10/21/2007, 10:47 PM
andydude Wrote:\( f(b, t) = \exp_b^{t+2}(-1) \) describes the function your assumption produces, i.e.:
- \( f(b, 0) = \exp_b^{2}(-1) = b^{b^{-1}} = b^{1/b} \)
- \( f(b, 1) = \exp_b^{3}(-1) = b^{b^{b^{-1}}} = b^{b^{1/b}} \)
- \( f(b, 2) = \exp_b^{4}(-1) = b^{b^{b^{b^{-1}}}} = b^{b^{b^{1/b}}} \)
and so on, while still allowing \( f(f(b, n), 1/n) = f(f(b, 1/n), n) = b \) and retaining the property \( \lim_{n \rightarrow \infty}f(b, n) = {}^{\infty}{b} \).
By the term orbit I'm refering to a term from dynamical systems where, given a point x, the sequence {x, f(x), f(f(x)), ...} is referred to as the orbit of f from x which is a way of referring to iteration without referring to the t in \( f^t(x) \). By using it this way, though, I'm slightly misusing it, since its a sequence, and not a function. Here I'm using it as a function \( f \ :\ t \rightarrow f^t(x) \), sorry if it was confusing. One of the reasons why I like the term 'orbit' so much is that it pairs nicely with iterate which, given t, is a function \( f \ : \ x \rightarrow f^t(x) \). I've discussed these terms here as well.
Andrew Robbins
Ok, I see. Still a pathological function though. It cannot be made continuous at 1. If we define \( {^0}e=e^{1/e} \) to conform with the limit of the tetraroots towards zero, then from the functional equation of tetration we must have:
\( {^1}e=e^{e^{1/e}} \).
However, the limit from the left of y=1 can be found by using the following:
\( lim_{n \to \infty}{^{(10^n-1)/10^n}}e=y \Leftrightarrow\\
e=lim_{n\to\infty}y^{^{1/(10^n-1)}y} \Leftrightarrow\\
e=y^{y^{1/y}} \)
Solving the latter with Maple, one gets \( y=2.025415088 \neq {^1}e=2.71828... \), so there is a discontinuity at y=1. In short, no matter how this function is defined at y=0 and y=1, it cannot be made continuous simultaneously at y=0 and y=1.
I prefer it defined as:
- \( {^0}e=1 \)
- \( {^1}e=e \)
- \( {^2}e=e^e \)
- \( \cdots \)
and let it do as it pleases in between, even if it ends up discontinuous.
