Non-recursive coefficient formulas. Can the Riemann mapping be constructed? - 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: Non-recursive coefficient formulas. Can the Riemann mapping be constructed? (/showthread.php?tid=650) Non-recursive coefficient formulas. Can the Riemann mapping be constructed? - mike3 - 06/04/2011 Hi. Here: http://math.eretrandre.org/tetrationforum/showthread.php?tid=611&pid=5749#pid5749 I gave an explicit, non-recursive coefficient formula for the coefficients of the regular Schroder function of $e^{uz} - 1$: $\chi_1 = 1$ $\chi_n = \frac{1}{n!}\ \sum_{{1 = m_1 < m_2 < \cdots < m_k = n} \atop {2 \le k \le n}}\ \prod_{j=2}^k \frac{u^{m_j-1}}{1 - u^{m_j-1}} \left{{m_j \atop m_{j-1}}\right},\ n\ >\ 1$. (the curly-brackets thing is a Stirling number of the 2nd kind) This is in a very "general" form -- e.g. the Bernoulli numbers can also be written in a form similar to this (as in the thread, I mention how it comes from a solution of a general kind of recurrence equation.). I wonder if this can be "specialized" to yield a more interesting formula, like how the Bernoulli numbers can be written as a double-sum. Note that this formula requires exponentially increasing numbers of terms, which makes it not suitable for computation (recurrence is best). But I'm more curious about its use for doing analysis of the generated function. Could this be used somehow to help get the coefficients of the Riemann mapping to build the full tetrational at, say, base $e$? Perhaps it can be simplified to one with less terms. Any thoughts?