08/03/2009, 03:24 PM
(08/02/2009, 10:34 PM)tommy1729 Wrote: 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.
Quote:and more general base b^1/b iteration is linked to iterations of b^x - 1.
Yes just set \( \tau(x)=b(x+1) \).
Quote: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.
