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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!

Mohsen Shahriari
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Golbez
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1 Answers1

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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.

ivan
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  • But this time，$f(\frac{2p}{q+1-p})$ should be calculated, which cannot be derived from the induction in $q$... – Golbez Jan 29 '13 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 Jan 29 '13 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... – Golbez Jan 29 '13 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 Jan 29 '13 at 15:19
  • I am stuck in proving $p=2n$ for hours. I will tried other techniques. Anyway, thank you for your advice! – Golbez Jan 29 '13 at 15:27