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A colleague and I are working on a problem and part of it comes down to evaluating the residue of a rational function. In particular, $$ \mathrm{Res} \left( z^{kn-1} \left( az^{m}+1 \right)^{-k}; r \right), $$ where $a$, $k$, $m$ and $n$ are positive integers satisfying $a \geq 2$ and $0<m<n$ and $r$ is any $m$-th root of $-1/a$.

The residue appears to have a nice form, $r^{kn}/m^{k}$ times a polynomial in $m$ and $n$ (of total degree $k-1$, it seems), and we have been able to prove this for $k=1$ and $2$ using series expansions, etc. But this becomes increasingly complicated and messy for larger $k$ and we have not been able to find any general pattern to help us along the way.

So our question is whether readers have seen residue problems for such rational functions or know of techniques that could help us to prove the value of this residue for any positive integer $k$.

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    $\begingroup$ By the substitution $w=z/r$ you can reduce to calculating the residue of $(w^n/(1-w^m))^k/w$ at $w=1$. That's tidier, but not obviously easier. $\endgroup$ – Neil Strickland Mar 10 at 14:36
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We want to calculate $$\rho(k,n,m)=\operatorname*{res}_{w=1}\left(\frac{w^n}{1-w^m}\right)^k\frac{dw}{w}. $$ If $kn$ is divisible by $m$ then it seems that $\rho(k,n,m)=-\binom{-k}{kn/m-k}/m$. This is because in this case the residues at all $m$'th roots of unity are the same, and the sum of those residues is minus the residue at $\infty$, which is easily calculated by the substitution $w=t^{-1}$ and the binomial expansion of $(1-t^m)^{-k}$. I have checked this in Maple for a range of cases. I don't know if this method can be adapted to the case where $kn$ is not divisible by $m$.

One can also check experimentally that the denominator and numerator of $\rho(k,n,m)$ are large, but their factorisation only involves fairly small primes $p$, certainly with $p<knm$. This typically indicates that the function can be expressed in terms of binomial coefficients and factorials, rather than general polynomials.

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  • $\begingroup$ That is really neat! Thanks so much for that! In principle, the same idea (looking at the residue at $\infty$) should work regardless of whether m divides kn or not, shouldn't it? $\endgroup$ – Alice Mar 10 at 20:43
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    $\begingroup$ If m does not divide $kn$ then the residue at $\omega r$ will be $\omega^p$ times the residue at $r$ for some $p\neq 0\pmod{m}$ so the sum of the residues at finite poles will be zero, as will the residue at $\infty$. There may be some more subtle way to adapt the argument. I tried a couple of things that did not work, but I did not spend a long time on it. $\endgroup$ – Neil Strickland Mar 10 at 21:50
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For $m=1$, this is the residue: $$\operatorname{Res} \left( z^{kn-1} \left( az+1 \right)^{-k}; -1/a \right)=\frac{(-1)^{k n-k} }{a^{kn}(k-1)!}\prod _{p=1}^{k-1} (k n-p).$$

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    $\begingroup$ Thanks for that, Carlo. I should have added to the question that we did work out $m=1$ already. $\endgroup$ – Alice Mar 10 at 14:49
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Let $l=kn-1$. Then we want to find $$R:=\operatorname{Res} \left(\frac{z^l}{ (1-(z/r)^m)^{k}};r\right)=\operatorname{Res}\left(\frac{(z+r)^l}{\bigl(1-(1+z/r)^m\bigr)^k};0\right).$$ This is the coefficient of $z^{-1}$ in the Laurent series expansion of $$\frac{(z+r)^l}{\bigl(1-(1+z/r)^m\bigr)^k}.$$ If we replace $z$ with $rz$, we divide the coefficient of $z^{-1}$ by $r$, so $$R=(-1)^k r^{l+1}[z^{-1}]\left(\frac{(1+z)^l}{\bigl((1+z)^m-1\bigr)^k}\right), $$ where $[z^i]$ extracts the coefficient of $z^i$.

We have $$ \frac{1}{\bigl((1+z)^m-1\bigr)^k}=(mz)^{-k}\left(\frac{mz}{(1+z)^m-1}\right)^k, $$ and since $(1+z)^m -1 = mz\bigl(1+\frac{1}{2}(m-1)z+\frac{1}{6}(m-1)(m-2)z^2+\cdots\bigr)$, we have $$\left(\frac{mz}{(1+z)^m-1}\right)^k=\sum_{i=0}^\infty P_i(m) z^i,$$ where $P_i(m)$ is a polynomial of degree $i$. Thus $$ \begin{aligned} R&=(-1)^k r^{l+1}m^{-k}[z^{k-1}] (1+z)^l \sum_{i=0}^\infty P_i(m) z^i\\ &=(-1)^k r^{l+1}m^{-k} \sum_{j=0}^{k-1} \binom{l}{j} P_{k-1-j}(m)\\ &=(-1)^k r^{kn}m^{-k} \sum_{j=0}^{k-1} \binom{kn-1}{j} P_{k-1-j}(m). \end{aligned} $$ The polynomials $P_i(m)$ are essentially ``higher-order degenerate Bernoulli numbers".

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  • $\begingroup$ thank you, Ira. That is nice too. $\endgroup$ – Alice Mar 16 at 20:27

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