We have the following expansion $$\pi\cot \pi z=\sum_{n\in\mathbb Z}\frac{1}{z-n},$$ if a suitable method of summation is used. Successively differentiating we further obtain $$\frac{\pi^2}{\sin^2\pi z}=\sum_{n\in\mathbb Z}\frac{1}{(z-n)^2}$$ and $$\pi^3\frac{\cos \pi z}{\sin^3\pi z}=\sum_{n\in\mathbb Z}\frac{1}{(z-n)^3}.$$ Hence we observe that $$\sum_{n\in\mathbb Z}\frac{1}{z-n}\cdot\sum_{n\in\mathbb Z}\frac{1}{(z-n)^2}=\sum_{n\in\mathbb Z}\frac{1}{(z-n)^3}.$$ On the other hand, $$\frac {\pi^4 }{3} \frac{1}{\sin^2 \pi z}=\sum_{n\in\mathbb Z}\frac{1}{(z-n)^4},$$ and so $$\sum_{n\in\mathbb Z}\frac{1}{z-n}\cdot\sum_{n\in\mathbb Z}\frac{1}{(z-n)^3}\neq\sum_{n\in\mathbb Z}\frac{1}{(z-n)^4}.$$

Let us call a triple of positive integers $(a,b,c)$ a Shimrod triple if $a+b=c$ and $$\sum_{n\in\mathbb Z}\frac{1}{(z-n)^a}\cdot\sum_{n\in\mathbb Z}\frac{1}{(z-n)^b}=\sum_{n\in\mathbb Z}\frac{1}{(z-n)^c}.$$ Therefore $(1,2,3)$ is a Shimrod triple, while $(1,3,4)$ is not. Are there Shimrod triples different from $(1,2,3)$? Can we prove identities of this type more directly?


1 Answer 1


Write $S_k = \sum_{n \in \mathbb{Z}} (z-n)^{-k}$, summations taken symmetrically in the Eisenstein sense. By uniform/normal convergence, we know that $S_k' = -kS_{k+1}$. We are interested in $S_aS_b - S_c$.

By various theorems on uniform/normal convergence $S_k$ has a Laurent expansion with principal part $z^{-k} $ at $z=0$. Consideration of the expansion of $S_1$ (namely $S_1 =\pi\cot{\pi z} = \frac{1}{z} + \sum_{m=1}^{\infty} -2\zeta(2m) z^{2m-1}$) implies that $S_{2k} - z^{2k} \sim 2\zeta(2k)$ for $a=2k$ even and $-(2k-1)\zeta(2k)z$ for $a=2k-1$ odd.

It follows immediately that $S_a S_b \sim z^{-a-b} $ as $z \to 0$, so certainly we must have $a+b=c$. Are there any others than $a=1,b=2$ and $c=a+b=3$? First check $S_a^2$. The Laurent expansion begins $$ z^{-2a}+2\alpha z^{-a}+\dotsb $$ or $$ z^{-2a}+2\alpha z^{1-a}+\dotsb. $$ $a \geq 1$, and for $a=1$, we know that $S_1^2 = S_2-\pi^2$, so it suffices to check $a>1$. But then $-a$ and $1-a$ are both negative, so there is no way to cancel these out.

So now suppose that $a>b \geq 1$. The beginning of the principal part of the expansion of $S_aS_b$ is given by one of the four cases $$ (z^{-a}+\alpha+O(z))(z^{-b}+\beta+O(z)) = z^{-a-b} + \beta z^{-a} + o(z^{a}) \\ (z^{-a}+\alpha z+O(z^2))(z^{-b}+\beta+O(z)) = z^{-a-b} + \beta z^{-a} + o(z^{a}) \\ (z^{-a}+\alpha+O(z))(z^{-b}+\beta z+O(z^2)) = z^{-a-b} + \beta z^{1-a} + \alpha z^{b} + o(z^{a}) \\ (z^{-a}+\alpha z+O(z^2))(z^{-b}+\beta z+O(z^2)) = z^{-a-b} + \beta z^{1-a} + o(z^{1-a}) $$ depending on the parity of $a$ and $b$. The first two of these are trivially disposed of. $a>1$, so we can also forget about the fourth. The third requires $a=b+1$ to be nontrivial. But $S_a S_{a+1}$ is the derivative of $(S_a)^2/(2a)$, so it suffices to check that we have no relation of the form $S_a^2 - S_{2a}+C$ except for $S_1^2 = S_2 - \pi^2$. But this again follows from the principal parts: there is a $2\alpha z^{-a}$ term or $2\alpha z^{1-a}$ that is not cancelled by the $S_{2a}$.

Hence the only possible identity of the given form is $S_1S_2=S_3$.

This is a general method that can be employed: we have the result

If $f$ is meromorphic and periodic with real period $L$, with finitely many poles $a_i$ in one period strip, and $f(z) $ converges to finite, but possibly different numbers as $\Im(z) \to \pm \infty$, then there are constants $c_{i,k}$ and $A$ so that $$f(z) = A+\sum_{i} \sum_{k=1}^{n_i} c_{i,k}\cot^k{\frac{z-a_i}{L}}$$

(Whittaker and Watson has a weaker version for simple poles, but the proof can be easily effected by making the principal parts in the sum match those in $f$'s Laurent series. The different is then a bounded analytic function, which is constant by Liouville.)

Differentiating $S_1^2+\pi^2=S_2$ gives $-2S_1S_2 = -2S_3$, which is your relation. Differentiating again gives $$3S_4 = S_2^2+S_1S_3 = S_2^2+S_1^2 S_2 = \pi^2+4\pi^2S_1^2+3S_1^4, $$ and so on. This type of argument, together with an induction, shows that any $S_k$ can be expressed as a polynomial in $S_1$ of degree $k$. This also means that there is an expansion of $f$ from the paragraph above of the form $$ f(z) = A' +\sum_{i} \sum_{k=1}^{n_i} c_{i,k}' S_k\left(\frac{z-a_i}{L}\right).$$

This was all worked out in detail by Eisenstein in the nineteenth century, but I don't have the references to hand. A modern account is given in Walker's Elliptic funtions: A constructive approach, which is astonishingly hard to get hold of considering that it's only 20 years old.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .