does tetration bring up a new number domain?

[bo198214]

It makes one wonder if the inverse of tetration would also create new numbers... I suspect it must've come up in this forum before, right?

It wasnt topic yet, so it is time to make it a topic!

In analogy to the demand that every polynomial shall have at least one zero, which led us to the algebraic numbers (or more sloppy to the complex numbers). The question here is whether every equation \( x[4]n=y \) has at least one solution \( x \) in the complex numbers.

By the striking correspondence of the function \( x^n \) on the reals compared with the function \( x[4]n \) on the positive reals, I would restrict the question to (in similarity to whether there is a complex solution of \( x^2=y \) for \( y<0 \)):

Is there a (complex) solution of \( x^x=y \) for \( 0<y<e^{-1/e} \) (which is the minimum of \( x^x \) at \( x=1/e \))?

The answer to this can be given with the lambertW function:
\( y=x^x=e^{x\ln(x)} \)
\( \ln(y)=x\ln(x) \) let \( x'=\ln(x) \)
\( \ln(y)=e^{x'}x' \)

We know that the Lambert W function (which is the inverse of \( f(x)=xe^x \)) is defined on any complex number, and so
\( x'=W(\ln(y)) \)
\( x=e^{W(\ln(y))} \).

Not only for every \( 0<y<e^{-1/e} \) but for every complex \( y \) the equation \( x^x=y \) has a solution.

I guess that is equally possible with the other \( x[4]n \) functions.
So, sorry guys, but tetration will not bring up any new numbers but is contained in the complex numbers (though it of course extends the algebraic numbers).