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\section*{Binomial coefficients}
\providecommand{\ddx}{\frac{d}{dx}}
\providecommand{\ddz}{\frac{d}{dz}}

\subsection*{Miscellaneous Curiosities}

\paragraph{Recall:}
\begin{itemize}\item For $n$ a non-negative integer and $r$ an integer,
\\	${n \choose r} =$ number of subsets of size $r$ of a set of size $n$
\mbox{}\\ \mbox{\hspace{2em}}$= \frac{n!}{n!(n-r)!} \quad$ if $0 \leq  r \leq  n$
\mbox{}\\ \mbox{\hspace{2em}}$= 0 \quad\quad$ else.

\item Pascal's triangle

\item Pascal's Formula: ${n \choose r} = {n-1 \choose r} + {n-1 \choose r-1}$

\item ${n \choose r} = {n \choose n-r}$

\item $\sum_{r=0}^n {n \choose r} = 2^n$

\item $(x+y)^n = \sum_{r=0}^n {n \choose r} x^{n-r}y^r$
\end{itemize}

\paragraph{Remark:}
\mbox{}\\    ${n \choose r} =$ number of paths from root of Pascal's triangle to the
    $(n,r)$ position.

\paragraph{Further identities:}
\begin{itemize}\item $k{n \choose k} = n{n-1 \choose k-1}$
        (immediate from ${n \choose k} = \frac{n(n-1)...(n-k+1)}{k(k-1)...1}$)

\item $(x+1)^n = \sum_{r=0}^n {n \choose r} x^r$

\item $0 = ((-1) + 1)^n = \sum_{r=0}^n {n \choose r} (-1)^r$;
\\        so alternating sum of binomial coefficients is 0;
\\        so sum of even coefficients = sum of odd coefficients $= 2^{n-1}$.
\end{itemize}

\paragraph{Yet further identities:}
\begin{enumerate}[(i)]\item ${n+1 \choose r+1} = \sum_{s=0}^n {s \choose r}$
\item $\sum_{r=0}^n {n \choose r}^2 = {2n \choose n}$
\item $\sum_{r=0}^n r {n \choose r} = n2^{n-1}$
\end{enumerate}
\paragraph{Proofs:}
\begin{enumerate}[(i)]\item Iteratively apply Pascal's formula:
\\        ${n+1 \choose r+1} = {n \choose r+1} + {n \choose r}$
\mbox{}\\ \mbox{\hspace{2em}}$= {n-1 \choose r+1} + {n-1 \choose r}$
\mbox{}\\ \mbox{\hspace{2em}}$= ...$
\mbox{}\\ \mbox{\hspace{2em}}$= {0 \choose r+1} + {0 \choose r} + ... + {n-1 \choose r}$
\mbox{}\\ \mbox{\hspace{2em}}$= {0 \choose r} + ... + {n-1 \choose r}$

        Alternative inductive proof:
\\        Easily holds for $n=r=0$.
\\        Suppose inductively it holds for smaller $n+r$.
\\        Then using Pascal's formula, we have:
\\        ${n+1\choose r+1} = {n \choose r+1} + {n \choose r}$
\mbox{}\\ \mbox{\hspace{2em}}$= \sum_{k=0}^{n-1} {k \choose r} + \sum_{k=0}^{n-1} {k \choose r-1}$
\mbox{}\\ \mbox{\hspace{2em}}$= \sum_{k=0}^{n-1} ({k \choose r} + {k\choose r-1})$
\mbox{}\\ \mbox{\hspace{2em}}$= \sum_{k=0}^{n-1} {k+1 \choose r}$
\mbox{}\\ \mbox{\hspace{2em}}$= \sum_{k=1}^{n} {k \choose r}$
\mbox{}\\ \mbox{\hspace{2em}}$= \sum_{k=0}^{n} {k \choose r}$

\item Consider diamonds in Pascal's triangle.

