Chủ đề này có 1 trả lời

[tpctnd]

Tìm tất cả các hàm số f thỏa mãn
$f^{2}(x)=f(x-y)f(y-x)+f(3y+x)f(3y-x)$

[ThEdArKlOrD]

If you mean $f^2(x)=f(x)^2$

Let $P(x,y)$ denote $f(x)^2=f(x-y)f(y-x)+f(3y+x)f(3y-x)$ for all $x,y \in \mathbb{R}$

$P(x,0)$ give us $f(x)^2=2f(x)f(-x)$,so $f(-x)^2=2f(x)f(-x)$ give us $f(x)^2=f(-x)^2$ for all $x\in \mathbb{R}$,

From $f(x)^2=2f(x)f(-x)$, suppose there exist $t$ such that $f(t)\neq 0$, we get $f(t)=2f(-t)$, but since $f(x)^2=f(-x)^2$ for all $x\in \mathbb{R}$, give $f(t)=f(-t)=0$, contradiction, so $f(x)=0$ for all $x\in \mathbb{R}$

 

If you mean $f^2(x)=f(f(x))$

Let $P(x,y)$ denote $f(f(x))=f(x-y)f(y-x)+f(3y+x)f(3y-x)$ for all $x,y \in \mathbb{R}$

$P(x,0)$ give us $f(f(x))=2f(x)f(-x)=f(f(-x))$ for all $x\in \mathbb{R}$ and $P(x,x)$ give us $f(f(x))=f(0)^2+f(4x)f(2x)$ for all $x\in \mathbb{R}$

Then $P(-y,y)$ give $f(f(-y))=f(-2y)f(2y)+f(2y)f(4y)$ for all $y\in \mathbb{R}$, so $f(0)^2+f(4x)f(2x)=f(f(x))=f(f(-x))=f(-2x)f(2x)+f(2x)f(4x)$ for all $x\in \mathbb{R}$

So $f(0)^2=f(-2x)f(2x)$ for all $x\in \mathbb{R}$, this give us $f(f(x))=f(f(-x))=2f(0)^2$ for all $x\in \mathbb{R}$

Then, $P(x,y)$ change to $2f(0)^2=f(0)^2+f(3y+x)f(3y-x)$ for all $x,y\in \mathbb{R}$, so $f(0)^2=f(3y)^2=f(0)f(6y)$ for all $y\in \mathbb{R}$

So $f(0)=0$ give $f(y)=0$ for all $y\in \mathbb{R}$ or $f(0)\neq 0$ give us $f(6y)=f(0)$ for all $y\in \mathbb{R}$

So $f$ is constant function and the rest is easy