Derivative of exp^[1/2] at the fixed point? sheldonison Long Time Fellow Posts: 684 Threads: 24 Joined: Oct 2008 12/23/2015, 04:39 PM (This post was last modified: 12/23/2015, 04:43 PM by sheldonison.) Let exp^[1/2](x) be the half iterate of exp(x), generated by Kneser's Riemann mapping. Is the derivative defined of exp^[1/2](L) defined at the fixed point of L~=0.318132 + 1.33724i, where exp(L)=L? From an old 2010 post#3..post#8 http://math.eretrandre.org/tetrationforu...4&pid=5400, we see that L is the closest singularity to the real axis for the half iterate, but it is a mild singularity, and that exp^[1/2](L)=L, so that exp^[1/2](L) is continuous at L. Is the derivative is also continuous at this singularity, and if so, then what is its value? If the derivative is continuous at the singularity, how many of the higher derivatives are also continuous? - Sheldon sheldonison Long Time Fellow Posts: 684 Threads: 24 Joined: Oct 2008 12/24/2015, 03:25 AM (This post was last modified: 12/29/2015, 11:02 PM by sheldonison.) (12/23/2015, 04:39 PM)sheldonison Wrote: Is the derivative is also continuous at this singularity, and if so, then what is its value? If the derivative is continuous at the singularity, how many of the higher derivatives are also continuous? In particular, there is a formal half iterate that is not real valued that can be developed at the fixed point L. $\exp(L+z) = L \cdot (1 + z + \frac{z^2}{2} + \frac{z^3}{6} + ...$ And the formal half iterate begins with $\exp^{0.5}(z+L) = L + \sqrt{L}z + \frac{L \cdot z^2}{2(L+\sqrt{L})} + ...$ Numerical experiments suggest that the derivative of Knesser's real valued half iterate is continuous at L, and the first derivative at the singularity at L is $\sqrt{L}$, and that the 2nd derivative may also match the formal half iterate. - Sheldon sheldonison Long Time Fellow Posts: 684 Threads: 24 Joined: Oct 2008 12/25/2015, 04:05 PM (This post was last modified: 12/25/2015, 04:24 PM by sheldonison.) (12/24/2015, 03:25 AM)sheldonison Wrote: Numerical experiments suggest that the derivative of Knesser's real valued half iterate is continuous at L, and the first derivative at the singularity at L is $\sqrt{L}$, and that the 2nd derivative may also match the formal half iterate. I think the key equations in understanding this behavior are the slog, the sexp, the Schroeder equation at L, and the Abel equation generated from the Schroeder equation, and the theta mapping from the Abel equation to the slog. Kneser's exp^{0.5}: $\exp^{0.5}(z) = \text{sexp}(\text{slog}(z) + 0.5)$ Here is the formal exp^{0.5} at L, generated from the formal Schroeder equation developed at the fixed point L $S(z)$ where: $S(\exp(z)) = S(z)\cdot L\;\;\;S(z+L)=z+a_2\cdot z^2 + a_3 \cdot z^3 ...$ Then the formal exp^{0.5} at L is exactly the same as: $S^{-1}\left( S(z) \cdot \sqrt{L} \right)$ The next step is to show the Abel equation, developed from the Schroeder equation; where $\alpha\left(\exp(z)\right) = \alpha(z)+1$ $\alpha(z) = \log_L\left(S(z)\right)$ And Kneser's slog(z) developed from the Abel equation is: $\text{slog}(z) = \alpha(z) + \theta\left(\alpha(z)\right)\;\;$ where $\theta(z)$ is a 1-cyclic function decaying to zero at $\Im \infty$ - Sheldon andydude Long Time Fellow Posts: 510 Threads: 44 Joined: Aug 2007 12/27/2015, 11:15 AM (12/24/2015, 03:25 AM)sheldonison Wrote: And the formal half iterate begins with $\exp^{0.5}(z+L) = L + \sqrt{L}z + \frac{L\cdot z^2}{2(L+\sqrt{L})} + ...$ Ok, so I replaced y with 1/2 and log(L) with L in the regular iteration power series to get this: $\exp^{1/2}(z + L) = L + \sqrt{L} z + \frac{\sqrt{L}}{2(1 + \sqrt{L})} z^2 + \frac{\sqrt{L} - 3L + 4L^{3/2} - 3L^2 + L^{5/2}}{6(1 + L)(1 - L)^2} z^3 + \cdots$ as expected it's the same power series. I wanted to highlight one of my findings in this paper (page 12) that is related but separate from this, which is a power series for $f^{1/x}(x)$ for any analytic function $f$ with a parabolic fixed point at 0. $f^{1/x}(x) = \frac{x}{1 - f_2} + \left(f_2 - \frac{f_3}{f_2}\right)\frac{\log(1 - f_2)}{(1 - f_2)^2} x^2 + \cdots$ Substituting in $f(z) = \exp_{\eta}(z) = \exp(z/e)$ we get $\exp_{\eta}^{1/z}(z) = e + 2(z-e) - \frac{2 \log(2)}{3e} (z - e)^2 + \frac{(1 + \log(4))^2}{18e^2} (z - e)^3 + \cdots$ which I realize is a different base, but still interesting. Using a similar technique, we might be able to find a comparable power series for $\exp^{1/z}(z)$, at which point $z=2$ would at the very least intersect with this function. sheldonison Long Time Fellow Posts: 684 Threads: 24 Joined: Oct 2008 12/27/2015, 11:40 PM (This post was last modified: 12/27/2015, 11:41 PM by sheldonison.) (12/27/2015, 11:15 AM)andydude Wrote: Ok, so I replaced y with 1/2 and log(L) with L in the regular iteration power series to get this: $\exp^{1/2}(z + L) = L + \sqrt{L} z + \frac{\sqrt{L}}{2(1 + \sqrt{L})} z^2 + \frac{\sqrt{L} - 3L + 4L^{3/2} - 3L^2 + L^{5/2}}{6(1 + L)(1 - L)^2} z^3 + \cdots$ as expected it's the same power series.Hey Andy, Thanks for your reply. Oops; I had a typo in my 2nd derivative which I fixed. I have a pari-gp program, that calculate the coefficients iteratively. Quote:I wanted to highlight one of my findings in this paper (page 12) that is related but separate from this, which is a power series for $f^{1/x}(x)$ for any analytic function $f$ with a parabolic fixed point at 0. $f^{1/x}(x) = \frac{x}{1 - f_2} + \left(f_2 - \frac{f_3}{f_2}\right)\frac{\log(1 - f_2)}{(1 - f_2)^2} x^2 + \cdots$ Substituting in $f(z) = \exp_{\eta}(z) = \exp(z/e)$ we get $\exp_{\eta}^{1/z}(z) = e + 2(z-e) - \frac{2 \log(2)}{3e} (z - e)^2 + \frac{(1 + \log(4))^2}{18e^2} (z - e)^3 + \cdots$ which I realize is a different base, but still interesting. The parabolic case is hugely interesting. I usually work with iterating $f(x)=\exp(x)-1$ which is equivalent to iterating base $\eta$. Anyway, the cool thing about the parabolic case is that the fixed point of zero for the fractional iterate is a singularity, and the formal power series is divergent at zero. References on mathoverflow: http://mathoverflow.net/questions/4347/f...ar-and-exp For the case at hand, $\exp(L+x)$, my new conjecture is that the first four derivatives are continuous, but the fifth derivative at the fixed point has a singularity. And the first four derivatives would match the first four derivatives of the formal half iterate at the fixed point. I'm still not totally comfortable it yet, so I haven't posted the justification. - Sheldon sheldonison Long Time Fellow Posts: 684 Threads: 24 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 andydude Long Time Fellow Posts: 510 Threads: 44 Joined: Aug 2007 12/29/2015, 10:51 PM (12/27/2015, 11:40 PM)sheldonison Wrote: Thanks for your reply. Oops; I had a typo in my 2nd derivative which I fixed. I have a pari-gp program, that calculate the coefficients iteratively. I don't think there was a typo, if you multiply the numerator and denominator by $\sqrt{L}$ and distribute, I think it's the same. tommy1729 Ultimate Fellow Posts: 1,859 Threads: 402 Joined: Feb 2009 12/30/2015, 01:27 PM Putting the issue in limit form (exp^[1/2](L+h i) - exp^[1/2](L-hi)) / h^n Where h is infinitesimal. Regards Tommy1729 sheldonison Long Time Fellow Posts: 684 Threads: 24 Joined: Oct 2008 12/31/2015, 11:02 AM (This post was last modified: 01/01/2016, 04:34 PM by sheldonison.) I updated some of the equations in post#6 So then we have a conjectured equation for the real valued Kneser half iterate in terms of the formal half iterate which is as follows edit: fixed typos $h_k(z) = h(z)\; +\; \sum_{n=1}^{\infty} \left( \sum_{m=0}^{\infty}c_{nm}\cdot(z-L)^{np+m} \right)\;\;\;\;p = \frac{2\pi i}{L} \approx 4.44695+1.05794i\;\;$ p is the pseudo period of sexp I think this is a complete form for the Kneser half iterate. Next I would like to calculate some of the $c_{10}, c_{11}, c_{12}...$ terms to test this equation out, as well as the $c_{20}$ term. The value and the first four derivatives of this equation are zero; the adder delta equation for the Kneser half iterate in terms of the formal half iterate. - Sheldon tommy1729 Ultimate Fellow Posts: 1,859 Threads: 402 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 »

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