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 Why bases 0-3.9... surprisingly remaining real valued after the logarithmic singularity at -2. I expect there is a unique solution for the complex plane, which can be developed using Kneser's Riemann mapping and the equivalent $\theta(z)$ mapping techniques we've invented ... which should be really interesting too! We expect this function to approach two of the function's other complex fixed points, and its conjugate, as $\Im(z)$ gets arbitrarily large, positive or negative. - Sheldon MorgothV8 Junior Fellow Posts: 18 Threads: 6 Joined: Dec 2012 04/15/2015, 09:02 AM Did anybody tried tetrations with negative bases, are there some special intervals here (like we have e^-e and e^(1/e)) on positive axis side? So far I see that there should be at least 4 differnt tetration classes: z < 0 e^-e > z > 0 e^(1/e) > z > e^-e z > e^-e And special cases: e^-e e^(1/e) and maybe 0? Is there any general summary of tetriation behavior in all those cases? Best Regards. Fuji GSW690III Nikon D3, Nikkors 14-24/2.8, 24/1.4, 35/2, 50/1.4, 85/1.4, 135/2, 80-200/2.8 tommy1729 Ultimate Fellow Posts: 1,370 Threads: 335 Joined: Feb 2009 04/15/2015, 02:51 PM (This post was last modified: 04/15/2015, 02:52 PM by tommy1729.) Plot exp(-exp(-x)) then you see why its not so popular. It has a Nice nonparabolic fix yet it is not nice. Notice the limits at - , + infinity. Regards tommy1729 JmsNxn Long Time Fellow Posts: 291 Threads: 67 Joined: Dec 2010 04/15/2015, 05:37 PM (This post was last modified: 04/15/2015, 05:38 PM by JmsNxn.) I have a conjecture I haven't been able to prove. I think the Fractional Calculus approach I took for hyper operators in between 1 and eta will work for bases inbetween $e^{-e} \le b < 1$ To be explicit: $(^z b) = \frac{1}{\G(1-z)} (\sum_{n=0}^\infty (^{n+1} b) \frac{(-1)^n}{n!} + \int_1^\infty (\sum_{n=0}^\infty (^{n+1} b) \frac{(-w)^n}{n!}) w^{-z}\,dw)$ This is of course if there exists an exponentially bounded solution. Since these bases do not induce a periodic tetration there requires more thought into how to solve this case. Constructing the Koenigs function $\psi$ of $b^z$ produces a function $\psi(b^z) = \lambda \psi(z)$ where $-1 < \lambda < 0$. Using regular Koenig iteration we can construct a tiny area about the fixed point $z_0$ which $\psi^{-1} (z)$ exists. If we can induce the Koenigs function such that $\exp_b^{\circ s+n} (z) = \psi^{-1}(\lambda^{s+n}\psi(z))$ is a well defined function for $|z - z_0 |<\delta$ and $n$ large enough so that this function is defined for all $\Re(s) > 0$ we can extend this to all $z$ in the immediate basin (including 1) using Fractional Calculus and similar techniques that I used here. This would give $\exp_b^{\circ s+n}(1) = (^{s+n} b)$, and then using more tricks $^s b$ can be recovered. The trick is of course ensuring that $\psi^{-1}(\lambda^{s+n}\psi(z))$ defined for $s$ in some $\Omega$ can be analytically continued to the right half plane, and then further is appropriately bounded. This would require analytically continuing the koenig function and as well its inverse. Unless someone knows if there exists a solution such that $|(^z b)| < M e^{\rho |\Re(z)| + \alpha |\Im(z)|}$ for $M,\rho, \alpha \in \mathbb{R}^+$ with $\alpha < \pi/2$, I see no other way of making this method work. However there is excitement, because certain holomorphic functions with a multiplier that is not real positive and between 0 and 1 can be complex iterated using fractional calculus. To give further motivation of this technique, consider the function $\phi_\lambda(\xi) = \lambda \xi + \sum_{n=2}^\infty a_n\xi^n$ for $0 < \lambda < 1$ We can iterate this using fractional calculus and for $\xi$ in the immediate basin of attraction, z in the right half plane, $\phi^{\circ z}_\lambda(\xi)$ is analytic in $\lambda$ and consequently the iterate must converge in the Fractional calculus transform for $\lambda$ in an epsilon neighbourhood of (0,1). The suspicion is that it will converge for all $0<|\lambda|<1$ on suitable $\phi$. I'm betting the exponential function is one of these suitable functions. sheldonison Long Time Fellow Posts: 638 Threads: 22 Joined: Oct 2008 04/16/2015, 05:09 PM (This post was last modified: 04/16/2015, 09:02 PM by sheldonison.) Conjecture: For all bases between $0, there are a pair of complex conjugate fixed points. Each of these fixed points has a Koenig solution, that can be used to generate a complex valued superfunction. Such a Koenig solution is entire, and never takes on the value of 0. The conjecture is the pair of Koenig complex superfunctions can be Kneser (Riemann) mapped (or equivalently) theta mapped, to a Tetration solution that is real valued between the singularity at -2, and infinity. Such