I'm trying to solve a exercise from a physics book. In that situation $x \gg a$ and $x > 0$.
I weren't able to solve, so I saw the solution: $$ \frac{1}{\sqrt{x^2 + a^2}}=\frac{1}{\sqrt{x^2(1+\frac{a^2}{x^2})}}=\frac{1}{|x| \sqrt{1 + \frac{a^2}{x^2}}}=\frac{1}{x} \left( 1+ \frac{a^2}{x^2} \right)^{-\frac{1}{2}} $$ Hence: $$ \frac{1}{x} \left( 1+ \frac{a^2}{x^2} \right)^{-\frac{1}{2}} = \frac{1}{x} \overbrace{ \left( 1 -\frac{a^2}{2x^2} \right)} = \frac{1}{x}-\frac{a^2}{2x^3} $$
I couldn't understand what had happened to $ \left( 1+ \frac{a^2}{x^2} \right)^{-\frac{1}{2}} $. What should I do?
So, in the generalized binomial theorem, we have
$$(1-x)^{-s} = \sum_{k=0}^\infty \binom{s+k-1}{k} x^k$$
Here, it seems $s=1/2$ and $x \mapsto -a^2/x^2$ giving
$$ \left( 1 + \frac{a^2}{x^2} \right)^{-1/2} = \sum_{k=0}^\infty \binom{k-1/2}{k} (-1)^k \frac{a^{2k}}{x^{2k}}$$
Note that since $x \gg a$, the relatively high-order terms have negligible contribution. While I wouldn't say strict equality applies, we would reasonably have, just by using the $k=0,1$ terms,
$$ \left( 1 + \frac{a^2}{x^2} \right)^{-1/2} \approx \binom{-1/2}{0} - \binom{1/2}{1} \frac{a^2}{x^2}$$
Note that, for these binomial coefficients, we define
$$\binom{\alpha}{k} = \frac{\alpha(\alpha - 1)(\alpha - 2) \cdots (\alpha - (k-1))}{k!}$$
where it equals $1$ for $k=0$.
Therefore
$$\binom{-1/2}{0} = 1 \qquad \binom{1/2}{1} = \frac 1 2$$
and
$$ \left( 1 + \frac{a^2}{x^2} \right)^{-1/2} \approx 1 - \frac 1 2 \frac{a^2}{x^2}$$