How do you use the binomial series to expand f(x) = (1+x^3)^(-1/2)?

Jan 29, 2018

${\left(1 + {x}^{3}\right)}^{- \frac{1}{2}} = 1 - {x}^{3} / 2 + \left(\frac{3}{8}\right) {x}^{6} - \left(\frac{5}{16}\right) {x}^{9} + \left(\frac{35}{128}\right) {x}^{12} - \ldots .$

Explanation:

It is in the form (a+b)^n where
a = 1, b = x^3, n = -(1/2).

The difference here is that the sum on the right doesn't have just n+1 terms, but infinitely many. This is called an infinite series.

In the binomial expansion formula for (1+x)^n = 1 +nx+ (n(n-1))/(2!)x^2 + ...

1+(-1/2)(x)^3 + ((-1/2)(-3/2) (x)^6) /(2!) + ((-1/2)(-3/2)(-5/2)x^9)/(3!) + ((-1/2)(-3/2)(-5/2)(-7/2)x^12)/(4!) + ....#

${\left(1 + {x}^{3}\right)}^{- \frac{1}{2}} = 1 - {x}^{3} / 2 + \left(\frac{3}{8}\right) {x}^{6} - \left(\frac{5}{16}\right) {x}^{9} + \left(\frac{35}{128}\right) {x}^{12} - \ldots .$

Jan 29, 2018

The answer is $= 1 - \frac{1}{2} {x}^{3} + \frac{3}{8} {x}^{6} - \frac{5}{16} {x}^{9} + o \left({x}^{9}\right)$

Explanation:

The binomial series is

${\left(a + b\right)}^{n} = {a}^{n} + \left(\begin{matrix}n \\ 1\end{matrix}\right) {a}^{n - 1} {b}^{+} \left(\begin{matrix}n \\ 2\end{matrix}\right) {a}^{n - 2} {b}^{2} + \left(\begin{matrix}n \\ 3\end{matrix}\right) {a}^{n - 3} {b}^{3} + .$

$= {a}^{n} + n {a}^{n - 1} b + \frac{n \left(n - 1\right)}{1 \times 2} {a}^{n - 2} {b}^{2} + \frac{n \left(n - 1\right) \left(n - 2\right)}{1 \times 2 \times 3} {a}^{n - 3} {b}^{3} + \ldots$

Here,

$a = 1$

$b = {x}^{3}$

$n = - \frac{1}{2}$

Therefore,

${\left(1 + {x}^{3}\right)}^{- \frac{1}{2}} = 1 + \left(- \frac{1}{2}\right) {x}^{3} + \frac{\left(- \frac{1}{2}\right) \left(- \frac{3}{2}\right)}{1 \cdot 2} {x}^{6} + \frac{\left(- \frac{1}{2}\right) \left(- \frac{3}{2}\right) \left(- \frac{5}{2}\right)}{1 \cdot 2 \cdot 3} {x}^{9} + \ldots$

$= 1 - \frac{1}{2} {x}^{3} + \frac{3}{8} {x}^{6} - \frac{5}{16} {x}^{9} + o \left({x}^{9}\right)$