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

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exp^[0.5](x + 2 pi i) - exp^[0.5](x)
MorgothV8 Offline
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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

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

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RE: exp^[0.5](x + 2 pi i) - exp^[0.5](x) - by MorgothV8 - 05/27/2015, 07:00 PM



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