Basel problem explained

The Basel problem is a problem in mathematical analysis with relevance to number theory, concerning an infinite sum of inverse squares. It was first posed by Pietro Mengoli in 1650 and solved by Leonhard Euler in 1734, and read on 5 December 1735 in The Saint Petersburg Academy of Sciences.^[1] Since the problem had withstood the attacks of the leading mathematicians of the day, Euler's solution brought him immediate fame when he was twenty-eight. Euler generalised the problem considerably, and his ideas were taken up more than a century later by Bernhard Riemann in his seminal 1859 paper "On the Number of Primes Less Than a Given Magnitude", in which he defined his zeta function and proved its basic properties. The problem is named after the city of Basel, hometown of Euler as well as of the Bernoulli family who unsuccessfully attacked the problem.

The Basel problem asks for the precise summation of the reciprocals of the squares of the natural numbers, i.e. the precise sum of the infinite series: $\sum_^\infty \frac = \frac + \frac + \frac + \cdots.$

The sum of the series is approximately equal to 1.644934.^[2] The Basel problem asks for the exact sum of this series (in closed form), as well as a proof that this sum is correct. Euler found the exact sum to be $/$ and announced this discovery in 1735. His arguments were based on manipulations that were not justified at the time, although he was later proven correct. He produced an accepted proof in 1741.

The solution to this problem can be used to estimate the probability that two large random numbers are coprime. Two random integers in the range from 1 to, in the limit as goes to infinity, are relatively prime with a probability that approaches $/$ , the reciprocal of the solution to the Basel problem.

Euler's approach

Euler's original derivation of the value $/$ essentially extended observations about finite polynomials and assumed that these same properties hold true for infinite series.

Of course, Euler's original reasoning requires justification (100 years later, Karl Weierstrass proved that Euler's representation of the sine function as an infinite product is valid, by the Weierstrass factorization theorem), but even without justification, by simply obtaining the correct value, he was able to verify it numerically against partial sums of the series. The agreement he observed gave him sufficient confidence to announce his result to the mathematical community.

To follow Euler's argument, recall the Taylor series expansion of the sine function $\sin x = x - \frac + \frac - \frac + \cdots$ Dividing through by gives $\frac = 1 - \frac + \frac - \frac + \cdots .$

The Weierstrass factorization theorem shows that the right-hand side is the product of linear factors given by its roots, just as for finite polynomials. Euler assumed this as a heuristic for expanding an infinite degree polynomial in terms of its roots, but in fact it is not always true for general

P(x)

.^[3] This factorization expands the equation into:

\begin\frac &= \left(1 - \frac\right)\left(1 + \frac\right)\left(1 - \frac\right)\left(1 + \frac\right)\left(1 - \frac\right)\left(1 + \frac\right) \cdots \\ &= \left(1 - \frac\right)\left(1 - \frac\right)\left(1 - \frac\right) \cdots\end

If we formally multiply out this product and collect all the terms (we are allowed to do so because of Newton's identities), we see by induction that the coefficient of is ^[4] $-\left(\frac + \frac + \frac + \cdots \right) = -\frac\sum_^\frac.$

But from the original infinite series expansion of, the coefficient of is . These two coefficients must be equal; thus, $-\frac = -\frac\sum_^\frac.$

Multiplying both sides of this equation by −² gives the sum of the reciprocals of the positive square integers. $\sum_^\frac = \frac.$

Generalizations of Euler's method using elementary symmetric polynomials

Using formulae obtained from elementary symmetric polynomials,^[5] this same approach can be used to enumerate formulae for the even-indexed even zeta constants which have the following known formula expanded by the Bernoulli numbers: $\zeta(2n) = \frac B_.$

For example, let the partial product for

\sin(x)

expanded as above be defined by

	S_n(x)
	x

	n
\prod\limits
	k=1

\left(1-

	x²
	k² ⋅ \pi²

\right)

. Then using known formulas for elementary symmetric polynomials (a.k.a., Newton's formulas expanded in terms of power sum identities), we can see (for example) that

