Question

If $f\left(x\right)\epsilon [1,2]$ when $x\epsilon R$ and for a fixed positive real number p, $f(x+p)=1+\sqrt{2f\left(x\right)-{f\left(x\right)}^2}$ for all $x\epsilon R$ then prove that f(x) is a periodic function .

Answer 1

$f(x+p)=1+\sqrt{2f\left(x\right)-f(x{)}^2}$

put $x=x+p$

$f(x+2p)=1+\sqrt{2f(x+p)-f(x+p{)}^2}$

$f(x+2p)=1+\sqrt{(1+\sqrt{2f\left(x\right)-f(x{)}^2}(1-\sqrt{2f\left(x\right)-f(x{)}^2}})$

$f(x+2p)=1+\sqrt{1-2f\left(x\right)+f(x{)}^2}=1+f\left(x\right)-1$

$f(x+2p)=f\left(x\right)$

hence $f\left(x\right)$ is a periodic function with period $=2p$