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 superfunctions of eta converge towards each other sheldonison Long Time Fellow Posts: 641 Threads: 22 Joined: Oct 2008 12/05/2012, 12:22 AM (This post was last modified: 12/05/2012, 06:44 PM by sheldonison.) (05/23/2011, 09:01 PM)sheldonison Wrote: ... Here, sexp(z) is the lower superfunction, with sexp(0)=1, and cheta(z) is the upper superfunction .... $\text{sexp}_{\eta}(z)=\text{cheta}(z+\theta(z)+ k )$. Where $\theta(z)$ is a 1-cyclic function, which quickly decays to zero as imag(z) increases. Then, the constant "k" is 5.0552093131039000 + 1.0471975511965977*I ...I made lots of minor updates and clarifications all over this reply. Apparently, when I posted this last year nobody noticed that the imaginary part of k=1.0471975511965977 is exactly Pi/3, which of course begs for an explanation! I didn't notice it either, until I started to work with the formal Abel series solution for iterates of $\exp(x)-1$, which is parabolic with a fixed point of zero. We start by noticing that solutions of $g(z)=\text{cheta}(z)=\exp^{[z]}_\eta(2e)$ are conjugate to solutions for $h(z)=(\exp(1)-1)^{[z]}$ so that $h(z) = \frac{g(z)}{e}-1$. So the two problems are trivially interchangeable. On mathstack, Will Jagy explained how to generate the formal abel function solution for the parabolic case. There are also papers by Baker on the abel function of exp(z)-1. Here is the formal solution for the abel function of exp(z)-1. I posted more terms below. $\alpha(\exp(z)-1)=\alpha(z)+1$ $\alpha(z) = \frac{-2}{z} + \frac{\log(z)}{3} + \frac{-z}{36} + \frac{z^2}{540} + \frac{z^3}{7776} + \frac {-71z^4}{435456} + \frac{8759z^5}{163296000} + O(z^6)$ $\alpha(z)=\text{cheta^{-1}(e(z+1))+k$ for $k\approx -2.025912$ The formal abel solution is only valid for real(x)>=0. It is also a divergent series, meaning that you need to truncate to some optimal number of terms for any particular value of z, but it is nonetheless very accurate. The 40 term series posted below was generated in pari-gp and is accurate to 32 decimal digits for |z|<=0.15. For larger values of z, iterate log(z+1) until the value is smaller than 0.15, and then evaluate the formal series. There is an analogous abel function for sexpeta. $\alpha_2(z)=\text{sexpeta^{-1}(e(z+1))+k$ for $k\approx 3.029297$ $\alpha_2(z) = \frac{-2}{z} + \frac{\log(-z)}{3} + \frac{-z}{36} + \frac{z^2}{540} + \frac{z^3}{7776} + \frac {-71z^4}{435456} + \frac{8759z^5}{163296000} + O(z^6)$ This nearly identical $\alpha_2(z)$ abel function is for the "sexpeta" superfunction of exp(z)-1. $\alpha_2^{-1}(z)$ approaches zero from the negative real numbers, as z goes to infinity. $\alpha_2(z)$ is only valid if real(z)<=0 . In this series, the log(z) term was replaced with log(-z), so that the abel function is real valued at the real axis for negative real numbers. For example, if we ignore the fact that $\alpha(z)$ is not valid for $\Re(z)<0$, then $\alpha(-0.01)\approx198.465+\frac{\pi i}{3}$, whereas $\alpha_2(-0.01)\approx198.465$. That $\pi i/3=\log(-1)/3$ difference between the two functions is exactly the imaginary part of the $\theta(z)$ constant term for the two superfunctions of $\exp_\eta(z)$, that I numerically calculated last year. If z=0.15i, than for the 40 term series posted below, both abel functions are valid, and should be accurate to >30 decimal digits. The two approximations differ by exactly $\frac{\pi i}{3}$. But, as |z| grows, the formal solution is no longer very accurate, and one must iterate exp(z)-1 for $\alpha_2(z)$, and iterate log(z+1) for $\alpha_2(z)$, until each |z| is a smaller number before evaluating the formal solution. This iteration leads to the two inverse abel functions (superfunctions) behaving very differently as z approaches the real axis. But as imaginary of z increases, the inverse of the two functions converge towards each other. $\lim_{z \to \Im \infty}\hspace{2 mm}\alpha(\alpha_2^{-1}(z))-z = \pi i/3$, which leads to the $\theta(z)$ function I calculated. My definition for theta is $\theta(z)=\alpha(\alpha_2^{-1}(z))-z$. The formal abel series solution may allow one to prove the exponential convergence as $\Im(z)$ increases, which is conjectured to be: $\alpha(\alpha^{-1}_2(z))-z=\pi i/3 + \sum_{n=1}^{\infty}a_n \exp(2n\pi zi)$. I've wanted to understand Ecalle cylinders for awhile .... If anyone wants the pari-gp code for parabolic abel solutions for the general case, for $f(x)=x+ax^2+...