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\begin{document}
\title{The Binomial Theorem}
%\subtitle{Notes for a colloquium}
\author{David Pierce}
\date{December 23, 2008\\
Edited July 6, 2018}
\publishers{Mathematics Department\\
Mimar Sinan Fine Arts University\\
Istanbul\\
\url{mat.msgsu.edu.tr/~dpierce/}\\
\url{polytropy.com}}

  \maketitle
\emph{I prepared these notes after teaching from \S~12.11 `The Binomial
  Series' and \S~12.12 `Applications of Taylor Polynomials' of
  Stewart's \emph{Calculus}, 5th ed., 2003.}

The Binomial Theorem learned in high-school is
  \begin{equation*}
    (1+x)^k=\sum_{n=0}^k\binom knx^n,
  \end{equation*}
where $k$ is a positive integer or $0$, and
\begin{equation*}
  \binom kn=
  \begin{cases}
    1,&\text{ if }n=0;\\
%\displaystyle\frac{k(k-1)\dotsm(k-n+1)}{n!}=
\displaystyle\frac kn\cdot
\displaystyle\frac{k-1}{n-1}
\dotsb
\displaystyle\frac{k-n+1}1,&\text{ if }n>0.
  \end{cases}
\end{equation*}
This definition of $\binom kn$ makes sense for \emph{all} real numbers
$k$.  In particular, if $k$ is an integer, and $0\leq k<n$, then
$\binom kn=0$.  Therefore, again if $k$ is a positive integer or $0$,
we have
\begin{equation}\label{eqn:series}
(1+x)^k=  \sum_{n=0}^{\infty}\binom knx^n.
\end{equation}
Since the series here has the form of a Maclaurin series, it must
\emph{be} the Maclaurin series for $(1+x)^k$, and its radius of
convergence is $\infty$.
In particular,
if $f(x)=(1+x)^k$, then
\begin{equation*}
  \frac{f^{(n)}(0)}{n!}=\binom kn.
\end{equation*}
This equation also holds for all real $k$ by direct computation:
\begin{align*}
  f'(x)&=k(1+x)^{k-1},\\
f''(x)&=k(k-1)(1+x)^{k-2},\\
&\hdots\\
f^{(n)}(x)&=k(k-1)\dotsm(k-n+1)(1+x)^{k-n}.
\end{align*}
So the series in~\eqref{eqn:series} is still the Maclaurin series for
$(1+x)^k$.  We ask:
\begin{compactenum}[(i)]
  \item\label{item:1}
what is the radius $R$ of convergence of this series?
\item\label{item:2}
when $\size x<R$, does~\eqref{eqn:series} hold?
\end{compactenum}
To answer~\eqref{item:1}, use the Ratio Test: if $a_n=\binom knx^n$, then
\begin{multline*}
\frac{a_{n+1}}{a_n}=
%  \frac{\binom k{n+1}x^{n+1}}{\binom knx^n}=
\frac{k(k-1)\dotsm(k-n)}{(n+1)!}\cdot
\frac{n!}{k(k-1)\dotsm(k-n+1)}\cdot\frac{x^{n+1}}{x^n}\\
=\frac{k-n}{n+1}\cdot x.
\end{multline*}
Since
\begin{equation*}
  \lim_{n\to\infty}\size{\frac{k-n}{n+1}\cdot x}=
\size x\cdot\lim_{n\to\infty}\size{\frac{-n+k}{n+1}}=\size x,
\end{equation*}
we have $R=1$.  So when $\size x<1$, then
\begin{equation*}
\sum_{n=0}^{\infty}\binom knx^n=g_k(x)
\end{equation*}
for some function $g_k(x)$.  But have we
$g_k(x)=(1+x)^k$?
This is question~\eqref{item:2}.  To answer it, note first that, if $n>0$, then
\begin{multline*}
  \binom{k-1}n
  =\frac{k-1}n\frac{k-2}{n-1}\dotsm\frac{k-n}1\\
=\frac{k-n}n\frac{k-1}{n-1}\dotsm\frac{k-n+1}1
=\frac{k-n}n\binom{k-1}{n-1},
\end{multline*}
so that
\begin{multline*}
\binom{k-1}n+\binom{k-1}{n-1}
=\left(\frac{k-n}n+1\right)\binom{k-1}{n-1}\\
=\frac kn\binom{k-1}{n-1}=\binom kn.
\end{multline*}
Therefore
\begin{multline*}
  (1+x)g_{k-1}(x)
=(1+x)\sum_{n=0}^{\infty}\binom{k-1}nx^n\\
=\sum_{n=0}^{\infty}\binom{k-1}nx^n
+\sum_{n=0}^{\infty}\binom{k-1}nx^{n+1}\\
=1+\sum_{n=1}^{\infty}\binom{k-1}nx^n
+\sum_{n=1}^{\infty}\binom{k-1}{n-1}x^n\\
=1+\sum_{n=1}^{\infty}\binom knx^n
=\sum_{n=0}^{\infty}\binom knx^n
=g_k(x).
\end{multline*}
Now we can compute
