Basel problem

The Basel problem is a mathematical problem that was unsolved for a long time and that was initially mainly dealt with by Basel mathematicians. It is a question of the sum of the reciprocal square numbers , i.e. the value of the series

${\displaystyle \sum _{n=1}^{\infty }{\frac {1}{n^{2}}}={\frac {1}{1^{2}}}+{\frac {1}{2^{2}}}+{\frac {1}{3^{2}}}+{\frac {1}{4^{2}}}+\cdots =1+{\frac {1}{4}}+{\frac {1}{9}}+{\frac {1}{16}}+\cdots .}$

It was solved in 1735 by Leonhard Euler, who found the series value . ${\displaystyle {\tfrac {\pi ^{2}}{6}}\approx 1{,}644934}$

Attempted solutions

In 1644 the Italian Pietro Mengoli asked himself whether this sum was converging, and if so, against what value, but could not answer this question. The Basel mathematician Jakob I Bernoulli found out about this problem a little later, but found no solution (1689). Several mathematicians then tried to answer the question, but were all unsuccessful. In 1726 Leonhard Euler , also a Basel mathematician and student of Jakob Bernoulli's brother Johann , began to deal with the problem. In 1735 he found the solution and published it in his work "De Summis Serierum Reciprocarum" .

Solutions

Euler's first solution

For his original solution, Euler considered the Taylor series of the cardinal sine function , so

${\displaystyle \mathrm {sinc} (x)={\frac {\sin(x)}{x}}=1-{\frac {x^{2}}{3!}}+{\frac {x^{4}}{5!}}-{\frac {x^{6}}{7!}}+-\cdots }$

and equated it with the product presentation of that function.

${\displaystyle {\frac {\sin(x)}{x}}=\prod _{n=1}^{\infty }\left(1-{\frac {x^{2}}{\pi ^{2}n^{2}}}\right)=\left(1-{\frac {x^{2}}{\pi ^{2}}}\right)\left(1-{\frac {x^{2}}{4\pi ^{2}}}\right)\left(1-{\frac {x^{2}}{9\pi ^{2}}}\right)\cdots =1-{\frac {x^{2}}{3!}}+{\frac {x^{4}}{5!}}-{\frac {x^{6}}{7!}}+-\cdots }$

When (hypothetically) multiplying the infinite product, he only considered those products which contain and . Since there is no other way that a term can contain a quadratic term, the two quadratic terms on the respective sides must be the same. ${\displaystyle 1}$ ${\displaystyle x^{2}}$

${\displaystyle -x^{2}\left({\frac {1}{\pi ^{2}}}+{\frac {1}{2^{2}\pi ^{2}}}+{\frac {1}{3^{2}\pi ^{2}}}+{\frac {1}{4^{2}\pi ^{2}}}+\cdots \right)=-{\frac {x^{2}}{3!}}=-{\frac {x^{2}}{6}}}$

and from this Euler deduced his solution:

${\displaystyle 1+{\frac {1}{2^{2}}}+{\frac {1}{3^{2}}}+{\frac {1}{4^{2}}}+\cdots ={\frac {\pi ^{2}}{6}}.}$

About a double integral

The proof of a double integral appears as an exercise in William J. LeVeque's 1956 textbook on number theory. In it, he writes about the problem: “I have no idea where this problem came from, but I'm pretty sure it is with did not originate from me. "

The representation is first obtained via the geometric row

${\displaystyle \zeta (2)=\int _{0}^{1}\int _{0}^{1}{\frac {\mathrm {d} x\mathrm {d} y}{1-xy}}.}$

Using a variable substitution and one arrives at ${\displaystyle u={\frac {y+x}{2}}}$ ${\displaystyle v={\frac {y-x}{2}}}$

${\displaystyle \zeta (2)=4\int _{0}^{\frac {1}{2}}\left(\int _{0}^{u}{\frac {\mathrm {d} v}{1-u^{2}+v^{2}}}\right)\mathrm {d} u+4\int _{\frac {1}{2}}^{1}\left(\int _{0}^{1-u}{\frac {\mathrm {d} v}{1-u^{2}+v^{2}}}\right)\mathrm {d} u,}$

where the inner integrals can be resolved with the help of the arctangent

${\displaystyle \zeta (2)=4\int _{0}^{\frac {1}{2}}{\frac {1}{\sqrt {1-u^{2}}}}\arctan \left({\frac {u}{\sqrt {1-u^{2}}}}\right)\mathrm {d} u+4\int _{\frac {1}{2}}^{1}{\frac {1}{\sqrt {1-u^{2}}}}\arctan \left({\frac {1-u}{\sqrt {1-u^{2}}}}\right)\mathrm {d} u.}$

With and you get the spelling ${\displaystyle g(u)=\arctan \left({\frac {u}{\sqrt {1-u^{2}}}}\right)}$ ${\displaystyle h(u)=\arctan \left({\frac {1-u}{\sqrt {1-u^{2}}}}\right)}$

${\displaystyle \zeta (2)=4\int _{0}^{\frac {1}{2}}g'(u)g(u)\mathrm {d} u+4\int _{\frac {1}{2}}^{1}h'(u)h(u)\mathrm {d} u=2\left[g(u)^{2}\right]_{0}^{\frac {1}{2}}+2\left[h(u)^{2}\right]_{\frac {1}{2}}^{1}={\frac {\pi ^{2}}{6}}.}$

Via a cotangent sum

Another proof uses the cotangent sum

${\displaystyle \sum _{n=1}^{m}\cot ^{2}\left({\frac {\pi n}{2m+1}}\right)={\frac {2m(2m-1)}{6}}.}$

This can be shown elementarily using Euler's identity .

Generalizations

Euler also generalized the problem. To this end, he examined the function later called Riemann's function

${\displaystyle \zeta (s)=\sum _{n=1}^{\infty }{\frac {1}{n^{s}}}={\frac {1}{1^{s}}}+{\frac {1}{2^{s}}}+{\frac {1}{3^{s}}}+{\frac {1}{4^{s}}}+\cdots }$

and found a general closed expression for all natural even-numbered arguments , viz ${\displaystyle s=2k}$

${\displaystyle \zeta (2k)=(-1)^{k-1}{\frac {(2\pi )^{2k}}{2(2k)!}}B_{2k},}$

where the -th represents Bernoulli's number . A general formula for odd natural arguments (see e.g. Apéry constant ) is not yet known. ${\displaystyle B_{2k}}$ ${\displaystyle 2k}$ ${\displaystyle s}$

Web links

• Leonhard Euler: De Summis Serierum Reciprocarum (Latin, English)
• Evaluating ζ (2) - 14 pieces of evidence for the value of ζ (2) compiled by Robin Chapman on a website of the University of Exeter (pdf, English; 181 kB)

literature

• C. Edward Sandifer: Euler's solution of the Basel problem — the longer story. Euler at 300, 105-117, MAA Spectrum, Math. Assoc. America, Washington, DC, 2007.
• Downey, Lawrence / Ong, Boon W. / Sellers, James A .: Beyond the Basel Problem: Sums of Reciprocals of Figurate Numbers. The College Mathematics Journal. Vol. 39, no. Nov. 5, 2008. P. 391-394