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+--- Thread: Series for x-tetra-x (/showthread.php?tid=276)
Series for x-tetra-x - andydude - 04/27/2009
I just found a nice "power series" for \( {}^{x}x \). It's based on regular iteration, and I took the regular iteration power series for \( \exp_x^y(z) \) and evaluated it at (z=1), then substituted (y=x) to obtain a power series for \( {}^{x}x \). At this point, I noticed there were lots of logarithms, or in other words, it seemed to be of the form \( \sum_{j\ge 0} \sum_{k\ge 0} A_{jk} x^j \ln(x)^k \) which is nice, but not very interesting. It was when I noticed that the power series expansion of \( x^x \) about 0 also produces this kind of series that I thought the two could be combined to make a new kind of series. So I tried doing some linear algebra change-of-basis stuff but I forgot how, so I used "undetermined coefficients" instead.
I want to discuss this more, but it will have to wait until this weekend. Here is the power series.
\(
\begin{tabular}{rl}
(x + 1) \uparrow\uparrow (x + 1)
& = 1 \\
& + x \\
& + x^2 (1 - x^x) \\
& + \frac{x^3}{2}(3 - 3 x^x) \\
& + \frac{x^4}{3!}(14 - 17 x^x) \\
& + \frac{x^5}{4!}(96 - 119 x^x - 12 x^{2x}) \\
& + \frac{x^6}{5!2}(1698 - 2013 x^x - 420 x^{2x}) \\
& + \frac{x^7}{6!4}(37448 - 41199 x^x - 15480 x^{2x}) \\
& + \cdots
\end{tabular}
\)
\begin{tabular}{rl}
(x + 1) \uparrow\uparrow (x + 1)
& = 1 \\
& + x \\
& + x^2 (1 - x^x) \\
& + \frac{x^3}{2}(3 - 3 x^x) \\
& + \frac{x^4}{3!}(14 - 17 x^x) \\
& + \frac{x^5}{4!}(96 - 119 x^x - 12 x^{2x}) \\
& + \frac{x^6}{5!2}(1698 - 2013 x^x - 420 x^{2x}) \\
& + \frac{x^7}{6!4}(37448 - 41199 x^x - 15480 x^{2x}) \\
& + \cdots
\end{tabular}
\)
[update]Fixed some coefficient signs[/update]
Andrew Robbins