08/31/2010, 08:33 PM
(08/31/2010, 07:34 PM)mike3 Wrote: For the variable-base case,Wow! Thanks Mike! Only matrices and statistics, "n choose k" formulas is one of my weak areas, so it might be a few days before I catch up.
\( a_n (L \log(B))^n = L \frac{B_n(\log(B), \log(B) 2! a_2, ..., \log(B) n! a_n)}{n!} \)
thus
\( a_n L^{n-1} \log(B)^n n! = B_n(\log(B), \log(B) 2! a_2, ..., \log(B) n! a_n) \)
\( a_n = \frac{B_n(\log(B), \log(B) 2! a_2, ..., \log(B) (n-1)! a_{n-1}, 0)}{(L^{n-1} \log(B)^n - \log(B)) n!} \).
I'm kind of intrigued at actually having a closed form for a real valued superfunction, like sqrt(2). I'm guessing that the number of terms required for convergence gets extremely large as z increases. From the terms I computed a couple of days ago, it looks like the terms are decreasing exponentially, which means the series acts like it has a singularity. However, since the regular superfunction is entire, we must have convergence to infinity, so the terms must eventually decrease faster than exponentially.
- Sheldon