Thread Rating:
• 0 Vote(s) - 0 Average
• 1
• 2
• 3
• 4
• 5
 slog_b(sexp_b(z)) How does it look like ? tommy1729 Ultimate Fellow Posts: 1,370 Threads: 335 Joined: Feb 2009 04/23/2014, 09:15 PM Forgive me if this is an old question or it could be easily derived from previous posts. Or if my memory fails me. But despite many posts on slog_b(z) and sexp_b(z) , I am somewhat puzzled by slog_b(sexp_b(z)). For clarity b is the base and in particular I am (mainly) intrested in bases larger than exp(1/2). Also to avoid confusion slog_b(sexp_b(z)) =/= z. This is the tetration analogue of log_b(exp_b(z)) =/= z. As often in the topic of tetration I feel " lured into deception ". By that I mean that many ideas pop up naturally but none of them seem very solid , intresting or true in retrospect. Also many variations on this question occur naturally , usually in the form of special cases and restrictions. Such as Re(z) > 0. Looking at changing values of b , rather than say z. Or considering different solutions to sexp and how that afffects things. A few notes : 1) analytic continuation is problematic since for REAL z , the equation does in fact hold ! ( and continuation of z = z of course ). This also makes me distrust Taylor series and related calculus. 2) my 2sinh fails here or is meaningless since it is not analytic ! Well at least I think so , because it is then imho not well defined for nonreal imput (z). 3) the chaotic nature of interations z , exp(z) , exp^[2](z) , ... is one of the reason I doubt my own intuition and I feel " lured into deception ". 4) I could probably link or rewrite almost all threads , questions and answers as a note here , since I find that most threads relate to this in one way or another ! 5) One of the " deceptions " was a " nonentire analytic function with all singularies at oo " what makes no sense of course. I however have no systematic way of doing things that automatically avoids such " nonsense ". Maybe Im not the only one who wonders about this. Some of my friends believe this requires new special functions to fully understand. I considered using higher dimensional numbers but that seems like running before you can walk. So I guess the complex plane is the place to start. regards tommy1729 sheldonison Long Time Fellow Posts: 641 Threads: 22 Joined: Oct 2008 04/23/2014, 10:46 PM (This post was last modified: 04/23/2014, 10:59 PM by sheldonison.) (04/23/2014, 09:15 PM)tommy1729 Wrote: ....I am somewhat puzzled by slog_b(sexp_b(z)). For clarity b is the base and in particular I am (mainly) intrested in bases larger than exp(1/2). Also to avoid confusion slog_b(sexp_b(z)) =/= z. This is the tetration analogue of log_b(exp_b(z)) =/= z.Hey Tommy, That is interesting actually. So, for $\log(\exp(z))=z+2n\pi i$ Is that what you have in mind? If so, there is a direct analog since tetration itself is Pseudo periodic, except, of course, as you approach the real axis, the Pseudo periodic part breaks down into approximately what I would call the theta mapping. The theta mapping is exactly 1-periodic. Consider the very similar 1-cyclic case, $\theta(z)=\text{Abel}_b(\text{sexp}_b(z))$, where the Abel function is from the Schroeder function fixed point. As imag(z) increases, theta goes to a constant. Of course, any solution of theta(z)+Period would be valid. And of course, there is a nasty singularity at the real axis, that is mathematically equivalent to the Kneser Riemann mapping. Before I go digging up an old plot of the 1-cyclic mapping for abel(sexp), is that close to what you are looking for? Another very interesting function is slog_a(sexp_b(z)), where a and b are different tetration bases, where both bases are greater than eta. Also, slog(2sinh(z)), is related. - Sheldon tommy1729 Ultimate Fellow Posts: 1,370 Threads: 335 Joined: Feb 2009 04/23/2014, 11:36 PM (This post was last modified: 04/23/2014, 11:40 PM by tommy1729.) Hi sheldon the pseudoperiodic property has been on my