thumb|upright=1.25|The Basel problem is analogous to the total [[apparent magnitude|apparent brightness of infinite identical point light sources on the number line viewed from the origin (top figure), compared to a single light source at position 1 (bottom)]]
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. 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:
<math display="block">\sum_{n=1}^\infty \frac{1}{n^2} = \frac{1}{1^2} + \frac{1}{2^2} + \frac{1}{3^2} + \cdots. </math>
The sum of the series is approximately equal to 1.644934. 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 <math display="inline">{\pi^2}/{6}</math> 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 <math display="inline">{6}/{\pi^2}</math>, the reciprocal of the solution to the Basel problem.
Euler's approach
Euler's original derivation of the value <math display="inline">{\pi^2}/{6}</math> 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
<math display=block> \sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \cdots </math>
Dividing through by gives
<math display=block> \frac{\sin x}{x} = 1 - \frac{x^2}{3!} + \frac{x^4}{5!} - \frac{x^6}{7!} + \cdots .</math>
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 <math>P(x)</math>. This factorization expands the equation into:
<math display=block>\begin{align}
\frac{\sin x}{x} &= \left(1 - \frac{x}{\pi}\right)\left(1 + \frac{x}{\pi}\right)\left(1 - \frac{x}{2\pi}\right)\left(1 + \frac{x}{2\pi}\right)\left(1 - \frac{x}{3\pi}\right)\left(1 + \frac{x}{3\pi}\right) \cdots \\
&= \left(1 - \frac{x^2}{\pi^2}\right)\left(1 - \frac{x^2}{4\pi^2}\right)\left(1 - \frac{x^2}{9\pi^2}\right) \cdots
\end{align}</math>
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
<math display=block> -\left(\frac{1}{\pi^2} + \frac{1}{4\pi^2} + \frac{1}{9\pi^2} + \cdots \right) = -\frac{1}{\pi^2}\sum_{n=1}^{\infty}\frac{1}{n^2}.</math>
But from the original infinite series expansion of , the coefficient of is . These two coefficients must be equal; thus,
<math display=block>-\frac{1}{6} = -\frac{1}{\pi^2}\sum_{n=1}^{\infty}\frac{1}{n^2}.</math>
Multiplying both sides of this equation by −<sup>2</sup> gives the sum of the reciprocals of the positive square integers.
<math display=block>\sum_{n=1}^{\infty}\frac{1}{n^2} = \frac{\pi^2}{6}.</math>
Generalizations of Euler's method using elementary symmetric polynomials
Using formulae obtained from elementary symmetric polynomials, 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:
<math display=block>\zeta(2n) = \frac{(-1)^{n-1} (2\pi)^{2n{2 \cdot (2n)!} B_{2n}. </math>
For example, let the partial product for <math>\sin(x)</math> expanded as above be defined by <math>\frac{S_n(x)}{x} = \prod\limits_{k=1}^n \left(1 - \frac{x^2}{k^2 \cdot \pi^2}\right)</math>. 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
<math display=block>
\begin{align}
\left[x^4\right] \frac{S_n(x)}{x} & = \frac{1}{2\pi^4}\left(\left(H_n^{(2)}\right)^2 - H_n^{(4)}\right) \qquad \xrightarrow{n \rightarrow \infty} \qquad \frac{1}{2\pi^4}\left(\zeta(2)^2-\zeta(4)\right) \\[4pt]
& \qquad \implies \zeta(4) = \frac{\pi^4}{90} = -2\pi^4 \cdot [x^4] \frac{\sin(x)}{x} +\frac{\pi^4}{36} \\[8pt]
\left[x^6\right] \frac{S_n(x)}{x} & = -\frac{1}{6\pi^6}\left(\left(H_n^{(2)}\right)^3 - 3H_n^{(2)} H_n^{(4)} + 2H_n^{(6)}\right) \qquad \xrightarrow{n \rightarrow \infty} \qquad \frac{1}{6\pi^6}\left(\zeta(2)^3-3\zeta(2)\zeta(4) + 2\zeta(6)\right) \\[4pt]
& \qquad \implies \zeta(6) = \frac{\pi^6}{945} = -3 \cdot \pi^6 [x^6] \frac{\sin(x)}{x} - \frac{2}{3} \frac{\pi^2}{6} \frac{\pi^4}{90} + \frac{\pi^6}{216},
\end{align}
</math>
