Three solutions to the Basel problem

The Basel problem was first posed in 1644 by Pietro Mengoli, author of the Novae quadraturae arithmeticae. It was to find the value of the sum

\displaystyle\sum_{n=1}^{\infty} \frac{1}{n^2} =  1 + \frac{1}{4} + \frac{1}{9} + \frac{1}{16} + \frac{1}{25} + \frac{1}{36} + \frac{1}{49} + \dots

whose terms are the reciprocals of the positive square numbers.

pietro
Pietro Mengoli

Mengoli’s problem remained intractable for a lifetime (91 years) before it was famously solved by Leonhard Euler in 1735, who showed that

\displaystyle\sum_{n=1}^{\infty} \frac{1}{n^2} = \frac{\pi}{6}

Later work by Bernhard Riemann defined the zeta function as

\zeta(s) = \displaystyle\sum_{n=1}^{\infty} \frac{1}{n^s}

where s \in \mathbb{C} is a complex number.

Thus the formula given by Euler is commonly written as \zeta(2) = \frac{\pi}{6}.

Proof 1:

We begin with Euler’s original proof of the theorem. Like all of Euler’s groundbreaking work it is a true jewel of the imagination.

Consider the MacLaurin expansion of \sin(\pi x).

\sin(\pi x) = \displaystyle\sum_{n=0}^{\infty} (-1)^{n+1} \frac{(\pi x)^{2n+1}}{(2n+1)!} = \pi x - \frac{(\pi x)^3}{3!} + \frac{(\pi x)^5}{5!} - \frac{(\pi x)^7}{7!} + \dots

The solutions to \sin(\pi x) = 0 occur when x \in \mathbb{Z}. Thus we may factorise \sin(\pi x) as

\begin{aligned} \displaystyle\sin(\pi x) & = \pi x \cdot \prod_{\substack{n \in \mathbb{Z} \\ n \neq 0}} \Big(1 - \frac{x}{n} \Big) \\ & = \displaystyle\pi x (1 - x)(1+x)\Big(1-\frac{x}{2}\Big)\Big(1+\frac{x}{2}\Big)\Big(1-\frac{x}{3}\Big)\Big(1+\frac{x}{3}\Big) \cdots \end{aligned}

The factor of \pi follows from the MacLaurin expansion given above. Collecting pairs of factors as the difference of two squares we have

\sin(\pi x) = \displaystyle \pi x (1 - x^2)\Big(1-\frac{x^2}{4}\Big)\Big( 1 - \frac{x^2}{9}\Big)\Big(1-\frac{x^2}{16} \Big)\Big(1-\frac{x^2}{25} \Big) \cdots

The coeffecient of x^3 in the MacLaurin expansion of \sin(\pi x) is -\frac{\pi^3}{3!} = -\frac{\pi^3}{6}. The coeffecient of x^3 in the product representation is

- \pi \displaystyle\Big(1+\frac{1}{4} + \frac{1}{9} + \frac{1}{16} + \frac{1}{25} + \frac{1}{36} + \dots \Big)

as the contributions to the coefficient of x^3 come from the \pi x factor multiplied with exactly one of the x^2 factors. Equating coefficients we have

- \pi \displaystyle\Big(1+\frac{1}{4} + \frac{1}{9} + \frac{1}{16} + \frac{1}{25} + \frac{1}{36} + \dots \Big) = -\frac{\pi^3}{6}

and thus

\displaystyle 1+\frac{1}{4} + \frac{1}{9} + \frac{1}{16} + \frac{1}{25} + \dots = \frac{\pi^2}{6}

as required.

Some technical issues are ignored in the proof above. Notably, how are we certain that the representation of the sine function as a product of linear factors is a permissible one? The necessary mathematical formalism to answer this question was provided by Weierstrass. See for example [4] which covers these details. Perhaps this treatment leaves some doubt as to the uniqueness of the series representation and the validity of equating coefficients in the final step of the proof. These are valid concerns which will not be dealt with here.

Proof 2:

apostolThe second proof is due to Apostol [1] and appeared in the in 1983 Mathematical Intelligencer. Apostol is a common noun to mathematical students everywhere owing to his seminal calculus textbook. Apostol’s proof has the advantage of being the work of a man rather than a magician. Apostol was lost to the world on May 8, 2016 at 92 years old. You may see for yourself (e.g in this lecture given in 2013) that he remained sharp until the end.

