10/06/2007, 01:04 PM
bo198214 Wrote:Though it contains a heap of errors:
- parabolic iteration works merely for the case \( b=e^{1/e} \). Jabotinsky's double binomial expansion works only in the parabolic case.
- Helm's method works generally for \( b>1 \) and yields real values. For \( 0<b<1 \) it obtains complex values.
- Woon's series was not designed for functions but for operators. I think it is equal to the parabolic case for \( w=1 \) and using a function instead of the operator.
- I think Jay's method works merely for \( b>e^{1/e} \).
- I think there are two solutions by Ioannis, one that is merely continuous but satisfies \( {}^{x}b = b^{\left({}^{(x-1)}b\right)} \) and one that is \( C^\infty \) and satisfies the condition only for integer \( x \).
One quick observation concerning some of the alternate approaches (like my first solution) considered above: I *think* (but I am not absolutely sure about it) that any method which defines \( {^y}x=f(x,y) \) for reasonable \( f(x,y) \), for \( y\in [0,1] \) could conceivably be extended to a \( C^{\infty} \), by using Andrew's method. For example, on my method for the first solution I define \( {^y}x=x^y \) for \( y \in [0,1] \). One could follow Andrew's impositions by requiring that \( lim_{y\to 1^-}D^{n}f(x,y)=lim_{y\to 1^+}D^{n}f(x,y) \). I haven't played around with my first solution to see if this works, but I suspect that it might for \( f(e,y) \). I don't think this approach will work if \( f(x,y) \) is linear in [0,1], because higher order derivatives will vanish.

