0
$\begingroup$

I want to calculate the residue of the following function without using to the development in series of Laurent.

$$f(z)=\frac{1}{z^2\,\sin(\pi z)}$$

$z=0$ is a triple pole

$$\mathrm{Res}[f,0]=\lim_{z->0}\frac{1}{3!}\frac{\mathrm{d^2} }{\mathrm{d} x^2}\left [ \frac{1}{z^2\,\sin(\pi z)} z^3\right ]$$

Usually in these cases I developed before making the limit the trigonometric function. Until now I had only considered simple poles in these cases and I always took the first term of the development. in $z=0$ $$\sin(\pi z)=\pi t-\frac{\pi^3t^3}{6}+o(t^4)$$

If I replace only the first term the result is $0$, and if I replace the second the result is right $\frac{\pi}{6}$. $$\mathrm{Res}[f,0]=\lim_{z->0}\frac{1}{3!}\frac{\mathrm{d^2} }{\mathrm{d} x^2}\left [ \frac{1}{z^2\,\pi z-\frac{\pi^3z^3}{6}} z^3\right ]= \frac{\pi}{6} $$

$$\mathrm{Res}[f,0]=\lim_{z->0}\frac{1}{3!}\frac{\mathrm{d^2} }{\mathrm{d} x^2}\left [ \frac{1}{z^2\,\pi z} z^3\right ]= 0 $$

why the two limits are different?

What is the rule? I need to replace up to the order of the pole or higher. Considering simple poles was fine always replace the first, this is why I make this assumption.

Someone can help me.

Thank you so much.

$\endgroup$
  • $\begingroup$ Are you willing to grant that $\sin z/z$ is analytic at $0$ if it takes the value $1$ there? $\endgroup$ – Lubin Apr 1 '17 at 23:54
  • $\begingroup$ Your question is unclear to me. $\endgroup$ – Zaid Alyafeai Apr 2 '17 at 1:32
  • $\begingroup$ I want to calculate the residue in $z=0$ using the formula with the limit of the multiple poles. $\endgroup$ – Stefano Barone Apr 2 '17 at 5:49
2
$\begingroup$

Your formula is not true : $$\mathrm{Res}[f,0]=\lim_{z->0}\frac{1}{2!}\frac{\mathrm{d^2} }{\mathrm{d} z^2}\left [ \frac{1}{z^2\,\sin(\pi z)} z^3\right ]$$

So : $$\frac{\mathrm{d^2} }{\mathrm{d} z^2}\left [ \frac{1}{z^2\,\sin(\pi z)} z^3\right ]=\frac{\mathrm{d^2} }{\mathrm{d} z^2}\frac{z}{\sin(\pi z)}=\frac{\mathrm{d} }{\mathrm{d} z}\frac{\sin(\pi z)-z^2\pi\cos(\pi z)}{\sin^2(\pi z)}=z (\pi^2 \csc^3(\pi z) + π^2 \cot^2(\pi z) \csc(\pi z)) - 2 \pi \cot(\pi z) \csc(\pi z)$$

So : $$\mathrm{Res}[f,0]=\lim_{z->0}\frac{1}{2!}[z (\pi^2 \csc^3(\pi z) + π^2 \cot^2(\pi z) \csc(\pi z)) - 2 \pi \cot(\pi z) \csc(\pi z)]$$

By using $\cot(\pi z)=\frac{1}{\pi z}-\frac{\pi z}{3}+o(z)$ and $\csc(\pi z)=\frac{1}{\pi z}+\frac{\pi z}{6}+o(z)$, we get : $$\lim_{z->0}\frac{1}{2!}[z (\pi^2 \csc^3(\pi z) + π^2 \cot^2(\pi z) \csc(\pi z)) - 2 \pi \cot(\pi z) \csc(\pi z)]=\frac{\pi}{6}$$

$$\mathrm{Res}[f,0]=\frac{\pi}{6}$$


You could also write : $$\frac{z}{\sin(\pi z)}=\frac{1}{\pi-\frac{\pi^3z^2}{6}+o(z^3)}$$

$$\frac{\mathrm{d^2} }{\mathrm{d} z^2}\frac{z}{\sin(\pi z)}=\frac{\mathrm{d^2} }{\mathrm{d} z^2}\frac{1}{\pi-\frac{\pi^3z^2}{6}+o(z^3)}=\frac{\pi}{3}+o(1)$$

So :

$$\mathrm{Res}[f,0]=\lim_{z->0}\frac{1}{2!}\frac{\mathrm{d^2} }{\mathrm{d} z^2}\left [ \frac{1}{z^2\,\sin(\pi z)} z^3\right ]=\lim_{z->0}\frac{1}{2!}(\frac{\pi}{3}+o(1))=\frac{\pi}{6}$$

$\endgroup$
  • $\begingroup$ Thank you so much for the very accurate answer. I wanted to avoid use derivatives and I prefer to write directly a series expansion. I just want to figure out how many terms of development I must put. Is right to reach the degree equal to the pole number? $\endgroup$ – Stefano Barone Apr 2 '17 at 19:51
  • $\begingroup$ You have to reach at least this degree but I can't ensure you that it will be enough sometimes you need to take a higher degree because you still could have a limit of the form $\frac{0}{0}$ , but I don't have an example in mind. $\endgroup$ – Jennifer Apr 2 '17 at 20:20
0
$\begingroup$

A pure Laurent series approach. Consider $$ \begin{aligned} \frac{1}{z^2\sin(\pi z)}&=\frac{1}{z^2}\frac{1}{\pi z-\frac{\pi^3z^3}{6}+O(z^5)}\\ &=\frac{1}{\pi z^3}\frac{1}{1-\frac{\pi^2z^2}{6}+O(z^4)}\\ &=\frac{1}{\pi z^3}\left[\frac{\pi^2z^2}{6}+O(z^4)\right]\\ &=\color\red{\frac{\pi}{6}}\frac{1}{z}+\boxed{O(z)}. \end{aligned} $$

All you need is to make sure that the remaining term has nothing to do with $1/z$, and that's why approximation $\sin(\pi z)\sim \pi z+O(z^3)$ fails.

$\endgroup$

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.