Question

Check whether the function defined by $f\left( x+\lambda \right)=1+\sqrt{2f\left( x \right)-{{f}^{2}}\left( x \right)},x\in \mathsf{\mathbb{R}},\lambda >0$ is periodic or not. If yes, then find its period. $$$$

Answer 1

We use the domain and range of the square root function to find the range of $f\left( x \right)$. We take $f\left( x+\lambda \right)-1=\sqrt{2f\left( x \right)-{{f}^{2}}\left( x \right)}$ and square both side, form a complete square ${{\left( f\left( x \right)-1 \right)}^{2}}$, replace $x$ with $x+\lambda$ and simplify to get squared expressions both side We take square root both side to get absolute values and use the range of $f\left( x \right)$ to get the period.

Complete step-by-step solution
We know that if the function $f$ is periodic then there exists nonzero constant $T$ such that
$f\left( x+T \right)=f\left( x \right)$
We are given the real valued function;
$f\left( x+\lambda \right)=1+\sqrt{2f\left( x \right)-{{f}^{2}}\left( x \right)},x\in \mathsf{\mathbb{R}},\lambda >0$
We know that the square root also returns non-negative values. So we have $\sqrt{2f\left( x \right)-{{f}^{2}}\left( x \right)}\ge 0$, then we have;

& f\left( x+\lambda \right)\ge 1+0 \\\ & \Rightarrow f\left( x+\lambda \right)\ge 1 \\\ \end{aligned}$$ Since square root is a strictly increasing function with given condition $\lambda >0$ we have; $$f\left( x \right)\ge 1....\left( 1 \right)$$ We know from the definition of square root function that the square root takes only non-negative values. So we have; $$\begin{aligned} & 2f\left( x \right)-{{f}^{2}}\left( x \right)\ge 0 \\\ & \Rightarrow f\left( x \right)\left( 2-f\left( x \right) \right)\ge 0 \\\ \end{aligned}$$ We multiply both side negative sign to have; $$\Rightarrow f\left( x \right)\left( f\left( x \right)-2 \right)\le 0.$$ We see from the above inequality that either we have $f\left( x \right)\ge 0,f\left( x \right)-2\le 0$ or $f\left( x \right)\le 0,f\left( x \right)-2\ge 0$. The latter deduction $f\left( x \right)\le 0,f\left( x \right)-2\ge 0$ is not possible since we have already obtained $f\left( x \right)\ge 1$. So we have; $$f\left( x \right)\ge 0,f\left( x \right)-2\le 0.....\left( 2 \right)$$ We take intersection of intervals or use wavy curve method to find the range of the functions $f\left( x \right)$ as $$f\left( x \right)\in \left[ 1,2 \right]$$ Now let us consider $$\begin{aligned} & f\left( x+\lambda \right)=1+\sqrt{2f\left( x \right)-{{f}^{2}}\left( x \right)} \\\ & \Rightarrow f\left( x+\lambda \right)-1=\sqrt{2f\left( x \right)-{{f}^{2}}\left( x \right)} \\\ \end{aligned}$$ We take square both sides to have; $$\Rightarrow {{\left( f\left( x+\lambda \right)-1 \right)}^{2}}=2f\left( x \right)-{{f}^{2}}\left( x \right)$$ Let us add $-1$ and $1$ in the right hand side of the above equation to get a complete square as; $$\begin{aligned} & \Rightarrow {{\left( f\left( x+\lambda \right)-1 \right)}^{2}}=2f\left( x \right)-{{f}^{2}}\left( x \right)+1-1 \\\ & \Rightarrow {{\left( f\left( x+\lambda \right)-1 \right)}^{2}}=1-\left( {{f}^{2}}\left( x \right)-2f\left( x \right)+1 \right) \\\ & \Rightarrow {{\left( f\left( x+\lambda \right)-1 \right)}^{2}}=1-{{\left( f\left( x \right)-1 \right)}^{2}}.....\left( 3 \right) \\\ \end{aligned}$$ Now let us replace $x$ with $x+\lambda $ in the above step and have; $$\Rightarrow {{\left\\{ f\left( x+2\lambda \right)-1 \right\\}}^{2}}=1-{{\left( f\left( x+\lambda \right)-1 \right)}^{2}}$$ We put the value of ${{\left( f\left( x+\lambda \right)-1 \right)}^{2}}$ in the right hand side of the equation obtained from (3). We have; $$\begin{aligned} & \Rightarrow {{\left\\{ f\left( x+2\lambda \right)-1 \right\\}}^{2}}=1-\left\\{ 1-{{\left( f\left( x \right)-1 \right)}^{2}} \right\\} \\\ & \Rightarrow {{\left\\{ f\left( x+2\lambda \right)-1 \right\\}}^{2}}=1-1+{{\left( f\left( x \right)-1 \right)}^{2}} \\\ & \Rightarrow {{\left\\{ f\left( x+2\lambda \right)-1 \right\\}}^{2}}={{\left( f\left( x \right)-1 \right)}^{2}} \\\ \end{aligned}$$ Let us take square root both side and have; $$\left| f\left( x+2\lambda \right)-1 \right|=\left| f\left( x \right)-1 \right|$$ Since the range of $f\left( x \right)$ is $\left[ 1,2 \right]$ $f\left( x+2\lambda \right)-1\ge 1-1=0$ and $f\left( x \right)-1\ge 1-1=0$. So we deduce the absolute values as; $$\begin{aligned} & f\left( x+2\lambda \right)-1=f\left( x \right)-1 \\\ & \Rightarrow f\left( x+2\lambda \right)=f\left( x \right) \\\ \end{aligned}$$ So the function is periodic and the period is $2\lambda$. $$$$ **Note:** We note that the least period if positive is called the fundamental period of function. Since We have used the identity $\sqrt{{{\left( f\left( x \right) \right)}^{2}}}=\left| f\left( x \right) \right|$ here with any real valued function $f\left( x \right)$. We must be careful when we are finding the range of $f\left( x \right)$ not $\left[ 0,2 \right]$. The square root function strictly increases in the domain $\left[ 0,\infty \right)$ because if $x > y $then $\sqrt{x} > \sqrt{y}$.