zero's of exp^[1/2](x) ? tommy1729 Ultimate Fellow Posts: 1,358 Threads: 330 Joined: Feb 2009 12/04/2010, 12:16 AM another brainstorm ... exp^[1/2](x) probably isnt an entire function. right ? i mean exp^[1/2](x) = sexp(slog(x)+1/2) and since both sexp and slog are not entire ... furthermore an entire function is of the form exp(f(x))*(x-a_0)...(x-a_n) where f(x) is entire too. but such a product identity is unlikely for exp^[1/2](x) , especially the exp(f(x)) part since an entire function grows faster than polynomial but f(x) grows slower hence a contradiction. but what if f(x) = constant ?? we still have the argument that sexp and slog are not entire. maybe there is a simple proof that exp^[1/2](x) is not entire. if it miraculously turns out to be entire it has to be of the form e^C *(x-a_0)...(x-a_n). eitherhow , already the zero's of exp^[1/2](x) interest me. in fact toying around a bit we find : exp^[1/2](x) = sexp(slog(x)+1/2) = 0 hence bye reversing : x = sexp(slog(0)-1/2) = exp^[-1/2](0) i know slog(x) has period 2pi i. hence the zero's of exp^[1/2](x) appear to be "exp^[-1/2](n*2pi i)" what reduces to exp^[-1/2](0). but things dont make sense , i must have made a mistake. first , there must be other zeros or the zeros are wrong : if exp^[-1/2] is entire then having only zero's exp^[-1/2](n*2pi i) or even just exp^[-1/2](0) cannot hold. now assume exp^[-1/2] is not entire. well log isnt , sexp isnt and slog isnt , so that makes more sense. i dont believe it only has one zero ? exp^[-1/2](0) = exp[1/2](log(0) + 2 n pi i) = sexp(slog(log(0))+1/2) hmm so ok , it seems exp^[-1/2](0) is indeed the only solution ? and it also seems 2 pi i is a period of exp^[1/2](x) , which we ofcourse actually already knew earlier. but wait a sec , if exp^[1/2] is periodic then we have as solutions : exp^[1/2](x) = 0 x = exp^[-1/2](0) + 2pi i. is that correct or are there others as well ? does this constitute a proof that exp^[1/2] is not entire ? JJacquelin Junior Fellow Posts: 5 Threads: 1 Joined: Oct 2010 12/07/2010, 08:53 AM Do you mean : exp^[1/2](x) = [exp(x)]^(1/2) = exp(x/2) ? sheldonison Long Time Fellow Posts: 633 Threads: 22 Joined: Oct 2008 12/07/2010, 01:03 PM (This post was last modified: 12/07/2010, 10:02 PM by sheldonison.) (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 tommy1729 Ultimate Fellow Posts: 1,358 Threads: 330 Joined: Feb 2009 12/08/2010, 12:32 AM (This post was last modified: 12/08/2010, 12:34 AM by tommy1729.) quote : The Period of the regular entire superfunction is 4.4470+1.0579i end quote. what is this number here ? the period of the kneser function before its riemann mapping right ? how is it computed ? i believe it was something along the line of 2pi i/ln(fixpoint) not ? i believe you made a typo and some 0.33 needs to be 0.3 ? thanks for the reply. regards tommy1729 mike3 Long Time Fellow Posts: 368 Threads: 44 Joined: Sep 2009 12/08/2010, 02:44 AM Here's a graph of $\exp^{1/2}(z)$, acquired via the Cauchy integral. Scale is from -5 to +5 on both real and imag axis:     sheldonison Long Time Fellow Posts: 633 Threads: 22 Joined: Oct 2008 12/08/2010, 10:16 AM (This post was last modified: 12/08/2010, 10:46 AM by sheldonison.) (12/08/2010, 02:44 AM)mike3 Wrote: Here's a graph of $\exp^{1/2}(z)$, acquired via the Cauchy integral. Scale is from -5 to +5 on both real and imag axis:very nice. Thanks Mike! I can see the cut points at L and conj(L). Your particular cutpoint choice don't show the singularity at -0.36237+iPi. The singularity at L is somewhat less intense, in that exp^[1/2](L) is defined and equals L. (12/08/2010, 12:32 AM)tommy1729 