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

Aug 13, 2017

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

#### Explanation:

The binomial series tell us that:

 (1+x)^n = 1+nx + (n(n-1))/(2!)x^2 +(n(n-1)(n-2))/(3!)x^3 + ...

And so for the given function:

$f \left(x\right) = \frac{1}{\sqrt{1 + {x}^{3}}}$
$\text{ } = {\left(1 + {x}^{3}\right)}^{- \frac{1}{2}}$

Then with $n = - \frac{1}{2}$, and replacing $x$ in the definition with ${x}^{3}$ we have:

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

$\text{ } = 1 - \frac{1}{2} {x}^{3} + \frac{\frac{3}{4}}{2} {x}^{6} - \frac{\frac{15}{8}}{6} {x}^{9} + \ldots$

$\text{ } = 1 - \frac{1}{2} {x}^{3} + \frac{3}{8} {x}^{6} - \frac{5}{16} {x}^{9.} . .$