Determine function such that $f\left(\sqrt{x_1^2+(f(x_1))^2}\right) = f\left(\sqrt{x_2^2+(f(x_2))^2}\right)$ for every $x_1,x_2$.

Question

Determine a numerical no constant function $f$ such that for all $x_1$, $x_2$ in its domaine of definition, the equality $$f\left(\sqrt{x_1^2+(f(x_1))^2}\right) = f\left(\sqrt{x_2^2+(f(x_2))^2}\right)$$ is verified. Can you give some special property of your solution $f$?

Answer 1

Let $g(x) = \sqrt{x^2 + f^2(x)}$. We need to show that $f(g(x_1)) = f(g(x_2))$, for all valid $x_1,x_2$. One way to approach that is to make $g$ constant, that is:

$$\sqrt{x^2 + f^2(x)} = c$$ $$x^2 + f^2(x) = c^2$$ $$f(x) = \sqrt{c^2 - x^2}$$

For some contant $c$. That will satisfy your original equation for all $x_1,x_2 \in [-c,c]$.

Answer 2

The domain of f is the set of x such that $\sqrt{x^2 + f^2(x)}\in\mathbb{R}$; besides, since the given equality is true for every $x_1, x_2$ in the domain of f, we must have $\sqrt{x^2 + f^2(x)}$= constant in order to f don't be constant. Hence $f(x) = \sqrt{c^2 - x^2}$ for an arbitrary constant c > 0, as Barry has answered above.

The special property of the function f is that it is its own inverse, $f\circ f(x) = x$ (it is an involution as can be easily proved).