Thread: Laurent series over different domains?

1. Laurent series over different domains?

Get the Laurent series expansion or f(z) = $\displaystyle \frac{1}{z+1} - \frac{4}{(z-2)^2}$ in the intervals;

(i) |z-1|<1 and
(ii) 1<|z-1|< 2

Now I don't have much trouble finding the Laurent series expansions but am not sure how to do it for specific intervals.
For example in (ii) of this problem i would get the Laurent expansions for $\displaystyle \frac{1}{z+1}$
for interval 1<|z-1|<2
and then get the expansions for $\displaystyle - \frac{4}{(z-2)^2}$
for 1<|z-1|<2,and add them all together. The textbook(correctly, just not sure why yet) just uses the expansion of $\displaystyle \frac{1}{z+1}$ for |z-1|<2 and adds to it the expansion of
$\displaystyle - \frac{4}{(z-2)^2}$ for 1<|z-1|. I'm not sure as to the procedure or the thinking behind getting Laurent expansions for these types of intervals
r1<|z-zo|< r2. Any help is much appreciated.

2. Re: Laurent series over different domains?

use this give you the exact expansion i think is the best posible
$\displaystyle -2 x+\frac{x^2}{4}-\frac{3 x^3}{2}-\frac{5 x^4}{4 (-2+x)^2}+\frac{x^5}{2 (-2+x)^2}+\frac{x^4}{1+x}$
kind regards

3. Re: Laurent series over different domains?

Originally Posted by punkstart
Get the Laurent series expansion or f(z) = $\displaystyle \frac{1}{z+1} - \frac{4}{(z-2)^2}$ in the intervals;

(i) |z-1|<1 and
(ii) 1<|z-1|< 2

Now I don't have much trouble finding the Laurent series expansions but am not sure how to do it for specific intervals.
For example in (ii) of this problem i would get the Laurent expansions for $\displaystyle \frac{1}{z+1}$
for interval 1<|z-1|<2
and then get the expansions for $\displaystyle - \frac{4}{(z-2)^2}$
for 1<|z-1|<2,and add them all together. The textbook(correctly, just not sure why yet) just uses the expansion of $\displaystyle \frac{1}{z+1}$ for |z-1|<2 and adds to it the expansion of
$\displaystyle - \frac{4}{(z-2)^2}$ for 1<|z-1|. I'm not sure as to the procedure or the thinking behind getting Laurent expansions for these types of intervals
r1<|z-zo|< r2. Any help is much appreciated.
(i) If we set $\displaystyle z+1=s \implies z=s-1$, then we have to find the Laurent expansion around s=0 of...

$\displaystyle f(s)= \frac{1}{s} - \frac{4}{(s-3)^{2}}$ (1)

Now we take into account that is...

$\displaystyle \frac{1}{(s-3)^{2}} = -\frac{d}{ds} \frac{1}{s-3}$ (2)

... and that for $\displaystyle |s|<3$ is...

$\displaystyle - \frac{1}{s-3}= \frac{1}{3}\ \frac{1}{1-\frac{s}{3}} = \frac{1}{3}\ \sum_{n=0}^{\infty} (\frac{s}{3})^{n}$ (3)

Deriving (3) and inserting the result in (1) we obtain...

$\displaystyle f(s)= \frac{1}{s} -\frac{4}{3} \sum_{n=1}^{\infty} \frac{n}{3^{n}}\ s^{n-1}$ (4)

... that is the result You are searching for. The (4) converges for $\displaystyle 0<|s|<3$ and setting $\displaystyle s=z+1$ You obtain the Laurent expansion of the original f(*) in z...

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$

4. Re: Laurent series over different domains?

Originally Posted by punkstart
Get the Laurent series expansion or f(z) = $\displaystyle \frac{1}{z+1} - \frac{4}{(z-2)^2}$ in the intervals;

(i) |z-1|<1 and
(ii) 1<|z-1|< 2

Now I don't have much trouble finding the Laurent series expansions but am not sure how to do it for specific intervals.
For example in (ii) of this problem i would get the Laurent expansions for $\displaystyle \frac{1}{z+1}$
for interval 1<|z-1|<2
and then get the expansions for $\displaystyle - \frac{4}{(z-2)^2}$
for 1<|z-1|<2,and add them all together. The textbook(correctly, just not sure why yet) just uses the expansion of $\displaystyle \frac{1}{z+1}$ for |z-1|<2 and adds to it the expansion of
$\displaystyle - \frac{4}{(z-2)^2}$ for 1<|z-1|. I'm not sure as to the procedure or the thinking behind getting Laurent expansions for these types of intervals
r1<|z-zo|< r2. Any help is much appreciated.
(ii) Setting $\displaystyle z-2=s \implies z=s+2$ Tou obtain that for $\displaystyle \s|<3$ is...

$\displaystyle \frac{1}{z+1}=\frac{1}{s+3}= \frac{1}{3}\ \frac{1}{1+\frac{s}{3}}= \frac{1}{3}\ \sum_{n=0}^{\infty} (-1)^{n} (\frac{s}{3})^{n}$ (1)

... so that is...

$\displaystyle \frac{1}{z+1} - \frac{4}{(z-2)^2}= - \frac{4}{(z-2)^2}+ \frac{1}{3}\ \sum_{n=0}^{\infty} (-1)^{n} (\frac{z-2}{3})^{n}$ (2)

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$