1968 IMO Problems/Problem 5

Problem 5

Let $f$ be a real-valued function defined for all real numbers $x$ such that, for some positive constant $a$, the equation \[f(x + a) = \frac{1}{2} + \sqrt{f(x) - (f(x))^2}\] holds for all $x$.

(a) Prove that the function $f$ is periodic (i.e., there exists a positive number $b$ such that $f(x + b) = f(x)$ for all $x$).

(b) For $a = 1$, give an example of a non-constant function with the required properties.

Solution

(a) Since \[f(x+a) \ge \frac{1}{2}\] is true for any $x$, and \[f(x+a)(1-f(x+a)) = \frac{1}{4} - (f(x)-(f(x))^2) = (\frac{1}{2}-f(x))^2\]

We have: \[f(x+2a) = \frac{1}{2} + \sqrt{(\frac{1}{2}-f(x))^2} = \frac{1}{2} + (f(x) - \frac{1}{2}) = f(x)\] Therefore $f$ is periodic, with $2a>0$ as a period.

(b) $f(x) = 1$ when $2n\le x < 2n+1$ for some integer $n$, and $f(x)=\frac{1}{2}$ when $2n+1\le x < 2n+2$ for some integer $n$.

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