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

Solution 1:

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, $b_8=\dfrac{127}{604800}$.

Solution 2:

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: $$\begin{align} \frac{1}{\sin(z)}&=\sum _{n=0}^{ \infty }{2{\it B}_{2n}\frac { (-1)^n\left( 1-2^{2n-1} \right) z^{2n-1}}{(2n)!}}\\ &=\frac{1}{z}+\frac{1}{6}\,z+{\frac {7 }{360}}\,{z}^{3}+{\frac {31}{15120}}\,{z}^{5}+... \tag{7} \end{align}$$

Solution 3:

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

EDIT: Solution here.