zero's of exp^[1/2](x) ?
#3
(12/07/2010, 08:53 AM)JJacquelin Wrote: Do you mean :
exp^[1/2](x) = [exp(x)]^(1/2) = exp(x/2)
?
He is using the notation, exp^[1/2] as the half iterate. exp^[1] would be the exponential function and exp^[-1] would be the inverse exponential function, or the logarithm function.

One of Tommy's questions is where are the zeros of the half iterate, and one of the zeros that Tommy points out is at sexp(-1.5), exp^[1/2](-0.6960247)=0. I think that might be the only zero, but I'm not sure. I don't understand the general behavior of the half iterate in the complex plane.

As to the other question, is the half iterate entire, it obviously is not, since the half iterate of sexp(-2.5)~=-0.36237+iPi would be the singularity. I made a graph of the half iterate, with real(x)=real(sexp(-2.5))=-0.36237 and imaginary varying from 0 to 0.9999*Pi*I, where the red is the real value of the half iterate, and the green is the imaginary of the half iterate. The singularity is pretty clear.
- Sheldon
   
added: a couple of questions. One question I have is what happens to the half iterate in the neighborhood of the fixed point L=0.3181+1.337i. slog(z) has a singularity at L. But does the half iterate have a singularity? I generated the half iterate following a path from 0.3 to 0.3+i*Pi, and then again following a path from 0.33 to 0.33*i*Pi. These pass nearby, on either side of the fixed point L. For both contours, I generated the slog, and then generated the half iterate via sexp(slog(z)+0.5). For the two cases, the two slogs follow very different paths in the complex plane, so I'm pretty sure the half iterate also has a singularity at L, although the half iterate of L itself is probably defined=L?

The half iterate following a path from 0.3 to 0.3+i*Pi, winds up at sexp(-2.248 ), which is on the real axis, as expected. But the half iterate following a path from 0.33 to 0.33+i*Pi isn't anywhere near the real axis, and winds up at sexp(2.2644+1.0858*I), which is a totally different place in the complex plane, and the imag(z) of the half iterate is neither zero, nor Pi. So the half iterates of these two numbers, 0.3+i*Pi and 0.33+i*Pi, have different values, depending on which side of the fixed point "L", the path follows in the complex plane.

half-iterate(0.33+i*Pi)=sexp(-2.2587)=-1.16627+i*Pi
half-iterate(0.33+i*Pi)=sexp(2.2644+1.0858*I)=-1.84495+2.6229*i

Another question, what if we used the entire superfunction, base e, which isn't real valued at the real axis? In this case, the half iterate would not be real valued at the real axis either. But would the half iterate be entire, or would it have singularities due to the slog singularities at the fixed point? The Period of the regular entire superfunction is 4.4470+1.0579i. The is also the pseudo periodicity of the sexp function, and the two different slog(0.33+i*Pi) in the example given, are approximately one period apart in the complex plane (4.5126+1.0858i). If we used the regular superfunction, instead of sexp, the two points would have been exactly one period apart, and the two half iterates would have been identical! So perhaps the fixed point would not cause a singularity in the half iterate when using the regular entire superfunction?
- Sheldon




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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