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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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    So for $y>0$ we have $f(y)=f((y+1)^2+1)-f(y^2+1)$. Since $2^x>0$ the functional equation tells you nothing about $f(y)$ for negative or zero $y$ anyway.2012-10-08
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    The function given by $f(y)-f((y+1)^2+1)+f(y^2+1)$ is analytic on $\mathbb R$ and is zero on $(0,\infty)$, hence is zero on all of $\mathbb R$. Especially, $t\mapsto f(t+\frac12)$ is odd, i.e. we have $f(y)=-f(-1-y)$ and $f(-\frac12)=0$.2012-10-08
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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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