Taylor series for $\frac{1}{\sqrt{(1-v^2)}}$

Question

I don't seem to get the answer the book "The Geometry of Spacetime" by Callahan does e.g. $1 + 1/2(v^2) + O(v^4)$ on Pg. $100$ and it is rather crucial for the ensuing discussion

Answer 1

The Taylor series of $(1+x)^\alpha$ (for all $\alpha\in\Bbb R$ and $x\in(-1,1)$ ) is the so called binomial series.

$$(1+x)^\alpha=\sum_{k=0}^\infty \binom \alpha kx^k$$

Where the generalized binomial is $\binom\alpha k=\frac{\alpha\cdot(\alpha-1)\cdots(\alpha-k+1)}{k!}$. In your case you are looking at $(1+x)^{-1/2}$ with $x=-v^2$.

Answer 2

It is simpler to consider the Taylor series $$(1-v)^{-1/2}=(1-0)^{-1/2}+\frac12(1-0)^{3/2}v+o(v)$$

and plug $v^2$, giving

$$(1-v^2)^{-1/2}=1+\frac12v^2+o(v^2).$$

Answer 3

The square root is locally Lipschitz (even differentiable) for positive arguments. Thus $\sqrt{a+\epsilon}=\sqrt{a}+O(\epsilon)$ for $a>0$. With that transform $$ \frac1{\sqrt{1-v^2}}=\frac{\sqrt{1+v^2}}{1+O(v^4)}=\sqrt{1+v^2+\frac{v^4}4}+O(v^4)=1+\frac{v^2}2+O(v^4) $$

This could be extended to higher orders, $$ \frac1{\sqrt{1-v^2}}=\frac{\sqrt{1+v^2+v^4}}{\sqrt{1-v^6}}=\sqrt{(1+\frac{v^2}2)^2+\frac{3v^4}4}+O(v^6)=...=1+\frac{v^2}2+\frac{3v^4}8+O(v^6) $$ etc.