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This is an extension to : Functions $f: \Bbb Q_{+} \to \Bbb Z$ such that $f(\frac{1}{x}) = f(x)$ and $(x+1)f(x-1)= xf(x)$

What can be said about functions $f : \Bbb Q^*_+ \to \Bbb Q$ such as $f(\tfrac1x)=f(x)$ and $f(x) = (1+\tfrac1x)f(x-1)$ for all $x$?

(the difference with the original question being the functions' codomain).

Community
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Alexandre Halm
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1 Answers1

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Every strictly positive rational can be transformed into $1$ by applying $x \mapsto x-1$ and $x \mapsto 1/x$.

This is immediate by induction on $p+2q$ : If $p/q > 1$ we can transform it into $p/q-1 = (p-q)/q$, and if $p/q < 1$ we can transform it into $q/p$. At every step, $p+2q$ decrease.

As a result, for every rational $x$ there is a rational $y(x)$ such that for every solution $f$ of the problem, $f(x) = y(x) f(1)$.

$y(x)$ itself has to be the solution of the problem that has value $1$ at $1$, and we know from the previous question that $y(p/q) = (p+q)/2$.

Therefore, for any solution $f$, we have $f(p/q) = (p+q)f(1)/2$ for all coprime $p$ and $q$. And every choice of $f(1) \in \Bbb Q$ gives a solution.

mercio
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  • "As a result, for every rational x there is a rational y(x) such that for every solution f of the problem, $f(x)=y(x)f(1)$." > I'm not sure I follow you. Since $f(x)$ is rational, $y(x) = f(x)/f(1)$ satisfies your equality trivially. You probably meant something else? – Alexandre Halm Apr 26 '15 at 13:19
  • yes but your choice of $y$ depends a priori on some solution $f$. – mercio Apr 26 '15 at 13:20
  • for example, take $x=2/3$. For every $f$, $f(2/3) = f(3/2) = (1+2/3)f(1/2) = (1+2/3)f(2) = (1+2/3)(1+1/2) f(1)$ so we have $y(2/3) = (1+2/3)(1+1/2)$. $y(x)$ is just the product of all the factors corresponding to the transformations that bring $x$ to $1$. – mercio Apr 26 '15 at 13:22
  • ah ok, misreading on my part. "we know from the previous question that $y(p/q)=(p+q)/2$" > why? – Alexandre Halm Apr 26 '15 at 13:22
  • because $p/q \mapsto (p+q)/2$ is a solution and the image of $1$ by this is $1$. Alternatively I could have just said that $f \mapsto f(1)$ is an isomorphism of $\Bbb Q$-vector spaces from the set of solutions to $\Bbb Q$. If you know $1$ (nonzero) solution, you know all of them. – mercio Apr 26 '15 at 13:24
  • ah yes of course, again I misread your sentence. Clear enough! – Alexandre Halm Apr 26 '15 at 13:29