Let $f\colon [0,1]\to[0,1]$ be given by $f(x) = 1-\sqrt{1-x^2}$, i.e., the increasing auto-homeomorphism of $[0,1]$ whose graph is a quarter circle centered at $(0,1)$. I am interested in what can be said about iterates of $f$, specifically:

  • Is there a closed-form expression (in function of $n,x$) for the value $f^{\circ n}(x)$ of the $n$-th iterate of $f$ when $n\in \mathbb{N}$ (or indeed $n\in \mathbb{Z}$, letting $f^{\circ(-1)}(x) = \sqrt{x(2-x)}$ stand for the reciprocal of $f$)?

  • Does there exist a $1$-parameter group $(f^{\circ\tau})_{\tau\in\mathbb{R}}$ of increasing auto-homeomorphisms of $[0,1]$ (meaning $f^{\circ\varsigma} \circ f^{\circ\tau} = f^{\circ(\varsigma+\tau)}$ for $\varsigma,\tau\in\mathbb{R}$) such that:

    • $f^{\circ n}$ coincides with the $n$-iterate of $f$ for $n\in\mathbb{N}$ (of course $1$ is enough here),

    • $1-f^{\circ\tau}(1-x) = f^{\circ(-\tau)}(x)$ for all $\tau,x$ (this holds for integer $\tau$),

    • for $x$ fixed, $f^{\circ\tau}(x)$ is a continuous function of $\tau$, and

    • for any $\tau$ we have $f^{\circ\tau}(x) = 2\,(\frac{x}{2})^{2^\tau} + o(x^{2^r})$ when $x\to 0$ (this too holds for integer $\tau$)

    ? (Note: I don't know if these conditions are enough to make such a family $(f^{\circ\tau})_{\tau\in\mathbb{R}}$ unique even if it exists.)

Ideally I'm hoping for an answer to both questions at once in the form of a closed-form expression of $f^{\circ\tau}(x)$, but I'd already be happy with an answer to either part.

Edit: as it's been suggested to me elsewhere, it seems that we can define $f^{\circ\tau}(x)$ by $\lim_{n\to+\infty} f^{\circ(-n)}(2\cdot(\frac{1}{2}f^{\circ n}(x))^{2^\tau})$: numerically this appears to work very well (and suggests that the last condition indeed makes the family unique), but it's not clear to me why this limit even exists.

Edit 2: taking into account Christian Remling's comments below, the gist of the problem is to compute the differential $\psi(x) := \lim_{\tau\to 0} \frac{1}{\tau}(f^{\circ\tau}(x)-x)$ of the family at $0$ (which Christian calls $-g$, I hope I got this right), from which we could then construct the flow. So combining this with the previous edit, maybe the essence of the question should be to determine the value of the following $\psi(x)$ candidate: $$\lim_{\tau\to 0} \frac{\left(\lim_{n\to+\infty} f^{\circ(-n)}\Big(2\cdot\big({\textstyle\frac{1}{2}}f^{\circ n}(x)\big)^{2^\tau}\Big)\right)-x}{\tau}$$ (e.g., for $x=\frac{1}{2}$ this appears to be approximately $-0.432465489$, but Google doesn't know this constant).

  • $\begingroup$ @ChristianRemling Ah yes! I made a large but odd number of sign mistakes. $\endgroup$
    – Gro-Tsen
    Commented Jun 29, 2023 at 15:29

2 Answers 2


Not really an answer, but maybe slightly more than a comment:

You want to embed the discrete dynamical system $f$ in a flow $\phi_t$ (so $\phi_1=f$), and the most straightforward way to attempt this is to just write down the ODE $\dot{x}=g(x)$ that produces $\phi_t$ and see what conditions $g$ must satisfy. If $H'=1/g$, then the solution is (implicitly) given by $H(\phi_t(x))-H(x)=t$. So in our situation $H$ must satisfy the functional equation $$ H\circ f - H = 1 \quad\quad\quad\quad (F) $$ In principle at least, we can easily describe the general solution: we fix two successive points $a$, $b=f(a)$ (for example $a=\sqrt{3}/2$, $b=1/2$), choose any (continuously differentiable, let's say) increasing $H$ on $[a,b]$ with $H(b)=H(a)+1$, and then the rest is forced on us by (F), by propagating the basic interval by $f$. Conversely, any $H$ obtained in this way will satisfy (F) and $g=1/H'$ will generate a flow with $\phi_1=f$.

Of course, any such $\phi_t$ will automatically be continous (in fact, differentiable) in $t$. However, the remaining two conditions seem tedious (though not impossible perhaps) to analyze in this way.

Your second condition is now equivalent to $H$ being odd about $1/2$, and one can write down more explicit versions, but I'm too lazy now to continue further. It appears that with my above choice of $a$, $b$, one only needs to verify this condition on $[\sqrt{3}/2, 1-\sqrt{3}/2]$, once it holds there, (F) will propagate it to the rest of the interval.

  • $\begingroup$ Thanks for making it very clear that the last condition is crucial to making the family unique (as it seems to). $\endgroup$
    – Gro-Tsen
    Commented Jun 29, 2023 at 11:55

The question is about the iterates of $\,f(x) := 1-\sqrt{1-x^2}\,$ defined on $[0,1].\,$ In particular, non-integer iterates $\,f^{\circ r}(x)\,$ where $r$ is a real number. The two fixed points of $\,f(x)\,$ are $0$ and $1$. use the symmetry equation

$$1-f^{\circ r}(1-x) = f^{\circ(-r)}(x) \tag1 $$

and expand the function $\,f(x)\,$ as a power series at $x=0$. First define the function

$$ g(y,m) := 1 + c_1(m) y + c_2(m)\frac{y^2}{2!} + \cdots+ \mathcal{O}(y^{M+1}) \tag2 $$

where the coefficients are assumed to be power series in $m$ to be determined later.

Define the approximation function

$$ F_r(x) := (x/2)^{2^r} 2g(x^2/8,2^r) \tag3 $$

such that for all integer $\,k\ge M\,$

$$ f^{\circ k}(x) = F_k(x) \tag4 $$

is true. The unique solution for the coefficients are

$$ c_1(m) = m,\; c_2(m) = 8m+m^2,\; c_3(m) = 104m+24m^2+m^3, \dots. \tag5 $$

Use this only for $\,r>0\,$ and use the symmetry equation $(1)$ for $\,r<0.\,$ Because the approximation function $\,F_r(x)\,$ gets better as $\,x\to 0^+\,$ the result is

$$ f^{\circ r}(x) = \lim_{k\to\infty} f^{\circ -k}( F_r (f^{\circ k}(x))) \tag6 $$

where $\,k>0\,$ is an integer, but I have no rigorous proof yet.

  • $\begingroup$ I'm confused as how your recursion for $g(y,m)$ is initiated: what is $g(y,1)$ for example? (You seem to define only $g(y,2^r)$, right?) Also, this seems to give a practical way to compute $f^{\circ\tau}$ numerically (extending the one contained in the first “edit“ of the question), but the question of why the limit in your last equation actually exists remains unclear (to me at least). $\endgroup$
    – Gro-Tsen
    Commented Jul 1, 2023 at 18:27
  • $\begingroup$ @Gro-Tsen Thanks for the comment! I will edit my answer. $\endgroup$
    – Somos
    Commented Jul 1, 2023 at 18:45

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.