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 base holomorphic tetration bo198214 Administrator Posts: 1,389 Threads: 90 Joined: Aug 2007 11/07/2009, 04:47 PM (11/06/2009, 11:29 PM)bo198214 Wrote: $g(x)=\ln(a) (e^x -1 )$, $\tau(x)=\frac{x}{\ln(b)}+a$, $\tau^{-1}(x)=\ln(b)(x - a)$, $g = \tau^{-1}\circ f\circ \tau$. $g(x)=\ln(a)x + \frac{\ln(a)}{2} x^2 + \frac{\ln(a)}{6} x^3 + \dots$ This gives the following unsimplified coefficients of $g^{\circ t}$: $ {g^{\circ t}}_1 = \mbox{lna}^{t}\\ {g^{\circ t}}_2 = \left(\frac{1}{2} \, \frac{{({(\mbox{lna})}^{t}^{2} - \mbox{lna}^{t})}}{{(\mbox{lna}^{2} - \mbox{lna})}}\right) \mbox{lna}\\ {g^{\circ t}}_3 = \left(\frac{1}{2} \, \frac{{(\frac{{({(\mbox{lna})}^{t}^{2} - \mbox{lna}^{t})} \mbox{lna}^{t}}{{(\mbox{lna}^{2} - \mbox{lna})}} - \frac{{({(\mbox{lna})}^{t}^{2} - \mbox{lna}^{t})} \mbox{lna}}{{(\mbox{lna}^{2} - \mbox{lna})}})}}{{(\mbox{lna}^{3} - \mbox{lna})}}\right) \mbox{lna}^{2} + \left(\frac{1}{6} \, \frac{{({(\mbox{lna})}^{t}^{3} - \mbox{lna}^{t})}}{{(\mbox{lna}^{3} - \mbox{lna})}}\right) \mbox{lna}$ One can see that the coefficients are polynomials in $\ln(a)^t$ with rational coefficients in $\ln(a)$. One needs to investigate whether $f^{\circ t}(1)=\tau\circ g^{\circ t}\circ\tau^{-1}(1)=a+\frac{1}{\ln(b)}\sum_{n=1}^\infty {g^{\circ t}}_n (\ln(b)(1 - a))^n$ is analytic in $b=e^{1/e}$ with $a=\exp(-W(-\ln(b)))$. I just wanted to unify the variables: with $a = \ln(a)/\ln(b)$ we can write: $b[4]t = f^{\circ t}(1)=\tau\circ g^{\circ t}\circ\tau^{-1}(1)=\frac{1}{\ln(b)}\left(\ln(a)+\sum_{n=1}^\infty {g^{\circ t}}_n (\ln(b) - \ln(a))^n\right)$, $\ln(b)\in (0,1/e)$, $\ln(a) = - W(-\ln(b))\in (0,1)$ or shorter, setting $x=\ln(b)$ and $y=\ln(a)$ $e^x [4] t = \frac{1}{x}\left(y+\sum_{n=1}^\infty {g^{\circ t}}_n (x - y)^n\right)$, $x\in (0,1/e)$, $y=-W(-x)\in (0,1)$. The thing is now that Lambert $W$ has a singularity at $-1/e$, i.e. if $x=\ln(b)$ approaches $1/e$. The question is whether this singularity gets compensated somehow by the infinite sum. I want to further simplify the formula: with $x = y e^{-y} = y/e^y$ $e^{ye^{-y}} [4] t = e^y\left(1+\sum_{n=1}^\infty {g^{\circ t}}_n (e^{-y} - 1)^n y^{n-1}\right)$, $y\in (0,1)$ where ${g^{\circ t}}_n$ are polynomials in $y^t$ with coefficients that are rational functions in $y$. I hope i dint put errors somewhere; « Next Oldest | Next Newest »

 Messages In This Thread base holomorphic tetration - by bo198214 - 11/05/2009, 02:12 PM RE: base holomorphic tetration - by Base-Acid Tetration - 11/05/2009, 10:50 PM RE: base holomorphic tetration - by mike3 - 11/06/2009, 04:15 AM RE: base holomorphic tetration - by mike3 - 11/06/2009, 11:58 AM RE: base holomorphic tetration - by bo198214 - 11/06/2009, 12:12 PM RE: base holomorphic tetration - by mike3 - 11/06/2009, 09:16 PM RE: base holomorphic tetration - by bo198214 - 11/06/2009, 11:29 PM RE: base holomorphic tetration - by mike3 - 11/07/2009, 12:23 AM RE: base holomorphic tetration - by bo198214 - 11/07/2009, 08:17 AM RE: base holomorphic tetration - by mike3 - 11/07/2009, 08:21 AM RE: base holomorphic tetration - by bo198214 - 11/07/2009, 09:55 AM RE: base holomorphic tetration - by bo198214 - 11/07/2009, 04:47 PM RE: base holomorphic tetration - by bo198214 - 11/08/2009, 05:39 PM RE: base holomorphic tetration - by mike3 - 11/08/2009, 08:27 PM RE: base holomorphic tetration - by mike3 - 11/08/2009, 08:25 PM RE: base holomorphic tetration - by bo198214 - 11/08/2009, 08:44 PM RE: base holomorphic tetration - by mike3 - 11/08/2009, 09:51 PM

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