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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:

alt text

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 folowing 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.

alt text

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.

alt text

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@Anixx: What do you mean by "probably smooth"? What is your heuristic? –  Alex B. Dec 1 '10 at 4:50
    
"probably smooth" - so you took the time to check that certain derivative values matched for each iterate? –  J. M. Dec 1 '10 at 4:59
    
No, I do not know. It would be interesting if somebody analyzed this question. –  Anixx Dec 1 '10 at 5:13
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2 Answers

A necessary and sufficient condition for continuity is that you have two end points $a < b$ such that $f(f(a)) = f(a)$ and similarly for $b$, in other words both $f(a)$ and $f(b)$ should be fixed points of $f$. So, take any continuous function on some interval such that the graph of this function cuts the line $y=x$ at $x_0$ on that interval, now extend it continuously to any interval $[a,b]$ containing the original one such that $f(a) = f(b) = x_0$. If you have two intersection points $x_0$ and $x_1$, then you can specify the function values at $a$ and $b$ to be different. This gives you an enormous supply of examples.

As for smoothness, I believe that in fact a necessary and sufficient condition is that $f$ is smooth on $(a,b)$, that $f'(a) = \pm\infty$ and that $f'(x_0)<0$, and similarly for $b$ (and possibly $x_1$, if that's how you constructed $f$). Indeed, $\frac{d}{dt}f(f(t))|_a = f'(f(a))f'(a)=f'(x_0)f'(a)$ which will also be $\pm\infty$ but with the opposite sign, if the above condition is satisfied, and the condition is clearly necessary. For the derivative of each higher iterate, you will have a corresponding number of $f'(x_0)$ in the product, so the sign will keep swapping. So yes, your curve seems to be smooth at the end points, thanks to arccos.

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Thank you, but what about smoothness in other points (not just $a$ and $b$)? For example, in points where the curve intersects itself? –  Anixx Dec 1 '10 at 5:41
    
And what about analiticity? Can we say that all integer iterates of this function are just different branches of the same analytic curve? –  Anixx Dec 1 '10 at 5:43
    
What do you mean by "where the curve intersects itself"? I took your question to mean "let's look at the original curve as parametrised by the interval $[0,1]$, say, and the iterate as parametrised by $[1,2]$ and so on, append them to each other to get something that is parametrised by $[0,\infty)$ and see whether all derivatives with respect to $t$ exist, where $t$ is the parameter". Is that not what you meant? –  Alex B. Dec 1 '10 at 5:54
    
No, it's more interesting than that. The suggested path traverses all the graphs on the interval [-1, 3], but not in the order you suggest, because the original paths aren't smooth. Rather, it appears that you can trace a curve that is $C^1$ at least by switching from one iterate of $f$ to another at singular points which conveniently meet. I have no idea how to find other examples, but for the purposes of creating a $C^1$ or smooth curve here, I notice the velocities don't match with the original parametrizations. To make them match a constant-speed path would be better. –  Elizabeth S. Q. Goodman Dec 1 '10 at 6:57
    
@Elizabeth arccos is smooth on $[-1,1]$ and cos is smooth everywhere, so $f$ is smooth on $[-1,3]$. Moreover, the values of $f$ in that interval also land in the same interval, so all iterates of $f$ are smooth there. Am I missing something obvious? –  Alex B. Dec 1 '10 at 7:32
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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.

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It seems that the free term equal to 1 is a special case, it is indeed not smooth at $x_1$. If we take free term slightly greater than 1 we will get a self-circuited loop at any iteration, while if we take the term slightly smaller than 1 we will get a more smooth looking curve without loops. –  Anixx Dec 1 '10 at 15:10
    
Regarding the first example it is still interesting if the curve smooth when going through blue, red and part of the yellow zones (before the yellow curve crosses the $x=x_1$ point). –  Anixx Dec 1 '10 at 15:28
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