and so on for subsequent coefficients of <math>[x^{2k}] \frac{S_n(x)}{x}</math>. There are other forms of Newton's identities expressing the (finite) power sums <math>H_n^{(2k)}</math> in terms of the elementary symmetric polynomials, <math>e_i \equiv e_i\left(-\frac{\pi^2}{1^2}, -\frac{\pi^2}{2^2}, -\frac{\pi^2}{3^2}, -\frac{\pi^2}{4^2}, \ldots\right), </math> but we can go a more direct route to expressing non-recursive formulas for <math>\zeta(2k)</math> 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
<math display=block>(-1)^{k}k e_k(x_1,\ldots,x_n) = \sum_{j=1}^k (-1)^{k-j-1} p_j(x_1,\ldots,x_n)e_{k-j}(x_1,\ldots,x_n),</math>
which in our situation equates to the limiting recurrence relation (or generating function convolution, or product) expanded as
<math display=block> \frac{\pi^{2k{2}\cdot \frac{(2k) \cdot (-1)^k}{(2k+1)!} = -[x^{2k}] \frac{\sin(\pi x)}{\pi x} \times \sum_{i \geq 1} \zeta(2i) x^i. </math>
Then by differentiation and rearrangement of the terms in the previous equation, we obtain that
<math display=block>\zeta(2k) = [x^{2k}]\frac{1}{2}\left(1-\pi x\cot(\pi x)\right). </math>
Consequences of Euler's proof
By the above results, we can conclude that <math>\zeta(2k)</math> is always a rational multiple of <math>\pi^{2k}</math>. In particular, since <math>\pi</math> and integer powers of it are transcendental, we can conclude at this point that <math>\zeta(2k)</math> is irrational, and more precisely, transcendental for all <math>k \geq 1</math>. By contrast, the properties of the odd-indexed zeta constants, including Apéry's constant <math>\zeta(3)</math>, 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:
<math display=block>\zeta(s) = \sum_{n=1}^\infty \frac{1}{n^s}.</math>
Taking , we see that is equal to the sum of the reciprocals of the squares of all positive integers:
<math display=block>\zeta(2) = \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 = \frac{\pi^2}{6} \approx 1.644934.</math>
Convergence can be proven by the integral test, or by the following inequality:
<math display=block>\begin{align}
\sum_{n=1}^N \frac{1}{n^2} & < 1 + \sum_{n=2}^N \frac{1}{n(n-1)} \\
& = 1 + \sum_{n=2}^N \left( \frac{1}{n-1} - \frac{1}{n} \right) \\
& = 1 + 1 - \frac{1}{N} \;{\stackrel{N \to \infty}{\longrightarrow\; 2.
\end{align}</math>
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 :
<math display=block>\zeta(2n) = \frac{(2\pi)^{2n}(-1)^{n+1}B_{2n{2\cdot(2n)!}.</math>
A proof using Euler's formula and L'Hôpital's rule
The normalized sinc function <math>\text{sinc}(x)=\frac{\sin (\pi x)}{\pi x}</math> has a Weierstrass factorization representation as an infinite product:
<math display=block>\frac{\sin (\pi x)}{\pi x} = \prod_{n=1}^\infty \left(1-\frac{x^2}{n^2}\right).</math>
The infinite product is analytic, so taking the natural logarithm of both sides and differentiating yields
<math display=block>\frac{\pi \cos (\pi x)}{\sin (\pi x)}-\frac{1}{x}=-\sum_{n=1}^\infty \frac{2x}{n^2-x^2}</math>
(by uniform convergence, the interchange of the derivative and infinite series is permissible). After dividing the equation by <math>2x</math> and regrouping one gets
<math display=block>\frac{1}{2x^2}-\frac{\pi \cot (\pi x)}{2x}=\sum_{n=1}^\infty \frac{1}{n^2-x^2}.</math>
We make a change of variables (<math>x=-it</math>):
<math display=block>-\frac{1}{2t^2}+\frac{\pi \cot (-\pi it)}{2it}=\sum_{n=1}^\infty \frac{1}{n^2+t^2}.</math>
Euler's formula can be used to deduce that
<math display=block>\frac{\pi \cot (-\pi i t)}{2it}=\frac{\pi}{2it}\frac{i\left(e^{2\pi t}+1\right)}{e^{2\pi t}-1}=\frac{\pi}{2t}+\frac{\pi}{t\left(e^{2\pi t} - 1\right)}.</math>
or using the corresponding hyperbolic function:
