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I tried to expand $\frac{z^2}{\sin(z)^{2}}$ using Taylor expansion, but the coefficient involved some limit of $\frac{0}{0}$ and was really difficult to calculate. (I tried to convince myself the constant term of the Taylor expansion is 1 because $\lim_{x\rightarrow 0}(\frac{z^2}{\sin(z)^{2}})$=1 and the coefficients of z with odd degree are $0$ because $\frac{z^2}{\sin(z)^{2}}$ is a even function). But how do I solve those coefficients explicitly...?

Thank you.

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You may just write, as $z \to 0$: $$\begin{align}\frac{1}{\sin^2z}&=\frac{1}{\left(z-\dfrac{z^3}{3!}+\dfrac{z^5}{5!}+\mathcal{O}(z^7)\right)^2}\\\\&=\frac{1}{z^2\left(1-\dfrac{z^2}{3!}+\dfrac{z^4}{5!}+\mathcal{O}(z^6)\right)^2} \\\\&=\frac{1}{z^2}\left( 1-2\times \left(-\frac{z^2}{3!}+\frac{z^4}{5!}\right)+3\times \left(-\frac{z^2}{3!}\right)^2+\mathcal{O}(z^6)\right)\\\\ &=\frac{1}{z^2}\left(1+\frac{z^2}{3}+\frac{z^4}{15}+\mathcal{O}(z^6)\right)\\\\ &=\frac{1}{z^2}+\frac{1}{3}+\frac{z^2}{15}+\mathcal{O}(z^4) \end{align} $$ and this gives the first terms of the Laurent series expansion of $\dfrac{1}{\sin^2z}$ near $z=0$.


To obtain the coefficient of order $n$, one may use the Bernoulli numbers, observing that $$ \frac{1}{\sin^2z}=-2i\:\partial_z \left(\frac{1}{e^{2iz}-1}\right) $$ then get

$$ \frac{1}{\sin^2z}=\sum_{n=0}^{\infty}\frac{(-1)^{n+1} 2^{2n}(2n-1) }{(2n)!}B_{2n}z^{2n-2},\quad 0<|z|<\pi. $$

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    $\begingroup$ How did you get the third equation from the second? $\endgroup$ – Cure Nov 11 '17 at 21:15
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    $\begingroup$ @Cure This comes from the Taylor series of $\dfrac{1}{(1+u)^2}$ as $u \to0$ which is equal to $1-2u+3u^2+O(u^3)$, then one puts $u=\frac{z^2}{3}+\frac{z^4}{15}+\mathcal{O}(z^6)$ as $z \to 0$. $\endgroup$ – Olivier Oloa Nov 11 '17 at 23:06

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