Binomial Theorem

\(
\require{AMSsymbols}
\def\lfz#1{\overset{\Large#1}{\,\circ\kern-6mu-\kern-7mu-\kern-7mu-\kern-6mu\bullet\,}}
\def\lfzraised#1{\raise{10mu}{#1}}
\def\laplace{\lfz{\mathscr{L}}}
\def\fourier{\lfz{\mathcal{F}}}
\def\ztransform{\lfz{\mathcal{Z}}}
\require{cancel}
\newcommand\ccancel[2][black]
{\color{#1}{\cancel{\color{black}{#2}}}}\)Consider
$$
f(x)=(a+x)^r\label{eq:axr}
$$

Recall the MacLaurin Series

$$
\begin{align}
f(x)&=\sum _{k=0}^{\infty }{\frac {f^{(k)}(0)}{k!}}\,x^{k}\nonumber
\end{align}\nonumber
$$

The \(k\)th derivative of equation \(\eqref{eq:axr}\)
$$
f^{(k)}(x)=r\,(r-1)\cdots(r-k+1)(a+x)^{r-k}
$$

Substitute \(x=0\) to find the derivatives at \(0\)
$$
f^{(k)}(0)=r\,(r-1)\cdots(r-k+1)\,a^{r-k}
$$

Apply the MacLaurin Series to equation \(\eqref{eq:axr}\)
$$
\begin{align}
f(x)=(a+x)^r&=\sum _{k=0}^{\infty }\frac{r\,(r-1)\cdots(r-k+1)}{k!}\,a^{r-k}\,x^k
\end{align}
$$

Isaac Newton generalized binomial theorem for \(r\in\mathbb{C}\)
$$
\shaded{(a+x)^r=\sum_{k=0}^{\infty}{r \choose k}\,a^{r-k}\,x^k,\quad\text{where }{r \choose k}=\frac{r\,(r-1)\cdots(r-k+1)}{k!}}\label{eq:newton}
$$

The binomial coefficient \({r \choose k}\)
$$
\begin{align}\frac{(r)_k}{k!}&=\frac{r(r-1)(r-2)\cdots (r-k+1)}{k(k-1)(k-2)\cdots1}\nonumber\\
&=\prod _{i=1}^{k}\frac{(r-(i-1))}{i}=\prod _{i=0}^{k-1}\frac{r-i}{i}
\end{align}
$$

The series converges for \(r\geq0\land r\in\mathbb{N}\), or for \(|x|\lt|a|\)
$$
\begin{align}
(a+x)^r&=\sum_{k=0}^{\infty}{r \choose k}\,a^{r-k}\,x^k\nonumber\\
&=a^r+r\,a^{r-1}\,x+\frac{r(r-1)}{2!}\,a^{r-2}\,x^2+\frac{r(r-1)(r-2)}{3!}\,a^{r-3}\,x^3+\cdots
\end{align}
$$

Binomial series

Consider equation \(\eqref{eq:newton}\) for \(a=1\), gives the Binomial series
$$
\shaded{(1+x)^r=\sum_{k=0}^{\infty}{r \choose k}\,x^k,\quad\text{where }{r \choose k}=\frac{r\,(r-1)\cdots(r-k+1)}{k!}}
$$

This series converges when

  • \(|x|\lt1\), converges absolutely for any complex number \(r\).
  • \(|x|\gt1\), converges only when \(r\) is a non-negative integer, what makes the series finite.

Special cases

1) where \(a=1\), converges for \(|x|\lt1\)
$$
\begin{align}
(1+x)^{r}&=\sum_{k=0}^{\infty}\frac{(r)_k}{k!}\,x^k\nonumber\\
&=1+r\,x+\frac{r(r-1)}{2!}\,x^2+\frac{r(r-1)(r-2)}{3!}\,x^3+\cdots
\end{align}
$$

2) the negative binomial series, converges for \(|x|\lt1\)

Apply the Negated Upper Index of Binomial Coefficient identity \({r \choose k}=(-1)^k{k-r-1 \choose k}\)
$$
\begin{align}
(a+x)^r&=\sum_{k=0}^{\infty}{r \choose k}\,a^{r-k}\,x^k\nonumber\\
&=\sum_{k=0}^{\infty}{k-r-1 \choose k}(-1)^k\,\,a^{r-k}\,x^k
\end{align}
$$

Substitute \(x\to -x\) and \(m\to -m\)
$$
\begin{align}
(a-x)^{-r}&=\sum_{k=0}^{\infty}{k+r-1 \choose k}(-1)^k\,\,a^{-r-k}\,(-x)^k\nonumber\\
&=\sum_{k=0}^{\infty}{k+r-1 \choose k}\cancel{(-1)^k}\,\,a^{-r-k}\,\cancel{(-1)^k}\,x^k\nonumber\\
&=\sum_{k=0}^{\infty}{k+r-1 \choose k}\,a^{-r-k}\,x^k
\end{align}
$$

For \(a=1\)
$$
\begin{align}
(1-x)^{-r}&=\sum_{k=0}^{\infty}{k+r-1 \choose k}\,x^k\nonumber\\
\end{align}
$$

—-old:
$$
\begin{align}
(1-x)^{-r}&=\sum_{k=0}^{\infty}{-r \choose k}\,(-x)^k=\sum_{k=0}^{\infty}\frac{(-r)_k}{k!}\,(-x)^k\nonumber\\[5mu]
&=1+r\,x+\frac{r(r+1)}{2!}\,x^2+\frac{r(r+1)(r+2)}{3!}\,x^3+\cdots
\end{align}
$$

3) where \(r=-1\) the geometric series, converges for \(|x|\lt1\)
$$
\begin{align}
(1+x)^{-1}&=\sum_{k=0}^{\infty}\frac{(-1)_k}{k!}\,x^k\nonumber\\
&=1-x+x^2-x^3+\cdots
\end{align}
$$

Integer exponents

1) when \(n=r\) is a positive integer, the binomial coefficients for \(k\gt r\) are zero, and the series terminates at \(n\)
$$
(a+x)^{n}=\sum _{k=0}^{n}{n \choose k}\,a^{n-k}\,x^{k}
$$ where $${n \choose k}=\frac{n!}{(n-k)!k!}
$$ for nonnegative \(n\) and \(k\), and \(0\leq k\lt n\)

2) when the exponent \(-n=-r\) is a negative integer, where \(|x|\lt a\)
$$
\begin{align}
(a+x)^{-n}&=\sum _{k=0}^{\infty}{-n \choose k}\,a^{-(n+k)}\,x^{k}\nonumber\\[5mu]
&=\sum _{k=0}^{\infty}(-1)^k{n+k-1 \choose k}\,a^{-(n+k)}\,x^{k}
\end{align}
$$

For \(a=1\), this simplifies to
$$
\begin{align}
(1+x)^{-n}
&=\sum _{k=0}^{\infty}{-n \choose k}\,x^{k}\nonumber\\[5mu]
&=1-nx+\frac{n(n+1)}{2!}x^2+\frac{n(n+1)(n+2)}{3!}x^3+\cdots
\end{align}
$$

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