7
$\begingroup$

If $f(\frac{x}{y})=\frac{f(x)}{f(y)} \, , f(y),y \neq 0$ and $f'(1)=2$ then $f(x)=$?

I am not sure where to begin, any hints on starting and steps is apreciated.

Thank you

$\endgroup$
4
  • $\begingroup$ Try and write the problem in the question, not in the title... $\endgroup$ Feb 16, 2013 at 10:38
  • $\begingroup$ if there is a problem in the font, then to make it clear, the derivative of the function at x=1 is 2. $\endgroup$ Feb 16, 2013 at 10:38
  • $\begingroup$ To Nicolas: this just gives the information that f(1)=1 (as f(x) cannot be 0) $\endgroup$ Feb 16, 2013 at 10:48
  • $\begingroup$ I could not read the post properly, now with the TeXing can. $\endgroup$ Feb 16, 2013 at 10:53

4 Answers 4

6
$\begingroup$

$f(x/y)=f(x)/f(y),\forall x,y$

  • if you take $x=y$ you obtain $f(1)=1$. As a consequence $f(1/y)=1/f(y)$.

  • $x=0$ and $f$ not constant implies $f(0)=0$.

  • if $a,b \neq 0$ then $f(ab)=f(a/(1/b))=f(a)/(f(1/b))=f(a)f(b)$. If for an element $x \neq 0$ we have $f(x)=0$ then $f$ would be constant, which contradicts $f'(1)=2$.

  • In particular you have $f(x^2)=(f(x))^2>0$.

  • Define for $x>0$ $g(x)=\log(f(e^x))$. Then $$g(x+y)=\log(f(e^xe^y))=\log(f(e^x))+\log(f(e^y))=g(x)+g(y)$$

  • This means that $g$ satisfies a Cauchy functional equation and it is continuous (since you assume $f$ to be differentiable at $1$). This means that there exists $a$ such that $g(x)=ax$.

  • This leads to: $\log(f(e^x))=ax \Rightarrow f(e^x)=(e^{x})^a \Rightarrow f(x)=x^a$ for $x>0$.

  • Now the condition $f'(1)=2$ implies that $a=2$ and $f(x)=x^2$ for $x>0$.

  • Because $f(x^2)=f(x)^2$ even for negative $x$ you have $f(-x)=\pm f(x)$.

$\endgroup$
4
  • $\begingroup$ Notice that your second line is not so rigorous: the inequality holds when both $f(y)$ and $y$ are not zero. And you seem to assume this to derive that $f(y)\neq 0 $ for $y\neq 0$. Just saying... $\endgroup$
    – awllower
    Feb 16, 2013 at 11:35
  • $\begingroup$ @awllower: If you choose $y\neq 0$ such that $f(y) \neq 0$ it's the same thing. $\endgroup$ Feb 16, 2013 at 11:45
  • $\begingroup$ OK. There is no problem. I am just suggesting that maybe it could be explaned further. $\endgroup$
    – awllower
    Feb 16, 2013 at 11:47
  • $\begingroup$ thanks for the contributions!!! $\endgroup$ Feb 16, 2013 at 13:32
3
$\begingroup$

Let $f(x)=x^2$. Then it is one.

For $f(x/y)=(x/y)^2=x^2/y^2=f(x)/f(y)$ and $f'(1)=2$.

Edit:
First we find that $f(1)=f(y/y)=f(y)/f(y)=1$. So, $$\begin{align}f(y^{-1}) &=f(1/y)=f(1)/f(y)=f(y)^{-1}\end{align}$$ and $$\begin{align} f(xy) &=f(x/y^{-1})=f(x)f(y) \,\,,\text{when}\,\,\,y\neq 0.\end{align}$$ That is, when restricted to $\mathbb R^*$, it is an endomomorphism of $\mathbb R^*$. Of course linear functions are all such endomorphisms. Notice that the word "linear" here means linear in the multiplicative case: $x\cdot x\cdot \cdot\cdot x=x^k$. Together with the derivative condition, ths gives us the answer.

Thanks.

$\endgroup$
4
  • $\begingroup$ please kind sir, have some spare words to go with that answer? $\endgroup$
    – jimjim
    Feb 16, 2013 at 11:13
  • $\begingroup$ Sorry if this appeared to be rude. $\endgroup$
    – awllower
    Feb 16, 2013 at 11:17
  • $\begingroup$ Thanks. I will keep that in mind. $\endgroup$
    – awllower
    Feb 16, 2013 at 11:19
  • $\begingroup$ x^2 solves the problem, but is it the only solution?? $\endgroup$ Feb 16, 2013 at 13:34
1
$\begingroup$

Hint For $x=y$ you get $f(1)=1$. What can you figure out from that and $f'(1)=2$?

$\endgroup$
0
-2
$\begingroup$

Linear functions satisfied this function-equation trivially.

Let $f(x)=2x$ then $f'(x)=2$ for all $x$ and $f\left(\frac{x}{y}\right)=\frac{f(x)}{f(y)}$.

That's all :-)

R

$\endgroup$
1
  • 4
    $\begingroup$ I don't think that would work- as f(x/y)=2(x/y) and f(x)/f(y)=x/y $\endgroup$ Feb 16, 2013 at 10:49

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .