### Vieta’s formulas and $\zeta(2)$

(This is Yan Sheng.)

At some point I came up with a new idea to prove the famous identity$$\zeta(2)\coloneqq\sum_{n=1}^\infty\frac1{n^2}=\frac{\pi^2}6.$$There is no shortage of existing proofs of this identity; see this math.SE question, or this collection of proofs compiled by Robin Chapman. I was inspired by this *illuminating* video, where 3Blue1Brown explains an elementary proof based on Euclidean geometry, by a physical interpretation involving systems of lighthouses. My new idea comes from trying to translate this proof into the complex plane.

## Vieta’s formulas

The main idea in the video is to get an expression for the sum of inverse squared distances from the origin to the vertices of a certain regular $n$-gon. We first observe that when viewed on the complex plane, the vertices of this $n$-gon are the roots of some polynomial of degree $n$, and the sum of inverse squares can be computed by Vieta’s formulas.

More explicitly, suppose that $\rho_1,\rho_2,\ldots,\rho_n$ are the roots of the polynomial$$a_nz^n+a_{n-1}z^{n-1}+\cdots+a_1z+a_0=0,$$with $a_n\neq0$ (so that there are $n$ roots) and $a_0\neq0$ (so the roots are nonzero), and we want to find the sum of inverse squares $\sum_j\frac1{\rho_j^2}$.

The astute reader might have noticed that this objective is different from the video, which computes $\sum_j\frac1{\lvert\rho_j\rvert^2}$ instead. We chose this reformulation because it is easier to handle algebraically.

If we had wanted to find the sum of squares $\sum_j\rho_j^2$ instead, we could use the following calculation:$$\begin{aligned} \sum_j\rho_j^2&=\sum_{j,k}\rho_j\rho_k-\sum_{j< k}\rho_j\rho_k-\sum_{j>k}\rho_j\rho_k\\ &=\left(\sum_j\rho_j\right)^2-2\sum_{j< k}\rho_j\rho_k\\ &\overset{(* )}=\left(-\frac{a_{n-1}}{a_n}\right)^2-2\frac{a_{n-2}}{a_n}\\ &=\frac{a_{n-1}^2-2a_na_{n-2}}{a_n^2},\qquad(\dagger) \end{aligned}$$where for $(*)$ we used Vieta’s formulas$$\sum_j\rho_j=-\frac{a_{n-1}}{a_n},\qquad\sum_{j< k}\rho_j\rho_k=\frac{a_{n-2}}{a_n}.$$To get the sum of inverse squares instead, we need to apply the above formula to some polynomial with roots $\frac1{\rho_1},\frac1{\rho_2},\ldots,\frac1{\rho_n}$. But this is exactly what the reciprocal polynomial does: if we reverse the order of the coefficients of a polynomial, the roots of the new polynomial are the reciprocals of the original roots. To show this, note that$$ \begin{aligned} a_n\rho_j^n+a_{n-1}\rho_j^{n-1}+\cdots+a_1\rho_j+a_0&=0\\ \implies a_n+\frac{a_{n-1}}{\rho_j}+\cdots+\frac{a_1}{\rho_j^{n-1}}+\frac{a_0}{\rho_j^n}&=0, \end{aligned}$$so the polynomial $a_0z^n+a_1z^{n-1}+\cdots+a_{n-1}z+a_n=0$ has roots $\frac1{\rho_j}$. Now the formula $(\dagger)$ applied to this polynomial gives$$\sum_j\frac1{\rho_j^2}=\frac{a_1^2-2a_0a_2}{a_0^2}.$$

## An Eulerian “proof”

In the video, a sequence of regular polygons is chosen such that the vertices tend to the sequence of odd integers on the real line. In our reformulation, we should thus choose a sequence of polynomials $f_n$ which tends to some power series $f$ with regularly spaced roots. For our purposes, we will choose$$ \begin{aligned} f(z)&=\frac{e^z-1}z\\ &=1+\frac12z+\frac16z^2+\cdots, \end{aligned}$$which has roots $2\pi ik$ for $k\in\mathbb Z\backslash{0}$.

If we pretend that our sum-of-inverse-squares formula works not just for polynomials, but for general power series, then applying it to $f$ gives$$ \begin{aligned} \sum_{k\in\mathbb Z\backslash\{0\}}\frac1{(2\pi ik)^2}&=\frac{(\frac12)^2-2(1)(\frac16)}{1^2}\\ -\frac2{4\pi^2}\sum_{k=1}^\infty\frac1{k^2}&=-\frac1{12}\\ \zeta(2)&=\frac{\pi^2}6, \end{aligned}$$which is a “proof” in the style of Euler that I think he might have enjoyed.