mind, but still I havent figured things out as much as I desire. As for " if im looking for what you say " : kinda. For me log(exp(z)) = z for -2pi z, where it is not 1-cyclic. (04/23/2014, 11:36 PM)tommy1729 Wrote: What is it about slog(2sinh(z)) ?? Does this relate to things said before ?I was thinking along the lines that there are a whole family of analytic iterated functions that grow super-exponentially, such that in the limit as z gets larger, $\lim_{z \to \infty} \text{slog}(f^z)=\text{slog}(f^{z+1})-1$ For example,iterated gamma function also falls into this category, as does 2sinh^z, as does sinh^z, and other bases for tetration. The conjecture is that for these functions, taking the limiting behavior of the slog(f^z)-z is a repeating 1-cyclic function, that is c infinity and nowhere-analytic, which I find fascinating. tommy1729 Ultimate Fellow Posts: 1,370 Threads: 335 Joined: Feb 2009 04/24/2014, 10:30 PM Well that seems alot like the base change to me at first glance. Afterall slog is similar to taking many logs and sexp is similar to taking many exponents. Kinda. Since many believe both 2sinh method and base change are C^oo but not analytic that makes sense too. Im not very confident about how to prove periodicity of functions that are not analytic ! As often " nonanalytic calculus " has less " tools " then " analytic calculus " so I find it harder. Has the periodicity for the base change been confirmed ? Or is it not ? I might have to look at base change again and maybe advise you the same... I think a proof might depend on stuff said about the base change. I had that idea once too. Not sure why I forgot about it. Maybe doubts. It seems like a nice thing to investigate. Although its a big deviant from the Original Problem since it is nowhere analytic ( though I see the connection ). I have some ideas to find places to read up on this, so when I have the time I might say more. regards tommy1729 tommy1729 Ultimate Fellow Posts: 1,370 Threads: 335 Joined: Feb 2009 04/24/2014, 10:33 PM I said : " Im not very confident about how to prove periodicity of functions that are not analytic ! " Plz do not respond with just simply saying something along the line of : " use fourier !! " without further explaination. ( that would be lame ! ) thank you. regards tommy1729 tommy1729 Ultimate Fellow Posts: 1,370 Threads: 335 Joined: Feb 2009 04/24/2014, 10:43 PM Btw @sheldon : very nice link. Very nice induction I love induction regards tommy1729 sheldonison Long Time Fellow Posts: 641 Threads: 22 Joined: Oct 2008 04/25/2014, 04:54 PM (This post was last modified: 04/25/2014, 10:50 PM by sheldonison.) (04/24/2014, 10:30 PM)tommy1729 Wrote: Im not very confident about how to prove periodicity of functions that are not analytic ! As often " nonanalytic calculus " has less " tools " then " analytic calculus " so I find it harder. Has the periodicity for the base change been confirmed ? Or is it not ?I haven't seen or written a rigorous proof of coo nowhere analytic functions, but I am working on a rigorous proof for a related coo nowhere analytic problem, iterated logarithm of x^^n. I still have some gaps, but I know the basic flow of the proof, and I will post it sometime in the next few months. As to your original question, For f=b^z, with fixed point L=b^L, $b^{L+\delta}=L\times(1+\delta\ln(L)), \lambda=\ln(L), \text{period}=\frac{2\pi i}{\ln(\lambda)}$ So, there if $\alpha(z)$ is the Abel function for f=b^z, then the iterated function from the Schroeder equation fixed point can be represented generically as follows $\alpha^{-1}(z) = f^z = L + \sum_{n = 1}^{\infty} a_n\exp( z \ln(\lambda))$ As $\Im(z)$ gets aribrarily large, the function can be well approximated by $\alpha^{-1}(z) \approx L+a_1\exp\(z \ln(\ln(L)))$. As imag(z) increases, $\text{sexp}_b(z)\approx \alpha^{-1}(z+k+n\text{Period})$, where we can solve for k easily enough. The key is that the pseudo period for sexp(z), 4.4+1.05i for base