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I am having trouble trying to expand this function using Laurent series, and finding the residue$$f(x)=\frac{1}{1+e^z}$$ If I replace $e^z$ with its series I get $$f(x)=\frac{1}{1+\sum_{n=0}^\infty\frac{z^n}{n!}}$$ and the expansion is $$\frac{1}{2}+\frac{1}{1+z}+\frac{1}{1+\frac{z^2}{n!}}+...$$ which is not useful in finding the residue at the singularity point(s) $z_0=(2k+1)\pi i$.

What would be a proper expansion, or is there another way to solve this?

edit:fixed summation formula layout

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  • $\begingroup$ What do you want exactly? The laurent series? The residue? Both? A closed formula for the laurent series or just a few terms? $\endgroup$ – Git Gud Jun 12 '14 at 11:45
  • $\begingroup$ I need to find the residue, and for that i thought I need a couple of terms from the expansion. The terms are not really necessary, if there is another way to find the residue. $\endgroup$ – Andrew Jun 12 '14 at 11:49
  • $\begingroup$ This answer of mine should help you. The first part of the answer gives you a formula to find the residue. The second part hints at how you can find the laurent series to find the residue. $\endgroup$ – Git Gud Jun 12 '14 at 11:50
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You can use the formula for finding residues of a pole $z_0$ of order $1$:

$$\text{Res}(f, z_0)=\lim \limits_{z\to z_0}\left[(z-z_0)f(z)\right].$$

The singularities here are the points $(2k+1)\pi i$, where $k$ ranges over the integers.

To use the above formula you need to first prove that these points are poles of order $1$, that is, you need to prove that these points are not removable discontinuities and you need to prove that $\lim \limits_{z\to (2k+1)\pi i}\left(\dfrac{z-(2k+1)\pi i}{e^z+1}\right)\in \mathbb C$. This limit actually is the same as in the above formula, so if you find it, you immediately find your residue.

One has $$\begin{align} \lim \limits_{z\to (2k+1)\pi i}\left(\dfrac{z-(2k+1)\pi i}{e^z+1}\right)&=\lim \limits_{w\to 0}\left(\dfrac{w}{e^{w+(2k+1)\pi i}+1}\right)\\ &=\lim \limits_{w\to 0}\left(\dfrac{w}{e^{w+\pi i}+1}\right)\\ &=\lim \limits_{w\to 0}\left(\dfrac{1}{e^{w+\pi i}}\right)\\ &=-1. \end{align}$$

Hence $\forall k\in \mathbb Z\left(\text{Res}(f,(2k+1)\pi i)=-1\right)$ which agrees with Wolfram Alpha.

In the above I used Cauchy's rule. If you want to avoid without having to resort to Laurent series, you can first prove that $\lim \limits_{w\to 0}\left(\dfrac{e^{w+\pi i}+1}w\right)=1$ and this you can do with the Taylor Series.

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  • $\begingroup$ Thank you for the help, but i still have one small issue, and this has been my problem from the start. How exactly do you prove that those singularity points are poles of order 1? $\endgroup$ – Andrew Jun 12 '14 at 12:35
  • $\begingroup$ I addressed that issue in my answer "You need to prove that these points are not removable discontinuities and you need to prove that $\lim \limits_{z\to (2k+1)\pi i}\left(\dfrac{z-(2k+1)\pi i}{e^z+1}\right)\in \mathbb C$". I took care of the limit, you just have to prove that they aren't removable discontinuities. $\endgroup$ – Git Gud Jun 12 '14 at 12:40
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As Git explained, you need to calculate $$\lim_{z\to (2k+1)\pi i}\dfrac{z-(2k+1)\pi i}{e^z+1} \,.$$ The fastest way of doing this is to observe that $$\lim_{z\to (2k+1)\pi i}\dfrac{e^z+1}{z-(2k+1)\pi i} \,.$$ is just the definition of the derivative of $e^z$ at $z=(2k+1)\pi i$. Thus $$\lim_{z\to (2k+1)\pi i}\dfrac{e^z+1}{z-(2k+1)\pi i}=e^{(2k+1)\pi i}=-1$$

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  • $\begingroup$ Nice. I usually am on the look out for these tricks. I missed it. $\endgroup$ – Git Gud Jun 12 '14 at 12:25
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Understanding you only want some coefficients to start: using Pari/GP

1/(1+exp(x)) + O(x^8)
%693 = 1/2 - 1/4*x + 1/48*x^3 - 1/480*x^5 + 17/80640*x^7 + O(x^8)

which is also $$ = {1 \over 2} - {1 \over 4} {x \over 1!} + {1 \over 8} {x^3 \over 3!} - {1 \over 4} {x^5 \over 5!}+ {17 \over 16} {x^7 \over 7!} - {31 \over 4} {x^9 \over 9!}+ {691 \over 8} {x^11 \over 11!} - ... $$

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