MathLabs

Problem 5

Let ff be a real-valued function defined for all real numbers xx such that, for some positive constant aa, the equation f(x+a)=12+f(x)−f(x)2f(x+a)=\frac12+\sqrt{f(x)-f(x)^2} holds for all xx. (a) Prove that ff is periodic. (b) For a=1a=1, give an example of a non-constant function with the required property.
Step 1 of 5: The equation forces f to be at least one-half
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In plain words

The recurrence never outputs a value below 12\frac12: it starts at 12\frac12 and adds a nonnegative quantity. Since every real input can be written as x+ax+a, this one-sided bound applies globally, not merely to a translated copy of the function.

f(x+a)=12+f(x)−f(x)2≥12f(x+a)=\frac12+\sqrt{f(x)-f(x)^2}\ge\frac12
Detailed analysis

The square root is nonnegative, so the defining equation gives f(x+a)=12+f(x)−f(x)2≥12f(x+a)=\frac12+\sqrt{f(x)-f(x)^2}\ge\frac12. Because x+ax+a ranges over all real numbers when xx does, this proves f(y)≥12f(y)\ge\frac12 for every real yy. In particular, every value of ff lies in the domain where the next square-root manipulation is valid.