Constructing the Cotangent Function

Foreshadowing

The point of this note is to go through some of the motions of the derivation of the Weierstrass elliptic function $\wp$, and the Eisenstein series $G_{2k}$, but in one dimension instead of two. So instead of summing a translated copy of $1/z^2$ at all of the points of a 2-dimensional lattice in $\C$, we define $f$ by summing a translated copy of $1/x$ at all of the points in a 1-dimensional lattice in $\R$, and instead of the Laurent series of $\wp$ featuring $G_{2k}$, we find a Laurent series whose coefficients feature $\zeta(2k)$.

By making a polynomial in terms of $f$ and $f'$, to eliminate the strictly rational parts of the Laurent series, we find a differential equation that $f$ satisfies: this is analogous to the differential equation \[ (\wp')^2 = 4\wp^3 - g_2 \wp - g_3 \] but simpler. We find that the "1-dimensional elliptic function" is just the cotangent function up to some constant factors.

A Periodic Function of Poles

Let's say we want to construct a function $f : \R \to \R$ that's periodic. We could try to do this by first picking an arbitrary function, and adding infinitely many copies of it translated to the left and right. That is, we could pick a function $g : \R \to \R$, and set \[ f = \sum_{n=-\infty}^{\infty} g(x-n)\] Then we'd find that for any $m\in \Z$, \[ f(x + m) = \sum_{n=-\infty}^{\infty} g(x + m -n) = \sum_{n=-\infty}^{\infty} g(x + m -n)\] \[ = \sum_{u+m=-\infty}^{\infty} g(x - u) = \sum_{u=-\infty}^{\infty} g(x - u) = f \] where $u = n-m$, whence $n = u+m$.

What functions $g$ might be suitable? Well, in order for the infinite sum to converge anywhere, we probably want $g$ to fall off in magnitude to the left and right, so that its "tail" doesn't accumulate too heavily on values of $f$ near the origin.

So let's just try $g = 1/x$. Although it blows up at the origin, it does fall off as $x$ gets big. This means we're deciding to define \[ f = \sum_{n=-\infty}^{\infty} {1\over x-n}\]

Laurent Series

Let's try to rearrange this expression as a power series in $x$. We can't really do anything with the $n=0$ case, as that blows up at the origin. So let's separate that out: \[ f = {1\over x} + \left(\sum_{n=1}^{\infty} {1\over x-n}\right) + \left(\sum_{n=-\infty}^{-1} {1\over x-n}\right)\] From the standard geometric series formula \[ {1\over 1-x} = \sum_{m=0}^\infty x^m \] we can deduce \[ {1\over x-n} = {-1\over n-x} = {-1/n\over 1-x/n} \] \[ = -(1/n) \sum_{m=0}^\infty (x/n)^m \] So we can plug this in to get \[ f = {1\over x} + \left(\sum_{n=1}^{\infty} -(1/n) \sum_{m=0}^\infty (x/n)^m \right) + \left(\sum_{n=-\infty}^{-1} -(1/n) \sum_{m=0}^\infty (x/n)^m\right)\] \[ = {1\over x} + \left(\sum_{n=1}^{\infty} -(1/n) \sum_{m=0}^\infty (x/n)^m \right) + \left(\sum_{n=1}^{\infty} (1/n) \sum_{m=0}^\infty (-1)^m (x/n)^m\right)\] \[ = {1\over x} + \left(\sum_{n=1}^{\infty} -(2/n) \sum_{m=0}^\infty (x/n)^{2m+1} \right) \] \[ = {1\over x} -2 \left(\sum_{n=1}^{\infty} \sum_{m=0}^\infty x^{2m+1}n^{-2m-2} \right) \] and now rearranging (recklessly setting aside issues of convergence:) \[ = {1\over x} -2 \left( \sum_{m=0}^\infty x^{2m+1} \sum_{n=1}^{\infty}n^{-2m-2} \right) \] \[f = {1\over x} -2 \left( \sum_{m=0}^\infty x^{2m+1} \zeta(2m+2) \right) \] And that's at least a Laurent series for $f$ centered at the origin. We can actually simplify this by using the fact that $\zeta(0) = -1/2$. That means \[f = -2 \sum_{m=-1}^\infty x^{2m+1} \zeta(2m+2) \]

