• 0 Vote(s) - 0 Average
• 1
• 2
• 3
• 4
• 5
 Kneser's Super Logarithm bo198214 Administrator Posts: 1,386 Threads: 90 Joined: Aug 2007 01/26/2010, 11:19 PM (This post was last modified: 01/26/2010, 11:20 PM by bo198214.) I can not yet follow you: (01/23/2010, 01:01 PM)sheldonison Wrote: $\chi(z)=\lim_{n\to\infty}( \exp^{\circ n}(f+f^{(z-n)})$, at least I think that is what is meant by $\chi(z)$. I think the formula is corrupt, the correct formula should be: $\chi(z)=\lim_{n\to\infty} \frac{\log^{\circ n}(z) -c}{c^{-n}}$. edit: oh now I see: you call the fixed point $f$! Please avoid, $f$ is reserved for functions. Kneser calls it $c$, in this thread I keep with his convention. edit: well now I also see that you mean the inverted Schröder function. As we can derive from the previous formula we get: $\chi^{-1}(z)=\lim_{n\to\infty}\exp^{\circ n}(z c^{-n} +c )$ No, this still is not your formula, I guess you mean the inverse of the Abel function $\psi(z)=\log_c(\chi(z))$ this would have the formula: $\psi^{-1}(z)=\lim_{n\to\infty}\exp^{\circ n}(c^{z-n} +c )$ which is finally your formula Quote:Is the real valued sexp at the real axis connected to the Schroeder equation, chi, by a complex 1-cyclic function, $\theta(z)-z$ $\text{sexp}_e(z)=\chi(\theta(z))$ The Abel function $\psi$ maps $H_0$ to some region $X_0$. But our final Abel function $\Psi$ shall map $H_0$ into the upper halfplane, correspondingly also $H_1=\exp(H_0)$ to $X_0+1$ and $H_{-1}=\log(H_0)$ to $X_0-1$. Thatswhy we consider the Riemann mapping $\rho$ that biholomorphically maps the union of the by integer translated $X_0$ to the upper halfplane. (though is this marginally different from Kneser's original approach I think it is better accessible.) Then $\Psi(z)=\rho(\psi(z))$ is the wanted Abel function (slog), and we obtain sexp as: $\operatorname{sexp}(z)=\psi^{-1}(\rho^{-1}(z))$ Quote:Where $\theta(z)$ is the Riemann mapping of contour of the limit of iterating the natural logarithm, starting with the real interval, 0..1 [color=#0000CD](which corresponds to sexp(z) from z=0 to z=1). Not sure what contour exactly you mean, but I think we are close already. Your theta should be the Riemann mapping that maps the upper halfplane to the union of the by integer translated images $\psi(H_0)$. Please continue from here with unified denotation. mike3 Long Time Fellow Posts: 368 Threads: 44 Joined: Sep 2009 01/27/2010, 06:28 PM Ah. What do you mean by the Fourier series "exponentially decays to a constant as i increases"? Do you mean the coefficients of the series (in which case how could it converge?)? Do you mean the behavior toward imaginary infinity (if so, does it decay to the two principal fixed points of exp)? sheldonison Long Time Fellow Posts: 631 Threads: 22 Joined: Oct 2008 01/27/2010, 07:51 PM (01/26/2010, 11:19 PM)bo198214 Wrote: .... Your theta should be the Riemann mapping that maps the upper halfplane to the union of the by integer translated images $\psi(H_0)$. Please continue from here with unified denotation.I appreciate the educational update on the notation. Your description in this post of Kneser's approach already makes a lot more sense, and I will continue working on it until I fully understand it. I also updated the notation in my reply (with the graph) using $\psi^{-1}(z)$ as sexp developed from the fixed point, and $\psi(z)$ as the Abel function (inverse sexp, based on the fixed point). - Shel sheldonison Long Time Fellow Posts: 631 Threads: 22 Joined: Oct 2008 01/27/2010, 08:30 PM (This post was last modified: 01/28/2010, 11:40 AM by sheldonison.) (01/27/2010, 06:28 PM)mike3 Wrote: Ah. What do you mean by the Fourier series "exponentially decays to a constant as i increases"? Do you mean the coefficients of the series (in which case how could it converge?)? Do you mean the behavior toward imaginary infinity (if so, does it decay to the two principal fixed points of exp)?I mean the 1-cyclic repeating function $(\theta(z)-z)$ exponentially decays to a constant as $\Im(z)$ increases. After the Riemann mapping, it's individual Fourier series terms all decay to zero as imaginary increases. The terms grow as we approach the real axis, and at the real axis, it has singularities at integer values. The complex sexp developed from the fixed point (the inverse Abel function), $\psi^{-1}(z)$ already goes to the fixed point as $\Im(z)$ increases, and as $Re(z)$ decreases. It would be nice to have a graph of $\psi^{-1}(z)$, which becomes super-exponential (complex only) as $\Re(z)$ increases, and as $\Im(z)$ decreases, and goes to the fixed point as $\Im(z)$ increases, and as $\Re(z)$ decreases. Finally, $\operatorname{sexp}_e(z)=\psi^{-1}(\theta(z))$, so as as $\Im(z)$ increases, the sexp_e(z) will converge to $\psi^{-1}(z+k)$, where k is a small constant. - Shel mike3 Long Time Fellow Posts: 368 Threads: 44 Joined: Sep 2009 01/28/2010, 08:52 PM (01/27/2010, 08:30 PM)sheldonison Wrote: (01/27/2010, 06:28 PM)mike3 Wrote: Ah. What do you mean by the Fourier series "exponentially decays to a constant as i increases"? Do you mean the coefficients of the series (in which case how could it converge?)