• 0 Vote(s) - 0 Average
• 1
• 2
• 3
• 4
• 5
 exp^[0.5](x + 2 pi i) - exp^[0.5](x) sheldonison Long Time Fellow Posts: 633 Threads: 22 Joined: Oct 2008 12/30/2015, 04:29 PM (This post was last modified: 12/30/2015, 04:31 PM by sheldonison.) (05/27/2015, 07:00 PM)MorgothV8 Wrote: (05/26/2015, 12:28 PM)tommy1729 Wrote: Im thinking about exp^[0.5](x + 2 pi i) - exp^[0.5](x). I have little time. But some ideas. regards tommy1729Hmmm can it be 2*pi*i periodic? Like exp? I can only compute that from my c++ tet(ate(x)+0.5) .... tet(ate(x)+1) --> exp(x) is 2*pi*i periodic, hmmm If the cut lines are drawn to the right from the logarithmic singularity at -0.36237+/-pi i, then exp^{0.5}(z) is 2pi i periodic. In this plot, the singularity at L,L* is drawn vertically, away from the real axis, but it is so slight that it is still not really visible, even where it intersects the pi i cut lines. This graph goes from +/-10 real, +/-10 imag, with grid lines every 5 units.     - Sheldon « Next Oldest | Next Newest »

 Messages In This Thread exp^[0.5](x + 2 pi i) - exp^[0.5](x) - by tommy1729 - 05/26/2015, 12:28 PM RE: exp^[0.5](x + 2 pi i) - exp^[0.5](x) - by MorgothV8 - 05/27/2015, 07:00 PM RE: exp^[0.5](x + 2 pi i) - exp^[0.5](x) - by sheldonison - 12/30/2015, 04:29 PM

Users browsing this thread: 1 Guest(s)