Consider the following integral

I = \displaystyle\int_0^1 \int_0^1 \frac{1}{1-xy}~ \,dx \,dy

We will show that I = \zeta(2). Considering the domain of integration (x,y) \in [0,1]^2, we may write \displaystyle\frac{1}{1-xy} as the geometric series

  \displaystyle\frac{1}{1-xy} = \sum_{n=0}^{\infty} (xy)^n

and substituting this into the double integral we have

zeta2

\begin{aligned} I & = \int_0^1 \int_0^1 \sum_{n=0}^{\infty} (xy)^n \,dx \,dy \\ & = \int_0^1 \sum_{n=0}^{\infty} y^n \Big[\frac{x^{n+1}}{n+1} \Big]_{x=0}^{x=1} \,dy \\ & = \int_0^1 \sum_{n=0}^{\infty} \frac{y^n}{n+1} \,dy \\ & = \sum_{n=0}^{\infty} \Big[ \frac{y^{n+1}}{(n+1)^2}\Big]_{y=0}^{y=1} \\ & = \sum_{n=0}^{\infty} \frac{1}{(n+1)^2} = \zeta(2) \end{aligned}

We now ‘double count’ by evaluating the integral in a different way. We rotate the coordinate axes by 45 degrees, so that the point (x,y) maps to (u,v) where

zeta_int

\begin{aligned} \begin{bmatrix} x \\ y \end{bmatrix} & = \begin{bmatrix} \cos(\frac{\pi}{4}) & -\sin(\frac{\pi}{4})\\ \sin(\frac{\pi}{4}) & \cos(\frac{\pi}{4}) \end{bmatrix}  \begin{bmatrix} u \\ v \end{bmatrix} \\ & = \displaystyle \frac{1}{\sqrt{2}}\begin{bmatrix} u-v \\ u+v \end{bmatrix} \end{aligned}

so that \displaystyle 1-xy becomes \displaystyle 1- \frac{u^2 - v^2}{2} = \frac{2 - u^2 + v^2}{2}.

The region of integration becomes the square with vertices (0,0), (\pm 1/ \sqrt{2},1/ \sqrt{2}), (0,\sqrt{2}) in the xy-plane.

Symmetry of the integral about the diagonal u = v allows us to write

\begin{aligned} \displaystyle I & = 2 \int_0^{1/ \sqrt{2}} \Big( \int_0^u  \frac{2}{2 - u^2 + v^2} \,dv \Big) \,du \\ & + 2 \int_{1/ \sqrt{2}}^{\sqrt{2}} \Big( \int_0^{\sqrt{2}-u} \frac{2}{2 - u^2 + v^2} \,dv \Big) \,du \end{aligned}

The reason for making this transformation is that is allows us to write the unknown integral in terms of a known formula for the inverse tangent function. Namely the identity

\displaystyle \int_0^x \frac{\,dt}{a^2 + x^2} = \frac{1}{a} \arctan \frac{x}{a}

Thus

\displaystyle \int_0^u  \frac{1}{2 - u^2 + v^2} = \frac{1}{\sqrt{2-u^2}} ~ \arctan \frac{u}{\sqrt{2-u^2}}

and

\displaystyle  \int_0^{\sqrt{2}-u}  \frac{1}{2 - u^2 + v^2} = \frac{1}{\sqrt{2-u^2}} ~ \arctan \frac{\sqrt{2}-u}{\sqrt{2-u^2}}

so that

\begin{aligned} \displaystyle I & = 4 \int_0^{1 / \sqrt{2}} \frac{1}{\sqrt{2-u^2}} \arctan \frac{u}{\sqrt{2-u^2}} \,du ~~ + ~~ 4 \int_{1/ \sqrt{2}}^{\sqrt{2}} \frac{1}{\sqrt{2-u^2}} \arctan \frac{\sqrt{2}-u}{\sqrt{2-u^2}} \,du \\ \end{aligned}

Let these two integrals be I_1 and I_2 respectively. Making the substitution u = \sqrt{2} \sin \theta in I_1 so that \,du = \sqrt{2} \cos \theta it follows that \frac{u^2}{2} = \sin ^2 \theta and by the Pythagorean identity \frac{u^2}{2} + \cos^2 \theta = 1  Hence \,du = \sqrt{2} \cos \theta = \sqrt{2 - u^2} and therefore \tan \theta = u / \sqrt{2-u^2}. Substituting this information into I_1 gives

\begin{aligned} I_1 & = \displaystyle 4 \int_0^{\pi/6} \theta \,d\theta \\ & = 4 \Big[ \frac{\theta^2}{2}\Big]_{\theta = 0}^{\theta = \pi /6} \\ & = 2 \Big( \frac{\pi}{6} \Big)^2 \end{aligned}

The second integral I_2 may be handled using a similar process. We make the change of coordinates u = \sqrt{2} \cos 2\theta so that

\begin{aligned} \,du & = - 2 \sqrt{2} \sin 2 \theta \,d\theta \\ & = -2 \sqrt{2} \sqrt{1 - \cos^2 2\theta} \,d\theta \\ & = -2 \sqrt{2 - u^2} \,d\theta \end{aligned}

Finally

\begin{aligned} \frac{\sqrt{2}-u}{\sqrt{2-u^2}} & = \frac{\sqrt{2}-\sqrt{2}\cos 2\theta}{\sqrt{2-2\cos^2 2\theta}} \\ & = \sqrt{\frac{1-\cos 2\theta}{1+\cos 2\theta }} \\ & = \sqrt{\frac{2\sin^2\theta}{2\cos^2\theta}} = \tan \theta \end{aligned}

Substituting the information above into the integral I_2, the calculation is reduced to

I_2 = \displaystyle 8 \int_0^{\pi/6} \theta \,d\theta = 4 \Big( \frac{\pi}{6}\Big)^2

And therefore

\zeta(2) = \displaystyle I_1 + I_2 = 6 \Big( \frac{\pi}{6} \Big)^2 = \frac{\pi^2}{6}

as expected.

