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Let $F(x)$ be a real-analytic function near $0$ ,with $0$ as one of its fixpoints and $f ' (0) > 1$.

$$F(x) = F \circ F \circ F^{-1} = \lim_{n \to \infty} F^{n} \circ F \circ F^{-n} = \lim_{n \to \infty} F^{n}(f'(0)\cdot F^{-n}(x)) $$

($*^n$ is $n$th composition)

How to prove this ?

Where does this expression $\lim_{n \to \infty} F^{n}(f'(0)\cdot F^{-n}(x)) $ converge ? I assume the radius of convergeance of $F^{-1}(x)$ places an upper boundary on where this can converge. But I assume this is also true for the radius of convergeance of $F(x)$. I assume this expression converges exactly in a circle centered at $0$ with radius $T$. Where $T$ is the smallest of the ROC of both $F(x)$ and $F^{-1}(x)$.

( I assume convergeance implies convergeance to the correct value ? )

I understand $f'(0) x$ is a good approximation for $F(x)$ near $0$ and that $F^{-n}(x)$ goes to $0$ but Im not convinced. I want proofs.

I assume l'Hopital can not be used here ??

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  • $\begingroup$ Mick copied from an equation with some details missing. $$g(z) = \lim_{n \to \infty} F^{[n]} (g'(0) \cdot F^{[-n]}(z))$$ here, $g'(0)$ refers to the derivative of the $\frac{1}{k}$ fractional iterate of interest, so that $(g'(0))^k = F'(0)$. In the example, F'(0)=4, and g'(0)=2, and we were calculating $g(z)= F^{0.5}(z)$ $\endgroup$ – Sheldon L Jan 15 '15 at 2:34
  • $\begingroup$ Another correction; although perhaps Mick wanted the form he posted; I'm not sure. I guess usually a similar equation is used for the Schroder equation; This is the form for the kth fractional iterate. $$g = F \circ g \circ F^{-1}$$ $\endgroup$ – Sheldon L Jan 15 '15 at 3:16
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Hint received from tommy1729 :

Plug in the koenigs function.

Brilliant !

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