\begin\left[x^4\right] \frac & = \frac\left(\left(H_n^\right)^2 - H_n^\right) \qquad \xrightarrow \qquad \frac\left(\zeta(2)^2-\zeta(4)\right) \\[4pt]& \qquad \implies \zeta(4) = \frac = -2\pi^4 \cdot [x^4] \frac +\frac \\[8pt]\left[x^6\right] \frac & = -\frac\left(\left(H_n^\right)^3 - 3H_n^ H_n^ + 2H_n^\right) \qquad \xrightarrow \qquad \frac\left(\zeta(2)^3-3\zeta(2)\zeta(4) + 2\zeta(6)\right) \\[4pt]& \qquad \implies \zeta(6) = \frac = -3 \cdot \pi^6 [x^6] \frac - \frac \frac \frac + \frac,\end

and so on for subsequent coefficients of

[x^2k]

	S_n(x)
	x

. There are other forms of Newton's identities expressing the (finite) power sums

	(2k)
H
	n

in terms of the elementary symmetric polynomials,

e_i\equiv

i\left(-	\pi²
	1²

	\pi²
	2²

	\pi²
	3²

	\pi²
	4²

,\ldots\right),

but we can go a more direct route to expressing non-recursive formulas for

\zeta(2k)

using the method of elementary symmetric polynomials. Namely, we have a recurrence relation between the elementary symmetric polynomials and the power sum polynomials given as on this page by

(-1)^k e_k(x_1,\ldots,x_n) = \sum_^k (-1)^ p_j(x_1,\ldots,x_n)e_(x_1,\ldots,x_n),

which in our situation equates to the limiting recurrence relation (or generating function convolution, or product) expanded as $\frac\cdot \frac = -[x^{2k}] \frac \times \sum_ \zeta(2i) x^i.$

Then by differentiation and rearrangement of the terms in the previous equation, we obtain that $\zeta(2k) = [x^{2k}]\frac\left(1-\pi x\cot(\pi x)\right).$

Consequences of Euler's proof

By the above results, we can conclude that

\zeta(2k)

is always a rational multiple of

\pi^2k

. In particular, since

\pi

and integer powers of it are transcendental, we can conclude at this point that

\zeta(2k)

is irrational, and more precisely, transcendental for all

k\geq1

. By contrast, the properties of the odd-indexed zeta constants, including Apéry's constant

\zeta(3)

, are almost completely unknown.

The Riemann zeta function

The Riemann zeta function is one of the most significant functions in mathematics because of its relationship to the distribution of the prime numbers. The zeta function is defined for any complex number with real part greater than 1 by the following formula: $\zeta(s) = \sum_^\infty \frac.$

Taking, we see that is equal to the sum of the reciprocals of the squares of all positive integers: $\zeta(2) = \sum_^\infty \frac = \frac + \frac + \frac + \frac + \cdots = \frac \approx 1.644934.$

Convergence can be proven by the integral test, or by the following inequality: $\begin \sum_^N \frac & < 1 + \sum_^N \frac \\ & = 1 + \sum_^N \left(\frac - \frac \right) \\ & = 1 + 1 - \frac \;\; 2.\end$

This gives us the upper bound 2, and because the infinite sum contains no negative terms, it must converge to a value strictly between 0 and 2. It can be shown that has a simple expression in terms of the Bernoulli numbers whenever is a positive even integer. With : $\zeta(2n) = \frac.$

A proof using Euler's formula and L'Hôpital's rule

sinc(x)=	\sin(\pix)
	\pix

has a Weierstrass factorization representation as an infinite product:

\frac = \prod_^\infty \left(1-\frac\right).

The infinite product is analytic, so taking the natural logarithm of both sides and differentiating yields $\frac-\frac=-\sum_^\infty \frac$

(by uniform convergence, the interchange of the derivative and infinite series is permissible). After dividing the equation by

and regrouping one gets

\frac-\frac=\sum_^\infty \frac.

