>[!abstract] >The **Basel problem** was posed by Italian mathematician Pietro Mengoli (1626–1686) in 1650 as follows: what is the *precise* summation (in closed form) of the reciprocals of the squares of the natural numbers, which is expressed as the following converging infinite series: > >$\sum_{n=1}^{\infty} \frac{1}{n^2} = \frac{1}{1^2} + \frac{1}{2^2} + \frac{1}{3^2} + \frac{1}{4^2} + \cdots = 1 + \frac{1}{4} + \frac{1}{9} + \frac{1}{16} + \cdots \approx 1.644934$ > >In 1734, Leonard Euler (1707–1783) found the exact sum to be $\pi^2 / 6$, which is a remarkable result considering the unexpected presence of $\pi$ and the unorthodox approach he followed. ## Context and motivation The problem was named after the Swiss town of Basel where resided two prominent mathematicians who studied the problem deeply, but unsuccessfully: the Bernoulli brothers. The brothers were inspired by William Leibniz’s recent discovery of a closed form value for the infinite series of the reciprocals of the *triangular* numbers: $\frac{1}{1} + \frac{1}{3} + \frac{1}{6} + \frac{1}{10} + \cdots = 2$ Leibniz’s result gave hope that other infinite series involving reciprocals could be solved. In particular, the square numbers were the next logical arrangement after the triangular numbers. Jacob Bernoulli published a treatise on the topic, while his brother Johan tutored students; among them were Leonard Euler, who soon achieved mathematical fame from solving the Basel problem. ## Connections to the Riemann zeta function and prime numbers The infinite series as written in the abstract above is the Dirichlet form of the Riemann zeta function $\zeta(s)$ for $s=2$, written as $\zeta(2)$. Its reciprocal, expressed as $1 / \zeta(2)$, is therefore equal to the reciprocal of Euler’s closed form solution, $6 / \pi^2$. It evaluates to approximately $61\%$ and is the probability (in the limit as $n \to \infty$) that two random integers greater than $1$ and up to $n$ are coprime (i.e., they only have $1$ as a common divisor, even if they themselves are not primes). ## Euler’s proof Euler’s approach was not grounded in algebraic theories known at the time, but it worked nonetheless. His initial intuition was to find a relationship between infinite series (which the Basel problem is) and roots of equations. Euler knew from the fundamental theorem of algebra that if you have a general polynomial $p(x)$ of degree $n$, coefficients $a_n$, known roots at $x = r_n$ such that $p(x) = 0$, and a known value at zero of $p(0) = 1$: $p(x) = a_{n}x^{n} + a_{n-1}x^{n-1} + a_{n-2}x^{n-2} + \cdots + a_{1}x^{0} + a_{0}$ then that polynomial can be factored into a *product* involving its roots: $p(x) = \prod_{1}^{n} (1 - \frac{x}{r_n}) = (1 - \frac{x}{r_1}) (1 - \frac{x}{r_2}) (1 - \frac{x}{r_3}) \cdots$ This product can be expanded simply using the elementary symmetric polynomials from Vieta’s formulas for a product over $n$ roots, which Euler was also very familiar with. The general form is: $\prod_{1}^{n} (1 - \frac{x}{r_n}) = 1 - (\sum{\frac{1}{r_n}})x + (\sum_{i \lt j}{\frac{1}{r_i r_j}})x^2 - \cdots$ In this expanded product, we can see that the coefficient of $x$ (the second term) is the negative sum of reciprocals of all roots: $a_{x} = -\sum{\frac{1}{r_n}} = -(\frac{1}{r_1} + \frac{1}{r_2} + \frac{1}{r_3} + \cdots)$ And the coefficient of $x^2$ (the third term in the expanded product) is the sum of all pairwise products of the reciprocals of roots: $a_{x^2} = \sum_{i \lt j}{\frac{1}{r_i r_j}} = \frac{1}{r_1 r_2} + \frac{1}{r_1 r_3} + \frac{1}{r_1 r_4} + \cdots$ The