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What is the Laurent series expansion of $\;\dfrac1{\left(x-1\right)^2\left(x-3 \right)}\,$ over $\;0<\left\lvert x-1\right\rvert < 2$ and $0<\left\lvert x-3 \right\rvert < 3$?

Vlad
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saikiran
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  • Use partial fractions. Expand it and then use standard binomial expansions.$(1+x)^{-1}$ – ANUPAM BISHT Oct 19 '16 at 09:00
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    This question is ill-posed: There are infinite Laurent series expansions, unless you specify where you want it to be centered. Moreover, even having specified the center is not enough, you have to specify with respect to which domain you'd like to compute the Laurent expansion, so that it converges there. – b00n heT Oct 19 '16 at 09:10
  • Chiming in with @b00nheT. The question is ill-posed. In addition to the center you need to specify a desired annulus of convergence. For example, with Laurent series centered at the origin you get one series converging when $|x|<1$ (a Taylor series actually), another converging when $1<|x|<3$, and yet another converging when $|x|>3$. If $x$ is a real variable, you need to, at the very least, specify a single point where you want the series to converge (and also the center). – Jyrki Lahtonen Oct 19 '16 at 09:13
  • It is possible that the criticisim about the center is all Greek to you. To clarify: are you looking for a series of the form $\sum_n a_n x^n$, where $n$ may or may not take negative integers as values also? That would be a Laurent series centered at the origin. If you want powers $(x-a)^n$, then the series would be centered at $a$. – Jyrki Lahtonen Oct 19 '16 at 09:18
  • Please edit the question to include such details. Also include your own thoughts. The point of this advice is to dispel the thoughts that you simply want somebody to do your homework for you instead of learning how to do it (which is what we seek to achieve in this site). Without such additions, your question is guaranteed to attract negative attention making your future participation on the site more problematic. – Jyrki Lahtonen Oct 19 '16 at 09:22

1 Answers1

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As I have mentioned in the comment section,We use Partial Fractions here.

\begin{equation} \frac{1}{(z-1)^2(z-3)} = \frac{-1}{4(z-1)} - \frac{1}{2(z-1)^2} +\frac{1}{4(z-3)} ---(1) \end{equation}

(i) Expanding in the region $0<|z-1|<2$

This means we are trying to expand around the point $z = 1$. Hence we will not disturb the first two terms in equation (1).

For the third term, we do the following.

$\frac{1}{4(z-3)} = \frac{1}{4(1+(z-4))} = \frac{(1+(z-4))^{-1}}{4}$.

(Using the expansion, $(1+x)^{-1} = 1-x+x^2-\ldots$)

$ \implies \frac{(1+(z-4))^{-1}}{4} = 1 - (z-4) +(z-4)^2 - \ldots $

Hence final answer should be $$ \boxed{= (1 - (z-4) +(z-4)^2 - \ldots) +\frac{-1}{4(z-1)} - \frac{1}{2(z-1)^2}} $$

Similarly for, (i) Expanding in the region $0<|z-3|<3$

This means we are trying to expand around the point $z = 3$. Hence we will not disturb the third term in equation (1).

For the first two terms,

$\frac{-1}{4(z-1)} = \frac{-1}{4((z-3)+2)} =\frac{-1}{8(\frac{z-3}{2}+1)}= \frac{-(\left(\frac{z-3}{2}\right)+1)^{-1}}{8} =\frac{-(1-(\frac{z-3}{2}) + \frac{(z-3)^2}{2}-\ldots)}{8} $

And we know that if we differentiate, $\frac{d}{dz}\frac{-1}{4(z-1)} = \frac{1}{4(z-1)^2}$

Hence to find the series of $\frac{1}{4(z-1)^2}$ , we differentiate the series of $\frac{-1}{4(z-1)}$ and multiply $-1$ to it

Add all the results for final answer.

ANUPAM BISHT
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