5
$\begingroup$

This question already has an answer here:

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:

enter image description here

Any ideas about how to attack this problem rigorously?

$\endgroup$

marked as duplicate by Simply Beautiful Art, Did, Joffan, Eclipse Sun, Community Oct 26 '17 at 23:02

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • 1
    $\begingroup$ @gt6989b Correct - by $q^{\circ 1/2}$, I mean any function such that $$q^{\circ 1/2}(q^{\circ 1/2}(x))=q(x)$$ $\endgroup$ – Frpzzd Oct 25 '17 at 23:10
  • 2
    $\begingroup$ 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 $x=0$... $\endgroup$ – Steven Stadnicki Oct 25 '17 at 23:16
  • 1
    $\begingroup$ This problem would be so much easier by replacing $x^2+1$ with $2x^2-1$ or $x^2-2$... $\endgroup$ – Jack D'Aurizio Oct 26 '17 at 3:06
  • 2
    $\begingroup$ @SimplyBeautifulArt: they are conjugated with a Chebyshev polynomial, so they have an elementary functional square root, like $2\cos\left(\sqrt{2}\arccos\frac{x}{2}\right).$ $\endgroup$ – Jack D'Aurizio Oct 26 '17 at 15:23
  • 1
    $\begingroup$ Can we not do the same with $\cosh$ and $\operatorname{arccosh}$? $\endgroup$ – Simply Beautiful Art Oct 26 '17 at 15:27
2
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ How might one show that $q^{\circ 1/2}$ cannot be analytic? $\endgroup$ – Frpzzd Dec 5 '17 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 $\exp(x)-1$). $\endgroup$ – Gottfried Helms Dec 10 '17 at 21:14

Not the answer you're looking for? Browse other questions tagged or ask your own question.