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exp^[0.5](x + 2 pi i) - exp^[0.5](x) - Printable Version

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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

tommy1729
Hmmm 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

tommy1729
Hmmm 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]