Taylor approximation of inverse square root

Question

Given the function $f(x)=\sqrt{1+mx+\mathcal{O}(x^2)}$ I am reading that $g(x) = \frac{1}{f(x)}$, the inverse square root, can be computed with first order Taylor approximation and take $g(x) = 1 - \frac{m}{2}x + \mathcal{O}(x^2)$. So given, $f'(x) = \frac{m+\mathcal{O}(x^2)}{2\sqrt{1+mx+\mathcal{O}(x^2)}}$, I started with the first order Taylor approximation of $f(x)\approx f(a) + f'(a)(x-a)+\mathcal{O}(x^2)$ in point $a$ but I do not know how to continue. Could you please give some help?

Answer 1

One may recall that, as $u \to 0$, $$ \frac1{\sqrt{1+u}}=1-\frac{u}2+O(u^2) $$ giving, as $x \to 0$, $$ g(x) = \frac{1}{\sqrt{1+\underbrace{mx+\mathcal{O}(x^2)}_{u}}}=1-\frac{mx}2+O(x^2). $$

Answer 2

It is easier to write $g(x)=\frac{1}{f(x)}=(1 + m x + O(x^2))^{-1/2}$. Doing the derivative on g(x) (not f(x)) gives you $g'(x)= -m/2 (1 + m x +O(x^2))^{-3/2} +O(x^2)$, so the Taylor expansion (you have a sign error and should read $h(x)=h(a)+h'(a)(x-a)+O(x^2))$ at $a=0$ is:

$g(x)=g(0) + g'(x) (x-0) + O(x^2)= 1 - \frac{m}{2} x + O(x^2)$