# Binomial theorem and series

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}