Exact and Unique solution for base e^(1/e)

[bo198214]

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:
\( g = \tau^{-1} \circ f \circ \tau \).

Iff \( f \) has a fixed point \( p \) then \( \tau^{-1} \circ f\circ \tau \) has the fixed point \( \tau^{-1}(p) \). The fixed points are merely mapped linearly. Also the regular iteration of the conjugate is equal to the conjugate of the regular iteration. So nothing new really occurs.

and more general base b^1/b iteration is linked to iterations of b^x - 1.

Yes just set \( \tau(x)=b(x+1) \).

what if b < e ?

we get again the multiple fixpoint problem.

yes, the number of fixed points of \( b^{x/b} \) and of \( b^x-1 \) is the same, by linear conjugation nothing new occurs.