From: kovarik@mcmail.cis.McMaster.CA (Zdislav V. Kovarik) Subject: Re: iterated exp() Date: 18 Jun 1999 15:06:59 -0400 Newsgroups: sci.math Keywords: Schroeder's equation, iterated functions In article <7kdhcf$bk43@mx2.hrz.uni-essen.de>, Knuth Rueddenklau wrote: :Hello, : :the complex exp() and ln() have a fixed point :f = exp(f) = ln(f) (f= approx. 0.3 + i*1.33) . :When starting with a point p in the complex :plane and iteratively taking p_(n+1)=ln(p_n) :the generated set of points spirals inwards :toward f, approaching it ever closer. : :My question is: Can one find a parametric :representation of this spiral, like :s(n) = exp^(n)(p) = p_n ? : :Thanks for answers, :Urs Schreiber Yes and no: "when in trouble, change the variable". Every function f which has a strictly attractive fixed point c and is analytic in a neighbourhood of that point (ln will do, near the point you mentioned) is locally analytically conjugate to multiplication by a constant q, namely by q = f'(c). In formulas, there exists an analytic invertible function g mapping a neighbourhood of 0 onto a neighbourhood of c (and let us denote G its inverse) such that f(z) = g(q * G(z)) near z = c, or f(g(y)) = g(q * y) near y = 0. This is called the Schroeder's equation (f known, g unknown, q=f'(c)). This g can be normalized to have g'(0)=1, and then it can be found as a power series, one coefficient at a time. It then follows that for the iterative sequence z_(n+1) = f(z_n) we have z_n = g( q^n * G(z_0)) and the spiral you mentioned is a conformal image of a logarithmic spiral. You can get more fixed points of the exponential by seeking fixed points of z |-> ln(z) + 2 * k * pi * i with k integer. More on iterations: Marek Kuczma: Functional Equations in a Single Variable Monografie Matematyczne 46, Warsaw 1968 and other books and articles mentioned there, or in books co-authored by Marek Kuczma. Cheers, ZVK(Slavek).