Almost-half iterate of $x^2+1$ Because I can't find a function $h:\mathbb R\mapsto\mathbb R$ with the property
$$h^{\circ 2}(x)=x^2+1$$
I'm looking for a function that almost has that property - that is, I would like to find a closed-form (and preferably elementary) function $h:\mathbb R\mapsto\mathbb R$ satisfying
$$\lim_{x\to\infty} (x^2+1-h(h(x)))=0$$
or, equivalently,
$$h(h(x))=x^2+1+\mathcal O(\epsilon(x))$$
where $\lim_{x\to\infty} \epsilon(x)=0$.
But I haven't been able to do this either. I've tried functions in the form $$|x|^{\sqrt 2}+C$$
but none of them have worked. Can anybody find such a function $h$?
 A: Starting with
$$
g(x) = x^\sqrt{2} + C
$$
we have
$$
\begin{align}
g(g(x)) &= \left(x^\sqrt{2} + C\right)^\sqrt{2} + C \\
&= x^2 \left(1 + Cx^{-\sqrt{2}}\right)^\sqrt{2} + C \\
&\approx x^2 \left( 1 + C\sqrt{2}x^{-\sqrt{2}} \right) + C \qquad \text{(binomial theorem)} \\
&= x^2 + C\sqrt{2}x^{2-\sqrt{2}} + C.
\end{align}
$$
It looks like we can get what we want if we replace $C$ with $x^{\sqrt{2}-2}/\sqrt{2}$.
Indeed, if
$$
h(x) = x^{\sqrt{2}} + \frac{1}{\sqrt{2}} x^{\sqrt{2}-2} \tag{$*$}
$$
then
$$
\begin{align}
h(h(x)) &= \left( x^{\sqrt{2}} + \frac{1}{\sqrt{2}} x^{\sqrt{2}-2} \right)^\sqrt{2} + \frac{1}{\sqrt{2}} \left( x^{\sqrt{2}} + \frac{1}{\sqrt{2}} x^{\sqrt{2}-2} \right)^{\sqrt{2}-2} \\
&= \left( x^{\sqrt{2}} + \frac{1}{\sqrt{2}} x^{\sqrt{2}-2} \right)^\sqrt{2} + O\!\left(x^{2(1-\sqrt{2})}\right) \\
&= x^2 \left( 1 + \frac{1}{\sqrt{2}} x^{-2} \right)^\sqrt{2} + O\!\left(x^{2(1-\sqrt{2})}\right) \\
&= x^2 \left[ 1 + x^{-2} + O\!\left(x^{-4}\right) \right] + O\!\left(x^{2(1-\sqrt{2})}\right) \\
&= x^2 + 1 + O\!\left(x^{2(1-\sqrt{2})}\right)
\end{align}
$$
as $x \to +\infty$.
A: Though neither elementary or of closed form, it is not difficult to numerically compute such a function. Notice that we have
$$h(x)^2+1=h(h(h(x)))=h(x^2+1)$$
$$h(x)=\sqrt{h(x^2+1)-1}$$
By iterating this using
$$h_0(x)=|x|^{\sqrt2}$$
$$h_{n+1}(x)=\sqrt{h_n(x^2+1)-1}$$
we converge to a functional square root of $x^2+1$, as demonstrated in this graph. Intuitively, $h_0$ is sufficiently accurate for large $x$ and $h_{n+1}$ approximates smaller arguments $(x)$ in terms of larger arguments $(x^2+1)$ of $h_n$.
