(05/04/2009, 04:23 AM)Tetratophile Wrote:(05/03/2009, 07:05 AM)bo198214 Wrote: Its regular if the iterates are differentiable (or at leas asymptotically differentiable) at the fixed point. This implies that \( (f^{[t]})'(x_0) = f'(x_0)^t \).
do you mean that the derivative of the function iterated \( t \) times at the fixed point is the derivative at the fixed point to the \( t^{th} \) power (multiplied t times)?
exactly. You see that is satisfied for natural numbered iterations:
\( (f^{[2]})'(x_0)=f'(f(x_0)) f'(x_0)= f'(x_0)^2 \)
\( (f^{[3]})'(x_0)= ... \)
\( (f^{[n+1]})'(x_0) = f'(f^{[n]}(x_0)) (f^{[n]})'(x_0) = f'(x_0) f'(x_0)^n = f'(x_0)^{n+1} \).
