Superlog determinant factored! - Printable Version +- Tetration Forum (https://math.eretrandre.org/tetrationforum) +-- Forum: Tetration and Related Topics (https://math.eretrandre.org/tetrationforum/forumdisplay.php?fid=1) +--- Forum: Mathematical and General Discussion (https://math.eretrandre.org/tetrationforum/forumdisplay.php?fid=3) +--- Thread: Superlog determinant factored! (/showthread.php?tid=381) Superlog determinant factored! - andydude - 11/07/2009 I will write the proof as a story, so as to make it more interesting, but first, some definitions. Let $\mathbf{A}[f]^{(z_0)}_n$ be the nth truncated Abel matrix of f expanded at $z_0$, and let $d_{nb}(z_0) = \det(\mathbf{A}[\exp_b]^{(z_0)}_n)$. As I wrote about in my 2005 paper, each nth solution is unique iff $d_{nb}(z_0)$ is nonzero. These solutions can be used as the coefficients of the superlog as $\text{slog}_b(z) = y_0 + \sum_{k=1} a_k (z - z_0)^k$. Let $f(b) \overset{n}{=} g(b)$ indicate that $f^{(k)}(1) = g^{(k)}(1)$ for all $0 \le k \le n$. In some textbooks, this is referred to as "f and g have contact of order n at 1". In order to keep the notation small, every occurance of this notation will refer to f and g as functions of b, not any other variable, and the expansion point will always be 1. Theorem. $\begin{tabular}{|c|} \hline{} \ \\ d_{nb}(z_0) \overset{n}{=} \frac{(F(b) + 1 - z_0)^n}{G(b)} \\ \ \\ \hline\end{tabular}$ Proof. So one day, I was playing around with the determinants $d_{nb}(z_0)$, and began differentiating them. I was bored, and stumbled upon the quotient: $\frac{d_{nb}'(z_0)}{d_{nb}(z_0)}$ and I noticed that the series expansions of this expression at b=1 increased with every approximation number n, so I then looked at the quotient: $ \frac{d_{nb}'(z_0)}{n d_{nb}(z_0)} \overset{n}{=} \sum_{k=0}^{\infty} \frac{(b-1)^k}{(z_0-1)^{k+1}} \sum_{j=0}^{k-1} c_{jk} z_0^j$ and to my surprise, the series expansions did not change at all with n, except for the coefficients of $(b - 1)^{n+1}$, which are beyond the claim of $f\overset{n}{=}g$. So after looking at the quotient a bit more, I saw that it was the logarithmic derivative of $d_{nb}(z_0)$, so I integrated both sides. $ \frac{1}{n}\ln(d_{nb}(z_0)) \overset{n}{=} \ln(z_0 - 1) + \sum_{k=0}^{\infty} \frac{(b-1)^k}{(z_0-1)^{k}} \sum_{j=0}^{k-1} c_{jk} z_0^j + \ln(-1) - g(b)/n$ The extra terms are to make up for the constant of integration, and for the integral of $\frac{1}{z_0 - 1}$ in the big summation on the right. So after some simplification, I realized that $ \sum_{k=0}^{\infty} \frac{(b-1)^k}{(z_0-1)^{k}} \sum_{j=0}^{k-1} c_{jk} z_0^j = f(b, z_0) = \ln\left(\frac{F(b)}{1 - z_0} + 1\right)$ or in other words, the $f(b, z_0)$ can be factored into a function of b and a function of $z_0$. Putting it all together: $ \begin{tabular}{rl} ln(d_{nb}(z_0)) & \overset{n}{=} n\ln(1 - z_0) + n f(b, z_0) - g(b) \\ d_{nb}(z_0) & \overset{n}{=} \exp(n\ln(1 - z_0) + n f(b, z_0) - g(b)) \\ & \overset{n}{=} (1 - z_0)^n \frac{\exp(n f(b, z_0))}{\exp(g(b))} \\ & \overset{n}{=} (1 - z_0)^n \frac{(\frac{F(b)}{1 - z_0} + 1)^n}{\exp(g(b))} \\ & \overset{n}{=} \frac{\left[(1 - z_0)(\frac{F(b)}{1 - z_0} + 1)\right]^n}{\exp(g(b))} \\ & \overset{n}{=} \frac{\left(F(b) + (1 - z_0))\right)^n}{\exp(g(b))} \\ & \overset{n}{=} \frac{\left(F(b) + (1 - z_0))\right)^n}{G(b)} \\ \end{tabular}$ where $ \begin{tabular}{rl} F(b+1) &= 0 + b + b^2 + \frac{3}{2} b^3 + \frac{7}{3} b^4 + 4 b^5 + \frac{283}{40} b^6 + \frac{4681}{360} b^7 + \cdots \\ G(b+1) &= 1 + b + \frac{5}{2} b^2 + \frac{35}{6} b^3 + \frac{173}{12} b^4 + \frac{2077}{60} b^5 + \frac{6079}{72} b^6 + \cdots \end{tabular}$ Although it is not as exciting as I hoped at first, it is still interesting. The formula can be verified because it asserts something that is true for all n, and so if it is true for $n < 20$ it will probably be true for more n. I have tried to use this formula to find information about base-e, but these functions don't converge out there. I'm thinking it has something to do with the fact that base-1 is divergent. I'm hoping that this kind of formula could help with the higher bases, but so far no such luck. I have also tried to find a similar formula for the same matrix with the first column replaced with (1, 0, 0, 0, ...) which represents the numerator of Cramer's rule for the first coefficient/derivative of superlog. One final remark. Notice that n has been isolated. This makes it obvious that $\lim_{n\to\infty} d_{nb}(z_0)$ diverges to infinity if $F(b) > z_0$ and converges to 0 if $F(b) < z_0$, but is this only for bases near 1? Andrew Robbins RE: Superlog determinant factored! - andydude - 11/09/2009 OMG, I just realized: $F(b) = {}^{\infty}b - 1$. RE: Superlog determinant factored! - bo198214 - 11/10/2009 (11/09/2009, 09:39 PM)andydude Wrote: OMG, I just realized: $F(b) = {}^{\infty}b - 1$. ui! RE: Superlog determinant factored! - andydude - 11/10/2009 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.