$f$ is a real valued function defined for all real numbers $x$ such that $0\le f(x)\le \dfrac12$ and for some fixed $a>0$, $f(x+a)=\dfrac12-\sqrt{f(x)-(f(x))^2}$ for all $x$. Show that the function is periodic.

My attempt:

I could not really make an attempt, I tried testing $a$ as the period of the function but that didn't really help. My next attempt was considering an arbitrary period and then bring it under the definition. But, that didn't help either. Please help. Thank you.


1 Answer 1


What about squaring both sides?

$$\left[f(x+a) - \frac 12 \right]^2= f(x) - f(x)^2 = \frac 14 - \left[ f(x) - \frac 12 \right]^2.$$

It follows that $$\left[f(x+2a) - \frac 12 \right]^2= \frac 14 - \left[ f(x+a) - \frac 12 \right]^2 = \left[ f(x) - \frac 12\right]^2.$$

Consequently $$ \left| f(x+2a) - \frac 12 \right| = \left| f(x) - \frac 12 \right|$$ but since $0 \le f \le \dfrac 12$ both arguments are nonpositive and you get $f(x+2a) = f(x)$.

  • 1
    $\begingroup$ now, that was unexpected and brilliant on your part. thank you. $\endgroup$
    – Swadhin
    Feb 26, 2015 at 15:00

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .