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\newcommand{\coproduct}{\coprod} \newcommand{\product}{\prod} \newcommand{\closure}{\overline} \newcommand{\integral}{\int} \newcommand{\doubleintegral}{\iint} \newcommand{\tripleintegral}{\iiint} \newcommand{\quadrupleintegral}{\iiiint} \newcommand{\conint}{\oint} \newcommand{\contourintegral}{\oint} \newcommand{\infinity}{\infty} \newcommand{\bottom}{\bot} \newcommand{\minusb}{\boxminus} \newcommand{\plusb}{\boxplus} \newcommand{\timesb}{\boxtimes} \newcommand{\intersection}{\cap} \newcommand{\union}{\cup} \newcommand{\Del}{\nabla} \newcommand{\odash}{\circleddash} \newcommand{\negspace}{\!} \newcommand{\widebar}{\overline} \newcommand{\textsize}{\normalsize} \renewcommand{\scriptsize}{\scriptstyle} \newcommand{\scriptscriptsize}{\scriptscriptstyle} \newcommand{\mathfr}{\mathfrak} \newcommand{\statusline}[2]{#2} \newcommand{\tooltip}[2]{#2} \newcommand{\toggle}[2]{#2} % Theorem Environments \theoremstyle{plain} \newtheorem{theorem}{Theorem} \newtheorem{lemma}{Lemma} \newtheorem{prop}{Proposition} \newtheorem{cor}{Corollary} \newtheorem*{utheorem}{Theorem} \newtheorem*{ulemma}{Lemma} \newtheorem*{uprop}{Proposition} \newtheorem*{ucor}{Corollary} \theoremstyle{definition} \newtheorem{defn}{Definition} \newtheorem{example}{Example} \newtheorem*{udefn}{Definition} \newtheorem*{uexample}{Example} \theoremstyle{remark} \newtheorem{remark}{Remark} \newtheorem{note}{Note} \newtheorem*{uremark}{Remark} \newtheorem*{unote}{Note} %------------------------------------------------------------------- \begin{document} %------------------------------------------------------------------- \section*{binomial theorem} \hypertarget{context}{}\subsubsection*{{Context}}\label{context} \hypertarget{arithmetic}{}\paragraph*{{Arithmetic}}\label{arithmetic} [[!include arithmetic geometry - contents]] \hypertarget{combinatorics}{}\paragraph*{{Combinatorics}}\label{combinatorics} [[!include combinatorics - contents]] \hypertarget{contents}{}\section*{{Contents}}\label{contents} \noindent\hyperlink{statement}{Statement}\dotfill \pageref*{statement} \linebreak \noindent\hyperlink{combinatorial_interpretation}{Combinatorial interpretation}\dotfill \pageref*{combinatorial_interpretation} \linebreak \noindent\hyperlink{pascals_triangle}{Pascal's triangle}\dotfill \pageref*{pascals_triangle} \linebreak \noindent\hyperlink{related_concepts}{Related concepts}\dotfill \pageref*{related_concepts} \linebreak \noindent\hyperlink{references}{References}\dotfill \pageref*{references} \linebreak \hypertarget{statement}{}\subsection*{{Statement}}\label{statement} For $k$ a [[natural number]] and $r$ a [[complex number]], define the [[falling factorial]] by \begin{displaymath} r^{\underline{k}} = r(r-1)\ldots (r-k+1), \end{displaymath} a polynomial of degree $k$ evaluated at $r$. If $r$ is a natural number, this expression vanishes for $k \gt r$. The binomial theorem may be stated thus: if $r$ is any complex number and ${|x|} \lt 1$, then \begin{displaymath} (1 + x)^r = \sum_{k \geq 0} \frac{r^{\underline{k}} x^k}{k!} \end{displaymath} where the left side may be formally defined as $\exp(r \cdot \log (1+x))$, taking the principal branch of the [[logarithm]] as defined by the power series \begin{displaymath} \log(1 + x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \ldots \end{displaymath} with radius of convergence equal to $1$. (A formal verification of the binomial theorem may be found at [[coinduction]].) Thus, if we define the \textbf{binomial coefficient} $\binom{r}{k}$ by the formula \begin{displaymath} \binom{r}{k} \coloneqq \frac{r^{\underline{k}}}{k!