The Herglotz Trick

29 Dec, 2023

An idea that can be used only once is a trick, many times, a method.

-Problems and Theorems in Analysis

This morning when I was having my routine coffee, I came across this interesting video, "how to build your own periodic function". The rational is fairly simple, to ensure that

f (x) = f (x + T)

where $T = 1$ for simplicity, a possible setting up would be let

f (x) = \sum_{k = - \infty}^{+ \infty} g (x + k)

Now this is a infinite serie, if it converges absolutely, we can rearrange the order of the summation:

f (x) = g (x) + \sum_{k = 1}^{\infty} (g (x + k) + g (x - k)) .

However to ensure the convergence, we at least require that $lim_{k \to \infty} g (x + k) = 0$ . Then comes the Herglotz trick:

Set $g (x) = \frac{1}{x}$ :

f (x) = \frac{1}{x} + \sum_{k = 1}^{\infty} (\frac{1}{x + k} + \frac{1}{x - k}) = \frac{1}{x} + \sum_{k = 1}^{\infty} \frac{2 x}{x^{2} - k^{2}} .

Show that $f (x) = π \cot π x$

$f (x) + f (- x) = 0$ ,
$f (x)$ is not defined on integers, but continuous otherwise,
$f (x)$ satisfies the functional equation:

f (\frac{x}{2}) + f (\frac{x + 1}{2}) = 2 f (x) .

Let $h (x) = f (x) - π \cot π x$ , due to (1), it is reasonable to let $h (n) = 0, n \in ℕ$ . Thus we have a continuous periodic function in $ℝ$ , and we only need to focus on one of the periods (that is a closed interval). Let $x_{0}$ be the point where $h (x) = m$ is maximised in $[0, 1]$ , by (3)

h (\frac{x_{0}}{2}) + h (\frac{x_{0} + 1}{2}) = 2 m

Thus, $h (x) \equiv m$ for $x \in [0, 1]$ , but notice that $h (0) = 0$ , we have

f (x) - π \cot π x \equiv 0,

f (x) = π \cot π x .

Thus a periodic function is created.

But, what if we make other decisions of $g (x)$ ? Notice that

- \sum_{n = - \infty}^{\infty} \frac{1}{(x + n)^{2}} = \frac{d}{d x} \sum_{n = - \infty}^{\infty} \frac{1}{x + n} = f^{'} (x) = - \frac{π^{2}}{\sin^{2} π x} .

Taking different $x^{- k}$ yields various derivatives of $π \cot π x$ . Also the above formula is particularly interesting as if $x = 0$ we will have

lim_{x \to 0} \frac{1}{x^{2}} + 2 \sum_{n = 1}^{\infty} \frac{1}{(x + n)^{2}} = {lim}_{x \to 0} \frac{π^{2}}{π^{2} x^{2} - \frac{1}{3!} π^{4} x^{4} .}

Exchange the limits, we have

\sum_{n = 1}^{\infty} \frac{1}{n^{2}} = {lim}_{x \to 0} \frac{1}{x^{2} - \frac{1}{6} π^{2} x^{4}} - \frac{1}{x^{2}} = \frac{π^{2}}{6} .