Maybe I was wrong to say Apostol was not a magician…

Proof 3

The final proof comes from a math.stackexchange post (and possibly elsewhere) where several other methods for showing \zeta(2) = \pi / 6 are also given. The proof relies on the expansion of the function f(x) = x^2 on the domain -\pi \leq x \leq \pi into its Fourier series.

\displaystyle \frac{a_0}{2} + \sum_{n=1}^{\infty} (a_n \cos nx + b_n \sin nx)

where

a_n = \displaystyle \frac{1}{\pi} \int_{-\pi}^{~ \pi} f(x) \cos (nx) \,dx

for n = 0,1,2,3, \dots and

b_n = \displaystyle \frac{1}{\pi} \int_{-\pi}^{~ \pi} f(x)\sin (nx) \,dx

for n = 1,2,3, \dots. We note immediately that b_n = 0 for n \in \mathbb{N} as f(x) is an even function, \sin(nx) is an odd function, and the product of an odd and even function is an odd function.

Calculating the initial coefficient

a_0 = \displaystyle \frac{1}{\pi}\int_{-\pi}^{~ \pi} x^2 \,dx = \frac{2}{\pi} \int_0^{\pi} x^2 \,dx = \frac{2\pi^2}{3}

For n = 1,2,3,\dots

\begin{aligned} a_n & = \displaystyle \frac{1}{\pi} \int_{-\pi}^{~ \pi} x^2 \cos (nx) \,dx \\ & = \frac{2}{\pi} \int_{0}^{\pi} x^2 \cos (nx) \,dx \end{aligned}

Integrating by parts twice gives (up to an additive constant)

\begin{aligned} \int x^2 \cos(nx) \,dx & = \frac{x^2 \sin(nx)}{n} - \int \frac{2x \sin(nx)}{n} \,dx \\ & = \frac{x^2 \sin(nx)}{n} + \frac{2x\cos(nx)}{n^2} - \int \frac{2 \cos (nx)}{n^2}\,dx \\ & = \frac{x^2 \sin(nx)}{n} + \frac{2x\cos(nx)}{n^2} - \frac{2 \sin (nx)}{n^3} \end{aligned}

As \sin(2\pi n) = 0 and \cos(2\pi n) = (-1)^n for integer values of n it follows that

\begin{aligned} a_n & = \displaystyle \frac{2}{\pi} \Big[\frac{2x \cos(nx)}{n^2} \Big]_0^{\pi} \\ & = \frac{4(-1)^n}{n^2} \end{aligned}

Thus we have the Fourier series expansion

\displaystyle f(x) = \frac{\pi^2}{3} + \sum_{n=1}^{\infty} \Big( \frac{4(-1)^n}{n^2} \cos(nx) \Big)

As f(\pi) = \pi^2

\begin{aligned} \displaystyle \pi^2 & = \frac{\pi^2}{3} + \sum_{n=1}^{\infty} \Big( \frac{4(-1)^n(-1)^n}{n^2} \Big) \\ & = \frac{\pi^2}{3} + 4 \sum_{n=1}^{\infty} \frac{1}{n^2} \end{aligned}

Thus \displaystyle \zeta(2) = \frac{\pi^2}{4} - \frac{\pi^2}{12} = \frac{\pi^2}{6}.

It is interesting that the evaluation occurs on the boundary of the domain. The final evaluation at x = \pi seems like a remarkable confluence. This method may also be used to calculate the values of \zeta(2n) for n \in \mathbb{N}. I still wonder what intuition guides these proofs. For now it will have to be enough that they are beautiful.

References

[1] Apostol, T. M. “A Proof That Euler Missed: Evaluating \zeta(2) the Easy Way.” Math. Intel. 5, 59-60, 1983.

[2] F. Beukers,  “A note on the irrationality of \zeta(2) and \zeta(3), Bull. Lon. Math. Soc. 11, 268-272, 1979.

[3] O’Connor, J.J. and Robertson, E.F. “Pietro Mengoli” http://www-history.mcs.st-and.ac.uk/Biographies/Mengoli.html

[4] Sullivan, B.W. “Numerous Proofs of \zeta(2) = \frac{\pi}{6}http://math.cmu.edu/~bwsulliv/MathGradTalkZeta2.pdf

 

 

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