Question

What is the binomial expansion of `(1-2x)^(1/3)`?

Answer 1

This is really a calculus problem. The expansion is an infinite series because the power is a fraction: `(1-2x)^(1/3)=1-2/3 x-4/9 x^2-40/81 x^3-160/243 x^4-704/729 x^5-cdots` for `|x|<1/2`. This could be used as an approximation for small `x`, such as `(1-2x)^(1/3) approx 1-2/3 x-4/9 x^2`.

Explanation:

The general form of the binomial theorem discovered by Newton, that works for any number `p`, can be written:

`(1+y)^{p}=1+py+(p(p-1))/(2!) y^2+(p(p-1)(p-2))/(3!)y^3+(p(p-1)(p-2)(p-3))/(4!)y^4+cdots` for `|y|<1`.

When `p` is a non-negative integer, the infinite series above "truncates" to a finite expansion that is the more familiar binomial theorem from precalculus. Such an expansion is then valid for all `y`.

For the given problem, `y=-2x` and `p=1/3`, so

`(1-2x)^(1/3)=1+(1/3)(-2x)+(1/3 * -2/3)/(2!)(-2x)^2+(1/3 * -2/3 * -5/3)/(3!)(-2x)^3+(1/3 * -2/3 * -5/3 * -8/3)/(4!)(-2x)^4+(1/3 * -2/3 * -5/3 * -8/3 * -11/3)/(2!)(-2x)^5+cdots`

`=1-2/3 x-4/9 x^2-40/81 x^3-160/243 x^4-704/729 x^5-cdots`

This is valid for `|2x|<1 leftrightarrow |x|<1/2`