Basel problem derivation


The basis for this derivation is the assumption that the properties of finite polynomials hold true for infinite series. First, consider the Taylor series expansion of the sine function:

sinx=n=0(-1)n(2n+1)!x2n+1=x-x33!+x55!-x77!+

Dividing through by x, we get:

sinxx=1-x23!+x45!-x67!+

The roots of the function sinx/x occur at x=nπ, where n=±1,±2,±3,. Let us assume that we can express this infinite series as a productPlanetmathPlanetmath of linear factors given by its roots, just as we do for finite polynomials:

sinxx =(1-xπ)(1+xπ)(1-x2π)(1+x2π)(1-x3π)(1+x3π)
=(1-x2π2)(1-x24π2)(1-x29π2).

If we multiply out this product and collect all the x2 terms, we get:

-(1π2+14π2+19π2+)=-1π2n=11n2

From the original infinite series of sinx/x the coefficient of x2 is -1/3!. Equating both expressions gives:

-16=-1π2n=11n2
n=11n2=π26

This was the method Euler used to solve the Basel problemMathworldPlanetmath. There are more rigorous and modern derivations, but this one is a simple, straight forward, and easily understood one.

Title Basel problem derivation
Canonical name BaselProblemDerivation
Date of creation 2013-03-22 18:40:50
Last modified on 2013-03-22 18:40:50
Owner curious (18562)
Last modified by curious (18562)
Numerical id 5
Author curious (18562)
Entry type Derivation
Classification msc 11A25