Let f be a real-valued function defined for all real numbers, such that for some a > 0 we have
f(x + a) = 1/2 + √(f(x) - f(x)2) for all x.
Prove that f is periodic, and give an example of such a non-constant f for a = 1.
Solution
Directly from the equality given: f(x+a) ≥ 1/2 for all x, and hence f(x) ≥ 1/2 for all x.
So f(x+2a) = 1/2 + √( f(x+a) - f(x+a)2 ) = 1/2 + √f(x+a) √(1 - f(x+a)) = 1/2 + √(1/4 - f(x) + f(x)2) = 1/2 + (f(x) - 1/2) = f(x). So f is periodic with period 2a.
We may take f(x) to be arbitrary in the interval [0,1). For example, let f(x) = 1 for 0 ≤ x < 1, f(x) = 1/2 for 1 ≤ x < 2. Then use f(x+2) = f(x) to define f(x) for all other values of x.
Solutions are also available in: Samuel L Greitzer, International Mathematical Olympiads 1959-1977, MAA 1978, and in István Reiman, International Mathematical Olympiad 1959-1999, ISBN 189-8855-48-X.
© John Scholes
jscholes@kalva.demon.co.uk
3 Oct 1998
Last corrected/updated 9 Oct 2003