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 2 of 5: Compute the product of the shifted value and its complement
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In plain words

The expression u(1−u)u(1-u) is designed to cancel the square root: the shift raises uu above 12\frac12, and the product with 1−u1-u converts the square-root term into a perfect square involving the original value.

f(x+a)(1−f(x+a))=14−(f(x)−f(x)2)=(12−f(x))2f(x+a)(1-f(x+a))=\frac14-(f(x)-f(x)^2)=(\frac12-f(x))^2
Detailed analysis

Let u=f(x+a)u=f(x+a). From the recurrence, u=12+f(x)−f(x)2u=\frac12+\sqrt{f(x)-f(x)^2}. Therefore u(1−u)=14−(u−12)2=14−(f(x)−f(x)2)=(12−f(x))2u(1-u)=\frac14-(u-\frac12)^2=\frac14-(f(x)-f(x)^2)=(\frac12-f(x))^2. Thus f(x+a)(1−f(x+a))=(12−f(x))2f(x+a)(1-f(x+a))=(\frac12-f(x))^2.