2
$\begingroup$

when dealing with Laurent series I am confused about how to compute negative coefficients.

Laurents theorem states (in my course at least) that if f is differentiable on $D(z_0, R)/ \{z_0\}$ R>0. then $\exists a_n\in \Bbb C s.t.f(\zeta)=\sum_{n=-\infty}^{\infty}a_n(\zeta-z_0)^n$, $\zeta \in D(z_0,R)/\{z_0\}$

where $a_n=1/2\pi i\int_{C(z_0,r)}\frac{f(z)}{(z-z_0)^{n+1}}dz$

combining this with cauchy's integral formula for derivatives gives $a_n=\frac{f^n(z_0)}{n!}$.

I tried to compute the Laurent series of $\frac{1}{z^2}$ but I dont understand how to do this for $a_{-1}$ , as $a_{-1}=f^{-1}(0)/-1!$ and I'm not sure how to compute this.

Does $f^{-1}(0)$ simply mean the inverse here ?

Does $-1!=-(1!)$, i.e. does $-n!=-(n!).$

$\endgroup$
  • $\begingroup$ You can use the Residue formula $\endgroup$ – Quoka Apr 22 '18 at 14:46
  • $\begingroup$ @MathUser_NotPrime We don't learn the residue formula until next year, I just want to know how to compute Laurent series which is the topic we're covering at the minute. $\endgroup$ – excalibirr Apr 22 '18 at 14:47
  • $\begingroup$ You are trying to find the Laurent series about what point? i.e. what is $z_0$ $\endgroup$ – Quoka Apr 22 '18 at 14:49
  • $\begingroup$ @MathUser_NotPrime I'm trying to find it at the singular point $z_0=0$ to to decide whether or not it's an essential singularity. $\endgroup$ – excalibirr Apr 22 '18 at 14:52
1
$\begingroup$

As mentioned in another answer, the Laurent series of $z^{-2}$ is simply $z^{-2}$. We now show it using your formula. First, note that $$ a_n = \frac{1}{2\pi i}\int_{C(0,1)}\frac{z^{-2}}{z^{n+1}}dz = \frac{1}{2\pi i}\int_{C(0,1)}z^{-3-n}dz $$ So if $-3-n \geq 0$, the $a_n = 0$ (by Cauchy's theorem). Thus, $a_n = 0$ for all $n\leq -3$. Now, if $n=-2$ then we may compute $$ a_{-2} = \frac{1}{2\pi i}\int_{C(0,1)}z^{-1}dz = \frac{1}{2\pi i}\int_0^{2\pi}\left(e^{i\theta}\right)^{-1}ie^{i\theta}d\theta =1 $$ Now, if $n \geq -1$ then \begin{align*} a_{n} = \frac{1}{2\pi i}\int_{C(0,1)}z^{-(n+3)}dz &= \frac{1}{2\pi i}\int_0^{2\pi}\left(e^{i\theta}\right)^{-(n+3)}ie^{i\theta}d\theta\\ &= \frac{1}{2\pi}\int_0^{2\pi}e^{-(n+2)i\theta} d\theta\\ &= \left.\frac{1}{2\pi}\frac{e^{-(n+2)i\theta}}{-(n+2)i}\right|_0^{2\pi} = 0 \end{align*}

$\endgroup$
  • $\begingroup$ this is perfect , I know how to compute them now :) $\endgroup$ – excalibirr Apr 22 '18 at 15:29
  • $\begingroup$ a small additional question the fact that $a_n=0$ for $\forall n \neq-2 \Rightarrow$ implies this is not an essential singularity and because the order of f at $z_0 < 0$ means its not removable. we do have though that that the order is negative means that we have a pole of order -2 at $z_0$ correct ? $\endgroup$ – excalibirr Apr 22 '18 at 15:37
  • $\begingroup$ At $z_0$ (which in this case is $0$) we have a pole of order 2, not $-2$. But yes, it is not an essential singularity nor is it removable $\endgroup$ – Quoka Apr 22 '18 at 17:05
1
$\begingroup$

The formula of $$ a_n=\frac{1}{2\pi i}\int_{C(z_0,r)}\frac{f(z)}{(z-z_0)^{n+1}}dz = \frac{f^n(z_0)}{n!}. $$ works only for $n > 0$. Note that $f^n(z)$ is the $n$-th derivative of $f$.

For $n \le 0$, it depends on the singularity of $f$ at $z_0$.

For example, if $f(x) = \frac{g(x)}{(x-x_0)^k}$ where $g(x)$ is holomorphic, then $$ a_n=\frac{1}{2\pi i}\int_{C(z_0,r)}\frac{g(x)}{(z-z_0)^{n+k+1}}dz. $$ Thus for $n > -(k+1)$, one can apply Cauchy's integral formula. For $n \le -(k+1)$, the integrand becomes analytic, which makes $a_n = 0$.

$\endgroup$
1
$\begingroup$

Cauchy's integral formula is for derivatives, that is, for things like $f^{(n)}(a)$, with $n\in\mathbb{Z}_+$. It doesn't make sense to try to apply it with negative $n$.

And the Laurent series of $\frac1{z^2}$ centered at $0$ is simply $z^{-2}$ or, if you prefer, $\sum_{n=-\infty}^{+\infty}a_nz^n$ with $a_{-2}=0$ and $a_n=0$ if $n\neq-2$.

$\endgroup$
  • $\begingroup$ How does one calculate the Laurent series step by step ? Like we can do for Taylor expansions . $\endgroup$ – excalibirr Apr 22 '18 at 15:03
  • $\begingroup$ @exodius That's too broad. Do you have a specific example in mind? $\endgroup$ – José Carlos Santos Apr 22 '18 at 15:05
  • $\begingroup$ say for example $\frac{1}{z^2+1}$ $\endgroup$ – excalibirr Apr 22 '18 at 15:07
  • $\begingroup$ @exodius You use the fact that$$\frac1{z^2+1}=\frac12\left(\frac1{1-\frac zi}+\frac1{1+\frac zi}\right),$$together with the fact that$$|z|<1\implies\frac1{1-z}=1+z+z^2+z^3+\cdots$$and that$$|z|>1\implies\frac1{1-z}=-z^{-1}-z^{-2}-\cdots$$ $\endgroup$ – José Carlos Santos Apr 22 '18 at 15:15
  • $\begingroup$ Ah I think I understand now, thank you for your detailed answer :) $\endgroup$ – excalibirr Apr 22 '18 at 15:30

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.