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

  1. #1
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    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
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  2. #2
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    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)))
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