Let $f(x, y)$ be a periodic function satisfying $f(x, y) = f(2x + 2y, 2y - 2x)$ for all $x, y$. Define $g(x) = f(2^x, 0)$. Then find the period of function $g$.
Step-by-Step Solution
Key Concept: Apply the transformation $f(x, y) = f(2x + 2y, 2y - 2x)$ iteratively to $g(x) = f(2^x, 0)$ by substituting appropriate values of $x, y$ such that the second argument cycles through specific values. The key is recognizing that successive applications of the transformation with properly chosen arguments will eventually return to $f(2^x, 0)$ after a finite number of steps.
Given $g(x) = f(2^x, 0) = f(2, 2+2.0, 2.0-2.2^x)$, we compute successive compositions to find the period. Starting with $g(x+1) = f(2^{x+1}, -2^{x+1})$ and simplifying through function compositions, we obtain $f(-2^{x+7}, 2^{x+7}) = f(0, 2^{x+9}) = f(2^{x+10}, 2^{x+10}) = f(-2^{x+12}, 0) = g(x+12)$. Therefore, the period of $g$ is $12$.
Correct Answer: 12