11/30/2007, 05:24 PM
bo198214 Wrote:I simply dont see how you get fixed points involved.
You have Carleman matrix \( B_b \) of \( b^x \) and you uniquely decompose the finite truncations \( {B_b}_{|n} \) into
\( {B_b}_{|n}=W_{|n} D_{|n} {W_{|n}}^{-1} \) and then you define
\( {B_b}^t = \lim_{n\to\infty} W_{|n} {D_{|n}}^t {W_{|n}}^{-1} \).
Where are the fixed points used?
In the usual eigen-decomposition, the diagonal consists of powers of \( f'(x_0) \) where \( x_0 \) is the fixed point. Maybe this was already answered.
Andrew Robbins

