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 Derivative of exp^[1/2] at the fixed point? sheldonison Long Time Fellow Posts: 641 Threads: 22 Joined: Oct 2008 12/29/2015, 10:25 PM (This post was last modified: 01/01/2016, 04:57 PM by sheldonison.) I wanted to post what I've found out so far; which is not complete, and explain why only four derivatives are defined for Kneser's half iterate at the fixed point of L. First, let S(z) be the Schroeder function, $\alpha(z)$ is the Abel function, and $h_f$ be the formal half iterate which is not real valued at the real axis. $h_f(z) = S^{-1}(\sqrt{L}\cdot S(z))$ $h_f(z) = \alpha^{-1}(\alpha(z)+0.5) = S^{-1}\left(L^{0.5 + \log_L(S(z))}\right)$ Then let $h_k$ be the real valued Kneser half iterate. Here's the closest I've gotten, where $\theta_h(z)$ is a new 1-cyclic function whose coefficients can be derived from the 1-cyclic $\theta(z)$ mapping used for the $\text{sexp}(z)=\alpha^{-1}(z+\theta(z))$. $h_k(z) = \alpha^{-1}(\alpha(z)+0.5) = S^{-1}\left(L^{0.5 + \log_L(S(z))+\theta_h\left(\log_L(S(z))\right)}\right)\;\;$ The constant term $a_0=0$ for $\theta_h(z)$, so that $\;\theta_h(z)=\sum_{n=1}^{\infty} a_n \cdot \exp(2n\pi i)\;\;$ $h_k(z) = S^{-1} \left(\sqrt{L} \cdot S(z) \cdot L^{\theta_h\left(\log_L(S(z))\right)}\right)\;\;$ Since $S(z)=(z-L)+\sum_{n=2}^{\infty}a_n\cdot(z-L)^n=(z-L)\left(1+\cdot\sum_{n=1}^{\infty}a_{n+1}\cdot(z-L)^{n}\right)$, then the $\log_L(S(z)) = \log_L(z-L) + \log_L\left( 1+\cdot\sum_{n=1}^{\infty}a_{n+1}\cdot(z-L)^{n}\right)\;\;$, a taylor series in (z-L). So what we have for the individual theta(z) terms is $\exp\left(2n\pi i \cdot \log_L(z-L) \right) \cdot \left( \sum_{m=0}^{\infty}b_{nm}\cdot(z-L)^{m}\right) \approx (z-L)^{(4.44695n+1.05794ni)} \cdot \left(\sum_{m=0}^{\infty}b_{nm}\cdot(z-L)^{m}\right)$ $p = \frac{2\pi i}{L} \approx 4.44695+1.05794i\;\;$ p is the pseudo period of sexp $L^{\theta_h\left(\log_L(S(z))\right)}= 1 + \sum_{n=1}^{\infty}\left(c_n \cdot (z-L)^{np} + c_{n1}\cdot(z-L)^{np+1} + c_{n2}\cdot(z-L)^{np+2} ... \right) \;\;\;$ multiplier for S(z); also a function of (z-L) $h_k(z) = h(z) +\sum_{n=1}^{\infty} \left( \sum_{m=0}^{\infty}d_{nm}\cdot(z-L)^{np+m} \right) \;\;\;$ update I think this is a complete form for the Kneser half iterate. I left out many details including all of details about how to derive $\theta_h(z)$ from $\theta(z)$. I haven't yet used these equation to calculate values for Kneser's half iterate in terms of the formal half iterate, and verify they match the expected values generated through other means. But the first 4 derivatives are zero at L, and the 5th derivative of $\approx k \cdot (z-L)^{(4.44695+1.05794i )}$ has a singularity at L. That's all for now; next I would like to calculate some of the $d_{10}, d_{11}, d_{12}$ terms to test this equation out. - Sheldon « Next Oldest | Next Newest »

 Messages In This Thread Derivative of exp^[1/2] at the fixed point? - by sheldonison - 12/23/2015, 04:39 PM RE: Derivative of exp^[1/2] at the fixed point? - by sheldonison - 12/24/2015, 03:25 AM RE: Derivative of exp^[1/2] at the fixed point? - by sheldonison - 12/25/2015, 04:05 PM RE: Derivative of exp^[1/2] at the fixed point? - by andydude - 12/27/2015, 11:15 AM RE: Derivative of exp^[1/2] at the fixed point? - by sheldonison - 12/27/2015, 11:40 PM RE: Derivative of exp^[1/2] at the fixed point? - by andydude - 12/29/2015, 10:51 PM RE: Derivative of exp^[1/2] at the fixed point? - by sheldonison - 12/29/2015, 10:25 PM RE: Derivative of exp^[1/2] at the fixed point? - by sheldonison - 12/31/2015, 11:02 AM RE: Derivative of exp^[1/2] at the fixed point? - by tommy1729 - 12/31/2015, 01:25 PM RE: Derivative of exp^[1/2] at the fixed point? - by sheldonison - 01/01/2016, 03:58 PM RE: Derivative of exp^[1/2] at the fixed point? - by tommy1729 - 12/30/2015, 01:27 PM

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