Continuous iteration from fixed points of base e

[andydude]

The formula you give is very interesting. You say you are unsure whether the denominators should be \( a_k \) or \( \log(a_k) \) (base e), so I'm going to assume the most general form of both of these:

\( f(z) = \sum_{k=0}^{\infty} \frac{\log(z - a_k)}{2 c_k} + \frac{\log(z - \overline{a})}{2 \overline{c_k}} \)

Then you show that \( \text{slog}_b(z) \approx f(z) \) where \( a_k \) are the fixed points of b, which I still have my doubts about. What I noticed recently while re-reading your posts is that since you are unsure about what the \( c_k \) are in the above formula, given that we kind of "know" the slog function with approximations, we can solve for \( c_k \) using a system of linear equations.

Using Mathematica, it is a very simple matter to show that:

\( f(z) = C_0 - \sum_{k=1}^{\infty} z^_k \sum_{j=0}^{\infty} \frac{1}{a_j^k c_j} \)

which naturally implies that, if \( \text{slog}_b(z) = f(z) \), then:

\( \left[\frac{\partial^n}{\partial z^n} \text{slog}_b(z) \right]_{z = 0} = n! \sum_{j=0}^{\infty} \frac{1}{a_j^k c_j} \)

(for n > 0), which is a system of linear equations, which can be solved for \( 1/c_j \). This could provide a way to determine if \( c_k \) should be just \( a_k \) or \( \log(a_k) \) numerically.

Andrew Robbins