\begin{multline*}
  g_k{}'(x)
  =\sum_{n=0}^{\infty}\binom knnx^{n-1}
=\sum_{n=1}^{\infty}\binom knnx^{n-1}\\
=\sum_{n=0}^{\infty}\binom k{n+1}(n+1)x^n
=k\sum_{n=0}^{\infty}\binom k{n+1}\frac{n+1}k\cdot x^n\\
=k\sum_{n=0}^{\infty}\binom{k-1}nx^n
=k\cdot g_{k-1}(x).
\end{multline*}
If we define
\begin{equation*}
  h(x)=\frac{g_k(x)}{(1+x)^k},
\end{equation*}
then $h(0)=1$, and
\begin{equation*}
  h'(x)=\frac{g_k{}'(x)(1+x)^k-k(1+x)^{k-1}g_k(x)}{(1+x)^{2k}}=0.
\end{equation*}
Hence $h$ is constant, and $h(x)=1$ whenever $\size x<1$, and then
$g_k(x)=(1+x)^k$.  Thus,~\eqref{eqn:series} holds
for all $x$, when $k$ is a positive integer or $0$, and when $\size
x<1$, otherwise.
For example,
\begin{multline*}
  \frac1{1-x}=\sum_{n=0}^{\infty}\binom{-1}n(-x)^n
  =\sum_{n=0}^{\infty}\frac{-1}n\frac{-2}{n-1}\dotsm\frac{-n}1(-1)^nx^n\\
  =\sum_{n=0}^{\infty}x^n,
\end{multline*}
as we expect.  Also
\begin{equation*}
  \sqrt{1+x}=\sum_{n=0}^{\infty}\binom{1/2}nx^n.
\end{equation*}
Since
\begin{equation*}
  \binom k{n+1}=\binom kn\frac{k-n}{n+1},
\end{equation*}
we have
\begin{align*}
  \binom{1/2}0&=1,&
  \binom{1/2}1&=\frac12,
\end{align*}
and then
\begin{gather*}
  \binom{1/2}2=\frac12\cdot\frac{-1/2}2=-\frac18,\\
  \binom{1/2}3=-\frac18\cdot\frac{-3/2}3=\frac1{16},\\
  \binom{1/2}4=\frac1{16}\cdot\frac{-5/2}4=-\frac5{128},
\end{gather*}
and generally
\begin{equation*}
  \binom{1/2}n=(-1)^{n+1}\cdot\frac{1\cdot 3\dotsm(2n-3)}{2^nn!}
\end{equation*}
when $n>1$.
Therefore
\begin{equation*}
  \sqrt{1+x}=1+\frac12x-\frac18x^2+\frac1{16}x^3-\frac5{128}x^4+\dotsb,
\end{equation*}
an alternating series when $x>0$.
In particular, 
\begin{equation*}
  \sqrt{1+x}\approx 1+\frac x2
\end{equation*}
when $\size x$ is small (this is the tangent-line approximation); to
be precise, if $0<x<1$, then
\begin{equation*}
0<\sqrt{1+x}-\left(1+\frac x2\right)\leq\frac{x^2}8.
\end{equation*}
For a better approximation, again when $0<x<1$,
\begin{equation*}
\size{\sqrt{1+x}-\left(1+\frac x2-\frac18x^2\right)}\leq\frac{x^3}{16}.  
\end{equation*}
In Einstein's special theory of relativity,
\begin{equation*}
  m=\frac{m_0}{\sqrt{1-v^2/\pc^2}},
\end{equation*}
where $m$ is the mass of an object with velocity $v$ whose rest mass
is $m_0$, and $\pc$ is the speed of light.  The kinetic energy $K$ of the
object is then given by
\begin{multline*}
K=
  (m-m_0)\pc^2
=m_0\pc^2\left(\frac1{\sqrt{1-v^2/\pc^2}}-1\right)\\
=m_0\pc^2\left(\left(1-\frac{v^2}{\pc^2}\right)^{-1/2}-1\right)\\
=m_0\pc^2\left(\sum_{n=0}^{\infty}
  \binom{-1/2}n\left(\frac{-v^2}{\pc^2}\right)^n-1\right)\\ 
=m_0\pc^2\sum_{n=1}^{\infty} (-1)^n
  \binom{-1/2}n\left(\frac{v^2}{\pc^2}\right)^n\\
=m_0\pc^2\left(\frac{v^2}{2\pc^2}+\frac{3v^2}{8\pc^2}+\dotsb \right).
\end{multline*}
In particular, for low velocities,
\begin{equation*}
  K\approx\frac12m_0v^2.
\end{equation*}
To be precise, by Taylor's Theorem,
\begin{equation*}
  f(x)=f(0)+f'(0)\cdot x+\frac12f''(t)\cdot x^2
\end{equation*}
for some $t$ between $x$ and $0$.  If $f(x)=(1+x)^{-1/2}$, then
$f''(x)=3(1+x)^{-5/2}/4$, so that, if $-1<x<0$, then also $x<t<0$, and
\begin{equation*}
  \size{f(x)-f(0)-f'(0)x}=\frac12\size{f''(t)x^2}<\frac{3x^2}{8(1+x)^{5/2}}.
\end{equation*}
In our case, letting $x=-v^2/\pc^2$, we have
\begin{equation*}
  0<K-\frac12m_0v^2
  <m_0\pc^2\cdot\frac{3(v^2/\pc^2)^2}{8(1-v^2/\pc^2)^{5/2}}.  
\end{equation*}

\end{document}