$, I could also post that. Also, I assume there is no equivalent formal solution for the superfunction, $\alpha^{-1}(z)$, for the parabolic case. The best reasonable approximation I could generate for the reciprocal of the superfunction of exp(z)-1 is: fixed typo, updated approximation with emperical error bounds $\frac{1}{\alpha^{-1}(z)} \approx \frac{-z}{2}+\frac{-1}{6}\log(\frac{-z}{2})+O(z^{-1})$ $\frac{1}{\alpha_2^{-1}(z)} \approx \frac{-z}{2}+\frac{-1}{6}\log(\frac{z}{2})+O(z^{-1})$ The $O(z^{-1})$ term seems to be $(\frac{\log(\frac{-z}{2})}{18}-\frac{1}{36})\times z^{-1}$ - Sheldon Code:First 30 terms, formal abel series term for exp(z)-1. log(z)/3 term also required a-1= -2 a0=   0 a1=  -1/36 a2=   1/540 a3=   1/7776 a4=  -71/435456 a5=   8759/163296000 a6=   31/20995200 a7=  -183311/16460236800 a8=   23721961/6207860736000 a9=   293758693/117328567910400 a10= -1513018279/577754311680000 a11= -1642753608337/3355597042237440000 a12=  3353487022709/1689531377909760000 a13= -11579399106239/40790114695249920000 a14= -254879276942944519/137219514685385932800000 a15=  13687940105188979843/14114007224782553088000000 a16=  215276054202212944807/100956663443150497382400000 a17= -2657236754331703252459529/1203529624071657866919936000000 a18= -146435111462649069104449/50302321749125019205632000000 a19=  715411321613253460298674267/135588231530708185101474201600000 a20=  16634646784735044775309724063/3702250880735601413534515200000000 a21= -104353470644496360229598950087621/7332274212470670094037711585280000000 a22= -1026800310866887782669304706891/145015557324117535367532380160000000 a23=  10532451718209319314810847524219487/239106170881428081691713129676800000000 a24=  426818206492321153424287945331450731/55748747292256998858987528725200896000000 a25= -209820349077359397909291778326518401351/1340114117602331703341046363586560000000000 a26=  525117796674628883106100578152841570958289/21674067658217791337645745152194510848000000000 a27=  196370501349536911290241763355698126325788423/308676831703848984325152590299385561088000000000 a28= -4655964318554330930550687915598236845144401499/14371804056954685277851548955241890185216000000000 a29= -9047134015490968185454900363573980634933739699733371/3082105239034217031261653931196399559670497280000000000 a30=  205360181531874254884259531693649741510468924878159/77138340242791861182855632315816451715891200000000000 a31=  0.0152680842325604720475463799792485 a32= -0.0208547307456560124435878815421207 a33= -0.0888426680278022904549764943201458 a34=  0.169545350486845480899257484416871 a35=  0.573829524409951440465435202028753 a36= -1.47178737132655819000458397068965 a37= -4.08073498995482601179218114027270 a38=  13.7938917842810175925883583879899 a39=  31.6710432837078833141996786165051 a40= -140.151276797120726823402361527161 a41= -265.538550008913692150238411154415 « Next Oldest | Next Newest »

 Messages In This Thread superfunctions of eta converge towards each other - by sheldonison - 05/23/2011, 09:01 PM RE: superfunctions of eta converge towards each other - by bo198214 - 05/23/2011, 09:49 PM RE: superfunctions of eta converge towards each other - by sheldonison - 05/23/2011, 11:26 PM RE: superfunctions of eta converge towards each other - by bo198214 - 05/24/2011, 01:06 PM RE: superfunctions of eta converge towards each other - by sheldonison - 05/24/2011, 02:18 PM RE: superfunctions of eta converge towards each other - by bo198214 - 05/24/2011, 02:27 PM RE: superfunctions of eta converge towards each other - by sheldonison - 05/30/2011, 06:06 AM RE: superfunctions of eta converge towards each other - by sheldonison - 05/25/2011, 04:03 AM RE: superfunctions of eta converge towards each other - by sheldonison - 05/27/2011, 09:36 AM RE: superfunctions of eta converge towards each other - by tommy1729 - 05/25/2011, 11:04 PM RE: superfunctions of eta converge towards each other - by bo198214 - 05/25/2011, 11:15 PM RE: superfunctions of eta converge towards each other - by tommy1729 - 05/26/2011, 12:10 PM RE: superfunctions of eta converge towards each other - by tommy1729 - 06/06/2011, 10:42 PM

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