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Under a post in this forum, I found a comment (https://math.stackexchange.com/a/46936/1173827) by Beni Bogosel that mentions the problem of finding all solutions for a function $f:\mathbb{R}\rightarrow\mathbb{R}$ with the property $f(x)=f(\cos x), \forall x \in \mathbb{R}$.

Out of interest, I began searching for a solution. Immediately the solution $f(x)=0$ came to mind. The next set of solutions is $f(x)=k, k\in\mathbb{R}$.

But now I wondered if there are non-constant solutions. I figure that the domain of $f$ must be confined to the interval [-1,1], since this is the output range of $\cos(x)$.

For my first attempt I considered if $f(x)=g_1(x)+g_2(x)$, where the new two functions are switch functions that just turn into the other if $x$ or $\cos(x)$ is the input. But this is just a rephrasing of the problem and did not get me far.

Then I remembered that the question under which I found this problem was about a solution for $x=\cos(x)$, so I decided to approach the issue from that direction.

When plugging $\cos(x)$ into $f(x)$, I get

$$f(\cos(x))=f(\cos(\cos(x)))$$

But from the defining property of $f(x)$ I know that

$$f(x) = f(\cos(x))=f(\cos(\cos(x)))$$

I can repeat this forever until I have an infinitely nested $\cos(x)$ inside of $f(x)$ so that

$$f(x) = f(\cos(\cos(\cos( \cdots ))))$$

I know that these nested $\cos(x)$ approach a constant with the approximate value $q\approx0.739085$. But would this not imply that $f(x)=f(q)=k$ again?

Is it true then that all solutions for $f(x)=f(\cos(x))$ are just constant? Does a non-constant solution truly not exist?

Space junk
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    When you invoke the limit of $\cos(\cos(....))$ being $q$ to say that $f(x)=f(q)$, you're using continuity. Think about what happens if $f$ is not given to be continuous. – whoisit Jan 14 '24 at 22:43
  • @whoisit I thought a bit about it, and I was able to find a solution for a non continuous $f$. I since $x=\cos(x)$ is true for the special number $q$, I made $f$ assume this value only when $x=q\approx 0.7391$ and nowhere else, with the rest being constant, so that $$f(x)= \begin{cases} q & \text{if } |x| = q \ k\in\mathbb{R} & \text{if } |x| \neq q \end{cases}$$ This is the only one I found. – Space junk Jan 15 '24 at 07:18
  • I also suspect that $$ f(x)=\begin{cases} q & \text{if} |x|=q+2\pi n, n\in\mathbb{N} \ k\in\mathbb{R} & \text{if} |x| \neq q+2\pi n \end{cases}$$ is a solution as well. – Space junk Jan 15 '24 at 07:35

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I think not.

I suspect you can partition $\mathbb R$ into infinitely many equivalence classes where $\cos(x)=y \implies x \sim y$.

Clearly $f(x)$ needs to be constant on any particular equivalence class, but it can take any real value for each equivalence class.

Henry
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    Here's some elaboration on why infinitely many classes are formed by the weakest such equivalence relation. We must have $x \sim y$ if $\cos^k(x) = \cos^l(y)$ for some non-negative integers $k$ and $l$. On the other hand, one can check that this condition is enough to give an equivalence relation. Now, one can fix $x = \alpha$ and look for all solutions $y$, where $k$ and $l$ are allowed to vary. Each pair $(k, l) \in \mathbb{N}^2$ gives countably many solutions, so the class as a whole is also countable. Since $\mathbb{R}$ is uncountable, there must be (uncountably) infinitely many classes. – Haran Jan 14 '24 at 22:51
  • Excuse my ignorance, but what do you mean with equivalence class? – Space junk Jan 16 '24 at 21:19
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    @Spacejunk See Wikipedia. Essentially a subset (or class) with elements which are equivalent under an equivalence relation which is reflexive ($x \sim x$) and symmetric ($x \sim y \implies y \sim x$) and transitive ($x \sim y \text{ and } y \sim z \implies x \sim z$). – Henry Jan 16 '24 at 23:38