slog_b(sexp_b(z)) How does it look like ? sheldonison Long Time Fellow Posts: 626 Threads: 22 Joined: Oct 2008 04/27/2014, 04:52 AM (This post was last modified: 04/27/2014, 11:03 AM by sheldonison.) (04/26/2014, 08:53 AM)tommy1729 Wrote: [quote='sheldonison' pid='6865' dateline='1398462526'] Hmmm If we take + instead of - : ..... I changed - to + to avoid log branches. regards tommy1729 Because sexp(z) is analytic in the upper and lower halves of the complex plane, there is no ambiguity in log branches when looking at sexp(z), sexp(z-1), sexp(z-2), sexp(z-3), so no need to change - to +. In addition, the Taylor series gets more and more chaotic as z increases, so its easier to take slog(sexp(z-n)) for smaller z, and as long as the branches are the same, it does not change the result, except by subtracting n, so stop when the branches first differ, and then take slog(sexp(z)+2npi i). Then develop an intuitive explanation for the behavior of this function locally at z, the point in question. In particular, one can always find a path towards the fixed point of L, for sexp(z) as imag(z) increases, and then along this path, we are taking slog(L+delta+2npi i), which seems to be sufficient to show this solution does not have the property that slog(sexp(z+1))=slog(sexp(z))+1, along the path of increasing imaginary(z). The relevant equation would be as for normal sexp(z), where slog(sexp(z))=z, $\lambda=\ln(L)$, then as Im(z) increases, $\text{slog}(L+\lambda\delta) = \text{slog(L+\delta)+1+O\delta$, where the trick would be coming up with a rigorous equation for the error term. If the branches do line up, then my conjecture would be that the above approximation holds, and the solution will always be a multiple of the period as imag(z) goes to infinity. In that case, the solution would have the property that $\text{slog}(\text{sexp}(z))=z+\theta_n(z)+n\text{period}$, where theta is a 1-cyclic function, which goes to zero as $\Im(z)$ goes to infinity, and theta is uniquely determined by n. - Sheldon tommy1729 Ultimate Fellow Posts: 1,358 Threads: 330 Joined: Feb 2009 04/27/2014, 03:27 PM (04/27/2014, 04:52 AM)sheldonison Wrote: $\text{slog}(\text{sexp}(z))=z+\theta_n(z)+n\text{period}$, where theta is a 1-cyclic function, which goes to zero as $\Im(z)$ goes to infinity, and theta is uniquely determined by n. - Sheldon Lets say z is close to a real > 0 , then for most z : slog(sexp(z)) = z. This suggests that $theta_n(z))-n\text{period}$ is almost Always flat and not differentiable. This theta seems different from the periodic base change theta , and from the periodic sexp theta (the one that changes one C^oo solution of tetration into another one ). In other words - and no offense - but that is a very very general answer ; a family of functions without much details or properties and not like any seen before ( I think ). Also Im still unsure about alot of things. I blame myself mainly but my confusions are not resolved yet. Lets say n : theta depends only on n. But what does n depend on ? On z I guess , but how ?? This is probably equivalent to asking when to switch branches ? Somehow I guess that relates to the case when interpolation of iterations exp^[a/b](x_0) for a complex x_0 and integer a,b leads to intersecting points. ( we interpolate fractions by density to get app. continu line of reals ). And switching branches seems related. Second : Why L as ' invariant ' ? Why not 2pi i ? afterall iterations of exp(z) + 2pi i seems to suggest branches that differ by 2 pi i !? Like exp(u+2pi i) = sexp(slog(u+2pi i)+1) = v. Seeming giving a branch difference of 2pi i. Said otherwise log(sexp(a)) = sexp(a-1) or sexp(a-1) + 2pi i. Hence a branch difference of 2pi i ? Then again a branch difference of 2pi i is the log , not the sexp. All quite confusing id say. Or does " period " mean 2pi i here ? ( for base e) Is it just me, or is tetration tricky ? (Sometimes I believe I like tricky stuff, hence my intrest in tetration and number theory.) And what the ideas z1 - z2 = L or z1 - z2 = 2pi i ? But some good news too. I think your conjecture from 4/24 is correct and provable. Or partially provable. ( btw I seem to remember mick mentioning such a conjecture on MSE , i believe from you. ( I know you have MSE account ) But I forgot some of his arguments about it , did you discuss it with him ? or did we write anything uselfull about it ? I probably should