!}=\dfrac{n}{1}=n.
\]


\textbf{(c)} For any number $n$, we have%
\[
\dbinom{n}{2}=\dfrac{n\left(  n-1\right)  }{2!}=\dfrac{n\left(  n-1\right)
}{2}.
\]


\textbf{(d)} For any number $n$, we likewise have%
\[
\dbinom{n}{3}=\dfrac{n\left(  n-1\right)  \left(  n-2\right)  }{6}.
\]


\textbf{(e)} For any $k\in\mathbb{N}$, we have%
\begin{align*}
\dbinom{-1}{k}  &  =\dfrac{\left(  -1\right)  \left(  -1-1\right)  \left(
-1-2\right)  \cdots\left(  -1-k+1\right)  }{k!}\\
&  =\dfrac{\left(  -1\right)  \left(  -2\right)  \left(  -3\right)
\cdots\left(  -k\right)  }{k!}=\dfrac{\left(  -1\right)  ^{k}\cdot1\cdot
2\cdot\cdots\cdot k}{k!}=\dfrac{\left(  -1\right)  ^{k}\cdot k!}{k!}=\left(
-1\right)  ^{k}.
\end{align*}


\textbf{(f)} We have%
\[
\dbinom{2}{\sqrt{2}}=0\ \ \ \ \ \ \ \ \ \ \left(  \text{since }\sqrt{2}%
\notin\mathbb{N}\right)
\]
but
\[
\dbinom{\sqrt{2}}{2}=\dfrac{\sqrt{2}\left(  \sqrt{2}-1\right)  }{2}\neq0.
\]

\end{example}

It would be nice to have the binomial coefficients all nicely tabulated; alas,
there are infinitely many of them, so the table would take a bit too much
space. Here is a table of the binomial coefficients $\dbinom{n}{k}$ for all
$n\in\left\{  -3,-2,-1,\ldots,6\right\}  $ and some of the $k\in\left\{
0,1,2,3,4,5\right\}  $ (this is a somewhat unusual table, since the values of
$n$ correspond to the rows, but the values of $k$ correspond to
\textbf{diagonals}, not columns):%
\[
\hspace{-2.5pc}\hspace{-0.3cm}%
\begin{tabular}
[c]{ll|ccccccccccccccccc}%
\vphantom{$g^b$} &  & $\phantom{20}$ & $\phantom{20}$ & $\phantom{20}$ &
$\phantom{20}$ &  &  &  &  &  &  & $\overset{k=0}{\swarrow}$ & $\phantom{20}$
& $\overset{k=1}{\swarrow}$ & $\phantom{20}$ & $\overset{k=2}{\swarrow}$ &
$\phantom{20}$ & $\overset{k=3}{\swarrow}$\\\hline
\vphantom{$g^b$} $n=-3$ & $\rightarrow$ &  &  &  &  &  &  &  &  &  & $1$ &  &
$-3$ &  & $6$ &  & $-10$ & \\
\vphantom{$g^b$} $n=-2$ & $\rightarrow$ &  &  &  &  &  &  &  &  & $1$ &  &
$-2$ &  & $3$ &  & $-4$ &  & \\
\vphantom{$g^b$} $n=-1$ & $\rightarrow$ &  &  &  &  &  &  &  & $1$ &  & $-1$ &
& $1$ &  & $-1$ &  & $1$ & \\
\vphantom{$g^b$} $n=0$ & $\rightarrow$ &  &  &  &  &  &  & $1$ &  & $0$ &  &
$0$ &  & $0$ &  & $0$ &  & \\
\vphantom{$g^b$} $n=1$ & $\rightarrow$ &  &  &  &  &  & $1$ &  & $1$ &  & $0$
&  & $0$ &  & $0$ &  & $0$ & \\
\vphantom{$g^b$} $n=2$ & $\rightarrow$ &  &  &  &  & $1$ &  & $2$ &  & $1$ &
& $0$ &  & $0$ &  & $0$ &  & \\
\vphantom{$g^b$} $n=3$ & $\rightarrow$ &  &  &  & $1$ &  & $3$ &  & $3$ &  &
$1$ &  & $0$ &  & $0$ &  & $0$ & \\
\vphantom{$g^b$} $n=4$ & $\rightarrow$ &  &  & $1$ &  & $4$ &  & $6$ &  & $4$
&  & $1$ &  & $0$ &  & $0$ &  & \\
\vphantom{$g^b$} $n=5$ & $\rightarrow$ &  & $1$ &  & $5$ &  & $10$ &  & $10$ &
& $5$ &  & $1$ &  & $0$ &  & $0$ & \\
\vphantom{$g^b$} $n=6$ & $\rightarrow$ & $1$ &  & $6$ &  & $15$ &  & $20$ &  &
$15$ &  & $6$ &  & $1$ &  & $0$ &  &
\end{tabular}
\ \ \ \ \ \ \ \
\]


\subsubsection{Fundamental properties of the binomial coefficients}

By looking closely at this table, we can spot several important properties of
binomial coefficients. Here, for one, is the explanation for the many $0$'s in
the right half of the table:

\begin{proposition}
\label{prop.binom.0}If $n\in\mathbb{N}$ and $k\in\mathbb{R}$ satisfy $k>n$,
then $\dbinom{n}{k}=0$.
\end{proposition}

\begin{proof}
Let $n\in\mathbb{N}$ and $k\in\mathbb{R}$ satisfy $k>n$. If $k\notin%
\mathbb{N}$, then $\dbinom{n}{k}=0$ by definition. So assume WLOG that
$k\in\mathbb{N}$. Then, the definition of $\dbinom{n}{k}$ yields%
\[
\dbinom{n}{k}=\dfrac{n\left(  n-1\right)  \left(  n-2\right)  \cdots\left(
n-k+1\right)  }{k!}.
\]
However, $k>n$ and $k\in\mathbb{N}$ entail that one of the factors
$n,n-1,n-2,\ldots,n-k+1$ in the numerator is $0$. So the whole product is $0$,
and with it $\dbinom{n}{k}$.
\end{proof}

\textbf{Warning:} Proposition \ref{prop.binom.0} does not hold if we drop the
$n\in\mathbb{N}$ requirement. For instance, $\dbinom{-1}{2}\neq0$ even though
$2>-1$.


\end{document}