From: edgar@math.ohio-state.edu (G. A. Edgar) Newsgroups: sci.math.symbolic Subject: Re: The correct definition of a strange function ... Date: Mon, 27 Jul 1998 09:35:21 -0400 In article <6peunq$dac$1@news00.btx.dtag.de>, tobi.kamke@t-online.de (Tobias Kamke) wrote: > The function f is defined as follows: > > f(x) = p * f(2x) 0 < x <= 0.5 > f(x) = p + (1 - p) * f(2x - 1) 0.5 < x < 1 > f(1) = 1 > f(0) = 0 > > 0 <= p <= 1 > > How can I calculate f((1/3)), f((2/5)), f((1/1984)), ... ? > > What can you say about this function? Is it monotone? What's the > derivation of f(x)? > > > Thanks, > > Tobi There is a unique continuous monotone function f with these properties. It is "singular" in the sense that its derivative is zero almost everywhere. Reference: Billingsley, PROBABILITY AND MEASURE (first edition) Example 31.1. This example goes back to Ces�ro (1906). For f(1/3) you get an equation: f(1/3) = p f(2/3) = p (p + (1-p)f(1/3)) and thus f(1/3) = p^2/(1-p+p^2) . Similarly for any rational: the base 2 expansion is eventually periodic, so you get an equation to solve. The process may be computerized. If I have not made a mistake, f(2/5) = p^2(2-p)/(1-p^2+2p^3-p^4) ; f(1/1984) = p^11/(1-p^4+p^5) . -- Gerald A. Edgar edgar@math.ohio-state.edu