09/04/2007, 10:54 AM
Gottfried Wrote:The half-iterate is (using 16 terms for the result)
\( \hspace{24} f^{\circ \frac{1}{2}}(b,1) \) = s^^(1/2) = 0.58983992+0.24626917*I
Thank you Gottfried! So the hypothesis of complex values for \( b<1 \) (and \( b>e^{-e} \)) is strongly supported by the matrix operator method (where this method quite precisely reflects the inutition behind fractional iteration, as it yields the expected results in all major application cases, as I will show in a subsequent post somewhen.)
@Jay
I nearly completely agree with you with the one exception that I dont think there is a direct relation between the development at the fixed points and the different branches of the \( \text{sexp} \). Because if this would be true there had to exist a fixed point such that regular iteration of \( \exp \) at this fixed point is real, which, I would guess, dont exist. And even the matrix operator method yields real values for \( b>\eta \).
For variety of views: The multiple branches of \( e^x \) (which are spirals in the dependency of a real \( x \) as Jay already mentioned: \( e^x=(e^{1+2\pi i k})^x = e^x e^{2\pi i k x} \)) can also be explained as in the following. If we take \( 1^{1/2} \) as the solutions of \( x^2=1 \) we have two solutions: +1 and -1. And generally if we take \( 1^{1/n} \) we have n distinct solutions on the unit circle with the arguments \( \alpha=k\frac{2\pi}{n} \), \( k=0..n-1 \). If we now consider \( 1^{\frac{m}{n}} \) we have n solutions if the fraction is cancelled. So we have defined \( e^x \) for all rational \( x \) as some sort of complex cloud. Now the different branches of \( e^x \) are all the possible continuous functions through this cloud!
In the same way there are n complex solutions for the nth iterative root of a powerseries of the form \( ax+bx^2+\dots \), \( |a|\neq 1 \). And we get infinitely many branches of a continuous (i.e. where the coefficients continuously depend on the iteration exponent) iteration.