## A dead end?

To make the argument rigorous, we need polynomials $f_n$ whose roots all lie on a circle, with $f_n\to f$ as $n\to\infty$; a very nice choice is given by$$ \begin{aligned} f_n(z)&=\frac{(1+\frac zn)^n-1}z\\ &=1+\frac{\binom n2}{n^2}z+\frac{\binom n3}{n^3}z^2+\cdots, \end{aligned}$$which has roots at $n(e^{2\pi ik/n}-1)$ for $n\nmid k$, ie. lying on a circle with radius $n$ centred at $z=-n$. As a sanity check, note that for fixed $k$, we have $n(e^{2\pi ik/n}-1)\to2\pi ik$; hence the roots of $f_n$ “tend pointwise” to the roots of $f$.

However, this is not enough to show that the sum-of-inverse-squares of the $n$ roots of $f_n$ tends to the sum-of-inverse-squares of the roots of $f$. This is where our argument falls short of the geometric proof, which had good control on the error of the finite approximation.

## A tour de force

To finish the argument, we have to abandon our initial hopes for a quick proof, and get some exact expressions from the sum-of-inverse-squares formula.

We fix $n$, and set $\omega=e^{2\pi i/n}$. Then the roots of $f_n$ are $\rho_{k,n}\coloneqq n(\omega^k-1)$ for $1\leq k\leq n-1$. Note that$$ \begin{aligned} \frac1n\left(\frac1{\omega^k-1}+\frac12\right)&=\frac1{2n}\frac{\omega^k+1}{\omega^k-1}\\ &=\frac1{2n}\frac{\omega^{k/2}+\omega^{-k/2}}{\omega^{k/2}-\omega^{-k/2}}\\ &=-\frac i{2n}\cot\frac{k\pi}n. \end{aligned}$$Hence$$ \begin{aligned} \frac1{\rho_{k,n}}&=\frac1{2n}\left(-1-i\cot\frac{k\pi}n\right)\\ \frac1{\rho_{k,n}^2}&=\frac1{4n^2}\left(1-\cot^2\frac{k\pi}n+2i\cot\frac{k\pi}n\right). \end{aligned}$$We can sum this over $k$, take the real part, and compare with the sum-of-inverse-squares formula to obtain$$ \begin{aligned} \frac1{4n^2}\sum_{k=1}^{n-1}\left(1-\cot^2\frac{k\pi}n\right)&=\frac{(\frac{\binom n2}{n^2})^2-2(1)(\frac{\binom n3}{n^3})}{1^2}\\ &=-\frac{(n-1)(n-5)}{12n^2}\\ \sum_{k=1}^{n-1}\cot^2\frac{k\pi}n&=\frac{(n-1)(n-2)}3\\ \sum_{k=1}^{(n-1)/2}\cot^2\frac{k\pi}n&=\frac{(n-1)(n-2)}6, \end{aligned}$$since $\cot^2(\pi-\theta)=\cot^2\theta$.

Now we are in the position to finish as in Cauchy’s proof of the identity, by invoking the following trick: adding 1 to each term above gives$$\sum_{k=1}^{(n-1)/2}\csc^2\frac{k\pi}n=\frac{(n-1)(n+1)}6.$$By the elementary inequality $\sin\theta<\theta<\tan\theta$ for $\theta\in(0,\frac\pi2)$, we have$$ \begin{gathered} \cot^2\theta< \frac1{\theta^2}< \csc^2\theta\\ \sum_{k=1}^{(n-1)/2}\cot^2\frac{k\pi}n< \sum_{k=1}^{(n-1)/2}\frac{n^2}{k^2\pi^2}< \sum_{k=1}^{(n-1)/2}\csc^2\frac{k\pi}n\\ \frac{\pi^2}6\frac{(n-1)(n-2)}{n^2}< \sum_{k=1}^{(n-1)/2}\frac1{k^2}< \frac{\pi^2}6\frac{(n-1)(n+1)}{n^2}, \end{gathered}$$and we finish by taking $n\to\infty$.

**Exercise**: Can we calculate $\zeta(4)\coloneqq\sum_{n=1}^\infty\frac1{n^4}$ similarly? How annoying is it going to be?

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