05/05/2009, 03:27 AM
(This post was last modified: 05/12/2009, 03:55 AM by Base-Acid Tetration.)
So, Andysus bo198214, \( \frac{d}{dx}\exp^{\circ x} (\pm W(-1)) = \exp'(\pm W(-1))^x = \exp(\pm W(-1))^x \)? Where do you get from this to derive your tetration?
Does \( \lim_{y \to \infty} {}^{\pm iy} e = \mbox{fixed point of logarithm}= \pm W(-1) \), as in Kouznetsov's, in your tetration?
Also, how does one get the differential equation \( \ln (a)\frac{({}^{x+1} a)'}{({}^x a)'}={}^x a \)?
Edit: Oh, it comes from the definition of tetration.
\( {}^{x} a = a^{{}^{x-1} a} \)
\( [{}^{x} a]' = ({}^{x-1} a)' a^{{}^{x-1} a} \, \ln a \)
\( \frac{({}^{x} a)'}{ \ln{a} \, ({}^{x-1} a)' } = {}^x a, \, {}^0 a=1 \)
Does \( \lim_{y \to \infty} {}^{\pm iy} e = \mbox{fixed point of logarithm}= \pm W(-1) \), as in Kouznetsov's, in your tetration?
Also, how does one get the differential equation \( \ln (a)\frac{({}^{x+1} a)'}{({}^x a)'}={}^x a \)?
Edit: Oh, it comes from the definition of tetration.
\( {}^{x} a = a^{{}^{x-1} a} \)
\( [{}^{x} a]' = ({}^{x-1} a)' a^{{}^{x-1} a} \, \ln a \)
\( \frac{({}^{x} a)'}{ \ln{a} \, ({}^{x-1} a)' } = {}^x a, \, {}^0 a=1 \)
