# functional equation

#### jacks

##### Well-known member
If $f$ is a Real valued function on the set of real no. such that for any real $a$ and $b$ and $f(af(b)) = ab$. Then $f(2012) =$

#### Sudharaka

##### Well-known member
MHB Math Helper
If $f$ is a Real valued function on the set of real no. such that for any real $a$ and $b$ and $f(af(b)) = ab$. Then $f(2012) =$
Hi jacks,

$$\mbox{By substituting, }a=1\mbox{ we get, }f[f(b)]=b\mbox{ for each }b\in\Re\,.\mbox{ Therefore the inverse of }f\mbox{ is itself.}$$

$$\mbox{Suppose there exist a real number }n\mbox{ such that, }f(n)=1\,.\mbox{ Then, }$$

$f\left[mf\left(\frac{1}{m}\right)\right]=1=f(n)\mbox{ where }m\in\Re\mbox{ and }m\neq 0$

$\Rightarrow mf\left(\frac{1}{m}\right)=n$

$\Rightarrow f\left(\frac{1}{m}\right)=\frac{n}{m}~~~~~~~(1)$

$\Rightarrow f\left(\frac{1}{m}\right)=f\left[nf\left(\frac{1}{m}\right)\right]$

$\Rightarrow \frac{1}{m}=nf\left(\frac{1}{m}\right)$

$\Rightarrow f\left(\frac{1}{m}\right)=\frac{1}{mn}~~~~~~~~(2)$

By (1) and (2);

$n=\pm 1$

Therefore $$n$$ can be $$1\mbox{ or }-1$$ depending on the function $$f$$. Hence,

If the function $$f$$ is defined such that, $$f(1)=1,$$

$f(2012)=f(2012f(1))=2012$

If the function $$f$$ is defined such that, $$f(-1)=1,$$

$f(2012)=f(2012f(-1))=-2012$

#### Jester

##### Well-known member
MHB Math Helper
Here's mine (same conclusion).

First I'll show that $f(1) \ne 0$

Sub $a = 0$ into the functional equation giving $f(0) = 0$

Then sub $a = b= 1$ giving $f(f(1)) =1$. If $f(1) = 0$ then$f(0) = 1$ but $f(0) = 0$ contradiction.

Now set $b = 1$ so $f(a f(1)) = a$. If we let $f(1) = k$ then we have $f(ka) = a$ or $f(x) = \dfrac{x}{k}$

Returning back to the original function equation gives

$f(af(b)) = f\left(\dfrac{ab}{k}\right) = \dfrac{ab}{k^2} = ab$ giving $k = \pm 1$

Thus, $f(x) = \pm x$