# Laurent series of $f(z)=\frac{1}{z(z-1)(z-2)}.$

Consider $$f(z)=\frac{1}{z(z-1)(z-2)}.$$

I want to determine the Laurent series in the point $z_0=0$ on $0<|z|<1$.

Partial decomposition yields:

$$f(z)=\frac{1}{z(z-1)(z-2)}=(1/2)\cdot (1/z) - (1/(z-1)) + (1/2)(1/(z-2)).$$

Is the general strategy now, to try to use the geometric series?

$(1/(z-1))=-(1/(1-z))=-\sum_{k=0}^\infty z^k$

$\displaystyle 1/(z-2)=-(1/2)\frac{1}{1-\frac{z}{2}}=-(1/2)\cdot\sum_{k=0}^\infty (z/2)^k$

So $f(z)=(1/2)\cdot (1/z)+(1/2)\cdot\sum_{k=0}^\infty z^k-(1/4)\cdot\sum_{k=0}^\infty (z/2)^k$ (*)

Some questions:

1) What is the difference between a Laurent and a Taylor series? I don't get it. It seems you calculate them the same.

2) Why didn't we write (1/z) as a series expansion too?

3) What makes the end result (*) to be a Laurent series?

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## 1 Answer

Partial fractions and geometric series give $$\begin{align} \frac1{(1-x)(2-x)} &=\frac1{1-x}-\frac1{2-x}\\ &=\frac1{1-x}-\frac12\frac1{1-x/2}\\ &=(1+x+x^2+x^3+x^4+\dots)\\ &-\left(\frac12+\frac14x+\frac18x^2+\frac1{16}x^3+\frac1{32x^4}+\dots\right)\\ &=\frac12+\frac34x+\frac78x^2+\frac{15}{16}x^3+\frac{31}{32}x^4+\dots \end{align}$$ Thus, the Laurent series for $\frac1{x(x-1)(x-2)}$ at $x=0$ is $$\frac1{x(x-1)(x-2)}=\frac1{2x}+\frac34+\frac78x+\frac{15}{16}x^2+\frac{31}{32}x^3+\dots$$ We could also expand the series at $x=1$. Let $y=x-1$ and then $$\begin{align} \frac1{x(x-1)(x-2)} &=\frac1{(y+1)y(y-1)}\\ &=-\frac1y\frac1{1-y^2}\\ &=-\frac1y-y-y^3-y^5-y^7-\dots\\ &=-\frac1{x-1}-(x-1)-(x-1)^3-(x-1)^5-(x-1)^7-\dots \end{align}$$

1. The Laurent series is much like the Taylor series except terms of negative degree are allowed.

2. We don't expand $\frac1x$ since there is no power series for $\frac1x$ at $0$ other than $\frac1x$.

3. Definition. It is a power series at a point, $x_0$ which can have both positive and negative powers of $x-x_0$.

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That's very good robjohn, thank you so much! Question on point 3: In my case, there are no negative powers of $z-0$, doesn't this fact change something? Do we still have a Laurent series, even if all powers are positive ($k$ starts at $0$ in both cases) – Chris Jul 24 '12 at 0:41

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@Chris: Yes, a Laurent series may have both positive and negative powers of $x-x_0$, but it need not have both. A Taylor series is a Laurent series, but we usually just call it a Taylor series unless it has negative powers of $x-x_0$. – robjohn♦ Jul 24 '12 at 0:44
Great, things are much clearer now. I was until now utterly confused about that! – Chris Jul 24 '12 at 0:49
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