$\DeclareMathOperator{\sign}{sign}$ Is there a way to rewrite $f(x)=\sign(x)\sqrt{\sqrt{1+x^2}-1}$ using (smooth) elementary functions?
As far as I can see the function seems infinitely differentiable, but is composed of two non-differentiable factors. Is there a way to write the same function (equal on $\mathbb{R}$), but composed using purely smooth functions? (So without $sign(\cdot)$.)
(This function was my attempt at making a function with linear behaviour near $x=0$, and $\mathcal{O}(\sqrt{x})$ behaviour for large $x$.)
How far I've got
I realised that the non-differentiable point comes into existence whenever you take the square root of a function for which its first two Taylor series terms are zero, i.e $\sqrt{g(x)}$ for $g \approx \mathcal{O}(x^2)$.
For example:
For $g(x) = 1 - \cos{2x}$, we have $$\sign(x)\cdot \sqrt{g(x)} = \sqrt{2}\sin{x}$$ on the domain $[-2\pi, 2\pi]$. This is due to the identity $\cos{2x} = 1 - 2\sin^2{x}$.
Thinking about it this way, the question thus becomes whether there exists an odd elementary function $h(x)$ for which $h(x)^2 = \sqrt{1+x^2}-1$.
$\DeclareMathOperator{\sign}{sign}$ I have found the answer before even posting the question; let me know if that is not allowed, but I found the answer quite interesting, so I thought I would share. Try it yourself before looking at the answer if you want!
The solution is $$h(x) = \frac{x}{\sqrt{\sqrt{1+x^2}+1}} = \sign(x)\sqrt{\sqrt{1+x^2}-1}$$
Inspired by my success with the sine/cosine identities, I tried a substitution $x = \sinh{u}$, and indeed, it works wonderfully! I found $h(x) = \sqrt{2}\sinh{\frac{u(x)}{2}}$, and from there a inverse substitution with some algebra (assisted by Wolfram Alpha) got me the answer in no time.
In hindsight, an easier (but harder to come up with, perhaps) way of finding the answer is by recognising $\sqrt{1+x^2}-1$ as a factor of $x^2$, along with its conjugate $\sqrt{1+x^2}+1$.