 exp^[0.5](x + 2 pi i) - exp^[0.5](x) - Printable Version +- Tetration Forum (https://math.eretrandre.org/tetrationforum) +-- Forum: Tetration and Related Topics (https://math.eretrandre.org/tetrationforum/forumdisplay.php?fid=1) +--- Forum: Mathematical and General Discussion (https://math.eretrandre.org/tetrationforum/forumdisplay.php?fid=3) +--- Thread: exp^[0.5](x + 2 pi i) - exp^[0.5](x) (/showthread.php?tid=1006) exp^[0.5](x + 2 pi i) - exp^[0.5](x) - tommy1729 - 05/26/2015 Im thinking about exp^[0.5](x + 2 pi i) - exp^[0.5](x). I have little time. But some ideas. regards tommy1729 RE: exp^[0.5](x + 2 pi i) - exp^[0.5](x) - MorgothV8 - 05/27/2015 (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 RE: exp^[0.5](x + 2 pi i) - exp^[0.5](x) - sheldonison - 12/30/2015 (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. [attachment=1217]