12/08/2010, 10:16 AM
(This post was last modified: 12/08/2010, 10:46 AM by sheldonison.)

(12/08/2010, 02:44 AM)mike3 Wrote: Here's a graph of , acquired via the Cauchy integral. Scale is from -5 to +5 on both real and imag axis:very nice. Thanks Mike! I can see the cut points at L and conj(L). Your particular cutpoint choice don't show the singularity at -0.36237+iPi. The singularity at L is somewhat less intense, in that exp^[1/2](L) is defined and equals L.

(12/08/2010, 12:32 AM)tommy1729 Wrote: .... i believe you made a typo and some 0.33 needs to be 0.3 ?You're welcome Tommy. Hopefully these plots will clarify. I'm going to plot two more plots showing the two different half iterates of 0.33+i*Pi, which depend on which side of "L" the path goes on. Here, the path on the right (in green) goes straight up from 0.33 to 0.33+i*Pi, but the path in red takes a parabolic arc from 0.33 to 0.33+i*Pi, passing by the other side of the fixed point of L. The path in green leads to one of the half iterates, and the path in red leads to the other half iterate. L=0.31813+i*1.33723, which is about 43% of the way up up, midway between the green and red lines. Mike's cutpoint follows the path in green.

thanks for the reply.

half-iterate(0.33+i*Pi)=sexp(2.2644+1.0858*I)=-1.84495+2.6229*i (green)

half-iterate(0.33+i*Pi)=sexp(-2.2587)=-1.16627+i*Pi (red)

Plot showing the half-iterate, from 0.33 to 0.33+i*Pi, for both the green and red paths.

(12/08/2010, 12:32 AM)tommy1729 Wrote: what is this number here ? the period of the kneser function before its riemann mapping right ?Correct, the Period before the Riemann mapping, which for base "e" is 2i*Pi/L=4.4470+1.0579i.

This plot shows the contour taken by the slog(z) in the complex plane for the green, and red slogs, each going on either side of the fixed point of L, leading to the two different values for the half iterate. The half iterate is sexp(slog(z)+0.5). The endpoints of these two paths are approximately, but not exactly, one Period apart. My guess, is that if we used the superfunction before the Riemann mapping, then the path taken by the two slogs would have been exactly one period apart, and then the two half iterates would have been identical, independent of which side of "L" the path took, and perhaps there would be no singularity at "L".

- Sheldon