Let $f$ be a function which maps $\mathbb{Q}^{+}\to\mathbb{Q}^{+}$. And it satisfies $$ \left\{ \begin{array}{l} f(x)+f\left(\frac{1}{x}\right)=1\\ f(2x)=2f(f(x)) \end{array}\right. $$ Show that $f\left(\frac{2012}{2013}\right)=\frac{2012}{4025}$.

I tried to prove that $f(x)=\frac{x}{x+1}$(which satisfies all these comditions), but failed.

So I tried induction:

Surely, $f(1)=\frac{1}{2}$ by $f(1)+f(1/1)=1$.

By $f(2)=2f\left(f(1)\right)=2f\left(\frac{1}{2}\right)$ ,we know $f\left(\frac{1}{2}\right)=\frac{1}{3},f(2)=\frac{2}{3}$.

Note that $f(1)=2f\left(f\left(\frac{1}{2}\right)\right)$, we have $f\left(\frac{1}{3}\right)=\frac{1}{4}$.

By $f(4)=2f(f(2))=2f\left(\frac{2}{3}\right)$ as well as $f\left(\frac{2}{3}\right)=2f\left(\frac{1}{4}\right)$, we know $f\left(\frac{1}{4}\right)=\frac{1}{5},f\left(\frac{2}{3}\right)=\frac{2}{5},f(4)=\frac{4}{5}.$

Unfortunately, these procedures seems to have no similarity, and the larger the denominator, the more complex the procedure is. Is there any hints or solutions? Thanks for attention!

  • 2
    $\begingroup$ Given that the numbers $2012$ and $2013$ occur, people might think this is from a competition which is not over yet. Perhaps it would be good to add where you got the problem from. $\endgroup$
    – user50407
    Jan 29 '13 at 10:22
  • $\begingroup$ @fs Sorry that I do not know the origin of the question either, or I will Google the answer.. $\endgroup$
    – Golbez
    Jan 29 '13 at 10:24

Use induction on $q$. There are gaps in the proof which I cannot fill yet.

First establish the hypothesis for $q=1$. We already know that $f(1)=1$ and $f(2)=2/3$. Let's assume that $f(x)=\frac{x}{x+1}$ for all integer $x<=2n$. Then we have: $ f(\frac{1}{2n})=\frac{1}{2n+1}; $ hence $2f(f(\frac{1}{2n}))=2f(\frac{1}{2n+1})$ or $f(\frac{1}{n}) = 2f(\frac{1}{2n+1})$ Hence we've proven it for $x=2n+1$.

I cannot yet establish the case $x=2n+2$ Once this is done we'll have established the case $q=1$.

Next establish the hypothesis for $q=2$. The first few cases are calculated by hand. Assume the hypothesis $f(p/2)=p/(p+2)$ holds for $p<=2n$ and we'll prove it for $p=2n+1$. Use that because of the induction we have: $\frac{2}{2n+1}=f(\frac{2}{2n-1})$. This may be enough to prove this step.

Now for the main induction step we will try to prove that $f(p/q)=p/(p+q)$ Note that: $p/(q+1)=f(p/(q+1-p))$ due to the induction. Apply $f$ to both sides, multiply by 2 and use the second functional equation to calculate $f(p/(q+1))$. That will prove the induction step.

  • $\begingroup$ But this time,$f(\frac{2p}{q+1-p})$ should be calculated, which cannot be derived from the induction in $q$... $\endgroup$
    – Golbez
    Jan 29 '13 at 12:17
  • $\begingroup$ It can, if your induction hypothesis is that the proposition holds for all $q_0\leq q$. As I said this means you'll need to establish the induction step for q=1 and q=2. $\endgroup$
    – ivan
    Jan 29 '13 at 12:28
  • $\begingroup$ emm.. for instance $p=1$ and calculate $q=6$,this time ,what I need is $f(2/7)$, which is hard to find... $\endgroup$
    – Golbez
    Jan 29 '13 at 13:41
  • $\begingroup$ You are right. I have modified the answer but it doesn't look good. I still think that some kind of funny induction may be possible. $\endgroup$
    – ivan
    Jan 29 '13 at 15:19
  • $\begingroup$ I am stuck in proving $p=2n$ for hours. I will tried other techniques. Anyway, thank you for your advice! $\endgroup$
    – Golbez
    Jan 29 '13 at 15:27

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