Question

Let $f$ be a real valued function defined for all real numbers $x$ such that for some positive constant $a$ the equation $f(x+a)=\frac{1}{2}+\sqrt{f\left(x\right)-\left(f\right(x){)}^2}$ holds for all $x$.If the function is periodic enter $1$, else enter $0$.

Answer 1

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

$f(x+a+a)$

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

$=\frac{1}{2}+\sqrt{\frac{1}{2}+\sqrt{f\left(x\right)-f(x{)}^2}-(\frac{1}{4}+f(x)-f(x{)}^2+\sqrt{f\left(x\right)-f(x{)}^2})}$

$=\frac{1}{2}+\sqrt{f(x{)}^2-f(x)+\frac{1}{4}}$

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

$=\frac{1}{2}+f\left(x\right)-\frac{1}{2}$

$=f\left(x\right)$

Hence

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

Thus $f\left(x\right)$ is a periodic function with period $2a$.