TITLE: Functions which form continuous curve with its own iterations
QUESTION [7 upvotes]: The following function 
$$f(x)=-2 \cos \left(\sqrt{2} \arccos \left(\frac{x-1}{2}\right)\right)+1$$
has interesting property to form a continuous curve with its own integer iterations. The following image illustrates this property:
      (source)
Here blue is $f(x)$, red is $f(f(x))$, yellow is $f(f(f(x)))$ and green is $f(f(f(f(x))))$. It seems that all these functions form a continuous, and, probably, smooth curve.
The question is what is the general criterion for a function to have such property. Can you point some more examples of functions with such property?
P.S. If to use the following function 
$$f(x)=-2 \cos \left(\sqrt{2} \arccos \left(\frac{x-1}{2}\right)\right)+1.1$$
the curve becomes as below with one continuous branch and numerous closed circular branches.
      (source)
For function
$$f(x)=-2 \cos \left(\sqrt{2} \arccos \left(\frac{x-1}{2}\right)\right)+0.9$$
the curve is completely continuous and seems to be smooth.
      (source)

REPLY [2 votes]: Playing with the first example, it appears that the answer is negative. Let $x_0$ be the fixed point of $f(x)=-\cos\left(\sqrt{2} \arccos\left(\frac{x-1}{2}\right) \right)+1$. Looking at the point $-1< x_1<0$ where $f(f(f(x_1)))=x_0$, we see that $f(f(x_1))=-1$ and the point where 
$f(x)=-1$ is $x=3$, so it's easy to see that $x_1=1+2\cos\left(\frac{\sqrt{2}\pi}{2}\right)$.
So, if I am understanding the question correctly, we want to know whether $\frac{df^{[4]}}{dx}(x_1^{-}) = \frac{df^{[3]}}{dx} (x_1^{+})$. I find that $\frac{df^{[4]}}{dx}(x_1^{-})=-\frac{8\sin(\sqrt{2}\pi)}{\sqrt{2-2\cos(\sqrt{2}\pi)}}$ whereas $\frac{df^{[3]}}{dx} (x_1^{+})=-\frac{4\sqrt{2}\sin(\sqrt{2}\pi)}{\sqrt{2-2\cos(\sqrt{2}\pi)}}$. 
(Here I've used $f^{[n]}$ for the nth iterate of $f$ and $f(x^{\pm})$ for the right/left hand limit of $f$.)
It seems to me the graphs are illustrations of two facts: 1.) if $f^{[n]}(X)=x_0$, then $f^{[n+1]}(X)=x_0$, and 2.) since $f$ is decreasing near the fixed point $x_0$, if $f^{[n]}(x)$ is increasing approaching an $X$ such that $f^{[n]}(X)=x_0$, then $f^{[n+1]}(x)$ will be decreasing near $X$ (and vice versa). But the matching of derivatives is not guaranteed.