Dmitrii Kouznetsov's Tetration Extension

[bo198214]

Bo, namely for \( b=\sqrt{2} \) my expression can be simplified to yours.

Yes you are right, again stupid mistake of mine. For \( b=\sqrt{2} \) the fixed point is \( a=2 \) and the period is \( \frac{2\pi i}{\ln(\ln(a))}=\frac{2\pi i}{\ln(\ln(2))}\approx -17.1431*I \).

Good, so seems quite as if tetration by regular iteration gives the same as tetration with your approach.

As it is too expensive to compute the contours, I give here pictures of the regular tetration as 3d plot, absolute value over the complex plane, here it is:

over -1-I*10 ... +1+I*10
   
over 0 ... 1+I*20
   

One can clearly see the period of -17.14 along the imaginary axis.

Btw. there are infinitely many analytic functions that have no singularity on right halfplane: If \( F \) has no singularity there, also \( F(x+c\sin(2\pi x)) \), \( 0<c<\frac{1}{2\pi} \), is a superexponential with no singularities on the right halfplane and any 1-periodic function \( p \) with \( p(x)>-x \) in \( 0<x\le 1 \) would do.

I disagree. At some smooth contour \( u \) in your strip, the contour \( v=u+c \sin(2\pi u) \) enters the left halfplane and crosses the cut \( v<-2 \). You have no need to evaluate any tetration in order to see it. Corresponding contour \( F(v) \) is not continuous.

You are right. As sin is not bounded in the real part on the strip it can go to arbitrary negative values.

Fig.2 is ready at
http://en.citizendium.org/wiki/TetrationAsymptoticParameters00. Try to run it.

ok, that works too (but as a c++ program, which I wasnt aware of).