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 exp^[0.5](x + 2 pi i) - exp^[0.5](x) tommy1729 Ultimate Fellow Posts: 1,365 Threads: 333 Joined: Feb 2009 05/26/2015, 12:28 PM Im thinking about exp^[0.5](x + 2 pi i) - exp^[0.5](x). I have little time. But some ideas. regards tommy1729 MorgothV8 Junior Fellow Posts: 18 Threads: 6 Joined: Dec 2012 05/27/2015, 07:00 PM (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 Fuji GSW690III Nikon D3, Nikkors 14-24/2.8, 24/1.4, 35/2, 50/1.4, 85/1.4, 135/2, 80-200/2.8 sheldonison Long Time Fellow Posts: 638 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 »

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