The Basel Problem: Euler’s Bravura Performance

The Basel problem was first posed by Pietro Mengoli, a mathematics professor at the University of Bologna, in 1650, the same year in which he showed that the alternating harmonic series sums to ${\ln 2}$. The Basel problem asks for the sum of the reciprocals of the squares of the natural numbers,

$\displaystyle \sum_{n=1}^\infty \frac{1}{n^2} = \frac{1}{1^2} + \frac{1}{2^2} + \frac{1}{3^2} + \frac{1}{4^2} + \cdots = \ ?$

It is not immediately clear that this series converges, but this can be proved without much difficulty, as was first shown by Jakob Bernoulli in 1689. The sum is approximately 1.645 which has no obvious interpretation.

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That’s Maths II: A Ton of Wonders

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For a century or so, the Basel problem outwitted some of the most brilliant mathematicians, including Leibniz and both Jakob and Johann Bernoulli. As the Bernoullis lived in Basel, and Euler had grown up there, the problem acquired the name of that city. It was solved by Leonhard Euler in 1734, and presented to the Imperial Academy of Sciences in St Petersburg the following year.

By solving the problem that had defeated so many other mathematicians, Euler achieved immediate fame. He found the exact sum to be ${\pi^2/6}$. His line of reasoning was ingenious, with arguments based on daring manipulations that were difficult to justify. However, some years later, he presented a more rigorous proof.

Euler did not rest there: he generalised the problem considerably, deriving the sums for several related series. The Basel series is a particular case of the much more valuable and important zeta function, which Bernhard Riemann used in his 1859 paper On the Number of Primes Less Than a Given Magnitude, and which is at the core of the Riemann hypothesis, the most important unsolved problem in mathematics.

Series expansion for ${\sin x}$

We are familiar from elementary calculus with the Taylor series expansion of the trigonometric functions. The sine expansion about ${x=0}$ is

$\displaystyle \sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \cdots \nonumber \ \ \ \ \ (1)$

We can get a visual idea of how the series converges by considering the partial sums. We truncate the sine series to get a sequence of polynomials:

$\displaystyle \begin{array}{rcl} T_1(x) &=& {x} \\ T_3(x) &=& {x - \frac{x^3}{3!} } \\ T_5(x) &=& {x - \frac{x^3}{3!} + \frac{x^5}{5!} } \\ T_7(x) &=& {x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} } \\ \cdots &=& \cdots \end{array}$

They approximate ${\sin x}$ better with increasing order. In the figure below, we show the expansions ${T_n(x)}$ up to order ${n=39}$, which approximates ${\sin x}$ over five wavelengths
(only positive ${x}$ is shown in the figure).

${\sin x}$ and 20 approximations to it. The function ${T_{39}(x)}$ fits ${\sin x}$ well over five wavelengths.

Product expansion for ${\sin x}$

We note that the zeros of the polynomials ${T_n(x)}$ do not correspond to the zeros of ${\sin x}$, which are ${\{ 0, \pm\pi, \pm 2\pi, \pm 3\pi, \dots \}}$. We can easily write a sequence of polynomials having zeros at these points:

$\displaystyle \begin{array}{rcl} P_1(x) &=& {x} \\ P_3(x) &=& {x \left(1-\frac{x^2}{\pi^2} \right)} \\ P_5(x) &=& {x \left(1-\frac{x^2}{\pi^2}\right)\left(1-\frac{x^2}{(2\pi)^2}\right)} \\ P_7(x) &=& {x \left(1-\frac{x^2}{\pi^2}\right)\left(1-\frac{x^2}{(2\pi)^2}\right)\left(1-\frac{x^2}{(3\pi)^2}\right) } \\ \cdots &=& \cdots \end{array}$

In the figure below, we show the expansions ${P_n(x)}$ up to order ${n=39}$ (only positive ${x}$ is shown). The products ${P_n(x)}$ approach ${\sin x}$ as the degree ${n}$ increases.

${\sin x}$ and 20 approximations to it. The products ${P_n(x)}$ approximate ${\sin x}$ better as ${n}$ increases.

