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

Sep 7, 2014

By binomial series,
1/{sqrt{1-x^2}}=sum_{n=0}^{infty}{1cdot3cdot5cdot cdots cdot(2n-1)}/{2^n n!}x^{2n}

Let us first find the binomial series for
$\frac{1}{\sqrt{1 + x}} = {\left(1 + x\right)}^{- \frac{1}{2}}$

Its binomial coefficient is
C(-1/2,n)={(-1/2)(-3/2)(-5/2)cdot cdots cdot(-{2n-1}/2)}/{n!}
by factoring out all $-$'s and $\frac{1}{2}$'s,
={(-1)^n[1cdot3cdot5cdot cdots cdot(2n-1)]}/{2^n n!}

So, we have the binomial series
1/{sqrt{1+x}}=sum_{n=0}^{infty}{(-1)^n[1cdot3cdot5cdot cdots cdot(2n-1)]}/{2^n n!}x^n

Now, we can find the binomial series for the posted function by replacing $x$ by $- {x}^{2}$.
$\frac{1}{\sqrt{1 - {x}^{2}}}$
=sum_{n=0}^{infty}{(-1)^n[1cdot3cdot5cdot cdots cdot(2n-1)]}/{2^n n!}(-x^2)^n
which simplifies to
=sum_{n=0}^{infty}{(-1)^n[1cdot3cdot5cdot cdots cdot(2n-1)]}/{2^n n!}(-1)^nx^{2n}
since ${\left(- 1\right)}^{n} \cdot \left(- {1}^{n}\right) = {\left(- 1\right)}^{2 n} = 1$,
=sum_{n=0}^{infty}{1cdot3cdot5cdot cdots cdot(2n-1)}/{2^n n!}x^{2n}