Binomial Series

The series expansions $\dfrac1{1-x}=1+x+x^2+\cdots$, $\dfrac1{1+x}=1-x+x^2-\cdots$ and $\dfrac1{1-2x}=1+2x+4x^2+\cdots$ can all be re-written using negative integer exponents: $\dfrac1{1-x}=(1-x)^{-1}$, $\dfrac1{1+x}=(1+x)^{-1}$, $\dfrac1{1-2x}=(1-2x)^{-1}$. This hints that $(1+x)^n$ might make sense for exponents $n$ well beyond the positive integers of Theorem 5.1 — and indeed it does, for any rational (even irrational) exponent.

Theorem 5.4 (Binomial Theorem for Rational Exponent). For any rational number $n$,

This is stated here without proof (the proof needs tools beyond this course); it is genuinely an infinite series when $n$ is not a non-negative integer (Theorem 5.1's finite expansion is the special case $n\in\mathbb N$, where the series terminates because $^nC_r=0$ once $r>n$).

Companion forms, obtained by substitution into Theorem 5.4:

1. Replacing $x\to-x$: $\ (1-x)^n = 1-nx+\dfrac{n(n-1)}{2!}x^2-\dfrac{n(n-1)(n-2)}{3!}x^3+\cdots$ $(|x|<1)$.
2. Replacing $n\to-n$: $\ (1+x)^{-n} = 1-nx+\dfrac{n(n+1)}{2!}x^2-\dfrac{n(n+1)(n+2)}{3!}x^3+\cdots$ $(|x|<1)$.
3. Replacing both $x\to-x,\ n\to-n$: $\ (1-x)^{-n} = 1+nx+\dfrac{n(n+1)}{2!}x^2+\dfrac{n(n+1)(n+2)}{3!}x^3+\cdots$ $(|x|<1)$.

Written with $n=p/q$ (to make the "rational number" explicit):

and similarly for $(1-x)^{p/q}$ with alternating signs.

Four expansions worth memorising outright (all $|x|<1$):