Calculate Laurent series for $1/ \sin(z)$

How can calculate Laurent series for

$$f(z)=1/ \sin(z) $$ ??

I searched for it and found only the final result, is there a simple way to explain it ?

Solutions Collecting From Web of "Calculate Laurent series for $1/ \sin(z)$"

Using the series for $\sin(z)$ and the formula for products of power series, we can get
$$
\begin{align}
\frac1{\sin(z)}
&=\frac1z\frac{z}{\sin(z)}\\
&=\frac1z\left(1-\frac{z^2}{3!}+\frac{z^4}{5!}-\frac{z^6}{7!}+\cdots\right)^{-1}\\
&=\frac1z\left(1+\frac{z^2}{6}+\frac{7z^4}{360}+\frac{31z^6}{15120}+\cdots\right)\\
&=\frac1z+\frac{z}{6}+\frac{7z^3}{360}+\frac{31z^5}{15120}+\cdots
\end{align}
$$


Using the formula for products of power series

As given in the Wikipedia article linked above,
$$
\left(\sum_{k=0}^\infty a_kz^k\right)\left(\sum_{k=0}^\infty b_kz^k\right)
=\sum_{k=0}^\infty c_kz^k\tag{1}
$$
where
$$
c_k=\sum_{j=0}^ka_jb_{k-j}\tag{2}
$$
Set
$$
c_k=\left\{\begin{array}{}
1&\text{for }k=0\\
0&\text{otherwise}
\end{array}\right.\tag{3}
$$
and
$$
a_k=\left\{\begin{array}{}
\frac{(-1)^j}{(2j+1)!}&\text{for }k=2j\\
0&\text{for }k=2j+1
\end{array}\right.\tag{4}
$$
Using $(2)$, $(3)$, and $(4)$, we can iteratively compute $b_k$.


For example, to compute the coefficient of $z^8$:
$$
\begin{align}
c_8=0
&=b_8-\frac16b_6+\frac1{120}b_4-\frac1{5040}b_2+\frac1{362880}b_0\\
&=b_8-\frac16\frac{31}{15120}+\frac1{120}\frac7{360}-\frac1{5040}\frac16+\frac1{362880}1\\
&=b_8-\frac{127}{604800}
\end{align}
$$
Thus, $b_8=\dfrac{127}{604800}$.

From the generating function of the Bernoulli polynomials:
$${\frac {t{{\rm e}^{xt}}}{{{\rm e}^{t}}-1}}=\sum _{n=0}^{\infty }{
\frac {{\it B}_{n}\left(x\right) {t}^{n}}{n!}} \tag{1}$$
and from:
$$ \frac{1}{\sin(z)}={\frac {-2i{{\rm e}^{-iz
}}}{{{\rm e}^{-2iz}}-1}} \tag{2}$$
we set $$t=-2iz,\quad x=\frac{1}{2} \tag{3}$$
in $(1)$ to get:
$${\frac {-2iz{{\rm e}^{-iz}}}{{{\rm e}^{-2iz}}-1}}=\sum _{n=0}^{
\infty }{\frac {{\it B_n} \left( \frac{1}{2} \right) \left( -2\,iz
\right) ^{n}}{n!}} \tag{4}$$
and we then use the known value that relates the polynomial evaluated at $\frac{1}{2}$, to the Bernoulli number:
$${\it B_n} \left( \frac{1}{2} \right) = \left( {2}^{1-n}-1 \right) {
\it B_n} \tag{5}$$
together with the fact that:
$${\it B}_{2n-1} =
\cases{1/2&$n=1$\cr 0&otherwise\cr} \tag{6}$$
and divide both sides of $(4)$ by $z$ to get:
$$\begin{align}
\frac{1}{\sin(z)}&=\sum _{n=0}^{
\infty }{2{\it B}_{2n}\frac { (-1)^n\left( 1-2^{2n-1} \right) z^{2n-1}}{(2n)!}}\\
&=\frac{1}{z}+\frac{1}{6}\,z+{\frac {7
}{360}}\,{z}^{3}+{\frac {31}{15120}}\,{z}^{5}+…
\tag{7}
\end{align}$$

Idea: $1=\sin z {1\over\sin z} =$ (known Taylor series)(unknown Laurent series).

EDIT: Solution here.