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 Derivative of exp^[1/2] at the fixed point? tommy1729 Ultimate Fellow Posts: 1,372 Threads: 336 Joined: Feb 2009 12/31/2015, 01:25 PM (12/29/2015, 10:25 PM)sheldonison Wrote: 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 $\exp\left(2n\pi i \cdot \log_L(z-L) \right) \cdot \left( \sum_{i=0}^{\infty}b_{ni}\cdot(z-L)^{i}\right) \approx \left(4.44695n+1.05794ni\right)^{z-L} \cdot \left( \sum_{i=0}^{\infty}b_{ni}\cdot(z-L)^{i}\right)\;\;$ $p = \frac{2\pi i}{L} \approx 4.44695+1.05794i\;\;$ p is the pseudo period of sexp $L^{\theta_h\left(\log_L(S(z))\right)}= \left(1 + p^{z-L}\cdot f_1(z-L) + (2p)^{z-L}\cdot f_2(z-L) +...\right)\;\;$ multiplier for S(z); also a function of (z-L) $h_k(z) = h(z) +\sum_{n=1}^{\infty}\left( (\frac{2 n \pi i}{L})^{z-L} \cdot \left( c_n + \sum_{i=1}^{\infty}c_{ni}\cdot(z-L)^{i}\right) \right) \;\;\;$ 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 $\approx \left(4.44695n+1.05794ni\right)^{z-L}$ has a singularity at L. That's all for now; next I would like to calculate some of the $c_{1}, c_{11}, c_{12}$ terms to test this equation out. The 5 th derivative of $\approx \left(4.44695n+1.05794ni\right)^{z-L}$ is equal to of $\approx \left(4.44695n+1.05794ni\right)^{z-L} ln(4.44 n + 1.05 n i)^5$ ?? No singularity ? Im sure you make sense , but it is not clear what you are doing to me. Regards Tommy1729 « 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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