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Which entire function $f\left(x\right)$ goes asymptotically to $\dfrac{e^{-x}}{x}$ as $x$ goes to infinity with $x$ positive? That is, $\left(e^{-x}/x \right)/f \left(x \right) \rightarrow 1$.

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Plenty of them. $(e^x-e^{-2x})/x$ is the simplest example. As a matter of fact, every continuous function $g(x)$ on $\mathbb R$ can be approximated by an entire function with arbitrary continuous precision $\varepsilon(x)>0$. Voting to close. –  fedja Feb 18 '13 at 12:39
    
You cannot say a lot about $f$ without additional assumption on the uniform growth at infinity. For instance if such a $f$ exists then any other entire function $f+g$, where $g$ is a tower of exponential of length greater than $1$ $g(x)=\exp(-\exp(\exp(\ldots(\exp x)\ldots)))$, will also satisfy your assumption. –  Loïc Teyssier Feb 18 '13 at 12:40
    
Dear fedja, Do you mean $\left(e^{-x} - e^{-2x} \right)/x$? Best, davwood83 –  davwood83 Feb 18 '13 at 13:48
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1 Answer 1

As mentioned in the comments, the asymptotic behavior of $f$ along the real axis doesn't really tell you anything about the function globally.

For example, your function $f$ can behave in any way you like as $x\to-\infty$.

Indeed, the following is implied by Arakeljan's (or Nersesjan's) approximation theorem: Let $A$ be any finite union of disjoint curves tending to infinity, and let $g:A\to\mathbb{C}$ and $\varepsilon:A\to(0,\infty)$ be continuous.

Then there is an entire function $f$ such that $|f(z)-g(z)|<\varepsilon(z)$ for every $z\in A$.

So, for example, let $A=(-\infty,0] \cup [1,\infty)$, let $g$ be the function $e^{-x}/x$ on $[1,\infty)$ and let $g$ be arbitrary on $(-\infty,0]$. Then you can find an entire function that approximates this function arbitrarily closely.

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