MHB Understanding the Laurent Series of $\frac{1}{z(z-1)(z-2)}$

[ognik]

Blundering on, this problem will help me confirm what I think I know ...

Find the Laurent series for $ f(z) = \frac{1}{z(z-1)(z-2)} = \frac{1}{2z}+\frac{1}{1-z} -\frac{1}{4}\frac{1}{1-\frac{z}{2}} $
I found this definition of the LS: $ f(z) = \sum_{-\infty}^{+\infty}{a}_{n}(z-{z}_{0})^n = \sum_{n=0}^{\infty}{a}_{n}(z-{z}_{0})^n + \sum_{n=1}^{\infty}{a}_{n}(z-{z}_{0})^{-n}$ Shouldn't the '+' be a '-'?

So singularities (poles) at 0, 1 & 2 , so I have 3 Annuli(r,R;centre) = Ann(0,1;(0)), Ann(1,2;(0)), Ann(2,infinity;(0))

1) Ann(0,1;(0))
0 < |z| : $ f(z) = \frac{1}{2z} - \sum_{n=1}^{\infty}{z}^{-n} - \frac{1}{4} \sum_{n=1}^{\infty} (\frac{2}{z})^n $ is the Laurent series part?
|z| < 1: $ f(z) = \frac{1}{2z} + \sum_{n=1}^{\infty}{z}^{n} + \frac{1}{4} \sum_{n=1}^{\infty} (\frac{z}{2})^n $ is the Taylor series part?

However, for the annulus enclosed by 0 < |z| < 1, we only need the Taylor series part above? The pole at z=0 is covered by the 1st term of the series, and the Taylor series covers |z| < 1?

What puzzles me is that the 3rd term is $ |\frac{z}{2}| < 1, IE. |z| < 2 $?
Appreciate all corrections and advice.

 

[Fernando Revilla]

I found this definition of the LS: $ f(z) = \sum_{-\infty}^{+\infty}{a}_{n}(z-{z}_{0})^n = \sum_{n=0}^{\infty}{a}_{n}(z-{z}_{0})^n + \sum_{n=1}^{\infty}{a}_{n}(z-{z}_{0})^{-n}$ Shouldn't the '+' be a '-'?

The coefficients $a_n$ are univocally determined, then you need not to write $-.$ Now, denote $$f_1(z)=\dfrac{1}{z},\quad f_2(z)=\dfrac{1}{1-z},\quad f_3(z)=\dfrac{1}{1-z/2}.$$ So, $$f(z)=\frac{1}{2}f_1(z)+f_2(z)-\frac{1}{4}f_3(z).$$ Using the well known theorem about the geometric series

$$f_1(z)=\frac{1}{z}\quad (|z|>0)$$ $$f_2(z)=\dfrac{1}{1-z}=\left \{ \begin{matrix} \displaystyle\sum_{n=0}^{+\infty}z^n\quad (0<|z|<1)\quad (S_1)\\-\dfrac{1}{z}\dfrac{1}{1-1/z}=-\displaystyle\frac{1}{z}\sum_{n=0}^{+\infty}\dfrac{1}{z^{n}}=- \sum_{n=0}^{+\infty}\dfrac{1}{z^{n+1}}\quad (1<|z|)\quad (S_2)\end{matrix}\right.$$ $$f_3(z)=\dfrac{1}{1-z/2}=\left \{ \begin{matrix} \displaystyle\sum_{n=0}^{+\infty}\frac{z^n}{2^n}\quad (0<|z|<2)\quad (S_3)\\-\dfrac{2}{z}\dfrac{1}{1-2/z}=-\displaystyle\frac{2}{z}\sum_{n=0}^{+\infty}\dfrac{1}{(2z)^{n}}=- \sum_{n=0}^{+\infty}\dfrac{1}{2^{n-1}z^{n+1}}\quad (2<|z|)\quad (S_4)\end{matrix}\right.$$ Then, $$f(z)=\left \{ \begin{matrix} \dfrac{1}{2z}+S_1-\dfrac{1}{4}S_3=\ldots &\text{if}& 0<|z|<1\\\dfrac{1}{2z}+S_2-\dfrac{1}{4}S_3=\ldots &\text{if}& 1<|z|<2\\ \dfrac{1}{2z}+S_2-\dfrac{1}{4}S_4=\ldots &\text{if}& 2<|z|<+\infty \end{matrix}\right.$$

 

[ognik]

I think you're using a slightly different approach to the one I used, but we seem to end up with basically the same terms, for example for your S2 I have $ - \sum_{n=1}^{\infty} \frac{1}{{z}^{n}} $ - because my lower limit (for LS terms) is n=1, they are the same are they not? I just found the method I used easier for me personally to remember and use.

I was looking first at the 1st annulus to make sure my approach was right, before doing the others - this seems to have been a bad idea :-)

The question then comes down to how to decide which annulus each term belongs, looking at your solution was most helpful - it confirms some of the terms can appear in more than 1 region.

Just to check though, please confirm my overall solution is:

$ f(z) = \frac{1}{2z} + \sum_{n=0}^{\infty}{z}^{n}- \frac{1}{4}\sum_{n=0}^{\infty} (\frac{z}{2})^n $, 0 < |z| < 1

$ = \frac{1}{2z} - \sum_{n=1}^{\infty}{z}^{-n}- \frac{1}{4}\sum_{n=0}^{\infty} (\frac{z}{2})^n $, 1 < |z| < 2

$ = \frac{1}{2z} - \sum_{n=1}^{\infty}{z}^{-n}- \frac{1}{4}\sum_{n=1}^{\infty} (\frac{2}{z})^n $, |z| > 2

 

[Fernando Revilla]

Just to check though, please confirm my overall solution is:

$ f(z) = \frac{1}{2z} + \sum_{n=0}^{\infty}{z}^{n}- \frac{1}{4}\sum_{n=0}^{\infty} (\frac{z}{2})^n $, 0 < |z| < 1

$ = \frac{1}{2z} - \sum_{n=1}^{\infty}{z}^{-n}- \frac{1}{4}\sum_{n=0}^{\infty} (\frac{z}{2})^n $, 1 < |z| < 2

$ = \frac{1}{2z} - \sum_{n=1}^{\infty}{z}^{-n}- \frac{1}{4}\sum_{n=1}^{\infty} (\frac{2}{z})^n $, |z| > 2

Right.

 

[ognik]

Much appreciated Fernando