How to compute the Laurent expansion of $\frac{1}{(\cos(z)-1)^4}$? I am trying to expand the Laurent series of:
$$\frac{1}{(\cos(z)-1)^4}$$
About $z=0$. If I simply expand $\cos(z)-1$, I get:
$$-\frac{z^2}{2}+\frac{z^4}{24}-\frac{z^6}{720}+\frac{z^8}{40320}-\frac{z^{10}}{3628800}+\dots$$
Which is not very useful when I substitute in the original expression, I'd get:
$$\frac{1}{\left(-\frac{z^{10}}{3628800}+\frac{z^8}{40320}-\frac{z^6}{720}+\frac{z^4}{24}-\frac{z^2}{2}+\dots\right)^4}$$
And I don't know what to do with that. I tried to first "expand" $1/x$ with the following substitution $x=1-y$ which yields:
$$1+y+y^2+y^3+y^4+\dots$$
And using that $y=x-1$, we obtain:
$$2- x + (1 - x)^2 + (1 - x)^3 + (1 - x)^4 +\dots$$
But that also didn't went too well. Can you help?
 A: A partial answer: Using $\cos(z)-1 = -2\sin^2(z/2)$ we have
$$
\frac{1}{(\cos(z)-1)^4} = \frac{1}{16} \left(\csc (\frac z 2)\right)^8
$$
where $\csc(z) = 1/\sin(z)$ is the cosecant whose power series expansion
$$
 {\begin{aligned}\csc x&=\sum _{n=0}^{\infty }{\frac {(-1)^{n+1}2\left(2^{2n-1}-1\right)B_{2n}}{(2n)!}}x^{2n-1}\\[5mu]&=x^{-1}+{\frac {1}{6}}x+{\frac {7}{360}}x^{3}+{\frac {31}{15120}}x^{5}+\cdots ,\qquad {\text{for }}0<|x|<\pi .\end{aligned}}
$$
is known in terms of the Bernoulli numbers.
You still have to raise that to the power of $8$, and I do not know if there is a nice closed form for the result. But it saves you from computing the reciprocal of a Taylor series.
A: Remark. I asked Maple for this
$$
16\,{z}^{-8}+{\frac{16}{3}}{z}^{-6}+{\frac{14}{15}}{z}^{-4}+{\frac{4}
{35}}{z}^{-2}+{\frac{2497}{226800}}+{\frac{317}{356400}}{z}^{2}+{\frac
{341749}{5448643200}}{z}^{4}+{\frac{11581}{2918916000}}{z}^{6}+{\frac{
1458367}{6351561216000}}{z}^{8}+{\frac{895337047}{72572361039360000}}{
z}^{10}+{\frac{7208739073}{11579677827379200000}}{z}^{12}+O \left( {z}
^{14} \right)  
$$
