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I'm trying to compute the Laurent series of $\frac{1}{\sin(z)}$ at $z_0=0$. From what I've seen on the internet this is given as

$f(z)=\frac{1}{z}+\frac{z}{3!}+\frac{7z^3}{360}+...$

My attempt:

$f(z)=\frac{1}{\sin(z)}$ can be rewritten as $f(z)=\frac{1}{z}\frac{z}{\sin(z)}$

We can express $\sin(z)$ as $\sin(z)=z-\frac{z^2}{3!}+\frac{z^4}{5!}+...$

$\therefore \frac{z}{\sin(z)}=(\frac{\sin(z)}{z})^{-1}=(1-\frac{z}{3!}+\frac{z^3}{5!}+...)^{-1}$

To get this to equal what I know it should, then we should have $(\frac{\sin(z)}{z})^{-1}=(1+\frac{z}{3!}+\frac{7z^3}{360}+...)$ but how can this be what do I do with my negative sign and how do I obtain the $\frac{7}{360}$ coefficient?

Bernard
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excalibirr
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  • Take the coefficients as unknown, multiply with that of $\mathrm{sinc}z$, then solve simultaneous equations. – J.G. Apr 25 '18 at 21:13
  • @ZacharySelk I don't feel as though this is a duplicate, as the step that was skipped in the accepted answer of that question is the point I'm asking for clarification on. – excalibirr Apr 25 '18 at 21:15
  • \begin{eqnarray} \sin(z)=z-\frac{z^{\color{red}{3}}}{3!}+\frac{z^{\color{red}{5}}}{5!}+\cdots \end{eqnarray} – Donald Splutterwit Apr 25 '18 at 21:19
  • See also the answer at https://math.stackexchange.com/a/4427590/945479. – qifeng618 Apr 15 '22 at 02:02

2 Answers2

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You are wrong about $\frac{\sin z}z$. Its Taylor series is$$1-\frac{z^2}{3!}+\frac{z^4}{5!}-\cdots$$Since it is an even function, the Taylor series of $\frac z{\sin z}$ is of the type $a_0+a_1z^2+a_2z^4+\cdots$. So, you have$$\left(1-\frac{z^2}{3!}+\frac{z^4}{5!}-\cdots\right)(a_0+a_1z^2+a_2z^4+\cdots)=1,$$which means that$$a_0+\left(a_1-\frac{a_0}{3!}+\right)z^2+\left(a_2-\frac{a_1}{3!}+\frac{a_0}{5!}\right)z^4+\cdots=1.$$This, in turn, means that$$a_0=1\text{, }\ a_1-\frac{a_0}{3!}=0\text{, }\ a_2-\frac{a_1}{3!}+\frac{a_0}{5!}=0$$and so on. And from this you can compute $a_0,a_1,a_2,\ldots$

José Carlos Santos
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1

Note the most direct way to obtain the expansion of $\dfrac1{\sin z}$ is division by increasing powers. I'll show how to obtain the first four terms:

Start from the expansion of $\sin z$ at order $7$:$$\frac1{\sin z}=\frac1{z-\dfrac{z^3}{6}+\dfrac{z^5}{120}-\dfrac{z^7}{5040}+o(z^7)}=\frac1z\cdot\frac1{1-\dfrac{z^2}{6}+\dfrac{z^4}{120}-\dfrac{z^6}{5040}+o(z^6)}$$ Now the division of $1$ by $1-\dfrac{z^2}{6}+\dfrac{z^4}{120}-\dfrac{z^6}{5040}$ goes as follows (we truncate the terms with degree more than $6$) :

enter image description here

Thus we obtain \begin{align} \frac1{\sin z} &= \frac1z \Bigl( 1 + \dfrac{z^2}{6} + \frac{7z^4}{360}- \frac{z^6}{4320} +o(z^6)\Bigr) \\[1ex] &=\frac1z + \dfrac{z}{6} + \frac{7z^3}{360}- \frac{z^5}{4320} +o(z^5) \end{align}

Bernard
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