Euler’s daring reasoning

For finite ${n}$ both ${T_n(x)}$ and ${P_n(x)}$ are polynomials of degree ${n}$, but ${T_n(x) \ne P_n(x)}$. However, in the limit ${n\rightarrow\infty}$, both the sequences tend to the sine function, so Euler assumed that

$\displaystyle \sum_{n=0}^{\infty}(-1)^{n} \frac{x^{2n+1}}{(2n+1)!} = x \prod_{n=1}^{\infty} \left( 1 - \frac{x^{2} }{(n\pi)^2} \right) \,. \ \ \ \ \ (2)$

He also had the brilliant idea that he could regard both the infinite sum ${T_\infty(x)}$ and the infinite product ${P_\infty(x)}$ as polynomials of infinite degree. Since they are both equal to the sine function, the coefficients of corresponding powers of ${x}$ in the sum and in the product must be equal.

The coefficient of ${x}$ in the Taylor series is ${1}$; the coefficient of ${x^3}$ is ${-1/3! = -1/6}$. To get corresponding terms in the infinite product, Euler expressed it as an infinite polynomial. This is best illustrated for a finite case: we expand ${P_7(x)}$,

$\displaystyle \begin{array}{rcl} P_7(x) &=& x \left(1-\frac{x^2}{\pi^2}\right)\left(1-\frac{x^2}{(2\pi)^2}\right)\left(1-\frac{x^2}{(3\pi)^2}\right) \\ &=& x - \left(\frac{1}{1^2} + \frac{1}{2^2} + \frac{1}{3^2} \right) \frac{x^3}{\pi^2} + \left(\frac{1}{1^2}\cdot\frac{1}{2^2} + \frac{1}{1^2}\cdot\frac{1}{3^2} + \frac{1}{2^2}\cdot\frac{1}{3^2} \right) \frac{x^5}{\pi^4} - \left(\frac{1}{1^2}\cdot\frac{1}{2^2}\cdot\frac{1}{3^2} \right) \frac{x^7}{\pi^6} \,. \end{array}$

so that the coefficient of ${x^3}$ is minus the sum of the first three terms of the Basel series. Euler assumed that this could be extended to the infinite case, yielding

$\displaystyle - \frac{1}{\pi^2}\left( \frac{1}{1^2} + \frac{1}{2^2} + \frac{1}{3^2} + \frac{1}{4^2} + \cdots \right) \,.$

Equating this to ${-1/6}$ and multiplying by ${-\pi^2}$, he obtained the remarkable result

$\displaystyle \sum_{n=1}^\infty \frac{1}{n^2} = \left( \frac{1}{1^2} + \frac{1}{2^2} + \frac{1}{3^2} + \frac{1}{4^2} + \cdots \right) = \frac{\pi^2}{6}$

The terms of the Basel series involve squares, whereas the sum, through ${\pi}$, is linked to the circle. Euler was pleasantly surprised with his result and wrote “quite unexpectedly … I have found an elegant formula depending on ${\pi}$.”

Returning to the equation ( 2 ) equating the infinite sum and product, Euler evaluated both sides for ${x=\pi/2}$. Since ${\sin x=1}$ for this argument, he obtained

$\displaystyle 1 = \frac{\pi}{2} \prod_{n=1}^{\infty} \left( 1 - \frac{1}{4 n^2} \right) = \frac{\pi}{2} \prod_{n=1}^{\infty} \left(\frac{2n-1}{2n}\right)\left(\frac{2n+1}{2n}\right) \,.$

This is equivalent to the result

$\displaystyle \frac{\pi}{2} = \frac{2}{1} \cdot \frac{2}{3} \cdot \frac{4}{3} \cdot \frac{4}{5} \cdot \frac{6}{5} \cdot \frac{6}{7} \cdot \frac{8}{7} \cdot \frac{8}{9} \cdots \,,$

an infinite product for ${\pi}$ that had been published in 1656 by John Wallis.

By further use of ( 2 ), Euler obtained several other sparkling results, which are described by Dunham (1990, Ch. 9). He also went on to derive the result we know as Euler’s Product or, as Derbyshire (2003) called it, the Golden Key, a topic to which we hope to return.

Sources

${\bullet}$ Derbyshire, John, 2003: Prime Obsession. Plume Books, 422pp. ISBN: 978-0-4522-8525-5.

${\bullet}$ Dunham, William, 1990: Journey through Genius: the Great Theorems of Mathematics. John Wiley & Sons. Published by Penguin, 1991, 300pp. ISBN: 978-0-1401-4739-1.