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 Carlson's theorem and tetration mike3 Long Time Fellow Posts: 368 Threads: 44 Joined: Sep 2009 08/19/2010, 08:43 AM (This post was last modified: 08/19/2010, 08:44 AM by mike3.) Hi. I found the following easy uniqueness theorem that characterizes the regular tetrational of the base $b = \eta = e^{1/e}$, and perhaps also the whole regular tetrational (with attracting fixed point) (though base-$\eta$ is particularly interesting since it seems that both the regular and non-regular (i.e. Kneser's, etc.) method approach the same tetrational at this base.), with some modification (of condition 4). The conditions are very simple and easy. Theorem: There is a unique complex function $F(z)$ satisfying 1. $F(z + 1) = \eta^{F(z)}$ 2. $F(0) = 1$ 3. $F(z)$ is holomorphic in the entire cut plane with real $z \le -2$ removed, 4. $\lim_{r \rightarrow \infty} F(re^{i\theta}) = e$ for all $\theta \ne (2n+1)\pi, n \in \mathbb{Z}$ (could be weakened simply to saying the limit exists) Proof: We use what is called Carlson's theorem. Assume $F(z)$ and $G(z)$ are two different solutions of the above conditions. Then consider $H(z) = F(z) - G(z)$. Carlson's theorem says if this function, in the right half-plane, vanishes at every nonnegative integer, and is bounded asymptotically by $O(e^{u|z|})$ for some $u < \pi$, then it vanishes everywhere. Conditions 1 and 2 imply that $F$ and $G$ are equal at every nonnegative integer, thus $H$ is zero there, and condition 4 implies the asymptotic bounding (if two functions $f(x)$ and $g(x)$ have a limit at a given point, then the difference $f(x) - g(x)$ does as well) because every function decaying to a fixed value will be bounded in the asymptotic by any exponential (can give proof here if needed to fill this out.). Thus $H(z) = 0$, so $F(z) = G(z)$ and we are done. QED. See: http://mathworld.wolfram.com/CarlsonsTheorem.html Indeed, this says that condition 3 can be weakened to just holomorphism in the right half-plane ($\Re(z) \ge 0$) and condition 4 to the function being of exponential type of at most $\pi$ in that same right half-plane. The modifications also provide the theorem characterizing the regular tetrationals for $1 < b < \eta$. tommy1729 Ultimate Fellow Posts: 1,491 Threads: 355 Joined: Feb 2009 08/19/2010, 10:56 PM and how does one construct this superfunction ? one could equally say that the radius = 0 for this parabolic iteration and all taylor series with radius 0 are identical hence ' uniqueness '. but that is of course not intresting. regards tommy1729 bo198214 Administrator Posts: 1,395 Threads: 91 Joined: Aug 2007 08/20/2010, 12:21 PM (This post was last modified: 08/20/2010, 12:26 PM by bo198214.) (08/19/2010, 08:43 AM)mike3 Wrote: I found the following easy uniqueness theorem that characterizes the regular tetrational of the base $b = \eta = e^{1/e}$, and perhaps also the whole regular tetrational (with attracting fixed point) (though base-$\eta$ is particularly interesting since it seems that both the regular and non-regular (i.e. Kneser's, etc.) method approach the same tetrational at this base.), Hm, you mean that the super-exponential is bounded on the positive real axis by e and is imaginary periodic and hence is (exponentially) bounded on the right halfplane ... This is realy a nice finding, Mike! I guess it can be generalized to arbitrary regular superfunctions as they are always of the form $\eta(\pm e^{\kappa z})$ for some function $\eta$ analytic at 0. sheldonison Long Time Fellow Posts: 683 Threads: 24 Joined: Oct 2008 08/20/2010, 05:08 PM (This post was last modified: 08/20/2010, 06:03 PM by sheldonison.) (08/19/2010, 08:43 AM)mike3 Wrote: Hi. I found the following easy uniqueness theorem that characterizes the regular tetrational of the base $b = \eta = e^{1/e}$, and perhaps also the whole regular tetrational (with attracting fixed point) (though base-$\eta$ is particularly interesting since it seems that both the regular and non-regular (i.e. Kneser's, etc.) method approach the same tetrational at this base.)Mike, When I was working with eta, I noticed the simmilarity between $\text{sexpUpper}_\eta(z)$ and $\text{sexpLower}_\eta(z)$. As I remember, in the complex plane, sexpLower(z) looked a lot like sexpUpper(z+i), especially as imag(i) increased. I'll try to dig up an old contour graph.... Anyway, there are two solutions at base eta, one, the regular superfunction developed from the fixed point of "e", that goes to infinity as real(z) increases. This is the sexpUpper(z) function which has no singularities, and real(z)>e for all z at the real axis. The sexpLower(z) function of eta has a singularity at sexp(z,z=-2), and approaches "e" as real(z) increases to infinity. What you are pointing out is that the sexpLower(z) is a conformal/Kneser map of the sexpUpper(z) which is pretty cool, since it can also be developed from the attracting fixed point at +real infinity=e. Neat! This conformal mapping of one solution into thte other would only