<math display=block>\frac{\pi \cot (-\pi i t)}{2it}=\frac{\pi}{2t}{i\cot (\pi i t)}=\frac{\pi}{2t}\coth(\pi t).</math>
Then
<math display=block>\sum_{n=1}^\infty \frac{1}{n^2+t^2}=\frac{\pi \left(te^{2\pi t}+t\right)-e^{2\pi t}+1}{2\left(t^2 e^{2\pi t}-t^2\right)}=-\frac{1}{2t^2} + \frac{\pi}{2t} \coth(\pi t).</math>
Now we take the limit as <math>t</math> approaches zero and use L'Hôpital's rule thrice. By Tannery's theorem applied to <math display="inline">\lim_{t\to\infty}\sum_{n=1}^\infty 1/(n^2+1/t^2)</math>, we can interchange the limit and infinite series so that <math display="inline">\lim_{t\to 0}\sum_{n=1}^\infty 1/(n^2+t^2)=\sum_{n=1}^\infty 1/n^2</math> and by L'Hôpital's rule
<math display=block>\begin{align}\sum_{n=1}^\infty \frac{1}{n^2}&=\lim_{t\to 0}\frac{\pi}{4}\frac{2\pi te^{2\pi t}-e^{2\pi t}+1}{\pi t^2 e^{2\pi t} + te^{2\pi t}-t}\\[6pt]
&=\lim_{t\to 0}\frac{\pi^3 te^{2\pi t{2\pi \left(\pi t^2 e^{2\pi t}+2te^{2\pi t} \right)+e^{2\pi t}-1}\\[6pt]
&=\lim_{t\to 0}\frac{\pi^2 (2\pi t+1)}{4\pi^2 t^2+12\pi t+6}\\[6pt]
&=\frac{\pi^2}{6}.\end{align}</math>
A proof using Fourier series
Use Parseval's identity (applied to the function ) to obtain
<math display=block>\sum_{n=-\infty}^\infty |c_n|^2 = \frac{1}{2\pi}\int_{-\pi}^\pi x^2 \, dx,</math>
where
<math display=block>\begin{align}
c_n &= \frac{1}{2\pi}\int_{-\pi}^\pi x e^{-inx} \, dx \\[4pt]
&= \frac{n\pi \cos(n\pi)-\sin(n\pi)}{\pi n^2} i \\[4pt]
&= \frac{\cos(n\pi)}{n} i \\[4pt]
&= \frac{(-1)^n}{n} i
\end{align}</math>
for , and . Thus,
<math display=block>|c_n|^2 = \begin{cases}
\dfrac{1}{n^2}, & \text{for } n \neq 0, \\
0, & \text{for } n = 0,
\end{cases}
</math>
and
<math display=block>\sum_{n=-\infty}^\infty |c_n|^2 = 2\sum_{n=1}^\infty \frac{1}{n^2} = \frac{1}{2\pi} \int_{-\pi}^\pi x^2 \, dx.</math>
Therefore,
<math display=block>\sum_{n=1}^\infty \frac{1}{n^2} = \frac{1}{4\pi}\int_{-\pi}^\pi x^2 \, dx = \frac{\pi^2}{6}</math>
as required.
Another proof using Parseval's identity
Given a complete orthonormal basis in the space <math>L^2_{\operatorname{per(0, 1)</math> of L2 periodic functions over <math>(0, 1)</math> (i.e., the subspace of square-integrable functions which are also periodic), denoted by <math>\{e_i\}_{i=-\infty}^{\infty}</math>, Parseval's identity tells us that
<math display=block>\|x\|^2 = \sum_{i=-\infty}^{\infty} |\langle e_i, x\rangle|^2, </math>
where <math>\|x\| := \sqrt{\langle x,x\rangle}</math> is defined in terms of the inner product on this Hilbert space given by
<math display=block>\langle f, g\rangle = \int_0^1 f(x) \overline{g(x)} \, dx,\ f,g \in L^2_{\operatorname{per(0, 1).</math>
We can consider the orthonormal basis on this space defined by <math>e_k \equiv e_k(\vartheta) := \exp(2\pi\imath k \vartheta)</math> such that <math>\langle e_k,e_j\rangle = \int_0^1 e^{2\pi\imath (k-j) \vartheta} \, d\vartheta = \delta_{k,j}</math>. Then if we take <math>f(\vartheta) := \vartheta</math>, we can compute both that
<math display=block>
\begin{align}
\|f\|^2 & = \int_0^1 \vartheta^2 \, d\vartheta = \frac{1}{3} \\
\langle f, e_k\rangle & = \int_0^1 \vartheta e^{-2\pi\imath k\vartheta} \, d\vartheta = \Biggl\{\begin{array}{ll} \frac{1}{2}, & k = 0 \\ -\frac{1}{2\pi\imath k} & k \neq 0, \end{array}
\end{align}
</math>
by elementary calculus and integration by parts, respectively. Finally, by Parseval's identity stated in the form above, we obtain that
<math display=block>
\begin{align}
\|f\|^2 = \frac{1}{3} & = \sum_{\stackrel{k=-\infty}{k \neq 0^{\infty} \frac{1}{(2\pi k)^2}+ \frac{1}{4}
= 2 \sum_{k=1}^{\infty} \frac{1}{(2\pi k)^2}+ \frac{1}{4} \\
& \implies \frac{\pi^2}{6} = \frac{2 \pi^2}{3} - \frac{\pi^2}{2} = \zeta(2).