? Do you mean the behavior toward imaginary infinity (if so, does it decay to the two principal fixed points of exp)?I mean the 1-cyclic repeating function $(\theta(z)-z)$ exponentially decays to a constant as $\Im(z)$ increases. After the Riemann mapping, it's individual Fourier series terms all decay to zero as imaginary increases. The terms grow as we approach the real axis, and at the real axis, it has singularities at integer values. The complex sexp developed from the fixed point (the inverse Abel function), $\psi^{-1}(z)$ already goes to the fixed point as $\Im(z)$ increases, and as $Re(z)$ decreases. It would be nice to have a graph of $\psi^{-1}(z)$, which becomes super-exponential (complex only) as $\Re(z)$ increases, and as $\Im(z)$ decreases, and goes to the fixed point as $\Im(z)$ increases, and as $\Re(z)$ decreases. Finally, $\operatorname{sexp}_e(z)=\psi^{-1}(\theta(z))$, so as as $\Im(z)$ increases, the sexp_e(z) will converge to $\psi^{-1}(z+k)$, where k is a small constant. - Shel So then it would seem to agree with the Kouznetsov function, then, wouldn't it, i.e. the $\operatorname{tet}_e(z)$ function developed this way decays to approximately $0.318 \pm 1.337i$ as $z \rightarrow \pm i \infty$? Hmm. Does this Kneser method work for other bases, too? Can it be used at a complex base, e.g. $2 + 1.5i$? sheldonison Long Time Fellow Posts: 631 Threads: 22 Joined: Oct 2008 01/28/2010, 10:08 PM (01/28/2010, 08:52 PM)mike3 Wrote: So then it would seem to agree with the Kouznetsov function, then, wouldn't it, i.e. the $\operatorname{tet}_e(z)$ function developed this way decays to approximately $0.318 \pm 1.337i$ as $z \rightarrow \pm i \infty$? Hmm. Does this Kneser method work for other bases, too?Yes, and Yes. I don't think anyone's tried it though. Quote:Can it be used at a complex base, e.g. $2 + 1.5i$? You could always generate the inverse Abel function from the fixed point for the complex base. But if f(z) is real valued, f(z+1) would have a complex value, so the Kneser mapping couldn't convert the Abel function into a real valued tetration. But it seems like it could generate an analytic complex base tetration where sexp(-1)=0, sexp(0)=1, sexp(1)=2+1.5i, sexp(2)=(2+1.5i)^(2+1.5i) and sexp(-2,-3,-4...)=singularity.... mike3 Long Time Fellow Posts: 368 Threads: 44 Joined: Sep 2009 01/29/2010, 06:43 AM (01/28/2010, 10:08 PM)sheldonison Wrote: (01/28/2010, 08:52 PM)mike3 Wrote: So then it would seem to agree with the Kouznetsov function, then, wouldn't it, i.e. the $\operatorname{tet}_e(z)$ function developed this way decays to approximately $0.318 \pm 1.337i$ as $z \rightarrow \pm i \infty$? Hmm. Does this Kneser method work for other bases, too?Yes, and Yes. I don't think anyone's tried it though. Quote:Can it be used at a complex base, e.g. $2 + 1.5i$? You could always generate the inverse Abel function from the fixed point for the complex base. But if f(z) is real valued, f(z+1) would have a complex value, so the Kneser mapping couldn't convert the Abel function into a real valued tetration. But it seems like it could generate an analytic complex base tetration where sexp(-1)=0, sexp(0)=1, sexp(1)=2+1.5i, sexp(2)=(2+1.5i)^(2+1.5i) and sexp(-2,-3,-4...)=singularity.... Yeah, that's what I'd be after, and I wouldn't expect it to be real-valued at the real axis (except at $z = -1$ and $z = 0$). What would a graph of this function $\operatorname{tet}_{2 + 1.5i}(x)$ look like along the real axis (two graphs, actually, one for real, one for imag)? Can this also be done with the Cauchy integral formula method as well? Note that we need to know which two fixed points it should decay to at $\pm i \infty$, and the presumed lack of conjugate symmetry (i.e. I expect that $\operatorname{tet}_{2 + 1.5i}(\bar{z})\ \ne\ \bar{\operatorname{tet}_{2 + 1.5i}(z)}$ in general, due to it being not real-valued at the real axis at most points) would seem to complicate attempts to try to run this one numerically. « Next Oldest | Next Newest »

 Possibly Related Threads... Thread Author Replies Views Last Post Is bugs or features for fatou.gp super-logarithm? Ember Edison 10 2,081 08/07/2019, 02:44 AM Last Post: Ember Edison A fundamental flaw of an operator who's super operator is addition JmsNxn 4 6,659 06/23/2019, 08:19 PM Last Post: Chenjesu Can we get the holomorphic super-root and super-logarithm function? Ember Edison 10 2,610 06/10/2019, 04:29 AM Last Post: Ember Edison Inverse super-composition Xorter 11 12,264 05/26/2018, 12:00 AM Last Post: Xorter The super 0th root and a new rule of tetration? Xorter 4 3,704 11/29/2017, 11:53 AM Last Post: Xorter Solving tetration using differintegrals and super-roots JmsNxn 0 1,835 08/22/2016, 10:07 PM Last Post: JmsNxn The super of exp(z)(z^2 + 1) + z. tommy1729 1 2,497 03/15/2016, 01:02 PM Last Post: tommy1729 Super-root 3 andydude 10 10,358 01/19/2016, 03:14 AM Last Post: andydude [split] Understanding Kneser Riemann method andydude 7 7,567 01/13/2016, 10:58 PM Last Post: sheldonison super of exp + 2pi i ? tommy1729 1 3,326 08/18/2013, 09:20 PM Last Post: tommy1729

Users browsing this thread: 1 Guest(s)