There is the nagging detail of picking a particular "starting point", a real constant greater than e. With further investigation of this function (and yes, I'm fairly certain it is a complex function, with no branch cuts, and I'm fairly certain without any singularities), I might someday find an intuitively "correct" starting point.
However, for the time being, I use \( e^{2} \) as my starting point. This has the useful property that -1, 0, and 1 gives very nice values:
\( \check{\eta}(-1) = e+e = 2e = e^{[2]}2 \)
\( \check{\eta}(0) = e \times e = e^2 = e^{[3]}2 \)
\( \check{\eta}(1) = e^e = {}^{2}e = e^{[4]}2 \)
(Note the bracketed notation is special operator, the name of which I forget, but which is fairly well discussed elsewhere in the forum.)
I suppose 2e would also make a good starting point (shifting the above special values to 0, 1, and 2), but I like e^2. It allows me to check that my first positive and negative iterate are correct, using easily remembered values.
Thus, without further ado, the cheta function is defined as:
\( \check{\eta}(z) = \exp_{\eta}^{\circ z}(e^2) \)
However, for the time being, I use \( e^{2} \) as my starting point. This has the useful property that -1, 0, and 1 gives very nice values:
\( \check{\eta}(-1) = e+e = 2e = e^{[2]}2 \)
\( \check{\eta}(0) = e \times e = e^2 = e^{[3]}2 \)
\( \check{\eta}(1) = e^e = {}^{2}e = e^{[4]}2 \)
(Note the bracketed notation is special operator, the name of which I forget, but which is fairly well discussed elsewhere in the forum.)
I suppose 2e would also make a good starting point (shifting the above special values to 0, 1, and 2), but I like e^2. It allows me to check that my first positive and negative iterate are correct, using easily remembered values.
Thus, without further ado, the cheta function is defined as:
\( \check{\eta}(z) = \exp_{\eta}^{\circ z}(e^2) \)
~ Jay Daniel Fox