# Is it possible to express the functional square root of the sine as an infinite product?

Cross-post from MSE.

It is known that the sine can be expressed as an infinite product: $$\sin(x) = x \prod_{n=1}^{\infty} \Big{(} 1 - \frac{x^{2}}{n^{2}{\pi}^{2}} \Big{)} .$$ We can define that functional square root of a function $$g(\cdot)$$ to be the function $$f(\cdot)$$ that satisfies $$f(f(x)) = g(x)$$. The square root of the sine function with respect to function composition has been discussed previously on MO on a number of occasions. For instance, here the formal power series is considered.

I wonder whether the functional square root of the sine also has an infinite product representation. If not, has any research been done on this question?

• You refer to the functional square root. Is it unique? – LSpice Feb 13 '20 at 11:17

Functional square root of sine is not unique, and it cannot be defined in the whole complex plane (according to a theorem of I. N. Baker). You can define it on $$(0,\pi)$$ with $$f(0)=0$$, for example. The resulting function is analytic in the component of Fatou set of sine adjacent to zero on the right (see, for example the first figure here). But this function $$f$$ has no zeros in its domain, therefore a representation as infinite product does not make much sense. I mean that $$\log f$$ can be defined and any breaking of this log into summands gives you an infinite product representation. Of course this answer is related to a true product, it does not exclude that there is some "formal" product representing this square root, whatever a "formal product" may mean.