Differential Equation

What's the derivative of $f$? It's \[f' = -2 \sum_{m=-1}^\infty (2m+1)x^{2m} \zeta(2m+2) \] Let's stare at the first few terms of these Laurent series: \[f = 1/x -2x\zeta(2) - 2x^3\zeta(4) - 2x^5\zeta(6) - \cdots \] \[f' = -1/x^2 -2\zeta(2) - 6x^2\zeta(4) - 10x^4\zeta(6) - \cdots \] They're almost ordinary power series: only one term of negative exponent exists in each. If we squared $f$, and added it to $f'$, those terms would cancel out. Let's investigate $f^2 - f'$, and see if it's an interesting function. To make that calculation a bit easier, let's consider \[-(1/2)x f = \sum_{m=0}^\infty x^{2m} \zeta(2m) \] so that \[(1/4)x^2 f^2 = \sum_{m=0}^\infty \sum_{p=0}^\infty x^{2m} \zeta(2m) x^{2p} \zeta(2p) \] \[ = \sum_{m=0}^\infty \sum_{k=0}^{m} x^{2m} \zeta(2k) \zeta(2(m-k)) \] So now we know \[ f^2 = 4\sum_{m=0}^\infty \sum_{k=0}^{m} x^{2m-2} \zeta(2k) \zeta(2(m-k)) \] \[= 4x^{-2}\zeta(0)^2 + 4 (\zeta(0)\zeta(2) + \zeta(2)\zeta(0)) + 4x^2 (\zeta(0)\zeta(4) + \zeta(2)\zeta(2) + \zeta(4)\zeta(0)) + \cdots \] \[= x^{-2} -4\zeta(2) - 4x^2(\zeta(4) - \zeta(2)^2) + \cdots\] Hence \[f^2 + f' = \] \[4\left(\sum_{m=0}^\infty \sum_{k=0}^{m} x^{2m-2} \zeta(2k) \zeta(2(m-k)) \right)-2 \left(\sum_{m=-1}^\infty (2m+1)x^{2m} \zeta(2m+2)\right) \] \[= 4\left(\sum_{m=0}^\infty x^{2m-2} \sum_{k=0}^{m}\zeta(2k) \zeta(2(m-k)) \right)-2 \left(\sum_{m=0}^\infty (2m-1)x^{2m-2} \zeta(2m)\right) \] \[= \sum_{m=0}^\infty x^{2m-2} 4\left(\sum_{k=0}^{m}\zeta(2k) \zeta(2(m-k)) \right)-2 (2m-1) \zeta(2m) \] Se if we define $c_m$ to be the coefficients of the power series \[ f^2 + f' = \sum_{m=0}^\infty c_m x^{2m-2} \] we find that \[c_m = 4\left(\sum_{k=0}^{m}\zeta(2k) \zeta(2(m-k)) \right)-2 (2m-1) \zeta(2m)\] and in the first few cases this works out to

$m$	$c_m$
$0$	$4\zeta(0)\zeta(0) + 2\zeta(0) = 0$
$1$	$4\zeta(0)\zeta(2) + 4\zeta(2)\zeta(0) - 2\zeta(2) = -6\zeta(2)$
$2$	$4\zeta(0)\zeta(4) + 4\zeta(2)\zeta(2) + 4\zeta(0)\zeta(4) - 2\cdot 3\zeta(4) = 0$
$3$	$4\zeta(0)\zeta(6) + 4\zeta(2)\zeta(4) +$ $ 4\zeta(4)\zeta(2) + 4\zeta(6)\zeta(0) - 2\cdot 5\zeta(6) = 0$
$4$	$4\sum_{k = 0}^4 \zeta(2k)\zeta(2(4-k)) - 2\cdot 7\zeta(8) = 0$
$5$	$4\sum_{k = 0}^5 \zeta(2k)\zeta(2(5-k)) - 2\cdot 9\zeta(10) = 0$
$6$	$4\sum_{k = 0}^6 \zeta(2k)\zeta(2(6-k)) - 2\cdot 11\zeta(12) = 0$

Strange! Above $m=1$, these coefficents all seem to be exactly zero. If that is true, we have that $f^2 + f'$ is a constant function, specifically \[f^2 + f' = -6\zeta(2)\] Knowing that $\zeta(2) = \pi^2/6$, this is \[f^2 + f' = -\pi^2\]

Solving the Differential Equation

So let's try to solve the more general differential equation \[f^2 + f' = -k^2\] \[f^2 + {df\over dx} = -k^2\] \[ {df\over dx} = -k^2-f^2\] \[ {df\over k^2+f^2} =-dx\] \[ \int {df\over k^2+f^2} = -x + C\] Substitute $f = k \tan u$: \[ \int {k \sec^2 u \,du\over k^2+ k^2\tan^2 u} = -x + C\] \[ \int {1\over k} du = -x + C\] \[ u/k = -x + C\] \[ \tan^{-1} (f/k)/k = -x + C\] \[ f = k \tan (k(C-x))\] By symmetry, we can determine that the constant $C$ must be $1/2$, so \[f = \pi \tan (\pi (C-x)) = \pi \cot (\pi x)\]

Exercise: Prove that in fact \[2\left(\sum_{k=0}^{m}\zeta(2k) \zeta(2(m-k)) \right) = (2m-1) \zeta(2m)\] when $m \ge 2$, which justifies the inference that $f^2 + f'$ is a constant function.