have written things down :/ ) regards tommy1729 tommy1729 Ultimate Fellow Posts: 1,358 Threads: 330 Joined: Feb 2009 04/27/2014, 03:39 PM I came to reconsider an old idea ; Use more than one index for the branches. In its simplest general case something similar to : $f^{-1}(f^{1}(z)) = z + f_n(z) + g_m(z) + h_{nm}(z)$ Where m,n are integers that depend on z. Apart from maybe double periodic functions Im not sure when this is useful ... regards tommy1729 sheldonison Long Time Fellow Posts: 626 Threads: 22 Joined: Oct 2008 04/28/2014, 03:12 AM (This post was last modified: 04/29/2014, 01:31 PM by sheldonison.) (04/27/2014, 03:27 PM)tommy1729 Wrote: (04/27/2014, 04:52 AM)sheldonison Wrote: $\text{slog}(\text{sexp}(z))=z+\theta_n(z)+n\text{period}$, where theta is a 1-cyclic function, which goes to zero as $\Im(z)$ goes to infinity, and theta is uniquely determined by n. - SheldonSecond : Why L as ' invariant ' ? Why not 2pi i ? afterall iterations of exp(z) + 2pi i seems to suggest branches that differ by 2 pi i !?Tommy, There's a lot of misconceptions in your post; you might want to go back and look at the threads where we went over the Kneser construction, which starts with the formal Schroeder equation. When you use the formal Schroeder equation to generate f^n, you get a solution that has a period of $\frac{2\pi i}{\ln(\lambda)}$. For base e, the period~=4.447+1.058i. A case you are familiar with is iterating 2sinh, where $\lambda=2; \; \lambda$ is the derivative of f(L+z) at the fixed point, where f(L)=L. For $f^{oz}$ for iterating $f=b^z, \; \lambda=\ln(L)$. For base e, L~=0.318+1.337i. For base e, $\lambda=\ln(L)=L$, though for base bases lambda<>L. For the inverse Abel function generated from the formal Schroeder function, then iterating $f^{o z} = L+ S(\lambda^z) \approx L+\lambda^z$, where $S=x+a_2 x^2 + a_3 x^3...$ is the formal inverse Schroeder function. For our case for b>e^(1/e), $f(z)=b^z$, and f^z approaches L as imag(z) increases, and as real(z) decreases. For our case, f^z is not real valued at the real axis. To get a real valued function, we do a Riemann/theta mapping to generate sexp(z). Because theta(z) very quickly decays to a constant as imag(z) increases, you can see the pseudo periodicity for sexp(z) as imag(z) increases in Dimitrii's complex plane plots, or my plot of the inverse Abel function and sexp. This link/picture also tries to show the incredible complexity of the singularity at z=0 for Abel(sexp(z)), for Kneser's sexp(z) solution, which Tommy alludes to in some of his comments. This last plot also shows the contours of theta(z)+z, which would be similar for slog(sexp(z)), n>0; see the n=1 period graph below. The program I have posted to calculate Kneser's sexp(z) solution starts with this inverse Abel function, and converges to the inverse Abel function as imag(z) increases. Finally, here is the contour graph for n=1, slog(sexp(-1.5+i1E-6..0.5+i1E-6))=z+theta1(z)+period, where the solution is 1-period away from the origin, for base e, for z just above the real axis, where $\Im(z)=i10^{-6}$. As you can see, the function is not "almost always flat", and instead resembles the theta(z) mapping for Kneser's solution, just as I predicted.     - hope this helps, Sheldon Levenstein tommy1729 Ultimate Fellow Posts: 1,358 Threads: 330 Joined: Feb 2009 04/28/2014, 09:01 PM (This post was last modified: 04/28/2014, 09:26 PM by tommy1729.) Dear Sheldon Its true that I have to read some old threads again. But I think there are a few misunderstandings in our communication. Forgive me for responding quicky without thinking deeply but time is Always against me. I think its easiest to consider base "e" only for the moment being. I have to think a bit more before I adress everything. But lets let me start with this one : ------------------------------------------------------------------------- Do you agree that sexp(z) + 2pi i is another branch of sexp(z) ?? ------------------------------------------------------------------------- I assume the logic of that idea is clear , even if its wrong. Like if exp(sexp(u)+2 pi i) = exp(sexp(u)) = sexp(u+1) then it seems logical to conclude sexp(z) + 2pi i is another branch of sexp(z). Also if sexp(z) + 2pi i is another branch of sexp(z) for some z near B, then by analytic continuation ( valid on a subset of a branch without singularities ) it must be so almost everywhere on that