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

Sep 27, 2014

This is a natural extension of raising a binomial to a whole number power:
${\left(a + b\right)}^{n} = {\sum}_{k = 0}^{n} \left(\frac{n}{k}\right) {a}^{n - k} {b}^{k}$ (sorry about the inner bar)
Where (n/k)=(n!)/(k!(n-k)!)=(n(n-1)…(n-k+1))/(k!)

So we can apply this to any exponent r even if r is an arbitrary real number.

(a+b)^r=sum_(k=0)^oo (r(r-1)…(r-k+1))/(k!) a^(r-k)b^k

Now put your info in, with $r = \frac{2}{3} , a = 1 , b = - x :$

(1-x)^(2/3)=sum_(k=0)^oo (2/3(2/3-1)…(2/3-k+1))/(k!) 1^(2/3-k)(-x)^k

=(1)^(2/3)+(2/3)/1(1)^(-1/3)(-x)^1+((2/3)(2/3-1))/2(1)^(-4/3)(-x)^2+…

=1-2/3x-1/9x^2+…

There's the start of the series; I dare you to compute the next two terms. Take the \dansmath challenge/!