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)=\frac12+\sqrt{f(x)-f(x)^2}

holds for all

x

. (a) Prove that

f

is periodic. (b) For

a=1

, give an example of a non-constant function with the required property.

Step 5 of 5: Verify the recurrence in both intervals

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In plain words

There are only two possible input states, and the recurrence swaps them deterministically: $1\mapsto\frac12$ and $\frac12\mapsto1$ . Checking these two transitions is therefore a complete verification, including interval endpoints because the intervals are half-open.

f(x+1)=\frac12+\sqrt{f(x)-f(x)^2}\quad\text{for the two cases }f(x)=1\text{ and }f(x)=\frac12

Detailed analysis

If $f(x)=1$ , then $\frac12+\sqrt{f(x)-f(x)^2}=\frac12$ , and $x+1$ lies in the next half-open interval where $f(x+1)=\frac12$ . If $f(x)=\frac12$ , then $\frac12+\sqrt{f(x)-f(x)^2}=\frac12+\sqrt{\frac14-\frac14}=1$ , and $x+1$ lies in the following interval where $f(x+1)=1$ . Thus the displayed recurrence holds for every real $x$ .