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 $h$?
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asked Dec 2, 2017 at 21:02
Franklin Pezzuti Dyer
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$\begingroup$ What is the domain of $h$? $\endgroup$orlp– orlp2017-12-02 21:06:03 +00:00Commented Dec 2, 2017 at 21:06
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$\begingroup$ @orlp Real numbers. I'll clarify in the question. $\endgroup$Franklin Pezzuti Dyer– Franklin Pezzuti Dyer2017-12-02 21:12:42 +00:00Commented Dec 2, 2017 at 21:12
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$\begingroup$ Negative reals too? Because if not you can quite easily construct a piecewise function. $\endgroup$orlp– orlp2017-12-02 21:13:46 +00:00Commented Dec 2, 2017 at 21:13
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$\begingroup$ @orlp Yes, negative reals as well. It seems that if you construct a function $h$ that works for positive reals, you could just take $h(|x|)$, and that would work for all reals. But I'm hoping for something nice (preferably continuous and differentiable). $\endgroup$Franklin Pezzuti Dyer– Franklin Pezzuti Dyer2017-12-02 21:16:11 +00:00Commented Dec 2, 2017 at 21:16
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1$\begingroup$ There a pretty interesting associated differential eqation: $$h(x) = x\cdot \frac{h'(x^2+1)}{h'(x)}$$ $\endgroup$orlp– orlp2017-12-02 22:42:33 +00:00Commented Dec 2, 2017 at 22:42
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2 Answers
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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 $x \to +\infty$.
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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 $h_n$.
answered Mar 28, 2020 at 18:55
Simply Beautiful Art
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$\begingroup$ I haven't seen this particularly simple construction before. Does it appear in the literature anywhere? It is not necessarily analytic at $x=0$ but enjoys good properties elsewhere (monotone derivative). I've tried to generalize it (e.g., to find a functional square root of exp$(x)$) without success so far. The fact that $h_{0}(x)$ is known exactly would appear to be crucial. $\endgroup$S. Finch– S. Finch2025-04-05 13:52:03 +00:00Commented Apr 5 at 13:52
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$\begingroup$ @S.Finch I don't know about functional equations, I'm sure it has. Another example about the case of finding an asymptotic equivalence at infinity and extrapolating it to finite values using a recurrence that defined small inputs in terms of large inputs would be Euler's limit definition of the Gamma function, which approximates $(n+x)!\sim n!n^x$ as $n\to\inf$. $\endgroup$Simply Beautiful Art– Simply Beautiful Art2025-04-09 02:02:14 +00:00Commented Apr 9 at 2:02
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$\begingroup$ I plan to cite this StackExchange posting soon in a preprint. I'd prefer to attribute this to you by name, but understand if privacy is paramount. It will be an anonymous citation unless you email me (see bottom of mit.edu/~sfinch for my address). I hope to hear from you. Thank you! $\endgroup$S. Finch– S. Finch2025-04-10 12:00:53 +00:00Commented Apr 10 at 12:00
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$\begingroup$ @S.Finch Thank you for the interest. I am not interested in the citation, but would not mind seeing the preprint if it's publicly available online. $\endgroup$Simply Beautiful Art– Simply Beautiful Art2025-04-14 03:03:39 +00:00Commented Apr 14 at 3:03
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$\begingroup$ At last the preprint is ready -- arxiv.org/abs/2504.19999 -- with much appreciation for your 2020 post. $\endgroup$S. Finch– S. Finch2025-04-29 02:17:56 +00:00Commented Apr 29 at 2:17