4
$\begingroup$

How can I find the principal part of the Laurent series for $f(z)=\dfrac{\pi^2}{(\sin \pi z)^2}$ centered at $k$ where $k \in \mathbb{Z}$.

I think there are two ways to do it either use the formula $a_k$ for Laurent coefficients or expand manipulate $\sin^2 \pi z$ and solve for the coefficients. I am not sure if any of these ways are efficient.

$\endgroup$
  • $\begingroup$ $f$ has period $1$, so the principal part at $k$ is the translation of the principal part at $0$. As $\dfrac{\sin (\pi z)}{\pi z} = 1 - O(z^2)$, we have $\dfrac{\pi z}{\sin (\pi z)} = 1 + O(z^2)$, whence $\dfrac{\pi^2z^2}{(\sin (\pi z))^2} = 1 + O(z^2)$ and hence $\dfrac{\pi^2}{(\sin (\pi z))^2} = \dfrac{1}{z^2} + O(1)$. $\endgroup$ – Daniel Fischer Jul 2 '16 at 12:21
3
$\begingroup$

By setting $z:=k+\varepsilon$, $k \in \mathbb{Z}$, $\varepsilon \to 0$, one has by the Taylor series expansion: $$ \sin^2 \pi z=\left(\sin (\pi k+\pi\varepsilon)\right)^2=\sin^2(\pi\varepsilon)=\pi^2 \varepsilon^2-\frac{\pi^4 \varepsilon^4}{3}+O(\varepsilon^6) $$ giving $$ f(z)=\frac{\pi^2}{\pi^2 \varepsilon^2-\frac{\pi^4 \varepsilon^4}{3}+O(\varepsilon^6)}=\frac1{\varepsilon^2(1-\frac{\pi^2 \varepsilon^2}{3}+O(\varepsilon^4))}=\frac1{\varepsilon^2}+\frac{\pi ^2}{3}+O(\varepsilon^2) $$ that is, as $z \to k$, $$ f(z)=\frac{\pi^2}{\sin^2 \pi z}=\frac1{(z-k)^2}+\frac{\pi ^2}{3}+O((z-k)^2), $$

the principal part is $$ \frac1{(z-k)^2}. $$

$\endgroup$
  • $\begingroup$ How do you know that the bottom portion converges uniformly? $\endgroup$ – adam Jul 1 '16 at 22:18
  • $\begingroup$ @adam Please see my edit, I've used that, as $u \to 0$, $\frac1{1-u}=1+u+O(u^2)$ giving $\frac1{1-\frac{\pi^2 \varepsilon^2}{3}+O(\varepsilon^4)}=1+\frac{\pi^2 \varepsilon^2}{3}+O(\varepsilon^4)$. Thanks. $\endgroup$ – Olivier Oloa Jul 1 '16 at 22:23
  • $\begingroup$ @Oliver Oloa I understand that but we need to know we have uniform convergence before we do that right? $\endgroup$ – adam Jul 1 '16 at 22:24
  • $\begingroup$ Yes, to prove it once for all. $\endgroup$ – Olivier Oloa Jul 1 '16 at 22:26
2
$\begingroup$

We can try the following using the well known power series for sine:

$$\sin\pi z=(-1)^k\sin\left[\pi(z-k)\right]\implies$$

$$ \sin^2\pi z=\sin^2\pi(z-k)=\left(\sum_{n=0}^\infty(-1)^n\frac{\pi^{2n+1}(z-k)^{2n+1}}{(2n+1)!}\right)^2=$$

$$=\left(\pi(z-k)-\frac{\pi^3(z-k)^3}{3!}+\ldots\right)\left(\pi(z-k)-\frac{\pi^3(z-k)^3}{3!}+\ldots\right)=$$

$$=\pi^2(z-k)^2-\frac{\pi^4}3(z-k)^4+\mathcal O((z-k)^6)\implies$$

$$\frac{\pi^2}{\sin^2\pi z}=\frac{\pi^2}{\pi^2(z-k)^2\left(1-\frac{\pi^2}3(z-k)^2+\mathcal O((z-k)^3)\right)}=$$

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

and the principal part is thus $\;\cfrac1{(z-k)^2}\;$ , which was expectable as we have double poles here.

$\endgroup$
  • 1
    $\begingroup$ I like this one! $\endgroup$ – adam Jul 1 '16 at 22:27

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.