Question

Let f(x, y) be a periodic function satisfying the condition $f(x,y)=f(2x+2y),(2y-2x)$ $\forall x,y\in R$. Now, define a function $g$ by $g\left(x\right)=f(2^x,0)$. Then, show $g\left(x\right)$ is a periodic function, and find its period .

Answer 1

Given $f(x,y)=f(2x+2y,2y-2x)$ (1)

$\Rightarrow f(2x+2y,2y-2x)=f\left[2\right(2x+2y)+2(2y-2x),2(2y-2x)-2(2x+2y\left)\right]$ [using Eq. (1)]

$\Rightarrow f(2x+2y,2y-2x)=f(8y,-8x)$

$\Rightarrow f(x,y)=f(2x+2y,2y-2x)=f(8y,-8x)$ (by (1))

Now again, since $f(x,y)=f(8y,-8x)$ .....(2)
$\Rightarrow f(8y,-8x)=f\left[8\right(-8x),-8(8y\left)\right]$ [using Eq. (2)]
$\Rightarrow f(8y,-8x)=f(-64x,-64y)$
$\Rightarrow f(x,y)=f(2x+2y,2y-2x)=f(8y,-8x)=f(-64x,-64y)$

Again since, $f(x,y)=f(-64x,-64y)$ .....(3)
$\Rightarrow f(-64x,-64y)=f(-64\times (-64x),-64\times (-64y\left)\right)$
$=f(2^{12}x,2^{12}y)$
$\Rightarrow f(x,y)=f(2^{12}x,2^{12}y)$ [using Eq. (3)]
$\Rightarrow f(2^x,0)=f(2^{12}{.2}^x,0)=f(2^{12+x},0)$ .....(4)

Given, $g(x,0)=f(2^x,0)$
$\Rightarrow$ $g(x,0)=f(2^x,0)=f(2^{12+x},0)$ [using Eq. (4)]
$\Rightarrow g(x,0)=g(x+12,0)$
Hence, $g\left(x\right)$ is periodic with period 12.