Infinite tetration and superroot of infinitesimal

[Ivars]

hej Andy

Thanks for careful examination and corrections. It seems so that h(infinitesimal) does not contain much new information, at least I do not feel so . But superroot is more interesting

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.

I am not sure You can apply limit to hyperreals. Infinitesimal I mean in this case is by definition smaller than any real, but NOT 0. So infinitesimal^(1/infinitesimal) = (infinitesimal^(infinity of the same grade) ) IS NOT 0. I would rather say it is e, or e^pi/2.

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

This is very interesting as well is it returns complex values from real numbers. I have to learn more about superroots.