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 Superlog determinant factored! andydude Long Time Fellow Posts: 509 Threads: 44 Joined: Aug 2007 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. « Next Oldest | Next Newest »

 Messages In This Thread Superlog determinant factored! - by andydude - 11/07/2009, 06:09 AM RE: Superlog determinant factored! - by andydude - 11/09/2009, 09:39 PM RE: Superlog determinant factored! - by bo198214 - 11/10/2009, 12:10 AM RE: Superlog determinant factored! - by andydude - 11/10/2009, 12:44 AM

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