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RE: using sinh(x) ? - sheldonison - 07/21/2010
(07/20/2010, 08:31 PM)tommy1729 Wrote: how do you know the superfunction of 2sinh is periodic in the imaginary direction ?It is interesting. The period is based on the limiting behavior of 2sinh at its fixed point of zero, where the slope=2. If -real(z) is large enough, then 2^z becomes an excellent approximation of the SuperFunction of 2sinh. The limit equation for the superfunction of 2sinh in the complex plane is: Since 2^z is periodic in the imaginary, with period = i*2pi/ln(2), then the SuperFunction of 2sinh is also periodic with that period. This leads to some interesting behavior (pointed out in earlier posts). The SuperFunction grows superexponentially negative at img(z)=i*pi/ln(2), with img(f)=0. At img(z)=0.5i*pi/ln(2), the SuperFunction has real(f)=0, and converges to an imaginary fixed point. - Sheldon RE: using sinh(x) ? - tommy1729 - 07/21/2010
sheldon , you write " if -real(z) is large enough ..." do you mean if real(z) is a large negative number ? but then it isnt periodic for all real(z) so its rather " pseudoperiodic ". then we approximate periodicity in the section with large negative real parts in the imaginary direction. weird , in comparison with kouznetsov sexp which approximates periodicity in the section with large imaginary parts in the real direction. thats the same property rotated by 90 degrees ?? hmm wouldnt by that logic almost every superfunction be periodic ?? i mean => af = f ' (0) * f(x) lim n-> oo af^[n] ( a^(z-n) ) real(z) << -10 since a^z is periodic in direction Q , with period i*2pi/ln(a) , then the superfunction of af is also periodic with that period. ... by analogue ... ???? sorry if im mistaken in advance. regards tommy1729 RE: using sinh(x) ? - sheldonison - 07/22/2010
(07/21/2010, 10:34 PM)tommy1729 Wrote: hmmMostly correct, with period i*2pi/ln(a). But not true for every superfunction. Knesser's solution for sexp_e is not periodic, but is pseudo periodic. It starts with the superfunction, which is complex valued at the real axis, and then does a conformal mapping, to put real values back on the real axis, along with a Schwarz reflection. For i(z)>=1i, sexp_e is fairly close to the (linearly shifted) complex valued superfunction. You can see the complex pseudo periodicity in the contour graphs in Kouznetsov's paper on the sexp_e. - Sheldon RE: using sinh(x) ? - tommy1729 - 07/28/2010
exp(x) plotted in red. ln ln ln 2sinh exp exp exp (x) plotted in black. this is 3 iterations of my formula for the approximation of exp(x). as you can see that is already pretty good. it is clear that for large x the approximation with my formula is better than for smaller x. so i plotted a section of negative x. regards tommy1729 RE: using sinh(x) ? - tommy1729 - 07/28/2010
as said before exp exp exp ... (z) does *not converge* in a neighbourhood of z when im(z) =/= 0. i believe my algoritm is coo on the reals and by analytic continuation / mittag leffler expansion we can construct " the " (imho i.e. "my") sexp. by * not converge* i mean it is chaotic , since of course exp exp ... (2) also diverges. this is so because e^(x + yi) = e^x ( cos y + sin y i ) and large y can become arbitrary close to a multiple of 2pi. although analytic continuation solves the problem - assuming it works and it has period 2pi i - i would like to investigate further the behaviour of the exp iteration. for instance , is it true that the neighbourhood of z always contains exactly one non-chaotic value ? call it hp. assuming we can find hp for every z as a limiting sequence going to zero in the correct way. is it then possible to extend my formula to : exp^[z1,z] = ln ln ln ... 2sinh^[z] exp exp exp ... ( z1 + hp(z1) ) regards tommy1729 RE: using sinh(x) ? - tommy1729 - 12/04/2010
to compute a taylor series , we need to know that it is justified. concretely that means we need to prove that , where D^m is the mth derivative with respect to x(*) , holds. ( with respect to z should give the same result , but seems harder at first sight ) analytic continuation preserves periodicity , which is 2pi i. that is also the reason why we apparantly cant extent this method to bases between eta and e^(1/2). ( hints : fourier , entire , think about it ) a further remark and actually request is that i would love to see a plot of all the solutions , including branches of exp(x) = t exp(exp(x)) = t ... exp^[n] = t where t = 0 +/- 1.895494239i is one of the nonzero fixpoint of 2sinh. it might give me new insight or ideas. thanks in advance. regards tommy1729 RE: using sinh(x) ? - tommy1729 - 12/08/2010
(12/04/2010, 11:09 PM)tommy1729 Wrote: a further remark and actually request is that i would love to see a plot of all the solutions , including branches of any volunteers ? RE: using sinh(x) ? - sheldonison - 01/17/2011