branch. A related argument is : slog(exp(z)) = slog(exp(z+ 2pi i)) = slog(z)+1 although this equation is not satisfied everywhere. [ more on this in another thread : ( http://math.eretrandre.org/tetrationforu...80#pid6880 ) although Im stuck there. Also the math below hints why its not possible that the equation holds everywhere ] Another idea is that the values (range) of one branch relates to another in a logical way : the fundamental branch that maps the positive reals to the positive reals and the solution L of exp(L) = L is at - oo , + oo i , - i oo. Or more accurately like you said ; approaches L as imag(z) increases, and as real(z) decreases. (ok I ignored the complex conjugate , im well aware of it ) The idea is that the solution L(2) that is different from L and satisfies exp(exp(L(2)) = L(2) cannot be on the same branch as the fundamental one. And thus values close to the reals or L are on the fundamental branch and values close to L(2) are on another. By analogue one can construct L(n) and have n branches. The reasons come from the slog rather than the sexp : slog(exp(z)) = slog(z)+1 this cannot hold for z=L slog(exp(exp(z)) = slog(z) + 2 ( by induction ) this cannot hold for z=L(2) I hope it is clear why ! This suggests that L(2) is not on the same branch because we use solutions to tetration that have a minimal amount of singularities for both slog and sexp. ( I assumed the L(n) do not form a natural boundary otherwise the idea is in trouble ) Notice I did not yet claim for instance that L(n) represents the n th branch. Or for instance that z + L(a) - L(b) is another branch. But these ideas need to be adressed. regards tommy1729 sheldonison Long Time Fellow Posts: 626 Threads: 22 Joined: Oct 2008 04/28/2014, 11:33 PM (This post was last modified: 04/28/2014, 11:43 PM by sheldonison.) (04/28/2014, 09:01 PM)tommy1729 Wrote: ------------------------------------------------------------------------- Do you agree that sexp(z) + 2pi i is another branch of sexp(z) ?? ------------------------------------------------------------------------- ....Like if exp(sexp(u)+2 pi i) = exp(sexp(u)) = sexp(u+1) then it seems logical to conclude sexp(z) + 2pi i is another branch of sexp(z). So for example, sexp(1)=e, is there a sense that sexp(0)=log(sexp(1))+2pi i = 1+2pi i? Not if we stay in the upper half of the complex plane, where there are no singularities. There are no degrees of freedom in picking the logarithmic branch for sexp(z)=log(sexp(z+1)), since we analytically continue, smoothly from sexp(z+1) to sexp(z). If sexp(1)=e, then we cannot analytically continue to sexp(0)=1+2pi i unless we circle a logarithmic singularity. So, can we get to a place on the Riemann surface where sexp(0)=1+2pi i? Well, it just so happens that sexp(z) has a logarithmic singularity at z=-2. And if you start at sexp(0)=1, and circle counterclockwise around the singularity, then you pass through the region between sexp(-3) and sexp(-2). In this segment, imag(z)=pi I in the upper half of the complex plane and imag(z)= -pi i from the lower half of the complex plane. So, we can go through this region, and then back to sexp(0). And then we are in a strange branch of sexp(z); which I will call sexp_l(z) for logarithmic. Now, sexp_l(0) = 1 + 2 pi i. sexp_l(1)=e + 2pi i. The real axis of sexp_l is not real valued, but real valued + 2pi i. In the sexp_l branch, the defining equation is sexp_l(z+1)= exp(sexp_l(z)-2 pi i) + 2 pi i. I'm quite sure this is not what Tommy has in mind ... but yeah, it just so happens to be right there on the Riemann surface. But other than that really weird branch of sexp(z), if you avoid the real axis less then or equal to -2, where all the singularities are, then nope, it has no meaning to talk about sexp(0)=1+2pi i. And if you are talking about that wierd branch, one should be crystal clear that you are referring to that part of the function where sexp(z+1)<>exp(sexp(z)). Now, back to the main branch. slog(1+2pi i) ~= 1.9433573+0.66346938i. But this is also completely unrelated to the concept of a branch of sexp(1). - Sheldon tommy1729 Ultimate Fellow Posts: 1,358 Threads: 330 Joined: Feb 2009 04/29/2014, 11:52 AM Agreed ! So the next question is : Does your n th theta function equal the result obtained by going through the ln^[n] branch of sexp in a similar manner as the previous post ? regards tommy1729 « Next Oldest | Next Newest »