Is there an analytic solution for such problem?

Question

Given function $$f_n(x) = \cos x - (\cos \cos x) + (\cos \cos \cos x) - (\cos \cos \cos \cos x) + \dots + (-1)^{n-1} \underbrace{ \cos \cos \dots \cos }_n x,$$ where $n \in \mathbb{N}$ and $\underbrace{ \cos \cos \dots \cos }_n$ means cosine of cosine of cosine and so on $n$ times, find value of $$\sup_{n \rightarrow \infty, x \in \mathbb{R}} \{f_k(x)\}^{n}_{k=1}$$

Answer 1

It has already achieved in the comments (as of 9/2/2023 by @mathon and @KenM) that the supremum occurs at $x=0$. In the following, we provide a rigorous proof of this conjecture, albeit much later than the original date of the question, to conclude it.

By definition, $f_n(x+2\pi)=f_n(x)$ and $f_n(-x)=f_n(x)$. Therefore, $$ \sup_{n \rightarrow \infty\\x \in \mathbb{R}} \{f_k(x)\}^{n}_{k=1} = \sup_{n \rightarrow \infty\\x \in [-\pi,\pi]} \{f_k(x)\}^{n}_{k=1} = \sup_{n \rightarrow \infty\\x \in [0,\pi]} \{f_k(x)\}^{n}_{k=1}. $$ Another property of $f_n(x)$ is that $f_n(\pi-x)=f_n(x)-2\cos x$. This implies that for every $x\in[0,\frac{\pi}{2}]$, $\cos x\ge 0$ and $f_n(\pi-x)\le f_n(x)$. Therefore, the supremum expression can be further simplified to $$ \sup_{n \rightarrow \infty\\x \in [0,\frac{\pi}{2}]} \{f_k(x)\}^{n}_{k=1}. $$ Now, we prove that $\underbrace{\cos(\cos(\cdots \cos}_\text{$n$ times}(x)))$ is decreasing over $[0,\frac{\pi}{2}]$ when $n$ is odd and increasing over $[0,\frac{\pi}{2}]$ otherwise. This can be done very simply without using any calculus. Since $\cos: [0,\frac{\pi}{2}]\to [0,1]$ is a decreasing function, for any $g:D\to R$ such that $R\subseteq [0,\frac{\pi}{2}]$, $\cos g(x)$ is increasing over $D$ if $g(x)$ is decreasing over $D$ and is decreasing over $D$ if $g(x)$ is increasing over $D$. Furthermore, the range of $\cos g(x)$ is a subset of $[0,\frac{\pi}{2}]$. Using this fact, since $\cos x$ is decreasing over $[0,\frac{\pi}{2}]$, $\cos\cos x$ is increasing over $[0,\frac{\pi}{2}]$, $\cos\cos\cos x$ is decreasing over $[0,\frac{\pi}{2}]$ and so forth.

This brings us to deducing that $f_n(x)$ is decreasing over $[0,\frac{\pi}{2}]$ for all values of $n$. The supremum then occurs at the smallest member of $[0,\frac{\pi}{2}]$, i.e. at $x=0$. It would be a small matter of time and effort to prove that $f_n(0)$ is maximum when $n=1$ $\blacksquare$

Answer 2

This is not a full answer, but should help you to derive bounds for the value in both directions.

Split the sum $$ \sum_{n=1}^\infty (-1)^{n-1} \cos^n(x) $$ into a finite part of leading terms with odd length and the remaining higher terms $$ \sum_{n=1}^\infty (-1)^{n-1} \cos^n(x) = \sum_{n=1}^{2N +1} (-1)^{n-1} \cos^n(x) + \sum_{n=2N+2}^\infty (-1)^{n-1} \cos^n(x). $$

The maximum of the first part ($f_{2N+1}$) is at $x=0$ and can be bounded in both directions. The supremum of the sequence is thus also attained at (or rather in a neighborhood of) $x=0$. This follows from the following approximation for the remaining terms (or more precisely, from a concrete bound that should be obtainaned from it).

For the remaining terms, observe that the sequence $\cos^n(x)$ converges to a unique fixpoint for all $x \in \mathbb{R}$, the unique solution $x_0$ of $\cos x = x$. The summands are thus almost equal up to their alternating signs. We "transform coordinates" and instead work with the summands $\pm \cos^n(x) - x_0$.

Let $r = |(\frac{d}{dx} \cos)(x_0)| = |\sin(x_0)|$. Then $r$ is the convergence order of $\cos^n(x) \to x_0$, i. e. $$ \Delta_n := |\cos^n(x) - x_0| \approx \Delta_0 r^n. $$ (This follows by a first-order Taylor approximation of the recurrence equation around $x_0$.)

For the remaining terms we thus get approximately \begin{align*} &\sum_{n = N + 1}^\infty (\Delta_{2n} - \Delta_{2n+1}) \\ = &\sum_{n = N + 1}^\infty (1-r) \Delta_{2n} \\ = &(1-r) \Delta_{2N + 2} \sum{n = 0}^\infty r^2 \\ = &\Delta_{2N+2} \frac{1-r}{1-r^2} \\ = &\Delta_{2N+2} \frac{1}{1+r}. \end{align*}

Concerning the question of an analytic form and whether the value is expressible in terms of $\pi$ I am inclined to say no to both. According to a calculation with Sage with 2000 bits of precision, the value has long since stabilized at $f_{10001}$ and differs from $\pi/2$ by about 3.9e-5.

Already the solution of $\cos(x) - x=0$ should be transcendental and not expressible in the standard transcendental functions, but such proofs are in general incredibly difficult.