15th Brazil 1993 Problem 5 Find a real-valued function f(x) on the non-negative reals such that f(0) = 0, and f(2x+1) = 3f(x) + 5 for all x.
Let I n be the interval [2 n -1, 2 n+1 -1) for n = 0, 1, 2, ... . Then the I n are disjoint and cover the non-negative reals. Also x → 2x+1 maps I n onto I n+1 . Thus f is determined by the values it takes on I 0 = [0,1). These can be arbitrary, but the simplest is to take f(x) = 0 on [0,1). Then we get f(x) = 5 on [1,3), 20 on [3,7), 65 on [7,15) and by a simple induction (5/2)(3 n -1) on I n . Thanks to Suat Namli . 15th Brazil 1993 © John Scholes jscholes@kalva.demon.co.uk 27 February 2004 Last corrected/updated 27 Feb 04