The "cheta" function

[jaydfox]

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:

\( \check{\eta}(-1) = e+e = 2e = e^{[2]}2 \)
\( \check{\eta}(0) = e \times e = e^2 = e^{[3]}2 \)
\( \check{\eta}(1) = e^e = {}^{2}e = e^{[4]}2 \)

I suppose that e^2 isn't a completely arbitrary starting point. In addition to having the nice properties from above, both 2e and e^2 have other reasons to recommend them as "preferred" starting points.

As Henryk pointed out a few days ago, there is a simple linear transform to get from the cheta function to the continuous iteration of \( e^{x}-1 \):

so tetration base e^1/e is linked to iterations of e^x - 1.

Yes, the link is a linear transformation \( \tau \).
If you set \( \tau(x)=e(x+1) \), and \( f(x)=e^{x/e} \) and \( g(x)=e^x-1 \) you have the relation:
\( g = \tau^{-1} \circ f \circ \tau \).

If you consider the iterator of e^x - 1, then x=1 seems a very logical starting point (since 0 makes no sense). Next, apply Henryk's tau function to x=1, and you get 2e.

Interestingly, there is another function we can use, one I prefer for reasons related to computing the base-change formula I mentioned. It is:

\( G(x) = e^{(x+1)} \)

If you apply this G function to x=1, you get e^2. Apply it to ln(2), the first negative (or "backward") iteration of exp(x)-1, and you get 2e. Apply it to e-1, the first positive (or "forward") iteration of exp(x)-1, and you get e^e.

So as I said, 2e and e^2 are both better candidates for a starting point than most other purely arbitrary numbers, but I don't see a strong reason to choose one over the other. I simply prefer e^2. It's useful computationally, and it lets me verify in my head that forward and backward iterations are correct.