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I'm hoping to give $\underset{z=0}{\text{Res}}\left[\csc^n(z)z^{n-j}\right]$ a nice form for nonnegative integers $n$ and $j$ but the only way I can think of is going through and evaluating $$\frac1{(j-1)!}\lim_{zā†’0}\left(\frac d{dz}\right)^{j-1}\csc^n(z)z^n.$$ I simply don't want to tangle myself up in that.

It may help to know that these residues are the coefficients $c_j$ in the series $$\csc^n(z)=\sum_{j=1}^{n}c_{j}\sum_{k=-\infty}^\infty\frac{(-1)^{nk}}{(z-\pi k)^{n-j+1}}$$ which I am trying to derive.

Spador Yedi
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  • Faa di Bruno's formula gives $$\csc^n z = 2^n n! \sum_{k \geq -n} B_{k + 2 n, n}(a_1, \ldots, a_{k + n + 1}) \frac {z^k} {(k + 2 n)!}, \ a_k = (-1)^{(k + 1)/2} (2^{k - 2} - 1) k B_{k - 1},$$ where $B_{n, k}$ are the Bell polynomials and $B_k$ are the Bernoulli numbers. ā€“ Maxim Dec 06 '20 at 22:46
  • @Maxim Thank you, I think that actually solves it. Definitely not as nice as I was hoping, but about as much as I expected ā€“ Spador Yedi Dec 06 '20 at 23:52

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