08/23/2007, 07:43 PM
Without lose of generality, set the fixed point of \( f(x) \) at zero. It is well known that \( D f^t(0)=f'(0)^t \). Consider that
\( D^2 f(g(x)) = f''(g(x)) g'(x)^2 + f'(g(x)) g''(x) \), then
\( D^2 f(f^{t-1}(x)) = f''(f^{t-1}(x)) (D f^{t-1}(x))^2 + f'(f^{t-1}(x)) D^2 f^{t-1}(x) \). But this gives a functional equation for a geometrical progression solved at \( x=0 \) by
\( D^2 f^t(0) = f''(0) \sum_{k=0}^{t-1} f'(0)^{2t-k-2} \).
\( D^2 f(g(x)) = f''(g(x)) g'(x)^2 + f'(g(x)) g''(x) \), then
\( D^2 f(f^{t-1}(x)) = f''(f^{t-1}(x)) (D f^{t-1}(x))^2 + f'(f^{t-1}(x)) D^2 f^{t-1}(x) \). But this gives a functional equation for a geometrical progression solved at \( x=0 \) by
\( D^2 f^t(0) = f''(0) \sum_{k=0}^{t-1} f'(0)^{2t-k-2} \).
Daniel