2 fixpoints , 1 period --> method of iteration series - 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: 2 fixpoints , 1 period --> method of iteration series (/showthread.php?tid=1107) 2 fixpoints , 1 period --> method of iteration series - tommy1729 - 12/21/2016 Some credit Goes to Gottfried. Assume a strictly increasing analytic function f(x) with fixpoints A and B such that B > A > 0 and F(x) > 0 for x > 0 And also the Period of the first fixpoint = the Period of the second fixpoint. ( notice this is possible because Period Q = Period - Q !!! Example f ' (A) = 1/ f ' (B) ) Now we construct a function that can be used as a kind of slog type , including if necc BRANCHES for the FUNCTIONAL inverse to make iterations of f. Call the slog type function T Then we have for a suitable C whenever f is not growing too fast : ( C depends on the growth of f and the derivate at the fixpoints ) T(x) = ... Rf^[-2] c^2 + Rf^[-1] C + R ( x ) + Rf(x)/C + R f^[2](x)/c^2 + ... Where Rf^[y] means R ( f^[y] (x) ) and R(x) = (x-a)(x-b). Now we split Up t(x) into t1(x) + t2(x) in the obvious way ( separate positive and negative iterates ) Such that we arrive at T(x) = t1(x) + t2(x) T(f(x)) = t1(x) C + t2(x) / C. So finally ( again for inverse use branches ) We get the slog type equation T( f(x)^[y] ) = t1(x) C^y + t2(x) C^{-y} As desired. Since the periods agree and we can use analytic continuation we have a solution that is analytic in the interval [A, +oo] , assuming that T is analytic in a nonzero radius. So we have a solution agreeing on 2 fixpoints !! Regards Tommy1729