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Suppose $f$ has an $n$-th order pole at $0$ and $g$ is holomorphic at $0$. Can I write the residue of the product "in terms of" the residue of $f$?

Since $$Res(fg)=\frac {1}{(n-1!)}\lim _{s\rightarrow 0}\left (\frac {d^{n-1}}{ds^{n-1}}\left \{ s^nf(s)g(s)\right \}\right )$$ and since the derivative here is $$\sum _{k=0}^{n-1}\left (\begin {array}{l}n-1\\ \hspace {4mm}k\end {array}\right )\frac {d^{n-1-k}}{ds^{n-1-k}}\left \{ s^nf(s)\right \} \frac {d^{k}}{ds^{k}}\left \{ g(s)\right \}$$ I can deduce $$Res(fg)=\frac {1}{(n-1)!}\sum _{k=0}^{n-1}\left (\begin {array}{l}n-1\\ \hspace {4mm}k\end {array}\right )(n-1-k)!g^{(k)}(0)\lim _{s\rightarrow 0}\left (\frac {d^{n-1-k}}{ds^{n-1-k}}\left \{ s^nf(s)\right \} \right )$$ which perhaps isn't too far from what I want? I'd like it if this last limit was "something like" the residue of $f$.

tomos
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  • The residue is the coefficient of $s^{-1}$ in the Laurent expansion around $s=0$. That $f$ has a pole of order $n$ means $f(s)= \sum_{m\ge -n} a_m s^m$ – reuns Dec 28 '19 at 01:05

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