1
$\begingroup$

This is kind of a reverse question.

A few years back I was presented with a functional equation problem, I don't remember it completely, and now I would appreciate the help of the math.SE hivemind to recreate it.

It concerned a function $f:\Bbb Q^+\to \Bbb Q$, with $\Bbb Q^+$ being the strictly positive rationals. The problem gave two identities for $f$, one of them being $f(x) = f(x^{-1})$. I can't for the life of me recall what the second one was. However, I believe it related $f(x)$ and $f(x+1)$. There is a possibility that $f(1) = 1$ was also included as a restraint.

What I do remember, though, is the solution. With $a, b\in \Bbb Z$ coprime, $f(\frac{a}{b}) = ab$. Also, that the two identities together is in some way related to the Euclidian algorithm.

Long story short, does anyone know what the second identity is / could be?

3 Answers 3

0

The 2nd condition can be:

$x \cdot f(x+1) = (x+1) \cdot f(x)$

This gives, $f(m) = m$ for all $m \in \mathbf{Z}$ when used with the fact $f(1) = 1$.

To extend this to positive rationals, we can use $f(x) = f(x^{-1})$ and a repeated use of Euclid's division algorithm (as you mentioned).

  • 0
    This looks like the right one to me. Thanks a lot.2012-10-13
1

This has nothing to do with $f(x+1)$, but what about: $f(x^2)=f(x)^2$

Edit: what about this one: $f(x+1)+f(x-1)=2\cdot f(x)$

  • 0
    @Henry oh true, I was still thinking of the $f$ in the OP which does obviously not work for $x=0$ :/ Thanks2012-10-12
0

If $a$ and $b$ are coprime then $a+b$ and $b$ are also coprime, so you get $ f\left(\frac{a}{b} + 1\right) = f\left(\frac{a+b}{b}\right) = (a+b)b = ab+b^2 \text{ if } a,b \text{ coprime} $ and $ f\left(\frac{b}{a} + 1\right) = f\left(\frac{b+a}{a}\right) = (b+a)a = ab+a^2 \text{ if } a,b \text{ coprime} $ and thus generally $ \frac{f(q^{-1} + 1) - f(q^{-1})}{f(q+1) - f(q)} = q^2 $