# What about the other $f$ such that $f(f(x)) = \sin(x)$?

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.

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

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 $-f_+(z) = f_-(z)$.

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