	OR: Given a set $S$ of size $2n$, arbitrarily split it into two sets
	$S_1,S_2$ of size $n$.
\\	Then an $n$-subset $S'$ of $S$ corresponds to
	the pair $(S' \cap S_1, S' \cap S_2)$.
\\	The pairs of subsets arising in this way are precisely those of sizes
	summing to $n$,
\\	so
\mbox{}\\ \mbox{\hspace{4em}}${2n \choose n} = \sum_{r=0}^n {n \choose r}{n \choose n-r} = \sum_{r=0}^n {n \choose r}^2$

\item Neat algebraic proof:
\\        $n (x+1)^{n-1} = \ddx(x+1)^n = \ddx \sum_{r=0}^n {n \choose r} x^r$
\mbox{}\\ \mbox{\hspace{2em}}$= \sum_{r=0}^n r {n \choose r} x^r$.
\\        This holds for all $x$; taking $x=1$ gives the result.

\end{enumerate}
\paragraph{Examples of (i):}
\begin{itemize}\item ${n \choose 1} = \sum_{s=0}^{n-1} {s \choose 0} = \sum_{s=0}^{n-1} 1 = n$
\item${n \choose 2} = \sum_{s=0}^{n-1} {s \choose 1} = \sum_{s=0}^{n-1} s =$ $n$th triangular number
\item${n \choose 3} = \sum_{s=0}^{n-1} {s \choose 2} =$ $n$th pyrimidal number
\end{itemize}


\subsection*{Multinomial theorem}

What is the coefficient $a_{r,s,t}$ of $x^ry^sz^t$ in the expansion of $(x+y+z)^n$?

Clearly $a_{r,s,t} \neq  0$ only if $r+s+t=n$.

$a_{r,s,t}$ is the number of ways of choosing $r$ $x$'s, $s$ $y$'s, and $t$
$z$'s from the $n$ factors $(x+y+z)$;
\\i.e. the number of strings like "xyzzyxyzzy" with this many of each letter;
\\i.e. the number of permutations of the multiset $\{ r*x, s*y, t*z \}$.

So as we saw before,
\mbox{}\\ \mbox{\hspace{2em}}$a_{r,s,t} = \frac{n!}{r!s!t!}$.

Write ${n \choose r\;s\;t}$ for this number.

Generalising to arbitrarily many variables, we have

\paragraph{Theorem:}
\mbox{}\\    $(x_1 + ... + x_t)^n = \sum_{n_i \geq  0,n_1+...+n_t=n}
        {n \choose n_1\;n_2\;...\;n_t} x_1^{n_1}x_2^{n_2}...x_t^{n_t}$

Here, ${n \choose n_1\;n_2\;...\;n_t} = \frac{n!}{n_1!...n_t!}$
are the \underline{multinomial coefficients} (only defined if $n_1+...+n_t=n$).

\paragraph{Note:}
\mbox{}\\    ${n \choose r} = {n \choose r\;n-r}$.

The number of terms in the multinomial expansion of $(x_1+...+x_t)^n$ is the
number of $n$-combinations with $t$ types in unlimited supply, which we saw is
\mbox{}\\ \mbox{\hspace{2em}}${n+t-1 \choose n}$.


\subsection*{Unnatural exponents: $(x+y)^\alpha$}

\paragraph{Theorem [Newton's Binomial Theorem]:}
\mbox{}\\    Let $\alpha$ be a real (or even complex) number.
\\    Suppose $0 \leq  |x| < |y|$.
\\    Then
\mbox{}\\ \mbox{\hspace{2em}}$(x+y)^\alpha = \sum_{k=0}^\infty {\alpha \choose k} x^k y^{\alpha-k}$

    where
\mbox{}\\ \mbox{\hspace{2em}}${\alpha \choose k} = \frac{\alpha(\alpha-1)...(\alpha-k+1)}{k!}$.

Note that for $\alpha$ natural, this agrees with our previous definition.

\paragraph{Proof (not on syllabus):}
\mbox{}\\    Dividing through by $y^\alpha$, sufficient to show that for $|z|<1$,
\mbox{}\\ \mbox{\hspace{2em}}$(1+z)^\alpha = \sum_{k=0}^\infty {\alpha \choose k} z^k$.

    We show this for complex $z$ with $|z|<1$.
\\    $(1+z)^\alpha = \exp(\alpha\log(1+z))$ for any choice of branch.
\\    This is holomorphic on the domain $|z|<1$,
\\    so the Taylor series at 0 converges to the value of the function on this domain.

