# A functional equation problem on $\mathbb{Q}^{+}$: $f(x)+f\left(\frac{1}{x}\right)=1$ and $f(2x)=2f\bigl(f(x)\bigr)$

Let $$f$$ be a function which maps $$\mathbb{Q}^{+}$$ to $$\mathbb{Q}^{+}$$ and satisfies $$\begin{cases} f(x)+f\left(\frac{1}{x}\right)=1\\ f(2x)=2f\bigl(f(x)\bigr) \end{cases}$$ 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 $$f(1)+f\left(\frac{1}{1}\right)=1$$.

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

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

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

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!

• 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.
– user50407
Commented Jan 29, 2013 at 10:22
• @fs Sorry that I do not know the origin of the question either, or I will Google the answer.. Commented Jan 29, 2013 at 10:24
• Commented Jan 1, 2022 at 7:58

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.

• But this time，$f(\frac{2p}{q+1-p})$ should be calculated, which cannot be derived from the induction in $q$... Commented Jan 29, 2013 at 12:17
• 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.
– ivan
Commented Jan 29, 2013 at 12:28
• emm.. for instance $p=1$ and calculate $q=6$,this time ,what I need is $f(2/7)$, which is hard to find... Commented Jan 29, 2013 at 13:41
• 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.
– ivan
Commented Jan 29, 2013 at 15:19
• I am stuck in proving $p=2n$ for hours. I will tried other techniques. Anyway, thank you for your advice! Commented Jan 29, 2013 at 15:27