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Given that $f:\Bbb R\to\Bbb R$ is a real-valued function. The iteration of $f$, say $f^n$, is defined here:

  • $f^0(x)=x$
  • $f^n(x)=f\left(f^{n-1}(x)\right)$ for any positive integer $n$.

For example, $f(x)=x+c\implies f^{n}(x)=x+nc$, and $f(x)=2x\implies f^{n}(x)=2^nx$.


Supposing that $f(x)$ is well-behaved enough, how can we explore the asymptotic value of $f^n(x)$?

A Concrete Example

For example, $f(x)=\sin x$, we can (very toughly to me) obtain that $\lim_{n\to\infty}\sqrt n\,f^n(x)=\sqrt 3$, therefore $$f^n(x)=\sqrt{\frac 3n}+o(n^{-1/2})$$ Can we obtain more details about $f^n(x)$?


Maybe we can suppose that $f(x)$ is analytic (at $x=0$), therefore we can rewrite $f(x)=\sum_{n\ge0}\alpha_nx^n$. Now can we obtain the details about the asymptotic value of $f^n(x)$?

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Why $\lim_{n\to\infty}\sqrt n\,f^n(x)=\sqrt{3}$? – M. Strochyk Sep 21 '12 at 16:25
@M.Strochyk It's quite hard to me. In my book, there's a function stated: $\phi_c(x)=1/\sqrt{x^{-2}+(\sqrt3c)^{-2}}=x-x^3/6c^2+o(x^3)$, and $\phi_{1-\epsilon}(x)<\sin x<\phi_{1+\epsilon}(x)$ holds in some neighborhood of $0$ with radius $r_\epsilon$. Since it's so hard for me to read that proof, I believe that there's a general way to cope with that stuff. – Frank Science Sep 21 '12 at 16:33
It seems that is not possible to determine asymptotic values of $f^n(x)$ without any additional assumptions on class of iterated functions – M. Strochyk Sep 21 '12 at 17:04
@M.Strochyk That's why I said well-behaved enough. I think there's a more general method, but it might not be too general. – Frank Science Sep 21 '12 at 17:22

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