(12/08/2010, 02:31 PM)tommy1729 Wrote:I am re-posting this reply after fixing a minor error. There is still some interest in the TommySexp solution (wikipedia talk), which as suggested previously, is infinitely differentiable, but probably nowhere analytic at the real axis. The goal of this post is to show the location of the singularities that limit the radius of convergence, for n=3 and n=4, to bolster the argument that the TommySexp function is nowhere analytic. The singularities occur wherever the 2sinh superfunction(z) has a value of 0 + 2nPi*I, where n is a positive integer.(12/04/2010, 11:09 PM)tommy1729 Wrote: a further remark and actually request is that i would love to see a plot of all the solutions , including branches Start with the n=3 case, renormalized with a bias=0.06783836607, so that TommySexp(0)=1. n=3 is sufficiently large to get more than double precision accurate results at the real axis, centered around z=0. Here is a plot of TommySexp(z) with a radius of 0.45, with z centered at the origin, for the n=3 case, which requires three logarithms. For this plot with red=real, and green=imaginary. For this plot, n=3, with z=0.45*exp(I*theta), plotting TommySexp(z). Inside this region of the complex plane, the TommySexp function is analytic, with no singularities. But now, we increase the sample radius merely 0.008, from 0.450 to 0.458. And now the TommySexp function becomes poorly behaved, because there are singularities inside the circle. Following this path to theta(Pi*I), the imag(TommySexp(-0.458 )) is no longer even zero, due to the singularities! This means the function's value has becomes dependent on which path around the singularity is chosen. Again, red is real, and green is imaginary. What's going on, is that there are singularities, so that the TommySexp function has a limited radius of convergence. Next, I post the first fifty singularities of TommySexp, for the n=3 case. The fourth singularity is closest to the origin (after iterating three logarithms), with an absolute value of 0.457, which is why the second plot of the TommySexp(z), centered at the origin, with a radius=0.458 is poorly behaved. Code: `First fifty singularities` For an example, consider the fourth singularity on the list. Let z= 2.285107737 + 0.3580209487*I superf2sinh(z)=25.13274123*I=8*Pi*I. superf2sinh(z+1)=0. Now iterate the logarithm of superf2sinh(z+1), three times. We get a singularity on the first iteration. The logarithm, on a path circling the singularity gives different results, depending on the path. The logarithm of the logarithm of a singularity is also a singularity, which explains the second graph. All of these particular singularities line on a contour line, where real(superf2sinh(z))=0. So, now I plot this curve (red), superimposing on it a radius (z=0.458 ) half circle, showing how the sampling circle now includes the first singularity. This curve also shows that for the n=4 case, the radius of convergence is much much smaller still. It turns out the 600 thousandth singularity is at z=4.0025+0.0347i, so the radius of convergence for n=4 drops to 0.035. Using this definition, as n (number of iterated logarithms used to generate TommySexp) grows larger, the radius of convergence for TommySexp(z) at the origin gets arbitrarily small. Also, as the singularities approach the real axis, they also get arbitrarily close together. Thus, even though it can be shown that all of TommySexp's derivatives at the real axis are continuous, it is nonetheless probably nowhere analytic, just like the base change sexp definition discussed in an earlier post. - Sheldon RE: using sinh(x) ? - tommy1729 - 01/17/2011
thanks for your reply sheldon. in fact my method is intended for the real line. for the complex plane it will not work * in its limit form *. i fact , i wont converge for most nonreal numbers * in its limit form *. ( fixpoints L and L* will work ) to sketch some of the reasons , apart from yours , exp exp exp ... exp(z) does not converge for the neighbourhood of any nonreal z. the identity like functions that commutes with * the limit form * is id(x) = log log ... id( exp exp ... ) however x = log log exp exp (x) but this does not hold for complex z z =/= log log exp exp (z) which has ofcourse great consequences ( again for the limit form ! ) i dont have much time , but i think i made some ideas clear. although slightly on different paths , i think our ideas will merge. i think i can show that my limit form can be transformed to work for all of z. later when i have more time. i had some intuition about those singularities , so it doesnt surprise me. on the other hand , we might be able to learn more about them , and i thank sheldon for the post and pics. regards tommy1729 ps : im still thinking about the base change too , despite not personally appealing to me and ( imho ? ) missing important properties ...[/font] RE: using sinh(x) ? - tommy1729 - 07/02/2011
i can now prove a positive radius in x when expanded at certain points and Coo for real z. i assume that can be strengthened to analytic in both x and z with some effort (and perhaps a good book). ( see lemma 1 * add link later * ) however i wont go into details here yet , im currently more intrested in other things about tetration and perhaps this should make a good paper. in particular i still dont know why 1.729 i ^1.729 i ^ ... 0.5 + 0.5 i gives a circle and that bothers me. |