Does the alternating composition of sines and cosines converge to a constant?

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!

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.

• 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 $f(x)=-x^3$. Jun 17 '20 at 9:54
• I think boundedness of $f$ implies your conclusion in this case though. Jun 17 '20 at 9:59

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

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

• 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? Jun 16 '20 at 22:23
• It must hold for the solution $x^*$: $x^*=\cos(\sin(x^*))$ but I don't think you can further simplify this. Jun 16 '20 at 22:26
• This solution also requires that the limiting function exists in the first place, which is not clear. Jun 17 '20 at 1:39
• 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. Jun 17 '20 at 10:30
• 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. Jun 17 '20 at 12:53

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 $$f$$.

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 $$x_0$$.