We make a change of variables (

x=-it

-\frac+\frac=\sum_^\infty \frac.

Euler's formula can be used to deduce that $\frac=\frac\frac=\frac+\frac.$ or using the corresponding hyperbolic function: $\frac=\frac=\frac\coth(\pi t).$

Then $\sum_^\infty \frac=\frac=-\frac + \frac \coth(\pi t).$

Now we take the limit as

approaches zero and use L'Hôpital's rule thrice. By Tannery's theorem applied to

\lim_\sum_^\infty 1/(n^2+1/t^2)

, we can interchange the limit and infinite series so that

\lim_\sum_^\infty 1/(n^2+t^2)=\sum_^\infty 1/n^2

and by L'Hôpital's rule

\begin\sum_^\infty \frac&=\lim_\frac\frac\\[6pt]&=\lim_\frac\\[6pt]&=\lim_\frac\\[6pt]&=\frac.\end

A proof using Fourier series

Use Parseval's identity (applied to the function) to obtain $\sum_^\infty |c_n|^2 = \frac\int_^\pi x^2 \, dx,$ where $\begin c_n &= \frac\int_^\pi x e^ \, dx \\[4pt] &= \frac i \\[4pt] &= \frac i \\[4pt] &= \frac i\end$

for, and . Thus, $|c_n|^2 = \begin\dfrac, & \text n \neq 0, \\0, & \text n = 0,\end$

and $\sum_^\infty |c_n|^2 = 2\sum_^\infty \frac = \frac \int_^\pi x^2 \, dx.$

Therefore, $\sum_^\infty \frac = \frac\int_^\pi x^2 \, dx = \frac$ as required.

Another proof using Parseval's identity

Given a complete orthonormal basis in the space

	2
L
	\operatorname{per

}(0, 1) of L2 periodic functions over

(0,1)

(i.e., the subspace of square-integrable functions which are also periodic), denoted by

\{e_i\}

	infty

	i=-infty

, Parseval's identity tells us that

\|x\|^2 = \sum_^ |\langle e_i, x\rangle|^2,

where

\|x\|:=\sqrt{\langlex,x\rangle}

is defined in terms of the inner product on this Hilbert space given by

\langle f, g\rangle = \int_0^1 f(x) \overline \, dx,\ f,g \in L^2_(0, 1).

We can consider the orthonormal basis on this space defined by

e_k\equive_k(\vartheta):=\exp(2\pi\imathk\vartheta)

such that

\langlee_k,e_j\rangle=

	1
\int
	0

e^2\pi\imathd\vartheta=\delta_k,j

. Then if we take

f(\vartheta):=\vartheta

, we can compute both that

\begin\|f\|^2 & = \int_0^1 \vartheta^2 \, d\vartheta = \frac \\\langle f, e_k\rangle & = \int_0^1 \vartheta e^ \, d\vartheta = \Biggl\

Notes and References

https://scholarlycommons.pacific.edu/euler-works/41/ E41 – De summis serierum reciprocarum
A013661. cs2.
A priori, since the left-hand-side is a polynomial (of infinite degree) we can write it as a product of its roots as $\begin\sin(x) & = x (x^2-\pi^2)(x^2-4\pi^2)(x^2-9\pi^2) \cdots \\ & = Ax \left(1 - \frac\right)\left(1 - \frac\right)\left(1 - \frac\right) \cdots.\end$ Then since we know from elementary calculus that
\lim_x

\sin(x)
x

=1

, we conclude that the leading constant must satisfy
A=1

.
In particular, letting
(2)
H
n

:=

n
\sum
k=1

k^-2

denote a generalized second-order harmonic number, we can easily prove by induction that
[x^2]

n
\prod \left(1-
k=1
x²
\pi²

\right)=-

(2)
H
n
\pi²

→ -

\zeta(2)
\pi²

as
n → infty

.
Cf., the formulae for generalized Stirling numbers proved in:

	(2)
H
	n

	n
\sum
	k=1

	\zeta(2)
	\pi²