    Since $\ddz (1+z)^\alpha = \alpha(1+z)^{\alpha-1}$ and $(1+0)^\alpha=1$,
\\    this gives
\mbox{}\\ \mbox{\hspace{2em}}$(1+z)^\alpha = ((1+z)^\alpha |_{z=0})\frac{z^0}{0!} +$
\mbox{}\\ \mbox{\hspace{4em}}$(\alpha(1+z)^{\alpha-1}|_{z=0})\frac{z^1}{1!} +$
\mbox{}\\ \mbox{\hspace{4em}}$(\alpha(\alpha-1)(1+z)^{\alpha-2}|_{z=0})\frac{z^2}{2!} + ...$
\mbox{}\\ \mbox{\hspace{2em}}$= \sum_{k=0}^\infty (\alpha(\alpha-1)...(\alpha-k+1))\frac{z^k}{k!}$
\mbox{}\\ \mbox{\hspace{2em}}$= \sum_{k=0}^\infty {\alpha \choose k} z^k$
\\    \qed

\paragraph{Examples:}
\begin{itemize}\item $\frac{1}{1+z} = (1+z)^{-1}$
\mbox{}\\ \mbox{\hspace{2em}}$= \sum_{k=0}^\infty {-1 \choose k} z^k$
\mbox{}\\ \mbox{\hspace{2em}}$= \sum_{k=0}^\infty \frac{(-1)*(-2)*...*(-k)}{k*(k-1)*...*1} z^k$
\mbox{}\\ \mbox{\hspace{2em}}$= \sum_{k=0}^\infty (-1)^k z^k$
\mbox{}\\ \mbox{\hspace{2em}}$= 1 - z + z^2 - z^3 + ...$

\item
\mbox{}\\ \mbox{\hspace{2em}}$\sqrt{37} = \sqrt{6^2+1} = 6\sqrt{1+1/36} = 6(1+1/36)^{1/2}$
\mbox{}\\ \mbox{\hspace{4em}}$= 6(\sum_{k=0}^\infty {1/2 \choose k} (1/36)^k)$

        Now for $k>0$,
\mbox{}\\ \mbox{\hspace{2em}}${1/2\choose k} = \frac{ \frac{1}{2}\frac{1-2}{2}...\frac{1-2(k-1)}{2} }{k!}$
\mbox{}\\ \mbox{\hspace{4em}}$= \frac{(-1)^{k-1} 1*3*5*...*(2k-3)}{2^k k!}$
\mbox{}\\ \mbox{\hspace{4em}}$= \frac{(-1)^{k-1} (2k-2)!}{2^k (2*4*...*(2k-2)) k!}$
\mbox{}\\ \mbox{\hspace{4em}}$= \frac{(-1)^{k-1} (2k-2)!}{2^{2k-1} (k-1)! k!}$
\mbox{}\\ \mbox{\hspace{4em}}$= \frac{(-1)^{k-1}}{k 2^{2k-1}} {2k-2 \choose k-1}$

        So
\mbox{}\\ \mbox{\hspace{2em}}$\sqrt{1+z} = 1 + \sum_{k=1}^\infty \frac{(-1)^{k-1}}{k 2^{2k-1}} {2k-2 \choose k-1} z^k$
\mbox{}\\ \mbox{\hspace{4em}}$= 1 + \frac{1}{2}z - \frac{1}{8}z^2 + \frac{1}{16}z^3 - \frac{1}{25}z^4 + ...$

        So $\sqrt{37} = 6(1+1/36)^{1/2}$
\mbox{}\\ \mbox{\hspace{2em}}$\approx 6(1 + 1/(2*36) - 1/(8*36^2) + 1/(16*36^3) - 1/(25*36^4))$
\mbox{}\\ \mbox{\hspace{2em}}$= 6.0828$

        (Error is very small: $(6*(1 + 1/(2*36) - 1/(8*36^2) + 1/(16*36^3) - 1/(25*36^4)))^2 = 36.99999992692224$)

\end{itemize}
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