a solution would be like tetration for bases>$\eta=\exp(1/e)$ analytic if $\Im(z)>0$ or $\Im(z)<0$ and at the real axis, for z>-2, with singularities at -2, -3, and other values <-3. For tetration bases $0, there is a primary real valued fixed point, with a negative $\lambda$ multiplier at the fixed point. For bases $0 the fixed point is repelling, but bifurcates to an attracting two-cycle. marraco and I have posted plausible real valued tetration solutions with Tet(0)=1 for base b=0.01. For bases>$\exp(-e)$ the fixed point is attracting, and the solution will slowly oscillate towards the fixed point, exponentially scaling as it does so. The initial seed for such a solution would be generated in much the same way as the earlier post for b=0.01; I haven't done the Kneser/theta mapping yet, from the complex pair of fixed points. For example, consider b=0.1, which has an attracting fixed point $\approx 0.399$, and a pair of repelling complex conjugate fixed points $-0.3018\pm1.981i$. Here is an initial seed, a 4th degree polynomial approximation roughly accurate between tet(-1) and tet(0), with boundary conditions tet(-1)=0, tet(0)=1, tet'(-1)=0, tet'(0)=0, and the $\ln_b(\text{tet}(x+0))\approx \text{tet}(-1+x)$ with a piecemeal approximation having a continuous 2nd derivative: $\text{tet}_{0.1}(z) \approx 1 -4.18324136x^2 -4.36648272*x^3 -1.18324136x^4$ Then this is what it looks like, at the real axis from -3.76 to 10. As $\Re(z) \to \infty, \; \text{tet}(z) \to L \approx 0.399\;\;\; 0.1^{L}=L$     And here is the function at $\Im(z)=0.1i$ from -$-4<\Re(z)<5$ Notice that the function would go to the secondary fixed point of $L2 \approx-0.3018-1.981i\;0.1^{L2}=L2$ as $\Re(z) \to -\infty$. For $\Im(z)=-0.1i$, it would go to the conjugate fixed point, $\approx-0.3018+1.981i$. So this graph is why I think there is a unique Kneser mapping to generate this Tetration function, with the boundary conditions that tet(-1)=0, tet(0)=1, tet'(-1)=0, tet'(0)=0, $\text{tet}(\Im(z)=\pm i\infty)\approx 0.3018\mp 1.981i\;\;$ (approaching Koenig's solution) The algorithm just sketched could work for all bases $0.     I also have some bonus questions. Is it possible that we derive the requirement that tet'(-1)=0 from the requirement that the Riemann mapping go to the pair of conjugate fixed points in the complex plane? btw, if Tet'(-1)=0, then it is trivial to show that tet'(0)=0, from the definition of the derivative of exp(f(x)) Also, this function is different than Tetration $\exp(-e)$,http://math.eretrandre.org/tetrationforu...0&pid=6748. Mike and I think that the function at this link is actually the Tetration function that is analytically connected to real base tetration for b>eta. What happens if we start with this new Tetration function, for 0-2, and converging to the two complex valued Koenig's solutions in the upper/lower halves of the complex plane. The uniqueness criteria and proofs that Henryk has written rely on the slog/Abel function (and the sexp function) both being analytic on a sickle, between the two fixed points. But since the slog/Abel function has singularities where sexp'(z)=0 ... I don't know how one might apply that proof technique to the solution at hand. - Sheldon tommy1729 Ultimate Fellow Posts: 1,370 Threads: 335 Joined: Feb 2009 04/17/2015, 09:15 PM (This post was last modified: 04/17/2015, 09:26 PM by tommy1729.) For bases b > 2 i think the following makes sense : For x > 0 : F " (x) > 0 F ' (x) > 0 And F ' (0) = b/2 What do you think ? Regards Tommy1729 marraco Fellow Posts: 93 Threads: 11 Joined: Apr 2011 04/17/2015, 10:22 PM (This post was last modified: 04/18/2015, 04:37 AM by marraco.) (04/17/2015, 08:26 AM)tommy1729 Wrote: I think $a_1=ln(a) a$ makes sense. Regards Tommy1729 (04/17/2015, 09:15 PM)tommy1729 Wrote: For bases b > 2 i think the following makes sense : For x > 0 : F " (x) > F ' (x)/2 F ' (x) > 0 And F ' (0) = b/2 What do you think ? Seems arbitrary to me. Do you care to expand? There is probably some value for a₁ that provides some benefit. For example, some values makes $\vspace{15}{^xa\approx ln_{N_a}(a^{a^{x-1}})}$ (which is right bracket tetration). Other values make $\vspace{15}{^xa\approx ln_{N_a}(^xsroot(a))}$. I don't like a₁=0, because it makes most bases look weird. For example, this is tetration base $\vspace{15}{e^{\frac{1}{e}}}$, when a₁=0 (up), and when there is no restriction in a₁ (down; It gives a₁=0,6118~~~~~) I find the smoothed curve more appealing, because I suspect that flow in pipes is governed by tetration. This is a semi empiric "Moody diagram" of flow in pipes. To me, it resembles tetration. When flow turns fully turbulent, it converges to a constant value. « Next Oldest | Next Newest »

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