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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 . . .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, and then the half iterate is 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 - 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 |