This question is inspired by the big MO question here; and also inspired by the big MO question here. The premise of this question requires a backdrop on fractional iteration; so I'll start slow.

Suppose we have a holomorphic function $\phi : G \to G$ where $G$ is an open and connected set. Assume that $\phi^{\circ n}(\xi) \to \xi_0$ as $n \to \infty$ where $\xi \in G$ and $\xi_0 \in G$. There then exists the Schroder function $\Psi: G \to \mathbb{C}$ where $\Psi(\phi(\xi)) = \phi'(\xi_0)\Psi(\xi)$; and $\Psi$ is non-trivial.

When searching for root functions of $\phi$; i.e. searching for functions $f = \sqrt[n]{\phi}:G \to G$ such that $\phi = f(f(\cdots(n\text{ times})\cdots f(\xi)$; it is a simple exercise that there are $n$ solutions. And that each solution is an analytic continuation of $\Psi^{-1}(\sqrt[n]{\phi'(\xi_0)}\Psi(\xi))$, from a tiny neighborhood about $\xi_0$ to all of $G$, for one of the $n$ values of $\sqrt[n]{\phi'(\xi_0)}$.

So since $\cos$, for example, is entire and about the immediate basin $I$ of its only real fixed point $\xi_0$, $\cos : I \to I$ and $\cos^{\circ n} \to \xi_0$; there equals two different functions $f_0,f_1 : I \to I$ such that $f_{01}(f_{01}(\xi)) = \cos(\xi)$. Interestingly enough, it can be showed neither of these $f_0$ and $f_1$ are real valued on the reals, proving there exists no analytic $f:\mathbb{R} \to \mathbb{R}$ such that $f(f(x)) = \cos(x)$.

Moving on, we come to the $\sin$ function that has a fixed point at zero. This is a vastly different scenario though. With this function there is no open set satisfying the above criterion; $\sin$'s fixed point at zero is neutral (its multiplier $\sin'(0) = 1$) and thus no neighborhood about zero where $\sin(\sin(...\sin(\xi) \to 0$. But nonetheless we have that $\sin(\sin(...\sin(\mathbb{R}) \to 0$.

Now, just like with $\cos$, there should be two solutions to the equation $f(f(x)) = \sin(x)$ for $ x \in \mathbb{R}$. In the first question I'm referencing, we're only talking about one of these solutions. In fact, they are solving for the solution where, $f'(0) = 1$. But technically there should also be a solution $f'(0) =-1$. Namely, by all lawful right, there should be a second solution that is actually decreasing in a neighborhood of $0$. By all rights, it'll probably look incredibly chaotic on $\mathbb{R}$!

To better explain this, assume that $0<\phi'(0) < 1$ and $\phi:\mathbb{R} \to \mathbb{R}$ where $\phi(\phi(...\phi(x) \to 0$ for all $x$ ($\phi$ analytic). Then there exists $f_+$ and $f_-$ where $f_{\pm} = \Psi^{-1}(\pm\sqrt{|\phi'(0)|}\Psi(x))$; and since both $\Psi$ and $\Psi^{-1}$ are $\mathbb{R} \to \mathbb{R}$ this means that $f_{\pm}:\mathbb{R} \to \mathbb{R}$. Obviously $-1 < f_{-}'(0) < 0$ which implies $f_{-}(x)$ is decreasing in a neighborhood of $0$.

Adding to this; I'll be rough here; taking $\phi_n = (1-\frac{1}{n})\sin(x)$; and solving for $f_{\pm}^{(n)} = \pm\sqrt{\phi_n}$ it can be shown that $f^{(n)}_{+} \to f_{+}$ where $f_{+}$ is presented in the original question (It is not the formal power series; but instead in the answers mentioned). However all the methods there fail at producing $f_{-}$; in fact they make no mention of it. Whereas, to me, it appears to be intuitive to assume $f_{-}^{(n)} \to f_{-}$; and that in fact there is a functional square root of $\sin$ that is decreasing in a neighborhood of zero.

Does there exist an analytic function $f_{-}: \mathbb{R}_{\neq 0} \to \mathbb{R}$ where $f_{-}(f_{-}(x)) = \sin(x)$ and $f_{-}'(0) =-1$?

Edit: I forgot to add that $f_{-}(x)$ need not be analytic at $0$. This is because $f_{-}$ isn't necessarily defined in a neighborhood of zero in the complex plane; where as it is more obvious that there is a neighborhood of $\mathbb{R}_{\neq 0}$ in the complex plane that $f_{-}$ is defined. This involves talking about Julia sets and such not, I'd like to avoid that as much as I can.

  • 1
    $\begingroup$ Erm... Isn't $f_+$ odd? If so, we can just put $f_-(x)=f_+(-x)$. Am I missing anything? $\endgroup$
    – fedja
    Feb 21, 2017 at 8:24

1 Answer 1


Clearly, if $f_+(z)$ is a functional square-root of $\sin(z)$, then so is $g(z) := -f_+(-z)$; moreover, they have the same derivative at the origin. Hence, your post implies that $g(z) = f_+(z)$, so $f_+(z)$ is an odd function.

It follows that $-f_+(z)$ is a functional square-root of $\sin(z)$, so $-f_+(z) = f_-(z)$.

  • $\begingroup$ I completely overlooked that. I can't believe it's that obvious. $\endgroup$
    – user78249
    Feb 21, 2017 at 22:11

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.