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

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

• Does this mean that we need to read p.100 of the book? – Wuestenfux Jul 13 at 15:10
• Please use [MathJax](math.meta.stackexchange.com/questions/5020/… – saulspatz Jul 13 at 15:12
• Use the definition: $f(v) = f(0) + f'(0) v + f''(0) v^2/2 + \ldots$. Compute $f(0)$, $f'(0)$ and $f''(0)$ and you have your result. – Winther Jul 13 at 15:12
• Thanks -- evaluating the results at v = 0 makes all the horrible stuff go away – luysii Jul 13 at 15:20

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 $$x=-v^2$$.

• If the OP cannot understand the expression in his/her text book, it is doubtful that this answer will be of much use. – Mark Viola Jul 13 at 15:16
• Sometimes one works in vain. – Gae. S. Jul 13 at 15:17
• @MarkViola , OP doesn't even know how to use MathJax. How is it that they got such a high reputation? – evaristegd Jul 13 at 15:26
• @evaristegd You, the OP and I have by no means a high reputation. Also, sound questions sometimes get upvoted. – Gae. S. Jul 13 at 15:31

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.

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

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