1
$\begingroup$

I was working on a problem of finding the Laurent series of $\frac{1}{z-3}$ that converges where $|z-4| > 1$

So I had one approach, let $u=z-4$ then:

$$\frac{1}{z-3} = \frac{1}{1+u} $$

$$ = \frac{1}{u} - \frac{1}{u^2} + \frac{1}{u^3}...$$

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

But this apparently incorrect.

The correct answer is found by noting:

$$ \frac{1}{z-3} = \frac{1}{z-4 + 1} = \frac{1}{z-4} \frac{1}{1 - \frac{-1}{z-4}} = -\frac{1}{(z-4)^2} + ... $$

Where did I go wrong?

$\endgroup$
  • 1
    $\begingroup$ Your first one is correct. $\frac {1}{z-3} = \frac {1}{z-4} - \frac {1}{(z-4)^2} + \cdots$ Try evaluating $z = 6\cdots \frac {1}{3} = \frac 12 - \frac 14 + \frac 18-\cdots$ $\endgroup$ – Doug M Apr 5 at 1:59
  • 1
    $\begingroup$ The Princeton review book says the first term is of order 2, and they make... you know what I think they are plain wrong here. Since their factorization introduces a first order term in either expansion direction $\endgroup$ – frogeyedpeas Apr 5 at 2:00
1
$\begingroup$

${1\over 1+x} = 1 - x + x^2 - x^3 + \cdots$ when $|x| < 1$.

This is derived from ${1 \over 1-x} = 1 + x + x^2 + \cdots$ with $|x| < 1$.

$\endgroup$
  • $\begingroup$ i.e. the first is incorrect because the Taylor expansion is wrong. $\endgroup$ – U2647 Apr 5 at 2:00
  • $\begingroup$ I’m not sure I understand. I’m using the formula for $|x|>1$ $\endgroup$ – frogeyedpeas Apr 5 at 2:03
  • $\begingroup$ @frogeyedpeas I believe that is the expansion of $\frac{1}{1+ {1\over u}}$ $\endgroup$ – U2647 Apr 5 at 2:11
  • $\begingroup$ You can also use this formula, but remember to divide $u$ on both sides. $\endgroup$ – U2647 Apr 5 at 2:12
  • $\begingroup$ It’s one of 2 expansions. $\frac{1}{1+x} = \frac{1}{x}-\frac{1}{x^2} + ...$ $\endgroup$ – frogeyedpeas Apr 5 at 2:12
1
$\begingroup$

$$ \begin{align} \frac1{z-4}\frac1{1+\frac1{z-4}} &=\frac1{z-4}\left(1-\frac1{z-4}+\frac1{(z-4)^2}-\dots\right)\\ &=\frac1{z-4}-\frac1{(z-4)^2}+\frac1{(z-4)^3}-\dots \end{align} $$ The series starts with $\frac1{z-4}$, not $\frac1{(z-4)^2}$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.