\end{align}
</math>
Generalizations and recurrence relations
Note that by considering higher-order powers of <math>f_j(\vartheta) := \vartheta^j \in L^2_{\operatorname{per(0, 1)</math> we can use integration by parts to extend this method to enumerating formulas for <math>\zeta(2j)</math> when <math>j > 1</math>. In particular, suppose we let
<math display=block>I_{j,k} := \int_0^1 \vartheta^j e^{-2\pi\imath k\vartheta} \, d\vartheta, </math>
so that integration by parts yields the recurrence relation that
<math display=block>
\begin{align}
I_{j,k} & = \begin{cases} \frac{1}{j+1}, & k=0; \\[4pt] -\frac{1}{2\pi\imath \cdot k} + \frac{j}{2\pi\imath \cdot k} I_{j-1,k}, & k \neq 0\end{cases} \\[6pt]
& = \begin{cases} \frac{1}{j+1}, & k=0; \\[4pt] -\sum\limits_{m=1}^j \frac{j!}{(j+1-m)!} \cdot \frac{1}{(2\pi\imath \cdot k)^m}, & k \neq 0. \end{cases}
\end{align}
</math>
Then by applying Parseval's identity as we did for the first case above along with the linearity of the inner product yields that
<math display=block>
\begin{align}
\|f_j\|^2 = \frac{1}{2j+1} & = 2 \sum_{k \geq 1} I_{j,k} \bar{I}_{j,k} + \frac{1}{(j+1)^2} \\[6pt]
& = 2 \sum_{m=1}^j \sum_{r=1}^j \frac{j!^2}{(j+1-m)! (j+1-r)!} \frac{(-1)^r}{\imath^{m+r \frac{\zeta(m+r)}{(2\pi)^{m+r + \frac{1}{(j+1)^2}.
\end{align}
</math>
Proof using differentiation under the integral sign
It's possible to prove the result using elementary calculus by applying the differentiation under the integral sign technique to an integral due to Freitas:
<math display=block>I(\alpha) = \int_0^\infty \ln\left(1+\alpha e^{-x}+e^{-2x}\right)dx.</math>
While the primitive function of the integrand cannot be expressed in terms of elementary functions, by differentiating with respect to <math>\alpha</math> we arrive at
<math display=block>\frac{dI}{d\alpha} = \int_0^\infty \frac{e^{-x{1+\alpha e^{-x}+e^{-2xdx,</math>
which can be integrated by substituting <math>u=e^{-x}</math> and completing the square. In the range <math>-2 < \alpha < 2</math> the definite integral reduces to
<math display=block>\frac{dI}{d\alpha} = \frac{2}{\sqrt{4-\alpha^2\left[\arctan\left(\frac{\alpha+2}{\sqrt{4-\alpha^2\right)-\arctan\left(\frac{\alpha}{\sqrt{4-\alpha^2\right)\right].</math>
The expression can be simplified using the arctangent addition formula and integrated with respect to <math>\alpha</math> by means of trigonometric substitution, resulting in
<math display=block>I(\alpha) = -\frac{1}{2}\arccos\left(\frac{\alpha}{2}\right)^2 + c.</math>
The integration constant <math>c</math> can be determined by noticing that two distinct values of <math>I(\alpha)</math> are related by
<math display=block>I(2) = 4I(0),</math>
because when calculating <math>I(2)</math> we can factor <math>1+2e^{-x}+e^{-2x} = (1+e^{-x})^2</math> and express it in terms of <math>I(0)</math> using the logarithm of a power identity and the substitution <math>u=x/2</math>. This makes it possible to determine <math>c = \frac{\pi^2}{6}</math>, and it follows that
<math display=block>I(-2) = 2\int_0^\infty \ln(1-e^{-x})dx = -\frac{\pi^2}{3}.</math>
This final integral can be evaluated by expanding the natural logarithm into its Taylor series:
<math display=block>\int_0^\infty \ln(1-e^{-x})dx = - \sum_{n=1}^\infty \int_0^\infty \frac{e^{-nx{n}dx = -\sum_{n=1}^\infty\frac{1}{n^2}.</math>
The last two identities imply
<math display=block>\sum_{n=1}^\infty\frac{1}{n^2} = \frac{\pi^2}{6}.</math>
Cauchy's proof
While most proofs use results from advanced mathematics, such as Fourier analysis, complex analysis, and multivariable calculus, the following does not even require single-variable calculus (until a single limit is taken at the end).