work for eta, since baseseta? - Sheldon mike3 Long Time Fellow Posts: 368 Threads: 44 Joined: Sep 2009 08/20/2010, 08:26 PM (This post was last modified: 08/20/2010, 08:28 PM by mike3.) (08/20/2010, 05:08 PM)sheldonison Wrote: When I was working with eta, I noticed the simmilarity between $\text{sexpUpper}_\eta(z)$ and $\text{sexpLower}_\eta(z)$. As I remember, in the complex plane, sexpLower(z) looked a lot like sexpUpper(z+i), especially as imag(i) increased. I'll try to dig up an old contour graph.... Anyway, there are two solutions at base eta, one, the regular superfunction developed from the fixed point of "e", that goes to infinity as real(z) increases. This is the sexpUpper(z) function which has no singularities, and real(z)>e for all z at the real axis. The sexpLower(z) function of eta has a singularity at sexp(z,z=-2), and approaches "e" as real(z) increases to infinity. What you are pointing out is that the sexpLower(z) is a conformal/Kneser map of the sexpUpper(z) which is pretty cool, since it can also be developed from the attracting fixed point at +real infinity=e. Neat! The two functions $\check{\eta}(z)$ (what you call "SexpUpper") and $^z \eta$ have similar asymptotic behavior, but along the real line, $\check{\eta}(z)$ grows toward infinity, and is chaotically-behaved (shows the "fractal structure") in a half-strip containing that line. It's quite a weird function, and because of the tetrational growth I don't see how Carlson's theorem implies anything about it. How does this show the $^z \eta$ is a Kneser map of $\check{\eta}(z)$? Quote:I wonder if Carlson's theorem can be used to say anything about bases >eta? - Sheldon Probably not. Those bases diverge, so the tetrational $^z b$ is not exponentially bounded in the right half-plane. mike3 Long Time Fellow Posts: 368 Threads: 44 Joined: Sep 2009 08/20/2010, 08:35 PM (08/20/2010, 12:21 PM)bo198214 Wrote: (08/19/2010, 08:43 AM)mike3 Wrote: I found the following easy uniqueness theorem that characterizes the regular tetrational of the base $b = \eta = e^{1/e}$, and perhaps also the whole regular tetrational (with attracting fixed point) (though base-$\eta$ is particularly interesting since it seems that both the regular and non-regular (i.e. Kneser's, etc.) method approach the same tetrational at this base.), Hm, you mean that the super-exponential is bounded on the positive real axis by e and is imaginary periodic and hence is (exponentially) bounded on the right halfplane ... This is realy a nice finding, Mike! Yeah, though for base $\eta$ there is no imaginary periodicity, but still, bounded on the right half plane (actually, the conditions you give that it should be bounded on the positive real axis and on the imaginary do not imply ("hence") it is bounded on the right half plane alone -- see, e.g. translations of the entire "upper regular iteration" along the imaginary axis). Quote:I guess it can be generalized to arbitrary regular superfunctions as they are always of the form $\eta(\pm e^{\kappa z})$ for some function $\eta$ analytic at 0. Yes, provided the fixed point is attracting and positive real. tommy1729 Ultimate Fellow Posts: 1,491 Threads: 355 Joined: Feb 2009 08/21/2010, 08:36 AM (08/20/2010, 08:35 PM)mike3 Wrote: Quote:I guess it can be generalized to arbitrary regular superfunctions as they are always of the form $\eta(\pm e^{\kappa z})$ for some function $\eta$ analytic at 0. Yes, provided the fixed point is attracting and positive real. i believe we need oo to be repelling and f^^n(z) converging for lim n-> oo and any z too. mike3 Long Time Fellow Posts: 368 Threads: 44 Joined: Sep 2009 08/21/2010, 08:08 PM (08/21/2010, 08:36 AM)tommy1729 Wrote: (08/20/2010, 08:35 PM)mike3 Wrote: Quote:I guess it can be generalized to arbitrary regular superfunctions as they are always of the form $\eta(\pm e^{\kappa z})$ for some function $\eta$ analytic at 0. Yes, provided the fixed point is attracting and positive real. i believe we need oo to be repelling and f^^n(z) converging for lim n-> oo and any z too. You mean $f^n(z)$, right? For "any" z seems too restrictive: $f(z) = \eta^z$, for example, does have many $z$-values for which its iteration diverges, but these do not show up in the range of the tetrational $^z \eta$. bo198214 Administrator Posts: 1,395 Threads: 91 Joined: Aug 2007 08/22/2010, 05:12 AM (08/20/2010, 08:35 PM)mike3 Wrote: Yeah, though for base $\eta$ there is no imaginary periodicity, but still, bounded on the right half plane (actually, the conditions you give that it should be bounded on the positive real axis and on the imaginary do not imply ("hence") it is bounded on the right half plane alone -- see, e.g. translations of the entire "upper regular iteration" along the imaginary axis). Ya, right of course it needs not only be bounded on the real axis but on the strip around the real axis of period width. But then how you prove the boundedness of the superexponential? « Next Oldest | Next Newest »

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