Binomial theorem and series

Coert Vonk

\(\)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 \(a=1\)

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