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

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.

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

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

andydude Wrote:Did you mean the 2nd super-root? or the infinite super-root? The infinite super-root is just , and 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.

andydude Wrote:The 2nd super-root (sometimes called the super-sqrt) is the inverse function of and since that function has a minimum at 1/e which gives the value . This means that the 2nd super-root is real-valued for , but would be complex-valued from that value down to zero. Using the 2nd super-root defined by the Lambert W function, 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.