% This is lecture no. 24 of MH1101/: Calculus II.

\chapter{The Binomial Series}%
\label{cha:the_binomial_series}

\section{Using the Binomial Theorem}%
\label{sec:using_the_binomial_theorem}

We can find the Maclaurin series for the function $f(x)=(1+x)^{k}$ where $k$ is a real number.
First, we consider expanding the general expression $(a+b)^{n}$, where $n$ is some positive integer.
The \textbf{Binomial Theorem} states that for any positive integer $n$:
\[
	(a + b)^{n} = \sum_{i=0}^{n} \binom{n}{i}a^{n-i}b^{i}
,\]
where $\ds{\binom{n}{i}}$ is the \textbf{binomial coefficient} defined by:
\[
	\binom{n}{i} = \frac{n!}{i!(n-i)!} = \frac{n(n-1)(n-2)\cdots(n-i+1)}{i!}
.\]
\begin{exmp}
	For $n=3$, we have:
	\[
		(a+b)^3 = \sum_{i=0}^{3} \binom{3}{i}a^{3-i}b^{i} = a^3 + 3a^2b + 3ab^2 + b^3
	.\]
\end{exmp}
This theorem is used exclusively in the case where $n$ in $(a+b)^{n}$ is strictly a positive integer.
When $n$ is not a positive integer, the theorem's conditions are not satisfied, however, the theorem is still useable.
First, we take a look at the Maclaurin series for $(1+x)^{k}$ where $k$ is any real number.
For notational convenience, we extend the definition of the binomial coefficient to any real number $k$:
\[
	\binom{k}{n} = \frac{k(k-1)(k-2)\cdots(k-n+1)}{n!}
.\]
By convention, $\ds{\binom{k}{0}=1}$.
\begin{exmp}
	Find the Maclaurin series for $f(x)=(1+x)^{k}$, where $k$ is any real number.
	\begin{sol}
		We first arrange the first few computations to deduce the general formula:
		\[
		\begin{aligned}
			f(x) & = (1+x)^{k}\\
			f^{\prime}(x) & = k(1+x)^{k-1}\\
			f^{(2)}(x) & = k(k-1)(1+x)^{k-2}\\
			f^{(3)}(x) & = k(k-1)(k-2)(1+x)^{k-3}\\
				   &\ \,\vdots\\
			f^{(n)}(x) & = k(k-1)\cdots(k-n+1)(1+x)^{k-n}\\
		\end{aligned}
		\quad
		\begin{aligned}
			f(0) & = 1\\
			f^{\prime}(0) & = k\\
			f^{(2)}(0) & = k(k-1)\\
			f^{(3)}(0) & = k(k-1)(k-2)\\
				   &\ \,\vdots\\
			f^{(n)}(0) & = k(k-1)\cdots(k-n+1)\\
		\end{aligned}
		\]
		Therefore, the Macluarin series of $f(x)=(1+x)^{k}$ is:
		\[
			\sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!}x^{n} = \sum_{n=0}^{\infty} \frac{k(k-1)\cdots(k-n+1)}{n!}x^{n} = \sum_{n=0}^{\infty} \binom{k}{n}x^{n}
		.\]
		Which is known as the \textbf{Binomial Series}.
	\end{sol}
\end{exmp}
\begin{note}
	If $k$ is a positive integer, then eventually, the terms in the binomial series will be 0.
	For other values of $k$, none of the terms will be 0, and so we can try the ratio test to determine its convergence:

	\bigbreak
	\noindent
	Let $a_n=\ds{\binom{k}{n}x^{n}}$.
	Then:
	\begin{align*}
		\left\lvert \frac{a_{n+1}}{a_n}\right\rvert & = \left\lvert \frac{k(k-1)\cdots(k-n+1)(k-n)x^{n+1}}{(n+1)!}\cdot \frac{n!}{k(k-1)\cdots (k-n+1)x^{n}}\right\rvert\\
							    & = \frac{\lvert k-n\rvert}{n+1}\lvert x\rvert\\
							    & = \frac{\left\lvert 1- \frac{k}{n}\right\rvert}{1+\frac{1}{n}}\lvert x\rvert
	\end{align*}
	Taking the limits of both sides we see that:
	\[
	\lim_{n \to \infty}\left\lvert \frac{a_{n+1}}{a_n}\right\rvert = \lvert x\rvert
	.\]
	Thus by the ratio test, the binomial series converges if $\lvert x\rvert<1$ and diverges if $\lvert x\rvert>1$.
\end{note}
\bigbreak
The following theorem states that the function $(1+x)^{k}$ is indeed equal to its Maclaurin series.

\begin{thrm}
	The Binomial Series.
	If $k$ is any real number and $\lvert x\rvert<1$, then:
	\[
		(1+x)^{k}= \sum_{n=0}^{\infty} \binom{k}{n}x^{n}
	.\]
\end{thrm}
\begin{exmp}
	Find the Maclaurin series for the function $f(x)=\frac{1}{\sqrt{4-x}}$.
	\begin{sol}
		We rewrite $f(x)$ in a form where we can apply the binomial series:
		\begin{align*}
			\frac{1}{\sqrt{4-x}} & = \frac{1}{\sqrt{4\left(1-\frac{x}{4}\right)}}\\
					     & = \frac{1}{2\sqrt{1-\frac{x}{4}}}\\
					     & = \frac{1}{2}\left(1-\frac{x}{4}\right)^{-\frac{1}{2}}
		\end{align*}
		Use the binomial series with $k=-\frac{1}{2}$, and with $x$ replaced by $-\ds{\frac{x}{4}}$:
		\begin{align*}
			\frac{1}{\sqrt{4-x}} & = \frac{1}{2}\sum_{n=0}^{\infty} \binom{-\frac{1}{2}}{n} \left(-\frac{x}{4}\right)^{n}\\
					     & = \frac{1}{2}\sum_{n=0}^{\infty} \binom{-\frac{1}{2}}{n}(-1)^{n}\frac{x^{n}}{4^{n}}\\
					     & = \frac{1}{2} + \frac{1}{2} \sum_{n=1}^{\infty} \binom{-\frac{1}{2}}{n}(-1)^{n}\frac{x^{n}}{4^{n}}
		\end{align*}
		We expand $\ds{\binom{-\frac{1}{2}}{n}}$, for $n\geqslant 1$, as follows:
		\begin{align*}
			\binom{-\frac{1}{2}}{n} & = \frac{\left(-\frac{1}{2}\right)\left(-\frac{3}{2}\right)\cdots\left(-\frac{2n-1}{2}\right)}{n!}\\
						& = (-1)^{n}\frac{1\cdot 3\cdot 5\cdots(2n-1)}{2^{n}n!}
		\end{align*}
		Substituting this back into equation, we get:
		\[
			\frac{1}{\sqrt{4-x}}=\frac{1}{2}+\frac{1}{2}\sum_{n=1}^{\infty} \frac{1\cdot 3\cdot 5\cdots(2n-1)}{8^{n}n!}x^{n}
		.\]
		From the binomial series theorem, we know that htis series converges when:
		\[
		\left\lvert -\frac{x}{4}\right\rvert < 1
		,\]
		that is, $\lvert x\rvert<4$, so the radius of convergence in this case is $R=4$.
	\end{sol}
\end{exmp}
