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I am reading "Elliptic Functions According to Eisenstein and Kronecker" and I am struggling with a particular formula derivation:

$$ \frac{1}{p^mq^n} = \sum_{h=0}^{m-1} \frac{n(n+1)...(n+h-1)}{h!p^{m-h}r^{n+h}} + \sum_{k=0}^{n-1} \frac{m(m+1)...(m+k-1)}{k!q^{n-k}r^{m+k}} $$

where $r = p + q$ and $m,n \in N $

I am aware that it can be done using successive differentiation and partial fractions but I would love some help with the working!

Reading this book is part of my advanced research unit (but I am still first year undergrad so please be nice!)

nowhere dense
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Hugh Entwistle
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1 Answers1

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The first term in the RHS can be written as $$ \frac{1}{p^m(p+q)^n}\sum_{h=0}^{m-1}\binom{n+h-1}{h}\left(\frac{p}{p+q}\right)^h \tag{1}$$ and by stars and bars the binomial coefficient $\binom{n+h-1}{h}$ is the coefficient of $x^h$ in $\frac{1}{(1-x)^n}$.
In a similar fashion the second term can be written as $$ \frac{1}{q^n(p+q)^m}\sum_{h=0}^{n-1}\binom{m+h-1}{h}\left(\frac{q}{p+q}\right)^h \tag{2}$$ where $\binom{m+h-1}{h}$ is the coefficient of $x^h$ in $\frac{1}{(1-x)^m}$. About $(1)$, we want to compute a partial sum of the Taylor series of $\frac{1}{(1-x)^n}$ evaluated at $x=\frac{p}{p+q}$. That is equivalent to computing the $(n-1)$-th derivative of $\frac{1-x^{m+n-1}}{1-x}$ (that is a polynomial with degree $m+n-2$) at $x=\frac{p}{p+q}$, while dealing with $(2)$ is equivalent to finding the $(m-1)$-th derivative of the same function at $x=\frac{q}{p+q}$.

If we compute such derivatives through Cauchy's integral formula we have an easy finish.

I think that there is also a combinatorial interpretation of such identity in terms of random walks (with steps always oriented towards north or east) along a $m\times n$ grid, but I have to think about it.

Jack D'Aurizio
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