I know the Taylor expansion of $\sin(z)$, but I still don’t understand how to expand it into the Laurent series. If I use the Taylor expansion $$\sin(z)=\sum_{n=0}^{\infty}\frac{(-1)^nz^{2n+1}}{(2n+1)!},$$ it still is centered at $0$. Maybe I need an example of it. Does $\sin\left(\frac{1}{z}\right)$ have a Laurent expansion centered at $0$?

  • $\begingroup$ Hello, there. What exactly do you know about Laurent expansions? $\endgroup$
    – Angel
    Nov 22 '21 at 13:37
  • $\begingroup$ The singularities of $\sin\left(\frac{1}{z}\right)$ occur at $z_n=\frac{1}{n\pi}$ for $n\in\mathbb{Z}$. $0$ is the accumulation point of $z_n$ because $\lim_{n\to\infty}z_n=\lim_{n\to-\infty}z_n=0$. So it is unclear to me that the Laurent series expansion centered at $0$ converges anywhere at all. However, this also seems off to me, so I must be missing something. It would be helpful if someone could tell me if I am correct or not. $\endgroup$
    – Angel
    Nov 22 '21 at 13:50
  • $\begingroup$ Wait, never mind. Obviously I am wrong. I was confusing the singularities with the zeroes. That was quite the brain fart. $\endgroup$
    – Angel
    Nov 22 '21 at 14:01

The series you have tried to write (look it up as it is not written correctly) converges for every complex value of $z$. So, if you replace $z$ with $\frac{1}{z}$ in that formula, you get a Laurent series, the one you are looking for, obviously converging for every $z$ except $z=0$. $z=0$ is called an essential singularity of $\sin{\left(\frac{1}{z}\right)}$. This type of series contains an infinite number of negative power terms.

  • $\begingroup$ Yes ,i agree with you on that .Can I think that the singularity of nature has no remnants?is it correct? $\endgroup$
    – Coisini
    Nov 22 '21 at 15:10
  • 1
    $\begingroup$ If by "remnant" you mean the residue, the series does have a residue because the residue is the $a_{-1}$ coefficient of the series. So what is the value of the residue here? $\endgroup$
    – user995139
    Nov 22 '21 at 15:26

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