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Let $f(x) > 0$ be a member of $C(a, {\infty})$, that is, the space of continuous functions from the real number a to $+{\infty}$ . Suppose further that $f$ tends to $0$ as $x$ tends to $+{\infty}$.

Is the following always true:

$\exists$ real numbers $a_n$ and a real number $b > a$ such that, $f(x) = \sum_{n=0}^{\infty} \frac{a_n}{x^{n}}$ for all $x > b$ ?

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That sounds incredibly unlikely to be true. For example if your $f$ is nowhere differentiable. – kahen Feb 1 '13 at 14:14
Not sure but what about $f(x) = \frac{1}{log(|x|)}$. Function f(1/x) does not have expansion at zero. – tom Feb 1 '13 at 14:18
up vote 3 down vote accepted

No, since you are asking for $g:x\to f(\frac1x)$ for $x>0$, extended by $g(0)=0$, to be given near $0$ by a power series, in other words to be an analytic function. But far from all continuous (or even indefinitely differentiable) functions are analytic. Try $f(x)=e^{-x}$, so that $g(x)=e^{-1/x}$ for a counterexample.

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Alright thanks for your answer. What if, I added the extra condition in the question that the area under the curve must be +inifnity (between any c > 0 and +infinity). – Adam Rubinson Feb 1 '13 at 14:30
Probably 1/log|x| or something does the trick. Maybe I am asking the wrong questions to what I want. – Adam Rubinson Feb 1 '13 at 14:43

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