Question

How do you use the binomial series to expand $\frac{1}{{(1-x^2)}^{{\displaystyle \frac{1}{2}}}}$?

Answer 1

Expand the given expression:

We can rewrite $\frac{1}{{(1-x^2)}^{{\displaystyle \frac{1}{2}}}}$ as ${(1-x^2)}^{\frac{-1}{2}}$.

As there is a negative index the only formula that can be used is

${\left(1+a\right)}^n=1+na+\frac{n\left(n-1\right)}{2!}{a}^2+\frac{n\left(n-1\right)\left(n-2\right)}{3!}{a}^3+...$

Here $n=-\frac{1}{2},a=-x^2$, so substitute these values in above formula to find the expansion of ${(1-x^2)}^{\frac{-1}{2}}$

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

$=1+\frac{1}{2}{x}^2+\frac{3}{8}{x}^4+\frac{15}{48}{x}^6+...$

Hence, the required expansion using binomial series is $1+\frac{1}{2}{x}^2+\frac{3}{8}{x}^4+\frac{15}{48}{x}^6+....$.