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Let $f(x)$ be a non-constant real-analytic function and for real $x$ it satisfies :

$f(2^x) = f(4^x + 2^{x+1} + 2) - f(4^x + 1)$

Before you ask if this simplifies by writing $2^x = y$ note that $2^x$ is never equal to $0$.

I think $f(x)$ is unique upto a multiplication constant $C$.

How to find $f(x)$ ?

What would make an excellent asymptotic to $f(x)$ ?

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    @ Mark Bennet : That is true. @ Hagen von Eitzen : Im sorry but if you say$f(y)$= "..." then f(y)-"..." is 0 by definition. I did not define $f$ for negative $y$. Considering those 2 I do not know what you mean.2012-10-08

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