by square root algorithm and long division (or by binomial theorem) it is simple matter to find $1/\sqrt{(1-x^2)} = (1+x^2/2 + 3x^4/8 + 5x^6/16 + ...)$ >
I am new to this kind of thing. Can someone explain what is the method (step by step) to convert this reciprocal of a redical to a series ? I know there is a formula of binomial expansion but how do u do it with a reciprocal and how to make it an infinite series. Thanks.
If $\alpha\in\Bbb r$ and if $x<1$, we have$$(1+x)^\alpha=1+\binom\alpha1x+\binom\alpha2x^2+\cdots,$$where$$\binom\alpha n=\frac{\alpha(\alpha-1)\cdots(\alpha-(n-1))}{n!}.$$Applying this (with $-x^2$ instead of $x$, which is not a problem, since $|x|<1\implies|-x^2|<1$) and $\alpha=-\frac12$, we have the equality that you're after.