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 $x\to 0$

  ? (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, but it's not clear to me why this limit even exists.

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).

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 $x\to 0$

  ? (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.

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 $x\to 0$

  ? (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, but it's not clear to me why this limit even exists.