zero's of exp^[1/2](x) ? sheldonison Long Time Fellow Posts: 683 Threads: 24 Joined: Oct 2008 12/10/2010, 08:35 PM (This post was last modified: 12/10/2010, 09:27 PM by sheldonison.) (12/08/2010, 01:09 PM)sheldonison Wrote: .... 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. - SheldonI'm fascinated by the "mild" singularity of exp^[1/2](L). L is the fixed point for base e, or ~0.318+1.337*I. It seems that the limiting value for exp^[1/2](L)=L from all directions in the complex plane. The branch singularity discontinuity goes to zero in the vicinity of L. Here is a parametric contour graph, where the inner red circle is a circular radius L+0.5*exp(i*z). The inner green circle is a radius of 1. The middle red circle is a radius of 1.5, and the outer green circle is a radius of 2. The outer red circle is a radius of 2.5. The branch cut-point jump is visible in the graph at radius 1.5, 2, and 2.5. The starting point for the circle is L-radius, going counter clockwise, which matches Mike's exp^[1/2] graph cutpoints.     This chart plots the absolute value of the cutpoint branch delta, after circling around the singularity, from radius 0.5 to radius 2.5. At a radius of 0.5, the singularity has a magnitude of 0.00038, which is pretty small. We are comparing exp^[1/2](L+r*exp(-Pi*i)) to exp^[1/2](L+r*exp(+Pi*i)), a difference of one rotation around the singularity.     This chart plots the absolute value of the singularity, from radius 0.02 to radius 0.5. At a radius of 0.02, the singularity has a magnitude of 1.4E-11, which is very small. So, at the singularity itself, it is likely that exp^[1/2](L)=L, from all directions. Mathmetically, circling counterclockwise around the singularity corresponds to adding approximately one Period to the slog. slog(L-r) becomes slog(L-r)+Period, where Period is the Period of the superfunction before the Riemann mapping. - Sheldon     mike3 Long Time Fellow Posts: 368 Threads: 44 Joined: Sep 2009 12/11/2010, 09:21 AM Hmm. What's so "fascinating" about the "mild singularity", anyway? I'd expect that precisely because $\exp^{1/2}$ is an iterate of $\exp$, that it would have fixed points wherever $\exp$ does, on the appropriate branch of course (note how $\log$ has the same fixed points as $\exp$, with different branches giving different fixed points.). On this branch, $L$ and $\bar{L}$ are fixed. Not sure how I could prepare a graph with the branch cut reoriented, however, because new branch points start to appear and I'm not at all sure how to handle all that cutting since there doesn't seem any easy way to "anticipate" what points will be branch points, anyway. tommy1729 Ultimate Fellow Posts: 1,493 Threads: 356 Joined: Feb 2009 02/17/2011, 11:56 PM ive been willing to post this before , but i was busy with number theory. although this might appear as a cheap spin-off , keep in mind that i have my clearly related " sinh method " for tetration. hence this brings us towards very similar questions : singularities and zero's of (2*sinh)^[1/2](z) ? since 2*sinh(z) has a nice unique fixpoint at 0 , i was hoping to get more results than just the analogues of the posts about the zero's and singularities of exp^[1/2](z). i also wonder how different / similar the plots look like. ( more formally , im intrested in the branch structures .. but i like nice pics too ) in fact , i think this may be more intresting than its somewhat random question appearance and might be 'key' in 'hacking' tetration. btw , Riemann surfaces are important and i wonder ; didnt Riemann write about tetration ? regards tommy1729 tommy1729 Ultimate Fellow Posts: 1,493 Threads: 356 Joined: Feb 2009 11/14/2012, 05:11 PM (This post was last modified: 11/14/2012, 05:15 PM by tommy1729.) Based upon ideas similar to my recent TPID 4 proof , and knesers sexp , I came close to a proof (imho) and now conjecture : There is a semistrip on the complex plane , parallel and symmetric to the real line , that includes $+oo$ with $-\pi/4 <$ $Im$ $< \pi/4$ such that exp^[1/2+eps](z) has no singularities for z in that strip. (of course eps > 0 and real as usual.) regards tommy1729 tommy1729 Ultimate Fellow Posts: 1,493 Threads: 356 Joined: Feb 2009 12/29/2015, 01:23 PM (12/08/2010, 02:24 PM)sheldonison Wrote: (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 I conjecture that - using mike3 branches - L and L* are the only singularities. Also i think 0.94 + 1.12 i is a singularity but on a nearby branch. Mike3 branches match very Well with the fake exp^[1/2]. So im intrested in the other branches ( and pics ) and also their fake analogues - although that fake should Maybe be discussed in another thread. I think it is justitied to search the sigularities by starting from re part and then going Up or down the branches. ( like sheldon did in part ) For negative real parts there are 3 lines to start with - using mike3 branches - , not sure how to decide. That's alot of " new " conjectures , although i had them for years actually. Just decided to post them now. Recently ( dec 2015 ) Sheldon started a thread about the singularity at L. I will only talk about that there. I wish you all a good 2016. Regards Tommy1729 « Next Oldest | Next Newest »