Basel Problem 2.0

Let $k=\frac{p}{q}$ . We're interested in the roots of $\frac{\tan x}{x}=k$ which we can work with more easily in the form $\sin x = kx\cos x$

Euler's original approach to the Basel problem was to consider a function's power series as an infinite polynomial, and manipulate it as such. (This has serious flaws, but a) it worked, b) Euler later proved the same result in different ways and c) the rigorous justification for the approach in this case was found later still.)

By Vieta, for a polynomial $a_0+a_1 x + a_2 x^2 + \cdots$ with roots $x_1,x_2, \cdots$ , we have $\sum \frac{1}{x_i^2} = \left( \frac{a_1}{a_0} \right)^2-2\frac{a_2}{a_0}$

Rewriting our equation in terms of power series, we have $\begin{aligned} \sin x &= kx\cos x \\ x-\frac{x^3}{6}+\cdots &= kx \left( 1-\frac{x^2}{2}+\cdots \right) \\ 1-\frac{x^2}{6}+\cdots &= k \left( 1-\frac{x^2}{2}+\cdots \right) \\ 1-k+(3k-1)\frac{x^2}{6}+\cdots &=0 \end{aligned}$

Applying the Vieta formula from above, the sum of the reciprocals of the squares of the roots will be $-2\frac{3k-1}{6(1-k)}$

Note that the sum in the problem is restricted to the positive reals. If $x$ is a root, so is $-x$ ; so the quantity above is double the required sum, and we get the equation $-\frac{3k-1}{6(1-k)}=1$ with solution $k=\frac53$ , so that $p+q=5+3=\boxed8$ .

For any positive real number $r > 1$ , consider the entire function $f_r(z) \; = \; r\cos z - \frac{\sin z}{z}$ Now $f_r(z) \; = \; (r - 1) - \tfrac16(3r - 1)z^2 + O(z^4) \hspace{2cm} z \to 0$ and so $0$ is not a root of $f_r$ , and thus the roots of $f_r(z)$ are just the nonzero solutions of $\tan z = rz$ . Since both $\sin z$ and $\cos z$ are entire functions of order $1$ at infinity, so is $f_r$ . Suppose that $X_r$ is the set of positive roots of the equation $\tan z = rz$ . Then $X_r$ contains one element in the interval $\big(n\pi,(n+\tfrac12)\pi\big)$ for each integer $n \ge 0$ . Thus $X_r$ is countable, and we can write $X_r = \{\alpha_{r,n} \,|\, n \in \mathbb{N}\}$ . Since it is certainly true that $\alpha_{rn} > (n-1)\pi$ , we deduce that $\sum_{n=1}^\infty \frac{1}{\alpha_{rn}^2} \; < \; \infty$ Using the Hadamard Factorisation Theorem, we deduce that $f_r(z) \; = \; e^{g(z)} \prod_{n=1}^\infty \left(1 - \frac{z^2}{\alpha_{rn}^2}\right)$ where $g(z)$ is a polynomial of degree at most $1$ . Since $f_r(z)$ is even, we deduce that $g(z)$ is contant, and hence $f_r(z) \; =\; (1 - r)\prod_{n=1}^\infty \left(1 - \frac{z^2}{\alpha_{rn}^2}\right) \; = \; (1 - r)\left[1 - \left(\sum_{n=1}^\infty \frac{1}{\alpha_{rn}^2}\right)z^2 + O(z^4) \right] \hspace{2cm} z \to 0$ and hence $\sum_{n=1}^\infty \frac{1}{\alpha_{rn}^2} \; = \; \frac{3r - 1}{6(r - 1)} \hspace{3cm} (\star)$ For this sum to be equal to $1$ , we must have $r = \tfrac53$ , and hence the solution is $5 + 3 = \boxed{8}$ .

It so happens that the all the roots of $\tan z = r z$ are real if $r > 1$ , while there are imaginary roots if $0 < r < 1$ , which is why formula $(\star)$ fails when $0 < r < 1$ . Well, it does not fail, but the infinite sum $\sum_n \alpha_{rn}^{-2}$ is over the imaginary roots as well as the real ones.

A similar problem appeared in The American Mathematical Monthly Vol. 93 (1986) which asks for the value of $S$ when $\tan{x}=x$ .

The value of $S$ in that problem turns out to be $\dfrac{1}{10}$ .

The general problem of finding $S$ is also dealt with, in the solution to the problem, which can be found here , in the section called 'A Rayleigh Popular Problem'.

Check out a similar problem here