half-iterates of x^2-x+1 - Printable Version +- Tetration Forum (https://math.eretrandre.org/tetrationforum) +-- Forum: Tetration and Related Topics (https://math.eretrandre.org/tetrationforum/forumdisplay.php?fid=1) +--- Forum: Mathematical and General Discussion (https://math.eretrandre.org/tetrationforum/forumdisplay.php?fid=3) +--- Thread: half-iterates of x^2-x+1 (/showthread.php?tid=787) half-iterates of x^2-x+1 - Balarka Sen - 03/22/2013 What is the analytic solution to f(f(x))=x^2-x+1? I am thinking about Ecalle's method . . . I don't think transforming this into Abel's equation would be of any use since it's mostly applicable where f has a attractive fixed points and transforming it into Schroeder's form and solving it by following Koening's line for 0 < |f'(z)| < 1. But x^2 - x + 1 has only one fixed point which is neutral. I think Taylor series maybe used since 1 is probably a fixed point of f but it doesn't seems like it would converge, would it? Can we analytically continue the taylor series then? Balarka . RE: half-iterates of x^2-x+1 - sheldonison - 03/22/2013 (03/22/2013, 08:19 PM)Balarka Sen Wrote: What is the analytic solution to f(f(x))=x^2-x+1? I am thinking about Ecalle's method . . . .... But x^2 - x + 1 has only one fixed point which is neutral. I think Taylor series maybe used since 1 is probably a fixed point of f but it doesn't seems like it would converge, would it?You are correct. If you develop the half iterate of a function at the parabolic fixed point, then it is a divergent series. see Will Jagy's comments at http://math.stackexchange.com/questions/208996/half-iterate-of-x2c. Your question is equivalent to f(f(x))=x^2+x, where now the parabolic fixed point is zero. For your case, $y^2-y+1$, the equivalent parabolic fixed point is 1 where y=x+1. Then there is a formal abel function solution given by Ecalle for $g(x)=x^2+x$. This abel function is actually a formally divergent series, but in practice it works quite well, especially for smaller values x, as x approaches zero, by simply iterating $g^{-1}(x)$ a few times before using the abel function, or if approaching from the negative reals, iterating g(z) a few times and also using the log(-x) in the abel function. $\alpha(g(x))=\alpha(x)+1$ $\alpha(x) = \log(x)+ \frac{-1}{x} + \frac{-x}{2}+ \frac{x^2}{3}+ \frac{-13x^3}{36}+ \frac{113x^4}{240}+ \frac{-1187x^5}{1800}+ \frac{877x^6}{945}+ \frac{-14569x^7}{11760}+ \frac{176017x^8}{120960}+ \frac{-1745717x^9}{1360800}+ \frac{88217x^{10}}{259875}+$ $ \frac{147635381x^{11}}{109771200}+ \frac{-3238110769x^{12}}{1556755200}+ \frac{-63045343657x^{13}}{23610787200}+ \frac{24855467017x^{14}}{1489863375}+ \frac{-20362710600601x^{15}}{817296480000}+ \frac{-13053665468881x^{16}}{380007936000}+ ...$ and then the half iterate is $f(x)=\alpha^{-1}(\alpha(x)+\frac{1}{2})$, and y=x+1 gets you back to your original equation. Since the abel function is a divergent series at x=0, then the half iterate would also be a divergent series at x=0. The half iterate generated at any center point other than zero would be a normal convergent analytic function. This is because there are really two different leau fatou flower petals with two different abel functions, one generated from x>0 fixed point (repelling), and one generated from x<0 (attracting), and you can't analytically continue from one to the other. For example, I generated the half iterate of 0.5i, from both flower petals. In one case, I iterated g(z) 40 times, to get an accurate abel function result. Then I added a half, and took the inverse abel function, and then iterated $g^{-1}$ 40 times to get the half iterate of 0.5i $\approx -0.1113+0.5335i$. In the other case, I did the reverse, starting with 0.5i, and iterating $g^{-1}$ 40 times, then I took the half iterate, and then I iterated g(z) 40 times to get a half iterate of 0.5i $\approx-0.1138+0.5246i$. The two results differed by nearly 0.01i. Had I started with 0.2i, the two results would be consistent to 4E-12. So, the half iterate centered at zero acts like an analytic function as long as z is close to zero, but as imag(z) gets bigger, the half iterate has two possible values that are increasingly different. You have to pick which leau flower petal you're interested in to get arbitrarily exact results. Hope that helps. - Sheldon RE: half-iterates of x^2-x+1 - tommy1729 - 04/30/2013 @Balarka I posted a method for half-iterates today. If you want to avoid matrices , fixpoints , abel- or super functions that might intrest you. regards tommy1729 "Truth is that what does not go away when you stop believing in it" tommy1729