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I want to find the asymptotic expansion of $e^{-x}$ as $x \rightarrow \infty$. I know that $\psi_n(x) = x^{-n}$ is an asymptotic sequence, and so I want my expansion to be in the form $\sum_{n=0}^{\infty} \frac{a_k}{x^k}$. I have no idea how to go about getting an expansion in this form. Taylor expansions are the only thing I can think of, but I'd need to be expanding $e^{\frac{-1}{x}}$ to get it in that form.

Taking the definition (below) is just giving me that all the coefficients must be zero.

That is, $\lim_{x \rightarrow \infty} \frac{e^{-x}-a_kx^{-k}}{x^{-(k+1)}} = -\infty$ unless $a_k = 0$. What am I doing wrong?

user112495
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  • Or is it something along the lines of saying that $e^{\frac{-1}{x}}$ as $x \rightarrow 0$ is the same as $e^{-x}$ as $x \rightarrow \infty$? – user112495 May 26 '17 at 20:13
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    What do you mean by "asymptotic expansion?" – Adam Hughes May 26 '17 at 20:14
  • @AdamHughes an expression of the form $\sum_{n=0}^{\infty} a_k\psi_k(x)$ as $x \rightarrow \infty$ (in this case) such that $(f(x)-\sum_{n=0}^{n}a_k\psi_k(x)) = O(\psi_{n+1}(x))$. And here I want to use $\psi_k(x) = x^{-k}$, which I've checked is an asymptotic sequence. – user112495 May 26 '17 at 20:16
  • So I want to determine the $a_k$s such that that is true. But I don't know how to get it in a form that will allow me to determine the coefficients. – user112495 May 26 '17 at 20:35
  • It seems you want a Laurent Series about infinity, but I'm really confused by your wording. What is the problem with just writing $e^{-1/x}$ using the Taylor Series for $e^x$, which already gives you a Laurent Series around every point? – Brevan Ellefsen May 26 '17 at 20:54
  • @BrevanEllefsenThe question as given says 'Find all terms of the asymptotic expansion of $e^{-x} \simeq \sum_{k=1}^{\infty} \frac{a_k}{x^k}$', where the definition of asymptotic expansion is the one stated above.

    But I want an expansion for $e^{-x}$, not $e^{-\frac{1}{x}}$. And when I expanded $e^{-\frac{1}{x}}$. it gave me a series that wasn't consistent with my definition of asymptotic expansion.

     – user112495 May 26 '17 at 21:00
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    You are doing absolutely nothing wrong. The asymptotic expansion with respect to the given sequence is identically $0$. That merely means that the function decays faster than any measurement you try to apply, so it is not noticeable at these scales near $\infty$, It doesn't mean that it should be identically $0$ itself if that's what bothers you. – fedja May 27 '17 at 04:29

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