For a proof using the residue theorem, see here.
History of this proof
The proof goes back to Augustin Louis Cauchy (Cours d'Analyse, 1821, Note VIII). In 1954, this proof appeared in the book of Akiva and Isaak Yaglom "Nonelementary Problems in an Elementary Exposition". Later, in 1982, it appeared in the journal Eureka, attributed to John Scholes, but Scholes claims he learned the proof from Peter Swinnerton-Dyer, and in any case he maintains the proof was "common knowledge at Cambridge in the late 1960s".
The proof
[[File:limit circle FbN.jpeg|thumb|The inequality<br>
<math>\tfrac{1}{2}r^2\tan\theta > \tfrac{1}{2}r^2\theta > \tfrac{1}{2}r^2\sin\theta</math><br>
is shown pictorially for any <math>\theta \in (0, \pi/2)</math>. The three terms are the areas of the triangle OAC, circle section OAB, and the triangle OAB.
Taking reciprocals and squaring gives<br>
<math>\cot^2\theta<\tfrac{1}{\theta^2}<\csc^2\theta</math>.]]
The main idea behind the proof is to bound the partial (finite) sums
<math display=block>\sum_{k=1}^m \frac{1}{k^2} = \frac{1}{1^2} + \frac{1}{2^2} + \cdots + \frac{1}{m^2}</math>
between two expressions, each of which will tend to as approaches infinity. The two expressions are derived from identities involving the cotangent and cosecant functions. These identities are in turn derived from de Moivre's formula, and we now turn to establishing these identities.
Let be a real number with , and let be a positive odd integer. Then from de Moivre's formula and the definition of the cotangent function, we have
<math display=block>\begin{align}
\frac{\cos (nx) + i \sin (nx)}{\sin^n x} &= \frac{(\cos x + i\sin x)^n}{\sin^n x} \\[4pt]
&= \left(\frac{\cos x + i \sin x}{\sin x}\right)^n \\[4pt]
&= (\cot x + i)^n.
\end{align}</math>
From the binomial theorem, we have
<math display=block>\begin{align}
(\cot x + i)^n
= & {n \choose 0} \cot^n x + {n \choose 1} (\cot^{n - 1} x)i + \cdots + {n \choose {n - 1 (\cot x)i^{n - 1} + {n \choose n} i^n \\[6pt]
= & \Bigg( {n \choose 0} \cot^n x - {n \choose 2} \cot^{n - 2} x \pm \cdots \Bigg) \; + \; i\Bigg( {n \choose 1} \cot^{n-1} x - {n \choose 3} \cot^{n - 3} x \pm \cdots \Bigg).
\end{align}</math>
Combining the two equations and equating imaginary parts gives the identity
<math display=block>\frac{\sin (nx)}{\sin^n x} = \Bigg( {n \choose 1} \cot^{n - 1} x - {n \choose 3} \cot^{n - 3} x \pm \cdots \Bigg).</math>
We take this identity, fix a positive integer , set , and consider for . Then is a multiple of and therefore . So,
<math display=block>0 = , </math>
where <math>v_n = 2n-1 \mapsto \{1,3,5,7,9,\ldots\}</math>.
See also
- List of sums of reciprocals
References
- .
- .
- .
- .
Notes
External links
- An infinite series of surprises by C. J. Sangwin
- From ζ(2) to Π. The Proof. step-by-step proof
- , English translation with notes of Euler's paper by Lucas Willis and Thomas J. Osler
- Robin Chapman, Evaluating (fourteen proofs)
- Visualization of Euler's factorization of the sine function
- (animated proof based on the above)