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zero's of exp^[1/2](x) ?
#9
(12/08/2010, 01:09 PM)sheldonison Wrote:
(12/08/2010, 01:03 PM)tommy1729 Wrote: -0.36237 + iPi ??

where does this come from ? and what is its closed form ?
from my earlier post, sexp(-2.5)~=-0.36237+iPi. Looking at Mike's graph again, the black "zero" is at -0.696. The cutpoint for L is not even visible at 0.318+1.34i, and only becomes visible around -1+1.34i. This gives somewhat of an idea as to how mild the singularity for the half iterate at L is, with a magnitude of a little less than 1 part in 100,000 in the immediate vicinity of L itself.
- Sheldon

then sexp(-3.5) , sexp(-4.5) , ... sexp(-(2n+1)/2)

should all have a singularity because sexp(x-1) = ln(sexp(x))

right ?
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Messages In This Thread
zero's of exp^[1/2](x) ? - by tommy1729 - 12/04/2010, 12:16 AM
RE: zero's of exp^[1/2](x) ? - by JJacquelin - 12/07/2010, 08:53 AM
RE: zero's of exp^[1/2](x) ? - by sheldonison - 12/07/2010, 01:03 PM
RE: zero's of exp^[1/2](x) ? - by mike3 - 12/08/2010, 02:44 AM
RE: zero's of exp^[1/2](x) ? - by sheldonison - 12/08/2010, 10:16 AM
RE: zero's of exp^[1/2](x) ? - by tommy1729 - 12/08/2010, 01:03 PM
RE: zero's of exp^[1/2](x) ? - by sheldonison - 12/08/2010, 01:09 PM
RE: zero's of exp^[1/2](x) ? - by tommy1729 - 12/08/2010, 01:39 PM
RE: zero's of exp^[1/2](x) ? - by sheldonison - 12/08/2010, 02:24 PM
RE: zero's of exp^[1/2](x) ? - by tommy1729 - 12/29/2015, 01:23 PM
RE: zero's of exp^[1/2](x) ? - by tommy1729 - 12/08/2010, 12:32 AM
RE: zero's of exp^[1/2](x) ? - by tommy1729 - 11/14/2012, 05:11 PM



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