The title pretty much gives it away already, I'm trying to find the difference between Laurent expanding $\frac{1}{1-x}$ around $x=1$ and $\frac{1}{1-e^x}$ around $x=0$. The first expansion does not work out very well, while the second one can be expanded quite easily. I find it odd that there is such a big difference between the two functions, and I was wondering if someone could give me some insights as to where this comes from.
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5The functions have a pole at $1$ resp. $0$, so it's not a Taylor expansion. Probably you're looking for a Laurent expansion? – Daniel Fischer Oct 28 '14 at 17:03
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Yes, you are correct, my apologies – user3183724 Oct 28 '14 at 17:11
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I may be missing something, but the Laurent expansion of $\frac{1}{1-x}$ around $1$ is pretty trivial: $-(x-1)^{-1}$. The second one is more complicated than the first. – Daniel Fischer Oct 28 '14 at 18:40
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Right, so I guess I should rephrase what I mean. I'm trying to approximate the expressions under certain limiting conditions (motivated from physics). Now the first one really just stays the way it is when expanding it, while the second one can be approximated to first order in x, for example. I was wondering why this cannot be done for the first, given that it is so similar to the second in these limiting cases. – user3183724 Oct 28 '14 at 19:33
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Thinking about it, the answer is probably just lies in the fact that even though e^0 = 1, e^x with x going to 0 behaves in a more complicated manner. – user3183724 Oct 28 '14 at 19:38