Infinite tetration and superroot of infinitesimal

[andydude]

h( infinitesimal) =?

I thought we already answered this question here. The problem is not that we can't find an answer for this. We can find h(0) (if we define it from the Lambert W function). We can also find \( h_O(0) = 0 \) and we can find \( h_E(0) = 1 \). So the problem is not that we can't find its value, the problem is that there are 3 extensions of h(x) beyond the interval \( [e^{-e}, e^{1/e}] \). Each extension has some very nice properties to it, and a reason for its existence. The "nice" extension is the one that is differentiable across \( e^{-e} \), which is the definition of h(x) in terms of the Lambert W function, or equivalently, as the inverse function of \( x^{1/x} \). Using this definition, h(0) = 0, and thats final, there is no oscillation. Asking about super-real numbers near zero is, at best, going to give a fourth extension of h(x), which is not going to bring us any closer to having single definition for h(x) near zero, as it is going to increase the number of definitions.

Superroot of infinitesimal = ?

Did you mean the 2nd super-root? or the infinite super-root? The infinite super-root is just \( \text{srt}_{\infty}(x) = x^{1/x} \), and \( \lim_{x \rightarrow 0} x^{1/x} = 0 \) which may answer your question. The 2nd super-root (sometimes called the super-sqrt) is the inverse function of \( x^x \) and since that function has a minimum at 1/e which gives the value \( (1/e)^{(1/e)} = 0.6922 \). This means that the 2nd super-root \( \text{srt}_{2}(x) = \frac{\log(x)}{W(\log(x))} \) is real-valued for \( x \ge 0.6922 \), but would be complex-valued from that value down to zero. Using the 2nd super-root defined by the Lambert W function, \( \lim_{x \rightarrow 0} \text{srt}_{2}(x) = -\infty \) which can be verified independently.

Andrew Robbins