e, is much larger than the period of the 1-cyclic kneser $\theta$ mapping, so the 1-cyclic mapping converges to the constant k much quicker than sexp(z) converges to L. $\text{sexp}_b(z)=\alpha^{-1}(z+ n\text{Period}+\theta(z))$ So $\alpha(\text{sexp}(z)) = z + \theta(z)$ If the slog(sexp) is well behaved as imaginary(z) gets larger, decaying to a constant +z, than the answer to Tommy's original question is as follows, noting that theta(z) is analytic in the upper half of the complex plane, with singularities at the real axis. $\text{slog}(\text{sexp}(z)) \approx z + k + n\text{Period} + \theta (z)$ - Sheldon sheldonison Long Time Fellow Posts: 641 Threads: 22 Joined: Oct 2008 04/25/2014, 10:48 PM (This post was last modified: 04/26/2014, 01:04 AM by sheldonison.) (04/25/2014, 04:54 PM)sheldonison Wrote: As to your original question, I found a much simpler way of thinking about it. For some cases, we conjecture that $f(z) =\text{slog}(\text{sexp}(z)) = z + \theta(z)$, where theta(z) is a well defined 1-cyclic function as imag(z) increases, but has singularities at the real axis. So, z2=slog(sexp(z1)). Now look at the sequence of points. If these are all equal to zero, then we can conjecture that we have a 1-cyclic solution, since sexp(z) goes to the fixed point of L as real(z) goes to minus infinity. sexp(z1-1)-sexp(z2-1) sexp(z1-2)-sexp(z2-2) .... sexp(z1-n)-sexp(z2-n) If they're not all equal, than there will be some lowest "n", for which $\text{sexp}(z1-n) + 2m\pi i = \text{sexp}(z2-n)$ At the point z1-n, we solve the function $f(z) =\text{slog}(\text{sexp}(z)+2m\pi i)$, which looks nothing like the 1-cyclic solution, due to the mismatching branches. The solution is only well behaved locally, near $f(z) =\text{slog}(\text{sexp}(z)+2m\pi i)$, where f(z) doesn't vary much at all as z increases to some medium sized imaginary number, and then z increases by a Period, all while f(z) hardly changes at all. This is clearly a very different function mostly uninteresting solution, than $f(z) =\text{slog}(\text{sexp}(z)) \approx z + \theta(z)$. - Sheldon tommy1729 Ultimate Fellow Posts: 1,370 Threads: 335 Joined: Feb 2009 04/26/2014, 08:53 AM (04/25/2014, 10:48 PM)sheldonison Wrote: (04/25/2014, 04:54 PM)sheldonison Wrote: As to your original question, I found a much simpler way of thinking about it. For some cases, we conjecture that $f(z) =\text{slog}(\text{sexp}(z)) = z + \theta(z)$, where theta(z) is a well defined 1-cyclic function as imag(z) increases, but has singularities at the real axis. So, z2=slog(sexp(z1)). Now look at the sequence of points. If these are all equal to zero, then we can conjecture that we have a 1-cyclic solution, since sexp(z) goes to the fixed point of L as real(z) goes to minus infinity. sexp(z1-1)-sexp(z2-1) sexp(z1-2)-sexp(z2-2) .... sexp(z1-n)-sexp(z2-n) If they're not all equal, than there will be some lowest "n", for which $\text{sexp}(z1-n) + 2m\pi i = \text{sexp}(z2-n)$ At the point z1-n, we solve the function $f(z) =\text{slog}(\text{sexp}(z)+2m\pi i)$, which looks nothing like the 1-cyclic solution, due to the mismatching branches. The solution is only well behaved locally, near $f(z) =\text{slog}(\text{sexp}(z)+2m\pi i)$, where f(z) doesn't vary much at all as z increases to some medium sized imaginary number, and then z increases by a Period, all while f(z) hardly changes at all. This is clearly a very different function mostly uninteresting solution, than $f(z) =\text{slog}(\text{sexp}(z)) \approx z + \theta(z)$. - Sheldon Hmmm If we take + instead of - : sexp(z1+1)-sexp(z2+1) sexp(z1+2)-sexp(z2+2) .... sexp(z1+n)-sexp(z2+n) Then by the locally analytic property it seems we have a periodic function sexp with period z2 - z1 ?? This implies z2 - z1 = L ? And that suggests slog(sexp(z)) = z + z2 - z1 = z + L ? Something is weird here. I changed - to + to avoid log branches. regards tommy1729 « Next Oldest | Next Newest »

Users browsing this thread: 1 Guest(s)