}, \end{displaymath} then we have \begin{displaymath} (y + x)^r = \sum_{k \geq 0} \binom{r}{k} y^{r-k}x^k \end{displaymath} whenever ${|\frac{x}{y}|} \lt 1$, i.e., whenever ${|y|} \gt {|x|}$. More precisely: for any fixed $y \neq 0$, this equation holds for any branch of the logarithm that we use to define $(y+x)^r$ as $\exp(r\log(y+x))$ over the domain $\{x: {|x|} \lt {|y|}\}$. \hypertarget{combinatorial_interpretation}{}\subsection*{{Combinatorial interpretation}}\label{combinatorial_interpretation} The special finitary case in which $i, j, n$ are positive integers, \begin{displaymath} (i + j)^n = \sum_{0 \leq k \leq n} \binom{n}{k} i^k j^{n-k} \end{displaymath} may be established combinatorially (or in ``bijective fashion'') as follows. Start by interpreting $(i + j)^n$ as the number of functions $f: N \to I \sqcup J$ from an $n$-element set, where $I, J$ have cardinalities $i, j$ respectively. By pulling back $f$ along each of the inclusions of $I, J$ into $I \sqcup J$, we get functions $f_I: f^{-1}(I) \to I$, $f_J: f^{-1}(J) \to J$. Here $f^{-1}(I)$ and $f^{-1}(J)$ are complementary subsets of $N$, say of cardinalities $k$ and $n-k$ respectively. (In effect, we are using the fact that the category of finite sets is an [[extensive category]].) Thus, $f$ determines and is uniquely determined by the following data \begin{itemize}% \item A subset $K (= f^{-1}(I))$ of $N$, \item A function $g (= f_I)$ of the form $K \to I$, \item A function $h (= f_J)$ of the form $N-K \to J$ \end{itemize} and by counting the number of such triplets $(K, g, h)$, we are led to the right-hand side of the previous displayed equation. Notice this gives a rigorous proof for the polynomial identity \begin{displaymath} (x+y)^n = \sum_{0 \leq k \leq n} \binom{n}{k} x^k y^{n-k} \end{displaymath} since a polynomial in $\mathbb{Z}[x, y]$ is the zero polynomial if it vanishes for all positive integer values substituted for $x$ and $y$. \hypertarget{pascals_triangle}{}\subsection*{{Pascal's triangle}}\label{pascals_triangle} The binomial coefficient polynomials $\binom{x}{k}$ (here $x$ is an indeterminate) satisfy the recurrence \begin{displaymath} \Delta \binom{x}{k} \coloneqq \binom{x+1}{k} - \binom{x}{k} = \binom{x}{k-1}, \qquad \binom{x}{0} \coloneqq 1, \binom{0}{k} = 0\; (k \neq 0) \end{displaymath} where the first two equations may also be written as \begin{displaymath} \Delta \frac{x^\underline{k}}{k!} = \frac{x^\underline{k-1}}{(k-1)!}, \qquad \frac{x^\underline{0}}{0!} = 1. \end{displaymath} The first equation may be viewed as the discrete analogue of the continuous [[derivative]] formula \begin{displaymath} \frac{d}{d x} \frac{x^k}{k!} = \frac{x^{k-1}}{(k-1)!}. \end{displaymath} The more familiar form of this recurrence, \begin{displaymath} \binom{x+1}{k} = \binom{x}{k} + \binom{x}{k-1}, \end{displaymath} may be interpreted combinatorially: a $k$-element subset $K$ of $X + 1$ is either entirely contained in $X$, or is determined by the $(k-1)$-element subset of $X$ that is $K \cap X$. \hypertarget{related_concepts}{}\subsection*{{Related concepts}}\label{related_concepts} \begin{itemize}% \item [[multinomial coefficient]] \item [[factorial]], [[double factorial]] \item [[Catalan number]] \end{itemize} \hypertarget{references}{}\subsection*{{References}}\label{references} See also \begin{itemize}% \item Wikipedia, \emph{\href{https://en.wikipedia.org/wiki/Binomial_theorem}{Binomial theorem}} \end{itemize} [[!redirects binomial coefficient]] [[!redirects binomial coefficients]] \end{document}