11/10/2009, 12:44 AM
So I guess I should rephrase the statement as:
\( d_{nb}(z_0) \overset{n}{=} \frac{({}^{\infty}b - z_0)^n}{G(b)} \)
which may provide a useful shortcut to showing whether or not intuitive/natural iteration even works. Suppose we make a Cramer's rule matrix \( \mathbf{E} \) with all the same entries of \( \mathbf{A}[\exp_b] \), except the first column is replaced with (1, 0, 0, 0, ...) so we can solve for \( \text{slog}_b'(z_0) \). Let \( c_{nb}(z_0) = \det(E) \), then if the limit exists, \( \text{slog}_b'(z_0) = \lim_{n\to\infty} \frac{c_{nb}(z_0)}{d_{nb}(z_0)} \) which is a rational polynomial mess.
If we take into account the above formula, then
\( \lim_{n\to\infty} \frac{c_{nb}(z_0)}{
({}^{\infty}b - z_0)^n} = \frac{\text{slog}_b'(z_0)}{G(b)} \)
which may provide a clear path to answering the question of convergence.
\( d_{nb}(z_0) \overset{n}{=} \frac{({}^{\infty}b - z_0)^n}{G(b)} \)
which may provide a useful shortcut to showing whether or not intuitive/natural iteration even works. Suppose we make a Cramer's rule matrix \( \mathbf{E} \) with all the same entries of \( \mathbf{A}[\exp_b] \), except the first column is replaced with (1, 0, 0, 0, ...) so we can solve for \( \text{slog}_b'(z_0) \). Let \( c_{nb}(z_0) = \det(E) \), then if the limit exists, \( \text{slog}_b'(z_0) = \lim_{n\to\infty} \frac{c_{nb}(z_0)}{d_{nb}(z_0)} \) which is a rational polynomial mess.
If we take into account the above formula, then
\( \lim_{n\to\infty} \frac{c_{nb}(z_0)}{
({}^{\infty}b - z_0)^n} = \frac{\text{slog}_b'(z_0)}{G(b)} \)
which may provide a clear path to answering the question of convergence.