# Math Help - Help with solutions to the functional equations f(x) = f(x^2 + 2012)

1. ## Help with solutions to the functional equations f(x) = f(x^2 + 2012)

Hey guys, I've never solved problems of these sorts and I'm not quite sure where to begin. The problem is given the following equation

$f(x) = f(x^2 + 2012)$

where the domain of f is defined to be the reals, how do I find the maximum of the expression $f(a) - f(b)$ for some a,b in reals?

I thought about finding a class of solutions to f(x) = f(x^2 + 2012), namely given the relation $(x,y) \in \sim \implies f(x) = f(y)$, and the function g such that $g(x) = x^2 + 2012$, one naive equivalence class containing x, $E_x$, is just the sequence of $\{x, g(x), g(g(x)), g(g(g(x))), \hdots, g^\infty(x)\} = \{g^n(x)\}_{n=0}^\infty$. However, because $g(x') = g(g(x))$ may have solutions that are not in ${g^n(x)}$, we have to also include those within $E_x$. At this point, I'm just a bit confused by the algebra involved and what I should do next. Any/all help would be greatly appreciated.

Thanks guys for looking at this

2. ## Re: Help with solutions to the functional equations f(x) = f(x^2 + 2012)

Okay, I think I've figured it out

First: Show that each of the equivalence classes are distinct. Because g(x) is strictly increasing, then it must be the case that $x \ne y \implies g(x) \ne g(y)$, then so must the sequence x, g(x), g(g(x)) be strictly increasing, meaning that $x \ne y \implies g(x) \ne y \implies g^2(x) \ne y \implies \hdots \implies x \ne g^2(y) \hdots$, which means that $E_x \cap E_y = \not 0$

Hence, we only need to look at the values of f(x) within the domain of (0,2012). The answer is then just $\max_{x \in [0,2012]}(max(f(x)) - min(f(x)))$