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September 27, 2010

My Six Favourite Formulas – #4

Filed under: mathematics — ckrao @ 2:07 pm

Here I will discuss another of my favourite formulas:

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

It boggles the mind how the sum of reciprocals of squared natural numbers could possibly be related to \pi! Interestingly, the reciprocal of this number, 6/\pi^2 \approx 0.6079, is the probability that two randomly selected positive integers are relatively prime. That is, choose two integers uniformly and independently between 1 and N, then as N tends to infinity the probability they share no common factor other than 1 is 6/\pi^2.

Finding the infinite sum was first posed by Pietro Mengoli in 1644 and is now known as the Basel problem. It stumped the great mathematicians of the next ninety years, ranging from Leibniz to Jacob Bernoulli. Finally in 1735 Euler became famous after being the first to find the answer, aged 28. He used his immense talent in numerical analysis (using what is now known as the Euler-Maclaurin formula) to conjecture the sum to be \frac{\pi^2}{6}, then this motivated him to use series formulas for \frac{\sin x}{x} to prove the result. We will outline three other proofs below, more proofs can be found in Robin Chapman’s collection of 14 proofs here.

Proof 1: Elementary proof, found on MathOverflow website here.

The idea is to sandwich the sum \displaystyle \sum_{n=1}^{N} \frac{1}{n^2} between two expressions which each converge to \frac{\pi^2}{6} as N \to \infty.

We start with the inequality

\displaystyle \sin x < x < \tan x, where 0 < x < \pi/2.

This can be seen by drawing a unit circle and comparing the areas of a sector with internal angle x radians (area x/2) , a triangle with unit sides and internal angle x (area \frac{1}{2}ab\sin C = \frac{1}{2}\sin x) and a right angle triangle with base length 1 and angle x (area \frac{1}{2}bh =\frac{1}{2}\tan x).

Next, let x_n = \frac{n}{2^N}\frac{\pi}{2}, n = 1,2, \ldots, 2^N - 1. Then from the above inequality,

\displaystyle \sum_{n=1}^{2^N-1}\frac{1}{\tan^2 x_n} < \sum_{n=1}^{2^N-1} \frac{1}{x_n^2} < \sum_{n=1}^{2^N-1} \frac{1}{\sin^2 x_n}........(*)

Now let us evaluate the right side of this inequality, denoted by S_N. We sum terms symmetric about \pi/4 and find

\begin{array}{lcl} \frac{1}{\sin^2 x_n} + \frac{1}{\sin^2 x_{(2^N-n)}}&=& \frac{1}{\sin^2 \frac{n}{2^N}\frac{\pi}{2}} + \frac{1}{\sin^2 \frac{2^N-n}{2^N}\frac{\pi}{2}}\\&=&\frac{1}{\sin^2 x_n} +\frac{1}{\sin^2 (\pi/2-x_n)}\\&=& \frac{\sin^2 x_n + \cos^2 x_n}{\sin^2 x_n \cos^2 x_n}\\&=&\frac{4}{\sin^2 2x_n}\\&=&\frac{4}{\sin^2 \frac{n}{2^{(N-1)}}\frac{\pi}{2}}.\end{array}


\begin{array}{lcl} S_N &=& \sum_{n=1}^{2^N-1} \frac{1}{\sin^2 x_n}\\&=&\sum_{n=1}^{2^{(N-1)}-1} \left(\frac{1}{\sin^2 x_n} + \frac{1}{\sin^2 x_{(2^N-n)}}\right)+\frac{1}{\sin^2 x_{2^{(N-1)}}}\\&=&\frac{1}{\sin^2 \pi/4} + \sum_{n=1}^{2^{(N-1)}-1} \frac{4}{\sin^2 \frac{n}{2^{(N-1)}}\frac{\pi}{2}}\\&=&2 + 4S_{N-1}.\end{array}

This means we have the recurrence relation S_N = 2 + 4S_{N-1} with the initial condition S_1 = 2 (initial terms 2, 10, 42, 170, …). Using mathematical induction or generating functions (z-transforms), one can show that this has the closed form solution S_N = 2(4^N-1)/3. Similarly,

\displaystyle \sum_{n=1}^{2^N-1}\frac{1}{\tan^2 x_n} = \sum_{n=1}^{2^N-1} \frac{1-\sin^2 x_n}{\sin^2 x_n} = \frac{2(4^N-1)}{3} - (2^N-1).

Then from (*),

\displaystyle \frac{2(4^N-1)}{3} - (2^N-1) < \sum_{n=1}^{2^N-1} \frac{1}{\left(\frac{n}{2^N}\frac{\pi}{2}\right)^2} < \frac{2(4^N-1)}{3}.

This implies

\displaystyle \frac{\pi^2}{4^{N+1}}\left(\frac{2(4^N-1)}{3} - (2^N-1)\right) < \sum_{n=1}^{2^N-1} \frac{1}{n^2}< \frac{\pi^2}{4^{N+1}}\frac{2(4^N-1)}{3}.

We see that both left and right sides tend to \frac{\pi^2}{6} as N \to \infty, and we conclude that \displaystyle \sum_{n=1}^{\infty} \frac{1}{n^2} = \frac{\pi^2}{6}.

Proof 2: Integration by clever change of variables, due to Calabi, Beukers and Kock.

