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I wonder whether all (composites of) trigonometric and inverse trigonometric functions with periodic functional iterations can be found. In order to specify what I mean by that, let's introduce some notation first.

Let $f^{[n]} (x)$ denote the function $f$ that has been iterated $n$ times with itself. So initially, we have $f(x) := f^{[1]}(x)$. Then we have $f^{[2]} (x)= (f \circ f)(x)$, $f^{[3]}(x) = (f \circ f \circ f)(x)$, etc. We call a function periodic if there is some positive integer $k$ such that $f^{[1+k]}(x) = f^{[1]}(x)$. If that is the case, the period of of the function $f(\cdot)$ is equal to $k$.

Now we turn to some examples of periodic (inverse) trigonometric functions.

Examples

  • Consider the function $f(x) := f^{[1]} (x)= \sin(\arccos(x)) = \sqrt{1-x^{2}}$. Then we obtain $f^{[2]} (x)= x$, and $f^{[3]} (x)= \sqrt{1-x^{2}} = f(x)$. Thus, the function $f(\cdot)$ is periodic, and has a period of $3 - 1 = 2$. The same applies to the function $v(x) = \cos(\arcsin(x))$.
  • Define the function $g(x) := g^{[1]}(x) = \tan( \cot^{-1}(x)) = \frac{1}{x}$. Then, $g^{[3]}(x) = g(x)$, so it's also periodic and has a period of $2$, too.
  • Let $h(x) := h^{[1]}(x) = \sin(\arctan(\cos(\arcsin(x)))) = \sqrt{1-x^{2}}$. Like the first example, it's periodic and has a period of two.

Counterexamples

  • Let $p(x) := p^{[1]}(x) = \cot(\arccos(x)) = \frac{x}{\sqrt{1-x^{2}}}$. Then, $g^{[n]}(x) = \frac{x}{\sqrt{1-nx^{2}}}$ for $n \geq 1$, so it is not periodic.
  • Consider $m(x) := m^{[1]}(x) = \cot(\arcsin(x)) = \frac{\sqrt{1-x^{2}}}{x}$. Then, $m^{[n]}(x) = \sqrt{ \frac{F_{n} - F_{n+1}x^{2}}{F_{n}x^{2} - F_{n-1}} \cdot (-1)^{n+1} } $ for $n >1$, where $F_{n}$ is the $n$'th Fibonacci number. Thus, it is not periodic.
  • Define $l(x) := l^{[1]}(x) = \cos(\arctan(x)) = \frac{1}{\sqrt{1+x^{2}}} $. In this case, $l^{[n]} (x) = \sqrt{\frac{F_{n} + F_{n-1}x^{2}}{F_{n+1} + F_{n}x^{2}}}$ for $n>1$. Again, not a periodic function.

This is what I've found so far. The list is not exhaustive.

Questions

  1. Can all periodic composites of trigonometric and inverse trigonometric functions be found?
  2. What are their periods? Are there any examples of functions with periods greater than two?
  3. Are there any periodic trigonometric functions that do not involve inverse trigonometric functions?

Note: let's discard trivial examples like $q(x) = \cos(\arccos(x)) = x$, which have a period of zero.

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  • $\begingroup$ Is there a reason to require periodic functions to return to the first iterate, rather than some other iterate? I don't know if it makes a difference. (For that matter, I'd prefer the base case of the iteration to be $f^{[0]} = \operatorname{id}$, but that's just a matter of taste.) $\endgroup$ Commented Sep 21, 2020 at 18:17
  • $\begingroup$ Also, what branches do you choose for the inverse functions? How do you ensure that your functions are iterable—e.g., do you allow a function that is periodic only on the subset of its domain where the iterate is defined, or do you insist that your functions be globally defined? $\endgroup$ Commented Sep 21, 2020 at 18:18
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    $\begingroup$ @LSpice w.r.t. your first comment: no, I did not have a specific reason in mind. It just seemed like the most natural thing to require. As for your second comment: the function does not have to be globally defined, it can be iterable only on the subset of the domain where its iterate is defined. $\endgroup$ Commented Sep 21, 2020 at 21:35

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