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I discovered the following question by accident, and found it interesting, but I only resolved it by brute force:

Problem: Let f: $\mathbb{R} \rightarrow \mathbb{R}$ have the property $(\pi)$ iff they satisfy the equation

$$f(s^2 + f(t)) = t + f(s)^2 $$

Determine the family of all functions with the property $(\pi)$.

Solution: If we can show $f(0) := y$ to be zero, then letting $t$ run over all real values, with $s = 0$, shows $f(f(x)) = x$ $\forall x \in \mathbb{R}$, from which it is clear that $\mathrm{Id}(x)$ has the property $(\pi)$.

Taking $s = 0, t = 0$ shows (1) $f(y) = y^2$. Taking $s = y, t = 0$ shows (2) $f(y^2 + y) = (f(y))^2$, so that substitution from (1) gives (3) $f(y^2 + y) = y^4$. However, letting $s = 0, t = y$ shows (4) $f(y^2) = y + y^2$, so that $f(f(y^2)) = y^4$ by substituting the LHS of (4) in (3), while letting $s= 0, t = y^2$ shows directly that (5) $f(f(y^2)) = 2y^2$. Equating (4) and (5) says $2y^2 = y^4$, so $y = -\sqrt{2}, 0,$ or $\sqrt{2}.$

Assume $y = \sqrt{2}.$ Letting $t = f(y^2), s = y$ says (6) $f(f(f(y^2)) + y^2) = f(3y^2)$ $ = f(y^2) + (f(y))^2$; but $s = \sqrt{2}y, t = y$ says (7) $f(3y^2) = (f(\sqrt{2}y))^2 + y = (f(y^2))^2 + y$. Since $(f(y))^2 = y^4$, $f(y^2) = y + y^2$, equating the RHS of (6) and (7) leaves

$$ f(y^2) + (f(y))^2 = (f(y^2))^2 + y \implies y + y^2 + y^4 = (y^4 + 2y^3 + y^2) + y \implies 2y^3 = 0$$

which is absurd. Assuming $y = - \sqrt{2}$ and setting $s = -\sqrt{2}y, t = y$ in (7) similarly shows that $y \neq - \sqrt{2}.$ We conclude that $f(0) = 0$.

That $\forall x \in \mathbb{R} (f(f(x)) = x)$ says that $f(x) = f^{-1}(x);$ this requires that $f$ be one-to-one. Since $f(0) = 0$, we have $t \neq 0 \implies f(t) \neq 0$; we can now show that $f$ is increasing. Pick $x, y \in \mathbb{R}, x < y,$ and set $\delta = y - x.$ By choosing $t = f(x), s = \sqrt{\delta}$, we have

$$f(\delta + f(f(x))) = f(\delta + x) = f(x) + f(\sqrt{\delta})^2 \implies f(x + \delta) - f(x) = f(\sqrt{\delta})^2$$ $$\therefore f(y) - f(x) > 0$$

since we established that $f(\sqrt{\delta}) \neq 0$. This proves that $f$ is increasing. Since, in fact, $\mathrm{Id}(x)$ is the only $f: \mathbb{R} \rightarrow \mathbb{R}$ which is both increasing and an involution, this says that $f(x) = x$ is the sole function with the property ($\pi$).


So several things:

1) What field(?) of study does this question come from? Is it relevant to quotidian mathematics?

2) This was tricky. It took a lot of messing around with various combinations to get the main result, and I had the uncomfortable feeling that I was being inefficient. (I filled up a page and a half of notebook!) Does anybody know any books/have any advice on how to improve on questions like these?

3) If this is its own field, what other fields are best connected with it? (i.e., furnish handy tools)

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1 Answer

up vote 6 down vote accepted

This is called a functional equation. They are very hard to solve in general, but there are books on the subject. You should not expect an elegant solution in general.

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Do functional equations come into play for working mathematicians? –  user1296727 Jul 26 '12 at 14:22
    
Certainly there are various functional equations (e.g. that of the Riemann zeta function) that are very important in mathematics. But the kinds of functional equations that appear on Olympiads are not, I think, terribly relevant to anything else. Mostly they are nice puzzles. (But to some extent this is true of most Olympiad problems.) I asked an MO question about this: mathoverflow.net/questions/53431/… –  Qiaochu Yuan Jul 26 '12 at 14:24
    
Heh, you and I had pretty much the same reaction to these questions. Thank you, however, for the link. –  user1296727 Jul 26 '12 at 14:30
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