Wrote: .... i believe you made a typo and some 0.33 needs to be 0.3 ? thanks for the reply.You're welcome Tommy. Hopefully these plots will clarify. I'm going to plot two more plots showing the two different half iterates of 0.33+i*Pi, which depend on which side of "L" the path goes on. Here, the path on the right (in green) goes straight up from 0.33 to 0.33+i*Pi, but the path in red takes a parabolic arc from 0.33 to 0.33+i*Pi, passing by the other side of the fixed point of L. The path in green leads to one of the half iterates, and the path in red leads to the other half iterate. L=0.31813+i*1.33723, which is about 43% of the way up up, midway between the green and red lines. Mike's cutpoint follows the path in green. half-iterate(0.33+i*Pi)=sexp(2.2644+1.0858*I)=-1.84495+2.6229*i (green) half-iterate(0.33+i*Pi)=sexp(-2.2587)=-1.16627+i*Pi (red)     Plot showing the half-iterate, from 0.33 to 0.33+i*Pi, for both the green and red paths.     (12/08/2010, 12:32 AM)tommy1729 Wrote: what is this number here ? the period of the kneser function before its riemann mapping right ?Correct, the Period before the Riemann mapping, which for base "e" is 2i*Pi/L=4.4470+1.0579i. This plot shows the contour taken by the slog(z) in the complex plane for the green, and red slogs, each going on either side of the fixed point of L, leading to the two different values for the half iterate. The half iterate is sexp(slog(z)+0.5). The endpoints of these two paths are approximately, but not exactly, one Period apart. My guess, is that if we used the superfunction before the Riemann mapping, then the path taken by the two slogs would have been exactly one period apart, and then the two half iterates would have been identical, independent of which side of "L" the path took, and perhaps there would be no singularity at "L". - Sheldon     tommy1729 Ultimate Fellow Posts: 1,358 Threads: 330 Joined: Feb 2009 12/08/2010, 01:03 PM (12/08/2010, 10:16 AM)sheldonison Wrote: (12/08/2010, 02:44 AM)mike3 Wrote: Here's a graph of $\exp^{1/2}(z)$, acquired via the Cauchy integral. Scale is from -5 to +5 on both real and imag axis:very nice. Thanks Mike! I can see the cut points at L and conj(L). Your particular cutpoint choice don't show the singularity at -0.36237+iPi. The singularity at L is somewhat less intense, in that exp^[1/2](L) is defined and equals L. -0.36237 + iPi ?? where does this come from ? and what is its closed form ? regards tommy1729 sheldonison Long Time Fellow Posts: 633 Threads: 22 Joined: Oct 2008 12/08/2010, 01:09 PM (This post was last modified: 12/08/2010, 01:33 PM by sheldonison.) (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 tommy1729 Ultimate Fellow Posts: 1,358 Threads: 330 Joined: Feb 2009 12/08/2010, 01:39 PM (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 ? sheldonison Long Time Fellow Posts: 633 Threads: 22 Joined: Oct 2008 12/08/2010, 02:24 PM (This post was last modified: 12/08/2010, 03:18 PM by sheldonison.) (12/08/2010, 01:39 PM)tommy1729 Wrote: then sexp(-3.5) , sexp(-4.5) , ... sexp(-(2n+1)/2) should all have a singularity because sexp(x-1) = ln(sexp(x)) right ?yes, but I don't know where they would be in the complex plane. sexp(-2.5)=-0.36237+iPi, and if you follow a path from -0.36237 to -0.36237+iPi, the singularity is right there (plotted the path earlier). But, for sexp(-3.5) = 1.1513+i1.6856, I'm not sure what the path would be in the complex plane. If I naively calculate slog(1.1513+i1.6856), I get 0.94439+i1.12428, which has no connection to the predicted singularity at exp^[0.5](sexp(-3.5)). - Sheldon « Next Oldest | Next Newest »