11/07/2009, 08:12 PM
(11/07/2009, 05:11 PM)bo198214 Wrote: But now there is still the question why the limit \( f(x) \) indeed satisfies
\( f(x+1)=b^{f(x)} \)?
Actually it is not the case but we can obtain something very similar.
I just read in [1], p. 31, th. 10, that we have the following limit for a function
\( f(x)=f_1 x+f_2x^2+f_3 x^3 + ... \) with \( 0<f_1<1 \):
\( \lim_{n\to\infty} \frac{f^{\circ n}(t)-f^{\circ n}(\theta)}{f^{\circ n}(t)-f^{\circ n+1}(t)}=w\frac{1-f_1^w}{1-f_1} \) for \( \theta = f^{\circ w}(t) \).
If we invert the formula we get
\( f^{\circ w}(t) = \lim_{n\to\infty} f^{\circ -n}\left(w\frac{1-f_1^w}{1-f_1}\left(f^{\circ n+1}(t)-f^{\circ n}(t)\right)+f^{\circ n}(t)\right) \)
In our case though we dont have f(0)=0 but there is some fixed point \( z_f \) of \( f \), \( f(z_f)=z_f \).
In this case however the formula is quite similar, the only change is that \( f_1=f'(z_f) \).
[1] Ecalle: Theorie des invariants holomorphes
