04/23/2009, 06:12 AM
The power series expansions of tetration \( {}^{z}a \) about the base are commonly discussed about (a=1), which produce very nice coefficients. However, I remembered that the same power series about other points made each coefficient depend on tetration, which isn't so bad if we already know one base. I'm not sure if this is useful for noninteger heights, because there are z's in the bounds for summation.
\(
\begin{tabular}{rl}
{}^{z}a
& = {}^{z}e \\
& +
\left[
\frac{1}{e} \sum_{k=1}^{z} \prod_{j=0}^{k} {}^{(z-j)}e
\right]
(a - e) \\
& +
\left[
\frac{1}{e^2}\sum_{k=1}^{z}
\left(
e(k-1) - 1 +
\sum_{j=0}^{k} \frac{1}{{}^{z-j}e}
\sum_{i=1}^{z-j}
\prod_{l=0}^{i} {}^{z-j-l}e
\right)
\prod_{j=0}^{k} {}^{z-j}e
\right]
(a - e)^2 \\
& + \cdots
\end{tabular}
\)
Pretty crazy...
Andrew Robbins
\(
\begin{tabular}{rl}
{}^{z}a
& = {}^{z}e \\
& +
\left[
\frac{1}{e} \sum_{k=1}^{z} \prod_{j=0}^{k} {}^{(z-j)}e
\right]
(a - e) \\
& +
\left[
\frac{1}{e^2}\sum_{k=1}^{z}
\left(
e(k-1) - 1 +
\sum_{j=0}^{k} \frac{1}{{}^{z-j}e}
\sum_{i=1}^{z-j}
\prod_{l=0}^{i} {}^{z-j-l}e
\right)
\prod_{j=0}^{k} {}^{z-j}e
\right]
(a - e)^2 \\
& + \cdots
\end{tabular}
\)
Pretty crazy...
Andrew Robbins
