Question

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

Answer 1

The above given function is a periodic function. A periodic function is a function that repeats its values at regular intervals. Or we can say a function with a graph that repeats identically from left to right. It can be represented by $f\left( x \right)=f\left( x+p \right)$. We have to find the period of the given function. The period is the horizontal distance required for a complete cycle of graph.

Complete step by step solution:
The given periodic function is:
$\Rightarrow f\left( x,y \right)=f\left( 2x+2y,2y-2x \right).......(1)$
Since it is a periodic function then it will repeat itself, then we can write is as:
$\Rightarrow f\left( 2x+2y,2y-2x \right)=f\left( 2\left( 2x+2y \right)+2\left( 2y-2x \right),2\left( 2y-2x \right)-2\left( 2x+2y \right) \right)$
Now simplify it
$\Rightarrow f\left( 2x+2y,2y-2x \right)=f\left[ 4x+4y+4y-4x,4y-4x-4x-4y \right]$
$\Rightarrow f\left( 2x+2y,2y-2x \right)=f\left( 8y,-8x \right)$
Now by equation (1), if we compare equation (1), and the above expression then we get
$\Rightarrow f\left( x,y \right)=f\left( 8y,-8x \right).....(2)$
Since it is a periodic function so we again repeat the process, then we get
$\begin{aligned} & \Rightarrow f\left( 8y,-8x \right)=f\left[ 8\left( -8x \right),-8\left( 8y \right) \right] \\\ & \Rightarrow f\left( 8y,-8x \right)=f\left( -64x,-64y \right) \\\ \end{aligned}$
Now if we compare eq. (1), eq. (2), and the above expression, then we get
$\Rightarrow f\left( x,y \right)=f\left( 2x+2y,2y-2x \right)=f\left( 8y,-8x \right)=f\left( -64x,-64y \right)$
Now from the above equation, we get
$\Rightarrow f\left( x,y \right)=f\left( -64x,-64y \right)........(3)$
Again we will do same procedure, then we get
$\Rightarrow f\left( -64x,-64y \right)=f\left( -64\left( -64x \right),-64\left( -64y \right) \right)$
We know we can write $64={{2}^{6}}$ , putting the value in above expression, then we get
$\Rightarrow f\left( -64x,-64y \right)=f\left( -{{2}^{6}}\left( -{{2}^{6}}x \right),-{{2}^{6}}\left( -{{2}^{6}}y \right) \right)$
Now, we know that ${{a}^{x}}.{{a}^{y}}={{a}^{x+y}}$, now applying this in above equation, then we get
$\Rightarrow f\left( -64x,-64y \right)=f\left( {{2}^{12}}x,{{2}^{12}}y \right)$
Now from equation (3), we get
$\Rightarrow f\left( x,y \right)=f\left( {{2}^{12}}x,{{2}^{12}}y \right)$
Now we have prove that the function $g\left( x \right)=f\left( {{2}^{x}},0 \right)$is periodic, so from the above the expression, we get
$\Rightarrow g\left( x \right)=f\left( {{2}^{x}},0 \right)=f\left( {{2}^{12}}{{2}^{x}},0 \right)=f\left( {{2}^{12+x}},0 \right).......\left( 4 \right)$
Given, $g\left( x,0 \right)=f\left( {{2}^{x}},0 \right)$
Now from equation (4), we get
$\begin{aligned} & \Rightarrow g\left( x,0 \right)=f\left( {{2}^{x}},0 \right)=f\left( {{2}^{12+x}},0 \right) \\\ & \Rightarrow g\left( x,0 \right)=g\left( x+12,0 \right) \\\ \end{aligned}$
Hence g(x) is a periodic function and its period is $12$.

Note: It is not hard to check the periodic function of any function. To determine the periodicity and period of a function, we follow some steps:
1 Put f(x+T)=f(x)
2 If there exists a positive number “T” satisfying equation in “1” and it is independent of “X”, then f(x) is periodic.
3 The least value of “T” is the period of the periodic function.