Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Find the functions $f:\Bbb R_*^+ \to \Bbb R_*^+$ such that:

$$f(x)f \left(\frac{1}{x}\right)=1$$

share|improve this question
    
What is $R_\ast^+$? –  Asaf Karagila Oct 24 '12 at 15:05
    
I presume it's better known, and less ambiguously written, as $\Bbb R_{> 0}$. –  Lord_Farin Oct 24 '12 at 15:06
    
Am I missing something? $f(x) = x$. –  copper.hat Oct 24 '12 at 15:07
    
Where's your proof that no other $f$ does the job? –  Lord_Farin Oct 24 '12 at 15:08
    
None. $f(x) = 1$ also works. As does $f(x) = \frac{1}{x}$. I guess the question is to characterize all such functions... –  copper.hat Oct 24 '12 at 15:08
show 2 more comments

2 Answers 2

up vote 1 down vote accepted

In fact this functional equation belongs to the form of http://eqworld.ipmnet.ru/en/solutions/fe/fe2111.pdf.

The general solution is $f(x)=\pm e^{C\left(x,\frac{1}{x}\right)}$ , where $C(u,v)$ is any antisymmetric function.

share|improve this answer
    
+1 for paul. I deserve a point for my comment imho :) –  mick Oct 29 '12 at 22:33
add comment

For any function $\tilde{f}:[1,\infty)\rightarrow \mathbb{R}_+^*$ such that $f(1)=1$, there is a unique extension $f:(0,\infty)\rightarrow \mathbb{R}_+^*$ defined for $x<1$ by $f(x)=1/\tilde{f}(1/x)$.

share|improve this answer
    
+1 very nice... –  copper.hat Oct 24 '12 at 15:47
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.