Let $f(x)$ be a function defined for all positive real numbers satisfying the conditions $f(x) 0$ for all $x 0$ and $f(x - y) = \sqrt{f(xy) + 1}$ for all $x y 0$. Determine $f(2009)$.

Question

Let $f(x)$ be a function defined for all positive real numbers satisfying the conditions $f(x) > 0$ for all $x > 0$ and $f(x - y) = \sqrt{f(xy) + 1}$ for all $x > y > 0$. Determine $f(2009)$.

jhannybean · Answer

@AravindG

jhannybean · Answer

Because 1 is a little bigger than 0?...

jhannybean · Answer

Confused!!!

dan815 · Answer

ah nvm scratch this stuff i dont think its the right way

beginnersmind · Answer

dan I think the math is just off. Use used f(2) = 2 :)

dan815 · Answer

oh lets use this we can use that induction step after all

beginnersmind · Answer

The start is to rewrite  $f(x - y) = \sqrt{f(xy) + 1}$ as
$$f(xy) = [f(x-y)]^2 - 1$$
In particular using that 2009 = 49-41=8

$$f(2009) = [f(8)]^2 - 1$$
So we're not looking for a closed formula. Just a few values like f(1) and f(2) would be enough.

beginnersmind · Answer

edit: I meant 2009 = 49*41 of course.

jhannybean · Answer

oohhh haha ok that makes a little bit more sense  lol

dan815 · Answer

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