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I've been doing a lot of research about functional half-iteration, and I posed the following question to myself:

Consider the function $q:\mathbb R\mapsto\mathbb R$ defined as $$q(x)=x^2+1$$ Does $q^{\circ 1/2}$ exist? Does a continuous $q^{\circ 1/2}$ exist? What about a differentiable $q^{\circ 1/2}$?

It seems to me that $q^{\circ 1/2}$ exists, but I don't know how to prove that it exists (I certainly can't find it, since it's probably not an elementary function). So far, I've proven that if it exists and is continuous, then it must be bounded between $x$ and $q(x)$. My intuition tells me that a differentiable solution probably exists... but it may not be differentiable at $x=0$. I've worked out an "almost-graph" of a possible solution, but it's far from a rigorous proof:

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Any ideas about how to attack this problem rigorously?

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    $\begingroup$ @gt6989b Correct - by $q^{\circ 1/2}$, I mean any function such that $$q^{\circ 1/2}(q^{\circ 1/2}(x))=q(x)$$ $\endgroup$ Commented Oct 25, 2017 at 23:10
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    $\begingroup$ You must have $\frac{d}{dx}g(g(x))|_{x=0}=g'(0)g'(g(0))=0$, so either $g'(0)=0$ or $g'(g(0))=0$; the latter is impossible if the function is convex (which seems a sensible restriction to impose) so you must have zero derivative (if it exists) at $x=0$, and so there should be a differentiable solution at $x=0$... $\endgroup$ Commented Oct 25, 2017 at 23:16
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    $\begingroup$ This problem would be so much easier by replacing $x^2+1$ with $2x^2-1$ or $x^2-2$... $\endgroup$ Commented Oct 26, 2017 at 3:06
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    $\begingroup$ @SimplyBeautifulArt: they are conjugated with a Chebyshev polynomial, so they have an elementary functional square root, like $2\cos\left(\sqrt{2}\arccos\frac{x}{2}\right).$ $\endgroup$ Commented Oct 26, 2017 at 15:23
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    $\begingroup$ Can we not do the same with $\cosh$ and $\operatorname{arccosh}$? $\endgroup$ Commented Oct 26, 2017 at 15:27

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I claim there is a continuous $f$ such that $f \circ f = q$.

Recursively define $a_0 = 0$, $a_1 = 1/2 < 1 = q(a_0)$, and $a_{n+2} = q(a_{n})$.
Now start by taking $f$ to be any continuous increasing function from $[0,a_1]$ onto $[a_1,a_2]$, and define $f$ on $[a_n,a_{n+1}]$ for $n \ge 1$ by $f(x) = t^2+1$ where $x = f(t)$. For negative $x$ we define $f(-x) = f(x)$.

If $f(x) \sim 1/2 + \alpha x^2$ near $x=0$, we'll want $f'(1/2) = 1/\alpha$ to make $f$ differentiable. Thus one possibility is $f(x) = 1/2 + (2-\sqrt{3}) x^2 + 4 \sqrt{3} x^4$ for $0 \le x \le 1/2$.

On the other hand, if I'm not mistaken an analytic $f$ seems not to be possible.

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  • $\begingroup$ How might one show that $q^{\circ 1/2}$ cannot be analytic? $\endgroup$ Commented Dec 5, 2017 at 0:18
  • $\begingroup$ @Nilknarf: when the taylor series has diverging coefficients such that the radius of convergence is zero (which is -for example- known to be the case for the half-iterate of $\exp(x)-1$). $\endgroup$ Commented Dec 10, 2017 at 21:14

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