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Let $f(x) = \cos(\sin(x))$ and let $c(f, n)(x)$ denote the function $\underbrace{f\circ f\circ...\circ f}_{n \text{ times}}$. For example, $c(f, 1)(x) = f(x)$, $c(f, 2)(x) = f(f(x))$ and so on.

My question is: does $c(f, n)(x)$ approach any constant function if $n \to +\infty$? I graphed this for some values of $n$ and the function seems to approach some value a little bit over $0.76$. Does anyone have any insight as to whether that is true? If so, what value is it approaching and why?

Any sort of help or material helps; this question has been stuck in my head for quite some time now! Thanks in advance!

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Yes—this is a question in dynamical systems, if you're looking for words to search with.

In this case, the equation $f(x)=x$ has exactly one fixed point (near $x=0.76817$), and at that fixed point we have $|f'(x)| \approx 0.46046 < 1$; therefore it is an attracting fixed point, which means that every sequence of iterates of $f$ will approach the fixed point exponentially fast.

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    $\begingroup$ This does not follow from what you have. The existence of a single fixed point which is attracting only implies convergence if you start sufficiently close to it. For example, consider the equation $f(x)=-x^3$. $\endgroup$ Jun 17 '20 at 9:54
  • $\begingroup$ I think boundedness of $f$ implies your conclusion in this case though. $\endgroup$ Jun 17 '20 at 9:59
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For the limit $f$ it must hold: $f(x)=\cos(\sin(f(x)))$. Taking the derivative (assuming $f$ differentiable) implies:

$f'(x)=-\sin(\sin(f(x)))\cos(f(x))f'(x)$

Impliing either $f'(x)=0$ or $1=-\sin(\sin(f(x)))\cos(f(x))$

The last equation can only hold if $f(x)=k\pi$ for $k\in \mathbb{Z}$ but this implies $\sin(\sin(f(x)))=0$, contradiction. So $f$ must be constant (or not differentiable).

You can show that a constant solution exists by Banach Fixpoint theorem on the sequence

$x_{n+1}=\cos(\sin(x_n))$

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  • $\begingroup$ Thank you! That makes a lot of sense! Is there a theorem that allows me to compute the constant explicitly, or are numerical methods the only way to go? $\endgroup$
    – Gauss
    Jun 16 '20 at 22:23
  • $\begingroup$ It must hold for the solution $x^*$: $x^*=\cos(\sin(x^*))$ but I don't think you can further simplify this. $\endgroup$
    – Philipp123
    Jun 16 '20 at 22:26
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    $\begingroup$ This solution also requires that the limiting function exists in the first place, which is not clear. $\endgroup$ Jun 17 '20 at 1:39
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    $\begingroup$ The first part was a necessary condition for the limit to exist. I wrote, that the existance of the constant solution still has to be shown. $\endgroup$
    – Philipp123
    Jun 17 '20 at 10:30
  • $\begingroup$ I used Maxima to numerically calculate the constant solution to 128 digits, then plugged the result into the Inverse Symbolic Calculator (wayback.cecm.sfu.ca/projects/ISC/ISCmain.html). It found no match. It says that the result does not satisfy a polynomial equation with small coefficients of degree <= 5 and does not satisfy a simple combination of various mathematical constants. $\endgroup$ Jun 17 '20 at 12:53
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This question can be settled by some elementary analysis. Note that $$\begin{aligned} I:=f(\mathbb R)&=\cos(\sin(\mathbb R))=\cos([-1,1])=[\cos(1),1]=[0.540,1],\\ f(I)&=\cos\left(\sin\left([0.540,\,1]\right)\right)\\ &=\cos([0.514,\,0.841])\\ &=[\cos(0.841),\,\cos(0.514)]\\ &=[0.666,\,0.871]\subset I. \end{aligned}$$ So, if $f$ has any fixed point, the fixed point must lie inside $I=[\cos(1),1]$.

Let $g(x)=f(x)-x$. Since $g(\cos(1))=0.330>0>-0.334=g(1)$, by the intermediate value theorem, $g$ has a zero on $I$, i.e. $f$ has a fixed point on $I$. As $g'(x)=-\sin(\sin(x))\cos(x)-1<0$, the fixed point of $f$ is also unique. Finally, on $I=[\cos(1),1]$, as $$|f'(x)|=| \sin(\sin(x))\cos(x)|\le|\sin(\sin(x))|\le|\sin(\sin(1))|=0.746<1,$$ the fixed point is attractive. Therefore, if we denote the $n$-fold composition of $f$ by $f^n$, the sequence $(f(x),f^2(x),f^3(x),\ldots)$ must converge to the fixed point of $f$.

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Denote $f(x)=\cos(\sin(x))$.

Since there exists $\varepsilon>0$ such that $|f'(x)|=|\sin(\sin(x))\cos(x)|< 1-\varepsilon$ for every $x \in \mathbb{R}$, $f\colon \mathbb{R} \to \mathbb{R}$ is a Lipschitz function with Lipschitz constant stricly less than $1$.

Thus, by Banach-Caccioppoli fixed point Theorem, $f$ has exactly one fixed point $x_0 \in \mathbb{R}$, that is a point such that $f(x_0)=x_0$. Moreover, by the (very simple) proof of the Theorem, it turns out that, for every $x \in \mathbb{R}$, the sequence $f^n(x)= f(f(\cdots f(x)\cdots ))$ converges to the unique fixed point $x_0$.

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