Using the result \displaystyle \int_0^1 x^{2r}\ dx = \frac{1}{2r+1},

\begin{array}{lcl} \sum_{r=0}^{\infty} \frac{1}{(2r+1)^2} &=& \sum_{r=0}^{\infty}\left(\int_0^1 x^{2r}\ dx \right)\left(\int_0^1 y^{2r}\ dy \right)\\&=&\int_0^1 \int_0^1 \sum_{r=0}^{\infty} (x^2y^2)^r\ dx\ dy\\&=& \int_0^1 \int_0^1 \frac{1}{1-x^2y^2}\ dx\ dy.\end{array}

Then make the remarkable substitution

\displaystyle (x,y) = \left(\frac{\sin u}{\cos v}, \frac{\sin v}{\cos u}\right).

We find that this sets up a bijective transformation between the unit square \{(x,y): 0 < x < 1, 0 < y < 1\} and the triangle

\displaystyle A = \{(u, v) : u > 0, v > 0, u + v < \pi/2\}.

(Note that \sin u/\cos v < 1 for 0 <u,v < \pi/2 implies \sin u < \sin\left(\frac{\pi}{2} - v\right) and so u < \pi/2 - v or u + v < \pi/2. The inverse map is \displaystyle (u,v) = \left(\arctan x\sqrt{\frac{1-y^2}{1-x^2}}, \arctan y\sqrt{\frac{1-x^2}{1-y^2}}\right).)

Also we find

\begin{array}{lcl} dx dy &=& d\left(\frac{\sin u}{\cos v}\right)\wedge d\left(\frac{\sin v}{\cos u}\right)\\&=& \left(\frac{\cos u}{\cos v}du + \frac{\sin u \sin v}{\cos^2 v}dv\right) \wedge \left(\frac{\cos v}{\cos u}dv + \frac{\sin v \sin u}{\cos^2 u}du\right)\\&=& \left( \frac{\cos u}{\cos v}\frac{\cos v}{\cos u} -\frac{\sin u \sin v}{\cos^2 v}\frac{\sin v \sin u}{\cos^2 u}\right)du dv\\&=&(1-x^2y^2)dudv.\end{array}

This means

\displaystyle \sum_{r=0}^{\infty} \frac{1}{(2r+1)^2} = \int\int_A du dv, where A is defined above, and has area \pi^2/8. Hence

\begin{array}{lcl}\sum_{n=1}^{\infty} \frac{1}{n^2}&=& \sum_{r=1}^{\infty} \frac{1}{(2r)^2} + \sum_{r=0}^{\infty} \frac{1}{(2r+1)^2} \\&=& \frac{1}{4}\sum_{n=1}^{\infty} \frac{1}{n^2} + \frac{\pi^2}{8}\end{array}

from which \displaystyle \sum_{n=1}^{\infty} \frac{1}{n^2} = \frac{\pi^2}{8}/\left(1-\frac{1}{4}\right) = \frac{\pi^2}{6}.

Proof 3: Proof by Fourier series

This is the proof seen in many textbooks, but requires a more advanced result than the previous two, namely Parseval’s theorem. We use the fact that the complex exponentials e_n(x) = \exp(2\pi i n x) where n \in \mathbb{Z} form a complete orthonormal basis in the set L^2[0,1] of square-integrable complex-valued functions on the unit interval [0,1]. This means such a function f on the interval [0,1] can be written as

\displaystyle f(x) = \sum_{n=-\infty}^{\infty} \alpha_n e_n(x),

where \alpha_n are Fourier coefficients which may be found by taking the inner product of both sides of the above with e_n:

\displaystyle \langle f, e_n \rangle = \sum_{m=-\infty}^{\infty} \alpha_m \langle e_m, e_n\rangle = \alpha_n.

In linear algebra terms, we are simply projecting f onto the basis vector e_n to find its component in that direction. Parseval’s theorem is simply a generalisation of Pythagoras’s theorem, stating that the square of the norm of f is the sum of the squared magnitudes of its components:

\displaystyle \|f\|^2 = \langle f,f \rangle = \sum_{n= -\infty}^{\infty} \left| \langle f,e_n\rangle \right|^2.

We now calculate this for f(x) = x, noting that \langle f,g \rangle = \int_0^1 f (x)\bar{g}(x)\ dx.

    • \displaystyle \langle f,f \rangle = \int_0^1 x^2\ dx = \frac{1}{3}
    • \displaystyle \langle f,e_0 \rangle = \int_0^1 x\ dx = \frac{1}{2}
    • For n \neq 0,

\begin{array}{lcl} \langle f,e_n \rangle &=& \int_0^1 xe^{-2\pi i n x}\ dx\\&=& \left[\frac{-xe^{-2\pi i nx}}{2\pi in} \right]_0^1 + \int_0^1 \frac{e^{-2\pi i n x}}{2\pi in}\ dx\\&=& \frac{-e^{-2\pi i n}}{2\pi i n} + \frac{1}{(2\pi i n)^2}\left(-e^{-2\pi i n} + 1\right)\\&=& \frac{-1}{2\pi i n}.\end{array}

Combining these results in Parseval’s theorem gives

\displaystyle \frac{1}{3} = \frac{1}{2^2} + \sum_{n \in \mathbb{Z}, n \neq 0} \frac{1}{4\pi^2 n^2} = \frac{1}{4} + \frac{1}{2\pi^2}\sum_{n=0}^{\infty} \frac{1}{n^2},

from which \sum_{n=1}^{\infty} \frac{1}{n^2} = 2\pi^2\left(\frac{1}{3} - \frac{1}{4}\right) = \frac{\pi^2}{6}, as desired.

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