brilliance of Euler was to recognize that if he knew all the roots of a function, he could apply Vieta’s formulas in reverse, and find the value of a sum of the form $\sum{\frac{1}{r_{n}^{2}}}$ from the coefficients of its power series. So what Euler needed was a function $f(x)$ that had simple roots (e.g., only integers, or multiples of a single constant) to keep the calculations tractable, and whose infinite series was already known so that the coefficients could be matched. Euler then had the intuition to choose the $\sin$ function, because it is the only familiar function whose roots follow a simple pattern: $x = 0, \pm \pi, \pm 2\pi, \pm 3\pi, \cdots$ The root at $x = 0$ is inconvenient, though, because it would make the factored form collapse trivially. So Euler first divided the function by $x$: $f(x) = \frac{\sin{x}}{x}$ which still preserves the property $f(0) = 1$ and retains all the other roots except $0$. Note that at this point, all the roots are of the form $\pm n \pi$. From here, Euler audaciously treated $\sin(x)$ as an ordinary polynomial with roots $\pm n \pi$, even though he was not (at the time) justified in doing so, given that $sin$ is a transcendental function, not a polynomial. Nevertheless, he applied the same polynomial factorization method as earlier and obtained: $f(x) = \frac{\sin(x)}{x} = (1 - \frac{x}{\pi}) (1 + \frac{x}{\pi}) (1 - \frac{x}{2\pi}) (1 + \frac{x}{2\pi}) \cdots$ which, after grouping the roots $+n\pi$ and $-n\pi$ into a single quadratic factor using the difference of squares, simplified to: $f(x) = \frac{\sin(x)}{x} = \prod_{n=1}^{\infty} (1 - \frac{x^2}{n^2 \pi^2})$ Notice here that the variable is $x^2$ and not $x$; however, we can substitute the numerator $x^2$ for another variable $u$ and the roots in the denominator $n^2 \pi^2$ for $r_n$: $f(x) = \prod_{n=1}^{\infty} (1 - \frac{u}{r_n})$ and it then becomes obvious that this is the same factorized polynomial form as before. Euler then applied Vieta’s formulas to expand this newly-constructed polynomial and identify its coefficients: $f(x) = \frac{\sin(x)}{x} = 1 - (\sum{\frac{1}{n^2 \pi^2}})x^2 + (\sum{\frac{1}{i^2 j^2 \pi^4}})x^4 - \cdots$ So the coefficient of $x^2$ (the second term) is: $a_{x^2} = -\sum_{n=1}^{\infty}{\frac{1}{n^2 \pi^2}}$ At this juncture, Euler needed to match this coefficient of $x^2$ with the same coefficient from $\frac{\sin(x)}{x}$. To do so, he first used the well-known Taylor series expansion of $\sin(x)$, which is part of the reason why he had chosen $sin$ in the first place: $\sin(x) = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \cdots$ and then divided both sides by $x$ (which is valid for $x \neq 0$) to find $\frac{\sin(x)}{x}$ again: $\frac{\sin(x)}{x} = 1 - \frac{x^2}{3!} + \frac{x^4}{5!} - \frac{x^6}{7!} + \cdots = 1 - \frac{x^2}{6} + \frac{x^4}{120} - \frac{x^6}{5040} + \cdots$ So the coefficient of $x^2$ in the Taylor series expansion is $-1/6$; at this point, the Basel problem was solved. All that was left to do was match both coefficients: $\sum_{n=1}^{\infty}{\frac{1}{n^2 \pi^2}} = -\frac{1}{6}$ which, multiplying both sides by $\pi^2$, gives $\sum_{n=1}^{\infty}{\frac{1}{n^2}} = \frac{\pi^2}{6}$ solving the Basel problem and launching Euler into prominence. >[!note] >Euler also used several more even coefficients of the same expanded polynomial, from $x^4$ onwards, to also solve for $\sum_{n=1}^{\infty} \frac{1}{n^4} = \frac{\pi^4}{90}$ >and a few more. >[!related] >- **North** (upstream): — >- **West** (similar): — >- **East** (different): — >- **South** (downstream): —