# explicit formula for coefficients of Laurent series

Give an explicit formula for the coefficients of the Laurent series on $A:=\{z: |z|>1\}$ for the function $g(z)=\frac{e^z}{z-1}$.

I know how to go about finding the Laurent series, using the geometric series for $\frac{1}{1-z}$. But how do I obtain an explicit formula for ALL the coefficients?.

I have seen some earlier posts,but I am not entirely clear on their work. Can we use the coefficient formula in the Laurent series expansion and then use Cauchy's formula?.

Any help is appreciated.

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You can at first determine the Laurent series for $$e^{z} = \sum_{m=0}^\infty \frac{z^m}{m!}$$ and $$\frac1{z-1} = \sum_{n=0}^\infty z^{-(n+1)} , \qquad |z|>1$$ independently.

To multiply two Laurent series, we can use this formula or simply calculate \begin{align}\frac{e^{z}}{z-1}= &\left(\sum_{m=0}^\infty \frac{z^m}{m!} \right) \left(\sum_{n=0}^\infty z^{-(n+1)}\right)\\ &=\sum_{m,n\in\mathbb{Z}}[0\leq m][0\leq n] \frac{z^{m-n-1}}{m!}\\ &=\sum_{k,m} [0\leq m][0\leq m-k-1] \frac{z^{k}}{m!}\\ &= \sum_{k=-\infty}^{-1} z^k \underbrace{\sum_{m=0}^\infty \frac1{m!}}_{e} + \sum_{k=0}^\infty z^k \sum_{m=1+k}^\infty \frac1{m!}. \end{align} with $k=m-n-1$ and where I used Iverson's bracket.

Edit: As Robert Israel pointed out, we can still express the last sum using the incomplete $\gamma$-function $$\gamma(n,x) = \int_0^x t^{n-1} e^{-t} dt = x^n (n-1)! e^{-x} \sum_{k=0}^\infty \frac{x^k}{(n+k)!} .$$ Setting $x=1$, we have with $n\geq 1$ $$\gamma(n,1) = (n-1)! \sum_{i=n}^\infty \frac{1}{i!}.$$ Thus, $$\frac{e^z}{z-1} = e \sum_{k=-\infty}^{-1} z^k+ \sum_{k=0}^\infty\frac{\gamma(k+1,1)}{k!} z^k.$$

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You may want to use the incomplete Gamma function to express $\sum_{m=1+k}^\infty 1/m!$ –  Robert Israel Dec 31 '12 at 10:13
@RobertIsrael: thanks for the hint, I added a few lines... –  Fabian Dec 31 '12 at 10:27
You left out the $z^k$ in the first sum of the last equation. –  Robert Israel Dec 31 '12 at 20:24
Thanks so much for your comments. I understand what you are doing. But what stops us from expanding e^z in powers of (z-1). –  Jack Dawkins Jan 1 '13 at 0:11
@user54755: Laurent expansion are always in rings about a centre. If you expand in powers of $z-1$ you get a laurent expansion in a ring centered at $1$. But the question obviously asks for a Laurent expansion in a ring centered at 0. –  Fabian Jan 1 '13 at 6:22

First observe $$\frac{1}{1-z}=-\frac1z\frac{1}{1-\frac1z}=\sum_{-\infty}^{n=-1}-z^n$$ and$$e^z=\sum_{k=0}^{\infty}\frac{z^k}{n!}$$ The general term of the first series is $a_n=0$, $n\ge 0$ and $a_n=-1$, $n<0$ while for the second $b_n=\frac1{n!}$ for $n\ge 0$ and $b_n=0$ for $n<0$. $$\sum_{n=-\infty}^{\infty}a_n\sum_{n=-\infty}^{\infty}b_n=\sum_{n=-\infty}^{\infty}c_n$$ where $$c_n=\sum_{k=-\infty}^{\infty}a_kb_{n-k}$$ Can you continue?

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Thanks for the help. But why can't I expand e^z in powers of (z-1)? –  Jack Dawkins Jan 1 '13 at 0:13
@user54755 Sorry for the late answer. As I wanted an expansion at $0$ I expanded $e^z$ in powers of $z$. Did you do what you wrote? –  Nameless Jan 1 '13 at 8:43
Yeah, I did. But if you expand around z=0 then the series for 1/(1-z) isn't valid there right? or am I missing something here? –  Jack Dawkins Jan 1 '13 at 8:57
@user54755 By expanding at $0$ I mean at $A=\left\{\left|z\right|>1\right\}$ which is centered at $0$ and the expansion is valid as in my answer. –  Nameless Jan 1 '13 at 9:00